Students' Forum - Career Point

XtraEdge for IIT-JEE 1 MARCH 2010

Dear Students,

All of us live and work within fixed patterns. These patterns and habits determine the quality of our life and the choices we make in life. There are a few vital things to know about ourselves. We should become aware of how much we influence others, how productive we are and what can help us to achieve our goals. It is important to create an environment which will promote our success. We should consciously create a system that would enable us to achieve our goals. Most of us live in systems which have come our way by an accident, circumstances or people we have met over a period of time. We are surrounded by our colleagues or subordinates who happened to be there by the fact of sheer recruitment earlier or later by the management. Our daily routines and schedules have been formed on the basis of convenience, coincidence, and the expectations of society and sometimes due to superstitions. The trick for success is to have an environment that helps in attaining our goals. Control your life. Make an effort to launch your day with a great start. A law of physics says that an object set in motion tends to remain in motion. It is the same thing with daily routine. To have a good start each morning will keep you upbeat during the day. If you begin the day stressed, you will tend to remain so that way. The best is to create a course of action or conditions where you are not hassled for being late for a meeting, worried about household affairs or distracted by happenings in the world.

Aim to be highly successful. Control the direction of your life. Not only should you start the day on a cheerful note but also continue to do so during the day. Keep yourself stimulated and invigorated during the entire day. Start your day with a purpose. Have a daily direction and trajectory of action. It will keep you on your course all day long. Throughout the day reinforce your positive values and your choices. Anything that helps you in maintaining your highest values and your most important priorities should be welcome. Be in control of your life and work. Create and sustain a wonderful environment filled with beauty, peace, inspiration and hope.

Plan your day in such a way that suits your plans objectives and makes you feel just right with the right amount of encouragement during the entire day. You should give a direction to your day and timing.

Presenting forever positive ideas to your success.

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Editor : Pramod Maheshwari

Worry is a misuse of imagination. Volume - 5 Issue - 9

March, 2010 (Monthly Magazine)

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Volume-5 Issue-9 March, 2010 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Key Concepts & Problem Solving strategy for IIT-JEE.

Know IIT-JEE With 15 Best Questions of IIT-JEE

Challenging Problems in Physics, Chemistry & Maths

Much more IIT-JEE News.

Mock Test IIT-JEE Paper 1 & Paper II with Solution

Mock Test AIEEE with Solution

Mock Test BIT SAT with Solution

Success Tips for the Month

• The difference between a successful person and others is not a lack of strength, not a lack of knowledge, but rather a lack of will.

• Footprints on the sands of time are not made by sitting down.

• To succeed, we must first believe that we can.

• The secret of joy in work is contained in one word - excellence. To know how to do something well is to enjoy it.

• Six essential qualities that are the key to success: Sincerity, personal integrity, humility, courtesy, wisdom, charity.

• Continuous efforts - not strength or intelligence - is the key to unlocking our potential.

• We can do anything we want to do if we stick to it long enough.

• The path to success is to take massive, determined action.

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 • Top medical honchos mull medical courses in IITs • Promise of IITs, from Mamata

IITian ON THE PATH OF SUCCESS 7 Mr. R. Madhavan

KNOW IIT-JEE 10 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 51 Mock Test IIT-JEE Paper-1 & Paper-2 Mock Test AIEEE Mock Test BIT SAT SOLUTIONS 90

Regulars ..........

DYNAMIC PHYSICS 17

8-Challenging Problems [Set# 11] Students’ Forum Physics Fundamentals Matter Waves, Photo-electric Effect Thermal Expansion, Thermodynamics CATALYST CHEMISTRY 32

Key Concept Purification of Organic Compounds Boron & Carbon Family Understanding: Inorganic Chemistry

DICEY MATHS 41

Mathematical Challenges Students’ Forum Key Concept Definite integrals & Area under curves Probability

Study Time........

Test Time ..........


Top medical honchos mull medical courses in IITs New Delhi: If the proposal of the Health secretary K Sujatha Rao comes into action, then the premier technology institutes of the country- the Indian Institutes of Technology (IITs) would soon see doctors passing out from the premier institute.

Rao, in order to gain the consensus of various top honchos from top ministries, directors of medical institutes and chairmen of medical councils, who have been invited in a meeting by the health secretary today, will determine whether allowing institutes like IITs to teach medicine will "help medical education or dilute its quality." Those called to attend include directors of All India Institute of Medical Sciences (AIIMS), PGI (Chandigarh), Sanjay Gandhi Post Graduate Institute (Lucknow), JIPMER, NIMHANS, National Institute of Communicable Diseases, National Institute of Paramedical Sciences and principal of CMC Vellore. This will be the first major meeting to discuss the issue.

Along with eminent doctors like Dr Ranjit Roychoudhury, Dr Devi Shetty, Dr Anupam Sibal and cardiologist Dr K Srinath Reddy, chairman of the Medical Council and the Nursing Council of India will also attend the meeting. The main pointers of the meeting would be to discuss major issues: if MBBS programme can be introduced at IITs, replacing the existing Medical Dental, Nursing and Pharma Councils of India to create the National Council for

Human Resource in Health (NCHRH) - as the overarching regulatory body and will address the medical personnel crunch in India.

An official from the health ministry said that, "We would like to know the experts views in order to understand the feasibility of the proposal, if starting medical courses in IITs will boost medical education in India. Once we have a consensus of all the experts we would send it to the Human Resource Development (HRD) Ministry for further consideration."

In about three years, some IITs like Kharagpur and Hyderabad are working upon to start medical schools in about three years. According to the Ministry officials, a Memorandum of Understanding (MoU) has been signed between IIT Kharagpur and University of California, San Diego, to set up a hospital, which will offer graduate, postgraduate, and research programmes in medicine and bio-medical engineering.

IIT Hyderabad has been expressing its interest to offer MD degrees in three years. HRD minister Kapil Sibal in a recent meeting with IIT directors, had asked them to expand their courses.

Promise of IITs, from Mamata Mamata Banerjee today promised a deal with the Union human resource development ministry to build “schools, colleges and IITs” on railway land.

The railway minister said she was in talks with the HRD ministry and a memorandum of understanding could be signed in a matter of days.

Railways to sign with IIT-Kharagpur for research Kolkata : Indian Railways will sign a Memorandum of Understanding (MoU) with the Indian Institute of Technology, Kharagpur (IIT-K) to promote research and development in the organisation, Railway Minister Mamata Banerjee said here on Wednesday. "We are going to sign a MoU with the IIT-K on February 13, 2010, to promote research and development activities for the railways," Banerjee said after inaugurating a computerised reservation counter in southern part of Kolkata. "We may float scholarships for IIT-K students to carry out research activities," she said. The minister announced two new trains for West Bengal. "Two new trains would be flagged off on February 13. One from Jhargram-Purulia and another Medinipur-Jhargram." She said the railways would also provide its land for setting up 372 diagnostic centres, 44 first-class hospitals and 88 second-class hospitals across the country.


"We'll provide land and the Union Health Ministry will develop the medical infrastructure on those lands. We have already signed a MoU with the Union Health Ministry to develop such health projects," Banerjee said. IANS

Indo-German Computer Centre inaugurated at IIT Delhi An Indo-German Max Planck Centre on Computer Science (IMPECS) was inaugurated at the Indian Institute of Technology, Delhi (IIT-D) here on Wednesday by Mr. Prithviraj Chavan, Minister of Science & Technology and Earth Sciences along with H.E. Mr. Horst Koehler, President of the Federal Republic of Germany. IMPECS will be engaged in collaborative basic research in Computer Science between Indian and German scientists and serve as a bridge between computer scientific communities from both sides. The Centre will act as a center of excellence for faculty and students from both sides. The research areas envisaged under the Indo-German Computer Centre would be Algorithms and Complexity, Database and Information Retrieval; Graphics and Vision and Networking. It would also engage researchers from institutes including Tata Institute of Fundamental Research, Mumbai; Indian Institute of Technology, Kanpur (IIT-K); Indian Institute of Technology, Bombay (IIT-B) and Indian Institute of Technology, Madras (IIT-M) - from the Indian side and Max Planck Institute for informatics (MP-INF), Saarbruecken from the German side. The Centre is expected to benefit both countries. The major benefits for India would be further strengthening of the research base in Computer Science that would develop expertise of highest caliber needed by academia and industry. Germany would benefit through improved collaboration with leading

Indian scientists and a highly visible outpost with highly professional environment and large pool of young talent. The Indo-German Centre on Computer Centre has been set up jointly by Department of Science & Technology (Govt. of India) and Max Planck Society of Germany under the over-all aegis of Indo-German S&T Cooperation programme for an initial duration of 5 years at a total cost of approximately Rs.12 crores from Indian side by DST and approximately 2 million Euros by Max Planck Society from German side. The concept of setting-up the Indo-German Centre on Computer Science in India was discussed during the visit of German Chancellor to India in October 2007. Dr. T. Ramasami, Secretary, Department of Science & Technology, Prof Peter Gruss, President Max Planck Society, Germany and Prof Surendra Prasad, Director, Indian Institute of Technology, Delhi were the guest of honour during the inauguration function.

'Spectacular years ahead in space' - former ISRO chief New Delhi: "The last 50 years of space have been fantastic while the next 50 years will be spectacular," remarked Prof. U R Rao, former chairman ISRO, delivering his lecture on "Challenges in Space" in the 97th Indian Science Congress. The space age began with the launch of Sputnik-I, 52 years ago from the former Soviet Union. Since then, plenty of satellites have been launched. The Cosmic Background Radiation Explorer (COBE) launched in 1989, confirmed the prediction made by the Big Bang theory. The Wilkinson Microwave Anisotropy Probe and recently launched Herschel and Plank have

contributed a lot to the study of the universe. Dr. Rao, enumerated nine great challenges in space as food security, energy security, environmental security, resource security, space security, space transportation, search for life, exploration of the universe and colonisation of Mars. "The per capita food productivity of India which is currently about 1.7 ton/ ha should be increased to about 4 tons/ha by 2050 to meet the growing food requirements. This can be done by initiating a new green revolution that requires the application of space technology along with biological inputs," he observed. "Space technology can be used for better meteorological forecasting which would help mitigate the consequences of disasters," he added. Dr. Rao stressed the importance of energy security for industrial expansion, agriculture and infrastructure growth. He explained with figures that per capita energy usage of India is far lower than other developed and developing nations like U.S (15 times more), EU (7.5 times more) and China (2.3 times more). "Space technology can play a significant role in coping with India's energy deficit by the better utilisation of energy resources as well as by learning the effects of global warming, carbon dioxide emission and so on," he added.

Three students bag Manmohan Singh scholarship at Cambridge New Delhi: Three students from Bangalore, Kolkata and Mumbai are set to receive the 2010 Manmohan Singh scholarship to fund their undergraduate studies at the University of Cambridge. For 2010, the scholarship will be given to Neal Duggal from Mallya Aditi International School, Bangalore, Jesika Haria from Dhirubhai Ambani International


School, Mumbai, and Rudrajit Banerjee from The Cambridge School, Kolkata. Neal Duggal and Jesika Haria have received conditional offers of places at St John's College and Emmanuel College, Cambridge, to study for degrees in Economics and Engineering respectively, while Rudrajit Banerjee has received an unconditional offer to study Natural Sciences at Christ's College, Cambridge, the university said in a statement on Tuesday. The Manmohan Singh Under-graduate Scholarship programme was established in 2009 in honour of India's Prime Minister who graduated from the Cambridge varsity with a first class in Economics in the late 1950s. Singh was also awarded an honorary doctorate by the university in 2006. The scholarship is awarded to students who have received an offer of a place at the University of Cambridge. Two of the three places offered by University of Cambridge are conditional on these students achieving specific grades. The scholarship programme will provide full funding, covering fees and maintenance for under-graduate study at the Cambridge. IANS

Srikanth Jagabathula, President of India gold medal winner. Internet connectivity in rural areas at cheap rates? Well, this could be a reality if Srikanth's dream comes true. Meet Srikanth Jagabathula, IIT's pride, the President of India gold medal winner for 2005-06 for scoring the highest marks among all batches at Indian Institute of Technology-Bombay. After an enviable stint at the IIT, Srikanth is all set to fly to the United States to pursue his studies at the prestigious Massachusetts Institute of Technology. After five years he plans to come back to Indian to start his own communications company.

As a kid he dreamt of becoming an engineer. Somehow he always thought an engineer's job would be very fascinating. He heard about the IIT when he was in the 7th standard. Since then IIT was his aim. After clearing his 10th class, he religiously worked towards cracking the IIT-Joint Entrance Examination. A rank of 38 at the IIT-JEE meant a smooth entry into IIT-Bombay. Srikanth, who hails from Hyderabad, was always a topper in school. Mathematics and physics were his favourite subjects, but he dreaded biology and chemistry.

World class university to be set up by Reliance To promote education and research in India, the Reliance Group plans to set up a ‘world-class’ university in India. The university, modeled on the lines of American universities such as The University of Pennsylvania will be set up either in Delhi or Mumbai. “It will be international in scale and in best practices, but with an Indian soul.

IITs successfully conduc-ted GATE online in two subjects New Delhi: Indian Institutes of Technology at Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras, Roorkee and the Indian Institute of Science Bangalore successfully conducted online Graduate Aptitude Test in Engineering (GATE) 2010 for two out of 21 papers yesterday. Examinations in two subjects, namely Textile Engineering and Fiber Science (TF), and Mining Engineering (MN) were conducted using computers by these institutes. About 1700 candidates were registered for these examinations which were conducted simultaneously in eight cities over two shifts. This experiment was conducted in GATE 2010 this year for the first time and depending upon the experience, online examination

might be repeated on a larger scale in subsequent years. The offline version of the exam in other 19 papers shall be conducted all over the country on Feb 14, 2010. Graduate Aptitude test in Engineering (GATE) is an all India examination administered and conducted jointly by Indian Institute of Science and seven Indian Institutes of Technology on behalf of the National Coordination Board - GATE, Department of Higher Education, Ministry of Human Resource Development (MHRD). Admission to postgraduate programmes with MHRD and some other government scholarship/ assistanceship in engineering colleges/institutes is open to those who qualify through GATE. GATE qualified candidates with Bachelor's degree in Engineering / Technology / Architecture or master's degree in any branch of Science / Mathematics / Statistics / Computer Applications are eligible for admission to master's / doctoral program in Engineering / Technology / Architecture as well as for doctoral programs in relevant branches of Science with MHRD or other government scholarship/assistantship.

Placements in full swing at IITs In just over a month, around 70% of the students at IIT have been placed. The final placements began on Dec 1, 2009 and will continue till March 2010. However, because of the good response from IT companies, the IITs hope that the placements might be wrapped early. However, last year due to economic slowdown, the IITs were able to place only 75-80 per cent of their student pool. Many students had to opt for higher studies or jobs in teaching. Barclays Bank made the highest offer of Rs 22 lakh at IIT-Kharagpur for placement at Singapore. IIT-Roorkee also achieved around.


R. Madhavan IITians who preferred to be different, rather than get into a corporate rat race. One of the most interesting themes at this year's Pan-IIT event was the session on rural transformation. IITians who have chosen an offbeat career hogged the limelight at the event. The star at the event was R Madhavan, an alumnus of IIT-Madras. This is Madhavan's success story as a farmer. . .

PASSION FOR AGRICULTURE

I had a passion for agriculture even when I was young. I don't know how my love for agriculture started. I only know that I have always been a nature lover. I used to have a garden even when I was a teenager. So, from a home garden, a kitchen garden, I gradually became a farmer! My mother used to be very happy with the vegetables I grew.

My family was against my ambition of becoming an agriculturist. So, I had to find a livelihood for myself. I wrote IIT-JEE and got selected to study at the Indian Institute of Technology, Madras. I enjoyed studying mechanical engineering. My intention was to transform what I study into what I love; mechanisation of farming. I felt the drudgery in farming is much more than in any other industry, and no one had looked into it.

I started my career at the Oil and Natural Gas Corporation (ONGC). My father refused to give me any money to start farming. So I asked the officials to let me work at the offshore sites, on the rigs. The advantage was that I could work on rigs for 14 days and then take 14 days off. I chose to work on the rigs for nine years, uninterrupted.

After 4 years, I saved enough money to buy six acres of land. I bought land at Chengelpet near Chennai. I chose that land because the plot had access to road and water. Back in 1989, a man in a pair of trousers aroused curiosity among the farming community. That was not the image of a farmer!

I became a full fledged farmer in 1993. It was tough in the beginning. Nobody taught me how to farm. There was no guidance from the gram sevaks or the University of Agriculture. I ran from pillar to post but couldn't find a single scientist who could help me. I burnt my fingers. My first crop was paddy and I produced 2 tonnes from the six acres of land, it was pathetic. When I lost all my money, my father said I was stupid. I told him, it didn't matter as I was learning. It was trial and error for me for three years. Until 1997, I was only experimenting by mingling various systems.

Mr. R. Madhavan Mechanical Engineering, IIT Madras

Success StoryThis article contains story of a person who get succeed after graduation from different IIT's


LEARNING FROM ISRAEL

In 1996, I visited Israel because I had heard that they are

the best in water technology. Take the case of corn: they

harvest 7 tonnes per acre whereas we produce less than a

tonne. They harvest up to 200 tonnes of tomatoes,

whereas here it is 6 tonnes, in similar area of land. I

stayed in one of the kibbutz, which is a co-operative farm

for 15 days. I understood what we do is quite primitive. It

was an eye opener for me. They treat each plant as an

industry. A plant producing one kilo of capsicum is an

industry that has 1 kilo output. I learnt from them that we

abuse water. Drip irrigation is not only for saving water

but it enhances your plant productivity. We commonly

practice flood irrigation where they just pump water. As

per the 2005 statistics, instead of 1 litre, we use 750 litres

of water.

DR. LAKSHMANAN

I met Dr Lakshmanan, a California-based NRI, who has

been farming for the last 35 years on 50-60,000 acres of

land. He taught me farming over the last one decade.

Whatever little I have learnt, it is thanks to him. I knew a

farm would give me much better returns in terms of

money as well as happiness. Working for money and

working for happiness are different. I work and get

happiness. What more do you need?

NO GUIDANCE IN INDIA

I said at one platform that we have to change the

curriculum of the agricultural universities. What they

teach the students is not how to farm, but how to draw

loans from a bank! What they learn cannot be

transformed to reality or to the villages. The problem in

the villages is not mentioned in the university. There is a

wide gap and it is getting worse.

MAKING PROFITS

After burning my fingers for four years, from 1997

onwards, I started making profits. Even though it took me

four years, I did not lose hope. I knew this was my path

ven though I didn't have any guidance from anyone.

In those days, communication was slow. Today, I can get

guidance from Dr Lakshmanan on Skype or Google Talk,

or through e-mail. I send him the picture of my problem

and ask his guidance. In those days, it took time to

communicate. There was no Internet or connectivity.

That was why it took me four years to learn farming.

Today, I would not have taken more than six months or

even less to learn the trick!

THE FARMING CYCLE

I started crop rotation after 1997. In August, I start with

paddy and it is harvested in December. I plant vegetables

in December itself and get the crops in February. After

that, it is oil seeds like sesame and groundnut, which are

drought-resistant, till May. During May, I go on trips to

learn more about the craft. I come back in June-July and

start preparations on the land to get ready for August. In

1999, I bought another four acres. My target is a net

income of Rs 100,000 per annum per acre. I have achieved

up to Rs 50,000.

I sell my produce on my own. I have a jeep and bring what

I produce to my house and sell from there. People know

that I sell at home. I don't go through any middle man. I

take paddy to the mill, hull it and sell it on my own. In the

future, I have plans to have a mill too. These days, people

tell me in advance that they need rice from me. I have no

problem selling my produce.

ENGINEERING HELPS IN FARMING

More than any other education, engineering helps in

farming because toiling in the soil is only 20 per cent of

the work. About 80 per cent of farming needs engineering

skills. Science is a must for any farming. I have developed a

number of simple, farmer-friendly tools for farming areas

like seeding, weeding, etc. as we don't have any tools for

small farmers. If I have 200 acres of land, I can go for food

processing, etc. My next project is to lease land from the

small farmers for agriculture. The village will prosper with

food processing industries coming there. My yield will also

be more with more land.

Don't compare yourself with anyone in this world. If you do so, you are insulting yourself. – Alen Strike


PHYSICS

1. A cart is moving along + x direction with a velocity of 4 m/s. A person on the cart throws a stone with a velocity of 6 m/s relative to himself. In the frame of reference of the cart the stone is thrown in y-z plane making an angle of 30º with vertical z axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from the branch of a tree by means of a string of length L. A completely inelastic collision occurs, in which the stone gets embedded in the object. Determine: [IIT-1997]

(i) The speed of the combined mass immediately after the collision with respect to an observer on the ground,

(ii) The length L of the string such that the tension in the string becomes zero when the string becomes horizontal during the subsequent motion of the combined mass.

Sol. θ = 30º, v = 6 m/s When the stone reaches the point Q, the component

of velocity in the +Z direction (V cos θ) becomes zero due to the gravitational force in the –Z direction.

Z

Q YVx=4m/s

V´

X

+Y

Vsinθ

Vx=4m t = 0 t = t

Vcosθ

+Z

P

V θ

L

The stone has two velocities at Q

(i) Vx in the +X direction (4 m/s) (ii) V sin θ in the + Y direction (6 sin 30º = 3 m/s) The resultant velocity of the stone V´ = 22

x )sinV()V( θ+

= 22 34 + = 5 m/s (i) Applying conservation of linear momentum at Q

for collision with an mass of equal magnitude m × 5 = 2m × V [Since the collision is completely inelastic the two

masses will stick together. V is the velocity of the two masses just after collision]

∴ V = 2.5 m/s (ii) When the string is undergoing circular motion, at

any arbitrary position

T – 2mg cos α = l

2mv2

Given that T = 0 when α = 90º

∴ 0 – 0 = l

2mv2 ⇒ v = 0

2mg cosα

∞ M

α

α 2mg sinα

2mg

Q

⇒ Velocity is zero when α = 90º, i.e., in the

horizontal position. Applying energy conservation from Q to M, we get

21 2mV2 = 2mgl ⇒ l =

g2V2

= 8.92

)5.2( 2

× = 0.318 m

2. Three particles A, B and C, each of mass m, are

connected to each other by three massless rigid rods to form a rigid, equilateral triangular body of side l. This body is placed on a horizontal frictioness table (x-y plane) and is hinged to it at the point A so that it can move without friction about the vertical axis through A (see figure). The body is set into rotational motion on the table about A with a constant angular velocity ω. [IIT-2002]

F →

x

l

ω

B C

A

y

(a) Find the magnitude of the horizontal force exerted

by the hinge on the body. (b) At time T, when the side BC is parallel to the

x-axis, a force F is applied on B along BC (as shown). Obtain the x-component and the y-component of the force exerted by the hinge on the body, immediately after time T.

KNOW IIT-JEE By Previous Exam Questions


Sol. The mass B is moving in a circular path centred at A. The centripetal force (mlω2) required for this circular motion is provided by F′. Therefore a force F′ acts on A (the hinge) which is equal to mlω2. The same is the case for mass C. Therefore the net force on the hinge is

Fnet = º60cos'F'F2'F'F 22 ++

Fnet = 21'F2'F2 22 ×+ = 3 F′ = 3 mlω2

X

l

l

B C

A

Y

60º

Fnet

F′ F′

l

F′ F′

(b) The force F acting on B will provide a torque to

the system. This torque is

F ×23l = Iα

F ×23l = (2ml2)α

⇒ α =lm

F43

The total force acting on the system along x-direction is

F + (Fnet)x This force is responsible for giving an acceleration ax

to the system.

l23

F

c.m

Therefore F + (Fnet)x = 3m(ax) c.m.

= 3mm4F Q ax = αr =

lmF

43 ×

3l =

4F

=4F3

∴ (Fnet)x = 4F

(Fnet)y remains the same as before = 3 mlω2.

3. A large open top container of negligible mass and uniform cross-sectional area A has a small holes of cross-sectional area A/100 in its die wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density ρ and mass m0. Assuming that the liquid starts flowing out horizontally through the hole at t = 0, Calculate : [IIT-2000]

(i) The acceleration of the container and (ii) The velocity when 75% of liquid drained out Sol. (i) Let at any instant of time during the flow, the

height of liquid in the container is x. The velocity of flow of liquid through small hole in the orifice by Torricelli's theorem is v = gx2 …(i) The mass of liquid flowing per second through the orifice = ρ × volume of liquid flowing per second

dtdm = ρ × gx2 ×

100A …(ii)

Therefore the rate of change of momentum of the system in forward direction

=dtdm × v =

100Agx2 ρ×× (From (i) and (ii))

The rate of change of momentum of the system in the backward direction = Force on backward direction = m × a where m is mass of liquid in the container at the instant t m = Vol. × density = A × x × ρ

P

v

x

∴ The rate of change of momentum of the system in the backward direction = Axρ × a By conservation of linear momentum

Axρ × a =100gxA2 ρ

⇒ a =50g

(ii) By toricell's theorem v′ = )h25.0(g2 × where h is the initial height of the liquid in the container m0 is the initial mass

∴ m0 = Ah × ρ ⇒ h =ρA

m0

∴ v′ = ρ

=ρ

××A2

gmAm

25.0g2 00


4. An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have equal cross-sectional area A. Atmospheric pressure is P0, and when the piston is in equilibrium, the volume of the gas is V0. The piston is now displaced slightly from its equilibrium position. Assuming that the system is completely isolated from its surroundings, show that the piston executes simple harmonic motion and find the frequency of oscillation. [IIT-1981]

Sol. Let the piston be displaced by a distance x.

Then γγ −

++=

+ )Axv(p

AMgpV

AMgp 0000

Q Initial pressure on the gas P1 = p0 + AMg

Final pressure on the gas P2 = p0 + AMg + p

V0 xA

P0

where p is the extra pressure due to which the

compression x takes place. Final volume of the gas V2 = V0 – Ax The above equation can be rearranged as

1 =γ

γ

+

−

++

00

00

VA

Mgp

)AxV(pA

Mgp=

γ

−

++

00

VAx1

AMgp

p1

⇒ 1 = 1 +

AMgp

p

0 +–

0VAxγ +

γ

+ 00

VAx

AMgp

p

Negligible as p and x are small

∴

AMgp

p

0 +=

0VAxγ

∴ p =0

0 VAx

AMgp γ

+

⇒ 0

0 VAx

AMgp

AF γ

+=

⇒ F =0

2

0 VxA

AMg

p γ

+

⇒ Ma =0

2

0 VxA

AMgp γ

+

⇒ a =MVxA

AMg

p0

2

0γ

+

Comparing it with a = ω2x we get

ω2 =MVxA

AMgp

0

2

0γ

+

∴ ω =MVxA

AMgp

0

2

0γ

+

If A

Mg is small as compared to p0 then

ω =MVAp

0

20γ

= 2πf

∴ f =π2

AMV

p

0

0γ

5. A hydrogen-like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between –0.85 eV and –0.544 eV (including both these values). [IIT-2002]

(a) Find the atomic number of the atom. (b) Calculate the smallest wavelength emitted in

these transitions. (Take hc = 1240 eV.nm. Ground state energy of hydrogen atom = – 13.6 eV)

Sol. (a) If x is the difference in quantum number of the states then x+1C2 = 6 ⇒ x = 3

n + 3

n

Smallest λ

Now, we have 2

2

n)eV6.13(z− = – 0.85 eV …(i)

and 2

2

)3n()eV6.13(z

+− = – 0.544 eV …(ii)

Solving (i) and (ii) we get n = 12 and z = 3 (b) Smallest wavelength λ is given by

λhc = (0.85 – 0.544)eV

Solving, we get λ ≈ 4052 nm.


CHEMISTRY

6. Compound A (C6H12O2) on reduction with LiAlH4 yields two compounds B and C. The compound B on oxidation gave D, which on treatment with aqueous alkali and subsequent heating furnished E. The latter on catalytic hydrogenation gave C. The compound D was oxidized further to give F which was found to be a monobasic acid (molar mass : 60.0 g mol–1). Deduce the structures of A, B, C, D and E.

[IIT-1990] Sol. The compound A is an ester. The equations involved

in the given reactions are as follows. The compound F is a monobasic acid (molar mass =

60 g mol–1). This may be represented as RCOOH. From the molar mass of F, it is evident that the molar mass of R is 15 g mol–1 [= (60 – 45) g mol–1]. Hence, the compound F is CH3COOH (ethanoic acid).

F is obtained by the oxidation of D. Hence, the compound D must be an aldehyde with the structure CH3CHO (ethanal). The compound D was obtained from the oxidation of B which must be an alcohol. Hence, the structure of B is CH3CH2OH (ethanol). D undergoes an aldol condensation (treatment with aqueous alkali) which subsequently gives E on heating. The reactions involved here are

2CH3CHO → KOH.aq

OH|

CHOCHCHCH 23

→heating CH3CH=CHCHO The reduction of E gives compound C. Hence, we

have

CH3CH=CHCHO → ]H[ )C(

OHCHCHCHCH 2223

Finally, the structure of A can be obtained from the two alcohols (CH3CH2OH and CH3CH2CH2CH2OH) produced on treating A with LiAlH4. Thus, we have

)A(

CHCOOCHCHCHCH 32223 → 4LiAlH

)C(

OHCHCHCHCH 2223 + )B(

OHCHCH 23

Thus, the reactions involved are as follows.

CH3CH2CH2COOCH2CH3 (A)

LiAlH4

CH3CH2OH(B)

CH3CH2CH2CH2OH(C)

[O]

CH3CHO → CH3COOH (D) (F)

aq KOH

CH3CHCH2OH

OHheating

CH3CH=CHCHO

[H]

Alternatively, the compound A may be

CH3COOCH2CH2CH2CH3

(butylacetate)

[O]

7. A metallic element crystallizes into a lattice

containing a sequence of layers of ABABAB ......... . Any packing of spheres leaves out voids in the

lattice. What percentage by volume of this lattice is empty space ? [IIT-1996]

Sol. A unit cell of hcp structure is a hexagonal cell, which is shown in figure (i) & (ii). Three such cells form one hcp unit.

For hexagonal cell, a = b ≠ c; α = β = 90º and γ = 120º. It has 8 atoms at the corners and one inside,

Number of atoms per unit cell = 88 + 1 = 2

Area of the base = b × ON [From fig.(ii)] = b × a sin 60º

= 23 a2 (Q b = a)

a γ b α

cβ

Figure (i) Volume of the hexagonal cell

= Area of the base × height = 23 a2.c

But c = 322 a


∴ Volume of the hexagonal cell

= 23 a2 .

322 a = a3 2

and radius of the atom, r = 2a

Hence, fraction of total volume or atomic packing factor

= cell hexagonal theof Volume

atoms 2 of Volume

60º

N b

O

a

figure (ii)

= 2a

r342

3

3π× =

2a2a

342

3

3

π×

= 23

π = 0.74 = 74%

∴ The percentage of void space = 100 – 74 = 26% 8. Using the data (all values are in kilocalories per mole

at 25ºC) given below, calculate the bond energy of C – C and C – H bonds.

0CombustionH∆ (ethane) = – 372.0

0CombustionH∆ (propane) = –530.0

∆Hº for C (graphite) → C(g) = 172.0 Bond energy of H – H = 104.0 0

fH∆ of H2O (l) = – 68.0

0fH∆ of CO2(g) = – 94.0 [IIT-1990]

Sol. Given that,

C2H6(g) + 27 O2(g) → 2CO2(g) + 3H2O(l)

∆H = – 372.0 kcal mol–1 ...(i) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ∆H = – 530.0 kcal mol–1 ...(ii) ∆H0 for C(gr) → C(g) = 172.0 kcal mol–1 Bond energy of H – H = 104.0 kcal mol–1 0

fH∆ of H2O(l) = –68.0 kcal mol–1

0fH∆ of CO2(g) = – 94.0 kcal mol–1

To find, 2C(g) + 6H(g) → C2H6(g) ∆H = ? ...(iii) 3C(g) + 8H(g) → C3H8(g) ∆H = ? ...(iv) and hence the bond energy of C – C and C – H bonds. We know that, 0

reactionH∆ = 0productH∆ – 0

reactant∆H From eq. (i), 0

reactant∆H = 2 × 0CO2

∆H + 3 × 0OH2

∆H – 0HC 62

∆H

or – 372.0 = 2 × (–94.0) + 3 × (–68.0) – 0HC 62

∆H

or 0HC 62

∆H = – 20 kcal mol–1 [for eq.(iii)] From eq. (ii) 0

reactant∆H = 3 × 0CO2

∆H + 4 × 0OH2

∆H – 0HC 83

∆H

or – 530.0 = 3 × (–94.0) + 4 × (–68.0) – 0HC 83

∆H

or 0HC 83

∆H = – 24 kcal mol–1 [for eq.(iv)]

0reactant∆H = Sum of bond energies of reactants

– Sum of bond energies of products From eq.(iii),

0reactant∆H = 2 × [C(s) → C(g)] + 6 ×

→ HH

21

2

– [1 × C – C + 6 × C – H] or – 20 = 2 × 172.0 + 3 × 104.0 – [1 × C – C + 6 × C – H] or [1 × C – C + 6 × C – H] = 676 kcal mol–1 ...(v) From eq. (iv),

0reactant∆H = 3 × [C(s) → C(g)] + 8 ×

→ HH

21

2

– [2 × C – C + 8 × C – H] or – 24 = 3 × 172.0 + 4 × 104.0 – [2 × C – C + 8 × C – H] or [2 × C – C + 8 × C – H] = 956 kcal mol–1 ...(vi) Solving eq. (v) and eq. (vi), we get Bond energy of C – C bond = 82 kcal mol–1 Bond energy of C – H bond = 99 kcal mol–1

9. The electrode reactions for charging of a lead storage battery are

PbSO4 + 2e– → Pb + SO42–

PbSO4 + 2H2O → PbO2 + SO42– + 4H+ + 2e–

The electrolyte in the battery is an aqueous solution of sulphuric acid. Before charging the specific gravity of the liquid was found to be 1.11 (15.7% H2SO4 by weight). After charging for 100 hours, the specific gravity of the liquid was found to be 1.28 (36.9% H2SO4 by weight). If the battery contained two litres of the liquid, calculate the average current used for charging the battery. Assume that the volume of the battery liquid remained constant during charging. [IIT-1972]

Sol. Volume of 100 g of 15.7% H2SO4 = 11.1

100 = 90.9 ml

90.9 ml of 15.7% H2SO4 contains = 15.7 g H2SO4 ∴ 2.0 L of 15.7% H2SO4 contains

= 9.90

0.210007.15 ××

= 345.43 g H2SO4

Volume of 100 g of 36.9% H2SO4 = 28.1

100 = 78.125 ml

78.125 ml of 36.9% H2SO4 contains = 36.9 g H2SO4 ∴ 2.0 L of 36.9% H2SO4 contains

=125.78

0.210009.36 ×× = 944.64 g H2SO4


Amount of H2SO4 formed after charging = 944.64 – 345.43 = 599.21 g The overall reaction is Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O 98 g H2SO4 (2 × 1 + 32 + 4 × 16) required = 1 F electricity

∴ 599.21 g H2SO4 requires = 98

21.599 F electricity

= 98

21.599 × 96500 C electricity

We know, Q = It Given that, t = 100 hrs = 100 × 3600 s

∴ I = tQ =

3600100989650021.599××

× A = 1.638 A

∴ Current used for charging battery = 1.638 ampere

10. Given the following standard free energies of formation at 25ºC : CO(g) = – 32.807,

CO2(g) = –94.260, H2O(g) = – 54.635, H2O (l) = – 56.69 kcal mol–1. (a) Find ∆Gº and the equilibrium constant Kp for the

reaction CO(g) + H2O(g) CO2(g) + H2(g) at 25ºC. (b) Find the vapour pressure of H2O at 25ºC. (c) If CO, CO2 and H2 are mixed so that the partial

pressure of each is 1.00 atm and the mixture is brought into contact with excess liquid H2O, what will be the partial pressure of each gas when equilibrium is attained at 25ºC. The volume available to the gases is constant. [IIT-1973]

Sol. Given that, 0

fG∆ (CO) = – 32.807 kcal mol–1

0fG∆ (CO2) = –94.260 kcal mol–1

0fG∆ [H2O (g)] = – 54.635 kcal mol–1

0fG∆ [H2O(l) = – 56.69 kcal mol–1

(a) For the reaction, CO(g) + H2O(g) CO2(g) + H2(g) 0

reactionG∆ = 0fG∆ (CO2) – 0

fG∆ [H2O(g)] – 0fG∆ (CO)

= –94.260 – (– 54.635) – (–32.807) = – 6.818 kcal = – 6818 cal Also, ∆G0 = – 2.303 RT log

1pK

∴ –6818 = –2.303 × 1.987 × 298 × log1pK

or log1pK = 5.00

or 1pK = Antilog 5.00 = 1.00 × 105 atm

(b) For the reaction, H2O(l) H2O(g)

0reactionG∆ = 0

fG∆ [H2O(g) ] – 0fG∆ [H2O(l)]

= – 54.635 – (– 56.69) = 2.055 kcal = 2055 cal Also, ∆G0 = –2.303 RT log

2pK

∴ 2055 = –2.303 × 1.987 × 298 log 2pK

or log2pK = – 1.507 = 493.2

or 2pK = Antilog 493.2 = 3.122 × 10–2

OH2P =

2pK = 3.122 × 10–2 atm

(c) CO(g) + H2O(l) CO2(g) + H2(g) Initial Pressure 1 1 1 Pressure at Equlibrium 1 – x 1 + x 1 + x

Hence, 3pK =

1pK . 2pK =

CO

HCO

PP.P

22

or 1.00 × 105 × 3.122 × 10–2 = x1

)x1)(x1(−

++

or x2 + 3124x – 3121 = 0

or x = a2

ac4bb 2 −±−

=12

)3121(14[)3124(3124 2

×−××−±−

= 0.9987

∴ PCO = 1 – x = 1 – 0.9987 = 1.3 × 10–3 atm

2HP = 2COP = 1 + x = 1 + 0.9987 = 1.9987 atm

MATHEMATICS

11. Complex numbers z1, z2, z3 are the vertices A, B, C respectively of an isosceles right angled triangle with right angle at C. Show that

(z1 – z2)2 = 2(z1 – z3) (z3 – z2) [IIT-1986] Sol. Since, ∆is right angled isosceles ∆. ∴ Rotating z2 about z3 in anticlock wise direction

through an angle of π/2, we get

C(z2) B(z3)

A(z1)

31

32

zzzz

−−

= |zz||zz|

31

32

−−

eiπ/2

where, |z2 – z3| = |z1 – z3| ⇒ (z2 – z3) = i(z1 – z3) squarring both sides we get, (z2 – z3)2 = – (z1 – z3)2 ⇒ z2

2 + z32 – 2z2z3 = –z1

2 – z32 + 2z1z3

⇒ z12 + z2

2 – 2z1z2 = 2z1z3 + 2z2z3 – 2z32 – 2z1z2

⇒ (z1 – z2)2 = 2(z1z3 – z32) + (z2z3 – z1z2)

⇒ (z1 – z2)2 = 2(z1 – z3) (z3 – z2) 12. The fourth power of the common difference of arithmetic

progression with integer entries is added to the product of any four consecutive terms of it. Prove that resulting sum is the square of an integer. [IIT-2000]


Sol. Let four consecutive terms of the A.P. are a – 3d, a – d, a + d, a + 3d which are integers. Again

P = (a – 3d) (a – d) (a + d) (a + 3d) + (2d)4 (by given condition) = (a2 – 9d2)(a2 – d2) + 16d4 = a4 – 10a2d2 + 9d4 + 16d4 = (a2 – 5d2)2 Now, a2 – 5d2 = a2 – 9d2 + 4d2 = (a – 3d)(a + 3d) + (2d)2 = I.I + I2 (given) = I2 + I2 = I2 = I (where I is any integer) Therefore, P = (I)2 = Integer

13. Evaluate : ∫ xsin)x2(cos 2/1

dx [IIT-1987]

Sol. I = ∫ xsin)x2(cos 2/1

dx

= ∫−xsin

xsinxcos 22dx = ∫ −1xcot2 dx

putting, cot x = sec θ ⇒ –cosec2x dx = sec θ tan θ dθ. We get,

I = ∫ −1xsec2 .)sec1(

tan.sec2 θ+−θθ dθ

= – ∫ θ+θθ

2

2

sec1tan.sec dθ

= – ∫ θ+θθ

3

2

coscossin dθ

= – ∫ θ+θθ−

)cos1(coscos1

2

2dθ

= – ∫ θ+θθ−θ+

)cos1(coscos2)cos1(2

22

= – ∫ θsec dθ + 2 ∫ θ+θ2cos1

cos dθ

= –log |sec θ + tan θ| + 2 ∫ θ−θ2sin2

cos dθ

= – log |sec θ + tan θ| + ∫ − 2t2dt , where sin θ = t

= – log |sec θ + tan θ| + 2.22

1 logθ−

θ+

sin2sin2 +c

= – log |cot x + 1xcot2 − |

+ xtan12

xtan12log2

12

2

−−

−+ + c

14. Show that

∫π 2/

0xdxsin)x2(sinf = ∫

π 4/

0dxxcos)x2(cosf2

[IIT-1990]

Sol. Let, I = ∫π 2/

0xdxsin)x2(sinf ...(1)

Then, I = ∫π 2/

0

−

π x2

2sinf .sin

−

π x2

dx

= ∫π 2/

0fsin 2x . cos x dx ...(2)

adding (1) and (2), we get

2I = ∫π 2/

0f(sin 2x) . (sin x + cos x) dx

= 2 ∫π 4/

0f(sin 2x) . (sin x + cos x) dx

= 22 ∫π 4/

0f(sin 2x) sin

π

+4

x dx

= 22 ∫π 4/

0f

−

π x4

2sin . sin

π

+−π

4x

4dx

= 22 ∫π 4/

0f(cos 2x) . cos x dx

∴ I = 2 ∫π 4/


Hence, ∫π 2/

0f(sin(2x)) . sin x dx

= 2 ∫π 4/


15. Find the area bounded by the curves, x2 + y2 = 25, 4y = |4 – x2| and x = 0 above the x-axis. [IIT-1987] Sol. Here, x2 + y2 = 25, 4y = |4 – x2| could be sketched as,

whose point of intersection could be obtained by y

5

5 4 2 O

–2–4

–5

x

x2 + 16

)x4( 2− = 25

⇒ (x2 + 24) (x2 – 16) = 0 ⇒ x = ± 4 ∴ Required area

= 2

−−

−−− ∫ ∫∫ dx

44xdx

4x4dxx25

2

0

4

2

224

0

2

= 2

+− −

4

0

12

5xsin

225x25

2x

–

−−

−

4

2

32

0

3x4

3x

41

3xx4

41 = 8+

π

425


Passage # 1 (Q. No. 1 to 4)

For the given phasor diagrams of AC circuit power factor seems to be same i.e. cos φ but the nature of the circuit is entirely different so to distinguish we use the following codes

V

I φ V

I φ

Phasor Diagram -1 Phasor Diagram -2 For circuit-1 Power factor cosφ (lagging)–Inductive nature For circuit-2 Power factor cosφ (leading)–Capacitive nature For a circuit shown in figure when V = 200 volt and f = 50 Hz then the voltmeter reading is zero.

V/f

R = 10Ω

V ~

L

C

When source frequency get varied then the power

factor becomes2

1 (lagging) and freq. f2 and2

1

(leading) at frequency f1 then.

1. Which frequency relation(s) is/are correct - (A) f > f1 > f2 (B) f < f1 < f2 (C) f1 < f < f2 (D) f2 < f < f1

2. Value of ∆f = f2 – f1 is -

(A) LR.

21π

(B) L2

R.21π

(C) LR2.

21π

(D) RR4.

21π

3. Value of watt less current when frequency is f1 (A) .Amp25 (B) .Amp210 (C) 10 Amp. (D) Zero Amp.

4. Value of charge on capacitor at frequency f (A) (5π)–1 cb (B) 10π cb (C) 10 cb (D) can not be calculated as the value of C is not known Passage # 2 (Q. No. 5 to 7)

A circular platform, 5.0m in radius, has pair of 600Hz sirens, mounted on posts at opposite ends of a diameter. The platform rotates with an angular velocity of 0.80 rad/s. A stationary listener is located at a large distance from the platform. The speed of sound is 350 m/ sec.

5. In situation the longest wavelength reaching the listener from the sirens, in cm, is closest to -

(A) 58.3 (B) 59.6 (C) 59.0 (D) 57.7 6. In situation, the highest siren frequency heard by the

listener in S.I. units, is closest to - (A) 605 (B) 607 (C) 611 (D) 609 7. In situation, the listener mounts on a bicycle and rides

directly away from the platform with a speed of 4.5 m/sec. The highest siren frequency heard by the listener, in SI units, is given by -

(A) 599 Hz (B) 585.6 Hz (C) 607 Hz (D) 614 Hz 8. Figure shows a pulse travelling in the x-direction in a

string stretched along x-axis. Then

distance (in meter)

y

x 1 2 3

(A) Acceleration of particle at x = 1 m is in +ve y-

direction (B) Velocity of particle at x = 1m is in-ve y-direction (C) Velocity of particle at x =1 m is zero (D) None of these

This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in physics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Dev Sharma Director Academics, Jodhpur Branch

Physics Challenging Problems

Solut ions wil l be published in next issue

Set #11


1. [D] As the magnetic field and area vector of the coil are in same direction so the magnetic flux passing through the coil is

0cosBs.S.B ==φ→→

)1)(r(B 2π=φ

Induced emf. atdNe φ

− n = 1 single turn coil

)rB(dtd)(

dtd

dtde 2π−=φ−=φ

−=

dtdBre 2π−= As B = B0 + B1 t2

[ ]t.B2.re 2π−= So tB20dtdB

1+=

e = – πr2 2(–2)t = 4π2.t e = 4πr2t ... (i)

Induced current

Rt.r4

Rei

2π==

as

π=

Rr

i = 4t ... (ii)

Induced electric field r2

t.r4r2

e.dist

eE2

ππ

=π

==

E = 2rt ... (iii)

Option (D) is correct, as induced electric field varies linearly with time for given value of radius.

2. [B] Induced current i = 4t So at t = 0 i = 0 3. [C] RMS value of current

)i......(dti

21dti

T1I

2

0

2T

0

2RMS ∫∫ ==

As )3/8(16038.16

3t16dt)t4(dti

2

0

32

0

22

0

2 =

−=

== ∫∫

.Amp

38

38.16.

21I

RMS==

4. [A] Induced charge 2

0

22

0

2

02t4tdt4idtq

=== ∫∫

coulomb80

24.4 =

−=

As 96500 C = 1 Faraday

So Faraday896500

1C8 ×= As x96500

1=

So induced charge = 8x faraday

5. [A] The forces working on metallic rod Fm = i L.B. Net force working on rod Fnet = Mg – i. LB

= LB.R

vBLMg −

=

RLvBMg

22−

a

Fm = i L.B.

b X

i

Fg = Mg

Fnet = R

LvBMgMa22

−=

a = acceleration of metallic rod ab

or acceleration = MR

LvBga22

−=

or acceleration kVgMR

LBvgdtdv 22

−=−

−=

Here

MRLBk

22=

dtkvg

dvkvgdtdv

=−

⇒−=

Integrating on both sides

∫ ∫=−

v

0

t

0

dtkvg

dv

tglnk1)kvgln(

k1t)kvgln(

k1 v

0=

−−−−⇒=−−

⇒

ktegkvgt

gkvgln

k1 −=

−⇒=

−−

Solution Physics Challenging Problems

Set # 10

8 Questions were Published in February Issue


⇒

ktkt e1gvke

gvk1 −− −=⇒=−

and v =

kg (1 – e–kt)

acceleration [ ] ktkt ekg)k(e

kg

dtdva −− =−−==

acceleration a = ge–kt

)e1(kgv kt−−= where

MRLBk

22=

6. [A] Acceleration a = ge–kt

When t = 1/k g37.e.ge.ga 1k1.k

=== −−

Acceleration a = 37 (maximum acceleration)

So time after which acceleration is 37% of maximum

acceleration is t = 1/k = MR/B2L2

7. [A] Velocity )e1(kgv kt−−= for t = 1/k

)velocityimum(max63.)e1(kgv 1 =−= −

Time after which velocity is 63% of maximum velocity is t = 1/k = MR / B2L2

8. [A] Area traversed by rod = x.L

x distance travelled down and in time t

So, Areal velocity = dtdA

= v.Ldtdx.L)xL(

dtd)A(

dtd

===

)e1(kg.L kt−−=

Areal velo.=

kgL (1 – e–kt)

Why can’t the Sun melt Snow?

There are some things in nature that have a great capacity to toss back or reflect a great deal of the sun’s light that falls on them. One of them is snow. Newly formed snow reflects about 90 per cent of the sunlight that falls upon it. This means that the sun is powerless to melt clean snow. And when snow does melt, it is not because of the sunlight. Snow does not melt on a spring day because of the sun’s heat. It melts because of the warm air from the sea.

After snow becomes ice, a different problem arises. Clean ice absorbs about two-thirds of the sunlight that hits it - but ice is transparent enough for the light to penetrate quite a long way (10 metres or more) before the absorption takes place.

It is remarkable what profound results follow from this simple property of transparency to sunlight. If, instead of penetrating deeply, the light were absorbed in a shallow surface layer of ice, the summer sun would quickly raise the temperature of the thin surface layer to melting point. And almost immediately, the water would run off. But when the sunlight penetrates a thick layer of ice before it can be absorbed, it cannot raise the temperature of the ice to melting point quickly enough. When the ice is very cold, the whole summer passes before any melting occurs at all. This is what happens today in the Antarctic, just as it must have happened in northern Europe during an Ice Age.

Just imagine, if by magic, ice were suddenly made opaque to light, the glaciers that exist today would melt away in a few years, raising the sea level by 60 metres or more. It would flood at least half the world’s population. Simply amazing how so much depends on so simple a physical property! Clouds toss back about 50 per cent of the light that hits them. Ice and deserts reflect 35 per cent. Land areas are generally a good deal lower in reflectivity - usually 10 to 20 per cent, depending on the nature of vegetation.

Oceans, which cover 71 per cent of the Earth’s surface, are least reflective of all - only about three per cent. That is why oceans appear dark in pictures of the Earth taken from the Moon or from artificial satellites. When all the sources of reflection are added together, our planet is found to turn back into space some 36 per cent of the solar radiation falling upon it.


1. Two circular rings A and B, each of radius a = 30 cm, are placed coaxially with their axes vertical s shown in Fig. Distance between centres of these rings is h = 40 cm. Lower ring A has a positive charge of 10 µC, while upper ring B has a negative charge of 20 µC. A particle of mass m = 100gm carrying a positive charge of q = 10 µC is released from rest at the centre of the ring A.

(i) Calculate initial acceleration of the particle.

(ii) Calculate velocity of particle when it reaches at the centre of upper ring B.

(g = 10 ms–2)

a

A

B

a

h

Sol. Since, Initially particle was at centre of lower ring A,

therefore, no force acts on the particle due to charge of this ring. At initial moment, two forces act on the particle :

(i) its weight mg =0.1 × 10 = 1 Newton (downwards)

(ii) force F exerted by the charge on ring B.

This force, F = q)ha(

hq4

12/322

2

0 +πε

where q2 = – 20 µC (charge on ring B)

∴ F = 5.76 Newton (upwards)

Let acceleration of the particle be a.

F

mg

ma

Considering its free body diagram (Fig.),

ma = F – mg

or a = 47.6 ms–2 Ans.(i)

When particle is released, it starts moving upwards. During its motion, work is done by electric forces acting on it. That work is used in two ways –

(i) to increase its gravitational potential energy by mgh and

(ii) to provide kinetic energy 21 mv2 to the particle.

But work done by electric field is W = q (V1 – V2)

where V1 is potential at centre of ring A

and V2 is potential at centre of ring B.

V1 = aq

41 1

0πε+

222

0 ha

q4

1

+πε= – 6 × 104 volt

V2 = .4

1

0πε 221

ha

q

++ .

41

0πε aq2 = – 42 × 104 volt

∴ W = q (V1 – V2) = 3.6 joule

But W = mgh + 21 mv2

∴ v = 8 ms–1. Ans. (ii)

2. In the circuit shown in Fig., C1 = 5 µF, C2 = 2.9 µF, C3 = 6 µF, C4 = 3 µF and C5 = 7 µF.

If in steady state potential difference between points A and B is 11 volt, calculate potential difference across C5.

A + –

C3

C4

C1 C2

C5

B

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' ForumPHYSICS


Sol. In the given circuit capacitors C4 and C5 are in series with each other while capacitor C2 is in parallel with this series combination. Then capacitors C3, C1 and above said combination are in series with each other.

When steady state is reached, no current flows through the circuit. To analyse the given circuit, it may be assumed that a charge q1 leaves the battery from its positive terminal and flows from A to B. This charge first reaches lower plate of capacitor C3. Hence, this plate becomes positively charged and upper plate negatively charged. Now charge q1 reaches the capacitor C1. Its left plate becomes positively charged and right plate negatively charged. Now charge q1 reaches the junction B. From where it gets divided into two parts. Let a charge q2 flow through series combination of capacitors C4 and C5. Then a charge (q1 – q2) flows through capacitor C2.

Hence, in steady state charges on different capacitors will be as shown in Fig.

+ –

C3

C4

C1 C2

C5

B C

D

A

E

q1 –

+ E

q1

+ –

(q1 – q2)+ –

q2+–

+ –

q2

Since terminal A is connected with positive terminal

of the battery, therefore, potential of A is higher than that of B.

It is given that potential difference (VA – VB) is equal to 11 volt,

∴ +3

1

Cq +

1

1

Cq = 11 or q1 =

31

31

CCCC11

+= 30 µC

Now applying Kirchhoff's voltage law on mesh BCDEB,

∴ +2

21

C)qq( − –

5

2

Cq –

4

2

Cq = 0 or q2 = 12.6 µC

Potential difference across C5 =5

2

Cq = 1.8 volt Ans.

3. System shown in Fig. consists of two large parallel metallic plates carrying current in opposite directions.

× × ×

Current density in each plate is j per unit width. Calculate

(i) magnetic induction in space between the plates and

(ii) force acting per unit area of each plate.

Sol. If a large plate carries a current which is uniformly distributed over its width, then a uniform magnetic field is established around it.

If a section of plate, which is normal to the direction of flow of current, is considered then it will be as shown in Fig.

× × ×

P

×

Q

S R B

B

l Fig. (1)

Let magnetic induction of the field induced due to current in one plate be B.

Considering a length l in the section as shown in Fig.(1) and applying Amperes's Circuital Law,

B. 2l = µ0 (lj) or B =21 µ0j

But there are two plates which carry equal current but in opposite directions. Therefore, magnetic fields due to these currents, in the space between the plates are unidirectional.

∴ Resultant magnetic field induction between the plates = 2B = µ0j Ans. (i)

Now consider an elemental width dx in the section of upper plate as shown in Fig.(2). This elemental width is similar to a long straight conductor carrying current di = jdx

× × ×

dx

Fig. (2)

Magnetic induction at this conductor due to current in

lower plate is B =21 µ0j (leftward)


Hence, force on this conductor, dF = B di per unit length

or dF =21 µ0j2 dx

per unit length

But area of unit length of the conductor considered = 1. dx = dx

∴ Force per unit area of upper plate =dxdF

=21 µ0j2 Ans. (ii)

4. Tyre of a bicycle has volume 2000 cm3. Initially, the

tube is filled to 0.75 of its volume by air at atmospheric pressure of P0 = 105 Nm–2. Area of contact of tyre with road is A = 24 cm2 when total mass of bicycle and its rider is m = 120 kg. Calculate the number of strokes of pump, which delivers v = 500 cm3 volume of air in each stroke, required to inflate the tyre.

Asume that area of contact of tyre with road remains unchanged. (g = 10 ms–2)

Sol. Atmospheric pressure, P0 = 105 Nm–2

Increase in pressure (due to weight of bicycle and its rider)

=A

mg =)1024(

101204−×

× Nm–2 = 5 × 105 Nm–2

∴ Final pressure of air in the tube,

P2 = 105 + (5 × 105) = 6 × 105 Nm–2

Final volume of air in the tube,

V2 = 2000 cm3 = 2 × 10–3 m3

Let temperature be T.

Finally, number of moles of air in the tube

n =RT

VP 22 =RT

)102()106( 35 −×××

Volume of these moles at atmospheric pressure,

V1 =0P

nRT = 5

35

10)102()106( −××× m3

= 12 × 10–3 m3

Initially, volume of air in tube (at atmospheric pressure) is V0 = 0.75 × 2000 cm3

= 1.5 × 10–3 m3

∴ Volume (at atmospheric pressure) of air to be pumped in is

∆V = V1 – V0 = 10.5 × 10–3 m3

Volume (at atmospheric pressure) of air pumped in each stroke is

v = 500 cm3 = 0.5 × 10–3 m3

∴ Number of strokes required

= 3

3

105.0105.10

×× = 21 Ans.

5. A ray of light travelling in air is incident at angle θ = 30º on a long rectangular slab of a transparent medium. The point of incidence is origin O of the co-ordinate system shown in Fig.

O

Airθ

x

y

The medium has a variable index of refraction µ (y)

given by µ = (0.25 + ky–2)1/2 where k = 1.0 m2. Calculate equation for trajectory of the ray in the medium.

Sol. If there are two transparent slabs having coefficients of refraction µ1 and µ2 and a ray is incident from air on first slab at angle i then it first refracts at interface of air and first slab as shown in fig.(1) and then at interface of two slabs. Let these angles of refraction be θ and r respectively.

i

(µ1) (µ2)

A θθ

r

Fig. (1)

Applying Snell's law at point A,

θsinisin = µ1 or sin θ =

1µisin … (i)


Refractive index of second slab with respect to first

slab = 1

2

µµ .

Now applying Snell's law at point B,

rsin

sin θ =1

2

µµ

Substituting value of sin θ from equation (1),

rsinisin = µ2.

This relation shows that if there are several refracting surfaces parallel to each other then Snell's law can be applied at two points also. In that case i is angle of incidence at one point, r is angle of refraction at the other point and µ is refractive index of that medium in which angle r is measured with respect to that medium in which angle i is measured.

Since, refractive index of given medium varies with y, therefore, it may be assumed that the given slab is composed of a large number of thin slabs having different refractive indices and refracting surfaces of all the slabs are normal to y-axis. Hence, angle of incidence and that of refraction are to be measured with y-axis.

O

Air θ

x

y

P α

(90

– α

)

Fig. (2)

Now consider a point P on trajectory of the refracted ray in medium as shown in fig. (2). Let inclination of tangent to the ray at this point with x-axis be α. Then angle of refraction is (90 – α).

Applying Snell's law at points P and O,

)90sin(

sinα−

θ =1µ

Substituting values of θ and µ,

αcos2

1 =2/1

2yk

41

+

or sec2 α = 4

+ 2y

k41

∴ 1 + tan2 α = 1 + 2yk4

But k = 1, therefore, tan α =y2

But tan α =dxdy

∴ y dy = 2 dx

Point O (x = 0, y = 0) and point P (x, y) lie on the trajectory of the ray. Hence, integrating above equation with these limits,

∫y

0dyy = ∫

x

0dx2

or y2 = 4x Ans.

SCIENCE TIPS

• Why is the cooling inside a refrigerator not proper when a thick layer of ice deposits on the freezer?

Because ice is a bad conductor of heat • Which type of computer is often found in small

business and in homes and classrooms ? The micro computer. It is the smallest and

the least costly type of computer • Out of joule, calorie, kilowatt and electron-volt

which one is not the unit of energy ? kilowatt • How does the atmospheric pressure vary with

height ? Atmospheric pressure P decreases with height

h above sea level. For an 'ideal' atmosphere at constant temperature P = P0 e–kh where k is a constant and P0 is the pressure at the surface

• How is r.m.s. velocity of gas molecules related to absolute temperature of the gas ?

vrms ∝ T • What are transducers ?

Devices which change signals from one form to another (e.g. sound to electrical) are called transducers

• Is polarization the property of all types of waves? No, it is property of only transverse waves


Matter Waves : Planck's quantum theory : Wave-particle duality - Planck gave quantum theory while explaining the

radiation spectrum of a black body. According to Planck's theory, energy is always exchanged in integral multiples of a quanta of light or photon.

Each photon has an energy E that depends only on the frequency ν of electromagnetic radiation and is given by :

E = hν .....(1) where h = 6.6 × 10–34 joule-sec, is Planck's

constant. In any interaction, the photon either gives up all of its energy or none of it.

From Einstein's mass-energy equivalence principle, we have

E = mc2 .....(2) Using equations (1) and (2), we get ;

mc2 = hν or m = 2chν .....(3)

where m represents the mass of a photon in motion. The velocity v of a photon is equal to that of light, i.e., v = c.

According to theory of relativity, the rest mass m0 of a photon is given by :

m0 = 2

2

cv1m −

Here, m = 2chν and v = c

Hence, m0 = 0 ....(4) i.e., rest mass of photon is zero, i.e., energy of

photon is totally kinetic. The momentum p of each photon is given by :

p = mc = 2chν × c =

chν =

ν/ch =

λh ......(5)

The left hand side of the above equation involves the particle aspect of photons (momentum) while the right hand side involves the wave aspect (wavelength) and the Planck's constant is the bridge between the two sides. This shows that electromagnetic radiation exhibits a wave-particle duality. In certain circumstances, it

behaves like a wave, while in other circumstances it behaves like a particle.

The wave-particle is not the sole monopoly of e.m. waves. Even a material particle in motion according to de Broglie will have a wavelength. The de Broglie wavelength λ of the matter waves is also given by :

λ = mvh =

ph =

mK2h

where K is the kinetic energy of the particle. If a particle of mass m kg and charge q coulomb

is accelerated from rest through a potential difference of V volt. Then

21 mv2 = qV or mv = mqV2

Hence, λ = mqV2h =

V34.12 Å

Photoelectric effect : When light of suitable frequency (electromagnetic

radiation) is allowed to fall on a metal surface, electrons are emitted from the surface. These electrons are known as photoelectrons and the effect is known as photoelectric effect. Photoelectric effect, light energy is converted into electrical energy.

Laws of photolectric effect : The kinetic energy of the emitted electron is

independent of intensity of incident radiation. But the photoelectric current increases with the increase of intensity of incident radiation.

The kinetic energy of the emitted electron depends on the frequency of the incident radiation. It increases with the increase of frequency of incident radiation.

If the frequency of the incident radiation is less than a certain value, then photoelectric emission is not possible. This frequency is known as threshold frequency. This threshold frequency varies from emitter to emitter, i.e., depends on the material.

There is no time lag between the arrival of light and the emission of photoelectrons, i.e., it is an instantaneous phenomenon.

Matter Waves, Photo-electric Effect

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Failure of wave theory : Wave theory of light could not explain the laws of

photoelectric effect. According to wave theory, the kinetic energy of

the emitted electrons should increase with the increase of intensity of incident radiation.

Kinetic energy of the emitted electron does not depend on the frequency of incident radiation according to wave theory.

Wave theory failed to explain the existence of threshold frequency.

According to wave theory there must be a time lag between the arrival of light and emission of photoelectrons.

Einstein's theory of photoelectric effect : Einstein explained the laws of photoelectric effect

on the basis of Planck's quantum theory of radiation.

Einstein treated photoelectric effect as a collision between a photon and an atom in which photon is absorbed by the atom and an electron is emitted.

According to law of conservation of energy,

hν = hν0 + 21 mv2

where hν is the energy of the incident photon; hv0 is the minimum energy required to detach the electron from the atom (work function or ionisation energy) and (1/2) mv2 is the kinetic energy of the emitted electron.

The above equation is known as Einstein's photoelectric equation. Kinetic energy of the emitted electron,

= 21 mv2 = h(ν – ν0) = hν – W

Explanation of laws of photoelectric effect : (a) The KE of the emitted electron increases with the

increase of frequency of incident radiation since W (work function) is constant for a given emitter. KE is directly proportional to (ν – ν0)

(b) Keeping the frequency of incident radiation constant if the intensity of incident light is increased, more photons collide with more atoms and more photoelectrons are emitted. The KE of the emitted electron remains constant since the same photon collides with the same atom (i.e., the nature of the collision does not change). With the increase in the intensity of incident light photoelectric current increases.

(c) According to Einstein's equation, if the frequency of incident radiation is less than certain minimum value, the photoelectric emission is not possible. This frequency is known as threshold frequency. Hence, the frequency of incident radiation below which photoelectric emission is not possible is known as threshold frequency or cut-off frequency. It is given by :

ν0 = h

mv)2/1(h 2−ν

On the other hand, if the wavelength of the incident radiation is more than certain critical value, then photoelectric emission is not possible. This wavelength is known as threshold wavelength of cut-off wavelength. It is given by :

λ0 = ]mv)2/1(h[

hc2−ν

(d) Since Einstein treated photoelectric effect as a collision between a photon and an atom, he explained the instantaneous nature of photoelectric effect.

Some other important points : Stopping potential : The negative potential

applied to the collector in order to prevent the electron from reaching the collector (i.e., to reduce the photoelectric current to zero) is known as stopping potential.

eV0 = 2.maxmv

21 = hν – W = h(ν – ν0)

Millikan measured K.E. of emitted electrons or stopping potentials for different frequencies of incident radiation for a given emitter. He plotted a graph with the frequency on x-axis and stopping potential on y-axis. The graph so obtained was a straight line as shown in figure.

ν0

Frequency of incident light

V0(s

topp

ing

pote

ntia

l)

Millikan measured the slope of the straight line

(=h/e) and calculated the value of Planck's constant.

I

Full intensity

75% intensity 50% intensity 25% intensity

V0– +

Potential difference


The intercept of V0 versus ν graph on frequency axis is equal to threshold frequency (ν0). From this, the work function (hν0) can be calculated.

Graphs in photoelectric effect : (a) Photoelectric current versus potential difference

graphs for varying intensity (keeping same metal plate and same frequency of incident light) : These graphs indicate that stopping potential is independent of the intensity and saturation current is directly proportional to the intensity of light.

I

(V0)1 – +

Potential difference (V0)2

ν2 ν1

ν2>ν1

(b) Photoelectric current versus potential difference

graphs for varying frequency (keeping same metal plate and same intensity of incident light) : These graphs indicate that the stopping potential is constant for a given frequency. The stopping potential increases with increase of frequency. The KE of the emitted electrons is proportional to the frequency of incident light.

A1 A2 A3

ν0

B1

B2

B3

Frequency

Stop

ping

pot

entia

l

(c) Stopping potential versus frequency graphs for

different metals : These graphs indicate that the stops is same for all metal, since they are parallel straight lines. The slope is a universal constant (=h/e). Further, the threshold frequency varies with emitter since the intercepts on frequency axis are different for different metals.

1. (i) A stopping potential of 0.82 V is required to stop

the emission of photoelectrons from the surface of a metal by light of wavelength 4000 Å. For light of wavelength 3000 Å, the stopping potential is 1.85 V. Find the value of Planck's constant.

(ii) At stopping potential, if the wavelength of the incident light is kept at 4000 Å but the intensity of light is increased two times, will photoelectric current be obtained? Give reasons for your answer.

Sol. (i) We have 1

hcλ

= eV1 + W

and 2

hcλ

= eV2 + W

⇒

λ

−λ 12

11hc = e(V2 – V1)

or h =

λ

−λ

−

12

1211e

)VV(e =

×−

××

−×

−−

−

778

19

1041

1031103

)82.085.1(106.1

= 6.592 × 10–34 Js (ii) No, because the stopping potential depends only

on the wavelength of light and not on its intensity. 2. A small plate of a metal (work function = 1.17 eV) is

plated at a distance of 2m from a monochromatic light source of wavelength 4.8 × 10–7 m and power 1.0 watt. The light falls normally on the plate. Find the number of photons striking the metal plate per square metre per second. If a constant magnetic field of strength 10–4 tesla is parallel to the metal surface, find the radius of the largest circular path followed by the emitted photoelectrons.

Sol. Energy of one photon = λhc = 7

834

108.4103106.6

−

−

×

×××

= 4.125 × 10–19 J Number of photons emitted per second

= 1910125.40.1

−× = 2.424 × 1018

Number of photons striking the plate per square metre per second

= 2

18

)2(14.3410424.2××

× = 4.82 × 1016

Maximum kinetic energy of photoelectrons emitted from the plate

Emax = λhc – W

= 4.125 × 10–19 – 1.17 × 1.6 × 10–19 = 2.253 × 10–19 J

Solved Examples


3. A monochromatic light source of frequency ν illuminates a metallic surface and ejects photoelectrons. The photoelectrons having maximum energy are just able to ionize the hydrogen atom in ground state. When the whole experiment is repeated with an incident radiation of frequency (5/6) ν, the photoelectrons so emitted are able to excite the hydrogen atom beam which then emits a radiation of wavelength 1215 Å. Find the work function of the metal and the frequency ν.

Sol. In the first case, Emax = Ionization energy = 13.6 eV = 21.76 × 10–19 J So, hν = 21.76 × 10–19 J ....(1) In the second case,

E'max = λhc

= 10

834

101215103106.6

−

−

×

×××

=16.3×10–19 J

So, 6h5ν = 16.3 × 10–19 + W ...(2)

Dividing Eq.(1) by Eq.(2)

56 =

W103.16W1076.21

19

19

+×

+×−

−

Solving, we get W = 11.0 × 10 – 19 J = 6.875 eV

From Eq.(1) ν = 34

1919

106.6100.111076.21

−

−−

×

×+×

= 5 × 1015 Hz 4. The radiation, emitted when an electron jumps from

n = 3 to n = 2 orbit in a hydrogen atom, falls on a metal to produce photoelectrons. The electrons from the metal surface with maximum kinetic energy are made to move perpendicular to a magnetic field of 1/320 T in a radius of 10–3 m. Find (i) the kinetic energy of electrons, (ii) wavelength of radiation and (iii) the work function of metal.

Sol. (i) Speed of an electron in the magnetic field,

v = m

Ber

Kinetic energy of electrons

Emax = 21 mv2 =

m2reB 222

= 2

3201

× 31

23219

101.92)10()106.1(

−

−−

××

××

= 1.374 × 10–19 J = 0.8588 eV (ii) Energy of the photon emitted from a hydrogen

atom

hν = λhc =

− 22 3

121

= 1.888 eV Wavelength of radiation,

λ = 19

834

106.1888.11031062.6

−

−

××

×××

= 6.572 × 10–7m = 6572 Å (iii) Work function of metal W = hν – Emax = 1.8888 – 0.8588 = 1.03 eV 5. X-rays are produced in an X-ray tube by electrons

accelerated through a potential difference of 50.0 kV. An electron makes three collisions in the target before coming to rest and loses half of its kinetic energy in each of the first two collisions. Determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms.

Sol. Initial kinetic energy of the electron = 50.0 keV Kinetic energy after first collision = 25.0 keV Energy of the photon produced in the first collision, E1 = 50.0 – 25.0 = 25.0 keV Wavelength of this photon

λ1 = 1E

hc = 319

834

100.25106.1103106.6×××

×××−

−

= 0.495 × 10–10 m = 0.495 Å Kinetic energy of the electron after second collision = 12.5 eV Energy of the photon produced in the second

collision, E2 = 25.0 – 12.5 = 12.5 keV Wavelength of this photon

λ2 = 2

hcλ

= 319

834

100.25106.1103106.6×××

×××−

−

= 0.99 × 10–10 m = 0.99 Å Kinetic energy of the electron after third collision = 0 Energy of the photon produced in the third collision,

E3 = 12.5 – 0 = 12.5 keV This is same as E2. Therefore, wavelength of this

photon, λ3 = λ2 = 0.99 Å.


Thermal Expansion :

.(a) When the temperature of a substance is increased, it expands. The heat energy which is supplied to the substance is gained by the constituent particles of the substance as its kinetic energy. Because of this the collisions between the constituents particles are accompanied with greater force which increase the distance between the constituent particles.

∆l = lα∆T ; ∆A = Aβ∆T ; ∆V = Vγ∆T or l' = l (1 + α∆T) ; A' = A(1 + β∆T) ; V' = V(1 + γ∆T) (b) Also ρ = ρ'(1 + γ∆T) where ρ' is the density at

higher temperature clearly ρ' < ρ for substances which have positive value of γ

* β = 2α and γ = 3α Water has negative value of γ for certain temperature

range (0º to 4ºC). This means that for that temperature range the volume decreases with increase in temperature. In other words the density increases with increase in temperature.

0 ml 5 ml 10 ml 15 ml 20 ml 25 ml 30 ml

If a liquid is kept in a container and the temperature

of the system is increased then the volume of the liquid as well as the container increases. The apparent change in volume of the liquid as shown by the scale is

∆Vapp = V(γ – 3α) ∆T Where V is the volume of liquid at lower temperature ∆Vapp is the apparent change in volume γ is the coefficient of cubical expansion of liquid α is the coefficients of linear expansion of the

container. Loss or gain in time by a pendulum clock with

change in temperature is ∆t = 21

α(∆T) × t

Where ∆t is the loss or gain in time in a time interval t ∆T is change in temperature and d is coefficient of

linear expansion. If a rod is heated or cooled but not allowed to expand

or contract then the thermal stresses developed

AF = γα∆T.

If a scale is calibrated at a temperature T1 but used at a temperature T2, then the observed reading will be wrong. In this case the actual reading is given by

R = R0(1 + α∆T) Where R0 is the observed reading, R is the actual

reading. For difference between two rods to the same at all

temperatures l 1α1 = l2α2. Thermodynamics

According to first law of thermodynamics q = ∆U + W

For an isothermal process (for a gaseous system) (a) The pressure volume relationship is ρV = constt. (b) ∆U = 0 (c) q = W (d) W = 2.303 nRT log10

i

fVV = 2.303 nRT log10

f

ipp

(e) Graphs T2 > T1

T2T1

P

V

P

T

V

T These lines are called isotherms (parameters at

constant temperature) For an adiabatic process (for a gaseous system)

(a) The pressure-volume relationship is PVγ = constt. (b) The pressure-volume-temperature relationship is

TPV = constt.

(c) From (a) and (b) TVγ–I = constt. (d) q = 0 (e) W = –∆U

Thermal Expansion, Thermodynamics

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


(f) ∆U = ncv∆T where cv = 1

R−γ

(g) W = 1

VpVp ffii−γ− =

1)TT(nR fi

−γ−

(h) Graphs

V

P

P

T

V

T Please note that P-V graph line (isotherm) is

steeper. For isochoric process

(a) P ∝ T (b) W = 0 (c) q = ∆U

(d) ∆U = nCv∆T where Cv = 1

R−γ

(e) Graphs

P

V

P

T

V

T For isobaric process

(a) V ∝ T (b) W = P∆V = P(Vf – Vi) = nR(Tf – Ti) (c) ∆U = nCv∆T (d) q = nCp∆T (e) Graphs

P

V

P

T

V

T For a cyclic process

(a) ∆U = 0 ⇒ q = W (b) Work done is the area enclosed in p-V graph.

For any process depicted by P-V diagram, area under the graph represents the word done.

Kirchoff's law states that good absorbers are good emitters also.

Problem solving Strategy : Thermal Expansion Step 1: Identify the relevant concepts: Decide

whether the problem involves changes in length (linear thermal expansion) or in volume (volume thermal expansion)

Step 2: Set up the problem using the following steps: Eq. ∆L = αL0∆T for linear expansion and Eq. ∆V = βV0∆T for volume expansion. Identify which quantities in Eq. ∆L = αL0∆T or

∆V = βV0∆T are known and which are the unknown target variables.

Step 3: Execute the solution as follows: Solve for the target variables. Often you will be

given two temperatures and asked to compute ∆T. Or you may be given an initial temperature T0 and asked to find a final temperature corresponding to a given length or volume change. In this case, plan to find ∆T first; then the final temperature is T0 + ∆T.

Unit consistency is crucial, as always. L0 and ∆L (or V0 ∆V) must have the same units, and if you use a value or α or β in K–1 or (Cº)–1, then ∆T must be in kelvins or Celsius degrees (Cº). But you can use K and Cº interchangeably.

Step 4: Evaluate your answer: Check whether your results make sense. Remember that the sizes of holes in a material expand with temperature just as the same way as any other linear dimension, and the volume of a hole (such as the volume of a container) expands the same way as the corresponding solid shape.

Problem solving strategy : Thermodynamics Ist Law Step 1: Identify the relevant concepts : The first law

of thermodynamics is the statement of the law of conservation of energy in its most general form. You can apply it to any situation in which you are concerned with changes in the internal energy of a system, with heat flow into or out of a system, and/or with work done by or on a system.

Step 2: Set up the problem using the following steps Carefully define what the thermodynamics system is. The first law of thermodynamics focuses on

systems that go through thermodynamic processes. Some problems involve processes with more than one step. so make sure that you identify the initial and final state for each step.

Identify the known quantities and the target variables.

Check whether you have enough equations. The first law, ∆U = Q – W, can be applied just once to each step in a thermodynamic process, so you will often need additional equations. These often

include Eq. ∫=2

1

V

V

dVpW for the work done in a

volume change and the equation of state of the material that makes up the thermodynamic system (for an ideal gas, pV = nRT).

Step 3: Execute the solution as follows : You shouldn't be surprised to be told that

consistent units are essential. If p is a Pa and V in m3, then W is in joules. Otherwise, you may want to convert the pressure and volume units into units of Pa and m3. If a heat capacity is given in terms of calories, usually the simplest procedure is to convert it to joules. Be especially careful with moles. When you use n = mtot/M to convert


between total mass and number of moles, remember that if mtot is in kilograms, M must be in kilograms per mole. The usual units for M are grams per mole; be careful !

The internal energy change ∆U in any thermodynamic process or series of processes in independent of the path, whether the substance is an ideal gas or not. This point is of the utmost importance in the problems in this topic. Sometimes you will be given enough information about one path between the given initial and final states to calculate ∆U for that path. Since ∆U is the same for every possible path between the same two states, you can then relate the various energy quantities for other paths.

When a process consists of several distinct steps, it often helps to make a chart showing Q, W, and ∆U for each step. Put these quantities for each step on a different line, and arrange them so the Q's, W's, and ∆U's form columns. Then you can apply the first law to each line ; in addition, you can add each column and apply the first law to the sums. Do you see why ?

Using above steps, solve for the target variables. Step 4: Evaluate your answer : Check your results for

reasonableness. In particular, make sure that each of your answers has the correct algebraic sign. Remember that a positive Q means that heat flows into the system, and that a negative Q means that heat flows into the system, and that a negative Q means that heat flows out of the system. A positive W means that work is done by the system on its environment, while a negative W means that work is done on the system by its environment.

1. A metallic bob weighs 50 g in air. It it is immersed

in a liquid at a temperature of 25ºC, it weighs 45 g. When the temperature of the liquid is raised to 100ºC, it weighs 45.1 g. Calculate the coefficient of cubical expansion of the liquid given that the coefficient of linear expansion of the metal is 2 × 10–6(ºC)–1.

Sol. Loss in weight in liquid at 25ºC = (50 – 45) = 5 gm Weight of liquid displaced at 25ºC = V25ρ25g ∴ 5 = V25ρ25g ...(1) Similarly, V100ρ100g = 50 – 45.1 = 4.9 gm ...(2) From eq.(1) & (2) we get,

9.4

5 = 100

25

100

25 .VV

ρρ

Now, V100 = V25(1 + γmetal × 75)= V25(1 + 3αmetal × 75) = V25(1 + 3 × 12 × 10–6 × 75) or V100 = V25(1 + 0.0027) = V25 × 1.0027 Also, ρ25 = ρ100(1 + γ × 75) where, γ = Required coefficient of expansion of the liquid

9.4

5 = 100

100

25

25 )751(0027.1V

Vρ

γ+ρ×

× =

0027.1751 γ+

or γ = 3.1 × 10–4 (ºC)–1

2. A one litre flask contains some mercury. It is found that at different temperature the volume of air inside the flask remains the same. What is the volume of mercury in flask ? Given that the coefficient of linear expansion of glass = 9 × 10–6(ºC)–1 and coefficient of volume expansion of mercury = 1.8 × 10–4 (ºC–1).

Sol. Let V = Volume of the vessel V' = Volume of mercury For unoccupied volume to remain constant increase

in volume of mercury should be equal to increase in volume of vessel.

∴ V' γm∆T = Vγg∆T or V' = m

gVγ

γ×

∴ V' = 4

6

108.110271000

−

−

×

×× = 150 cm3

3. A clock with a metallic pendulum gains 6 seconds each day when the temperature is 20ºC and loses 12 seconds each day when the temperature is 40ºC. Find the coefficient of linear expansion of the metal.

Sol. Time taken for one oscillation of the pendulum is

T = gL2π or T2 = 4π2 ×

gL .....(1)

Partially differentiating, we get

2T∆t = 4π2 × gL∆ .....(2)

Dividing (2) by (1), we get

TT∆ =

L2L∆ =

L2tL∆α = t

21

∆α

where ∆t is the change in temperature. Now, One day = 24 hours = 86400 sec Let t be the temperature at which the clock keeps

correct time. At 20ºC, the gain in time is

6 = 21

α × (t – 20) × 86400 ....(3)

At 40ºC, the loss in time is

12 = 21

α× (40 – t) × 86400 ...(4)

Dividing (4) by (3), we have

6

12 = 20t

t40−

−

which gives t = 3

80 ºC.

Using the value in equation(3), we have

6 = 21 × α ×

− 20

380 × 86400

which gives α = 2.1 × 10–5 perºC

Solved Examples


4. A piston can freely move inside a horizontal cylinder closed from both ends. Initially, the piston separates the inside space of the cylinder into two equal parts each of volume V0, in which an ideal gas is contained under the same pressure p0 and at the same temperature. What work has to be performed in order to increase isothermally the volume of one part of gas η times compared to that of the other by slowly moving the piston ?

Sol. Let volume of chambers changes by ∆V. According to the problem, the final volume of left chamber is η times final volume of right chamber.

∴ V0 + ∆V = η(V0 – ∆V)

or ∆V = 0V11

+η−η

P0,v0,T0 P0,v0,T0

As piston is moved slowly therefore, change in

kinetic energy is zero. By work-energy theorem, we can write

Wgas in right chamber + Wgas in left chamber + extAgentW = ∆KE

extAgentW = (Wgas(R) + Wgas(L))

We know that in isothermal process, work done is given by

W = nRT ln

i

fVV

∴ Work done by gas in left chamber (WL)

= P0V0 ln

∆+

0

0V

VV= P0V0 ln

+ηη

12

Similarly, work done by gas in right chamber (WR)

= P0V0 ln

∆−

0

0V

VV= P0V0 ln

+ηη

12

extAgentW = –P0V0 ln

+ηη

12 – P0V0 ln

+ηη

12

= P0V0 ln 2

41

η+η

5. A smooth vertical tube having two different sections is open from both ends equipped with two pistons of different areas figure. Each piston slides within a respective tube section. One mole of ideal gas is enclosed between the pistons tied with a non-stretchable thread. The cross-sectional area of the upper piston is ∆S greater than that of the lower one. The combined mass of the two pistons is equal to m. The outside air pressure is P0. By how many kelvins must the gas between the pistons be heated to shift the pistons through l.

P0

P0

Sol. Let A1 = Cross section of upper piston A2 = Cross section of lower piston T = Tension in the string P = Gas pressure m1 = Mass of upper piston m2 = Mass of lower piston Now, consider FBD of upper piston

PA1 m1g

P0 A1

From equilibrium consideration of upper piston we get, P0A1 + T + m1g = PA1 Similarly, consider FBD of lower piston

PA2

m2g P0 A2

T

∴ P0A2 + T = m2g + PA2 Eliminating T, we get

P = P0 + 21

21AA

g)mm(−

+

According to problem m = m1 + m2 and ∆S = A1 – A2

∴ P = P0 + S

mg∆

Now, PV = RT

or P∆V = R∆T or ∆T = R

VP∆

But ∆V = (A1 – A2)l = ∆S. l

∴ ∆T =

∆+

SmgP0 ∆S.l

ll

l


Qualitative Analysis : Qualitative analysis of an organic compound involves

the detection of various elements present in it. The elements commonly present in organic compounds are carbon, hydrogen, oxygen, nitrogen, halogens, sulphur and sometimes phosphorus.

Detection of Carbon and Hydrogen : This is done by heating the given organic compound

with dry cupric oxide in a hard glass test tube when carbon present is oxidised to carbon dioxide and hydrogen is oxidised to water.

C + 2 CuO →∆ CO2 + 2Cu

2H + CuO →∆ H2O + Cu Carbon dioxide turns lime water milky. Ca(OH)2 +

)Cfrom(2CO →

)Milky(3CaCO + H2O

Water condenses on the cooler parts of the test tube and turns anhydrous copper sulphate blue.

)white(4CuSO + 5H2O →

)Blue(4CuSO .5H2O

Detection of Nitrogen, Sulphur and Halogens : Nitrogen, sulphur and halogens in any organic

compound are detected by Lassaigne's test. Preparation of Lassaigne's Extract (or sodium extract): A small piece of sodium is gently heated in an

ignition tube till it melts. The ignition tube is removed from the flame, about 50–60 mg of the organic compound added and the tube heated strongly for 2–3 minutes to fuse the material inside it. After cooling , the tube is carefully broken in a china dish containing about 20–30 mL of distilled water. The fused material along with the pieces of ignition tube are crushed with the help of a glass rod and the contents of the china dish are boiled for a few minutes. The sodium salts formed in the above reactions (i.e., NaCN, Na2S, NaX or NaSCN) dissolve in water. Excess of sodium, if any, reacts with water to give sodium hydroxide. This alkaline solution is called Lassaigne's extract or sodium extract. The solution is then filtered to remove the insoluble materials and the filtrate is used for making the tests for nitrogen, sulphur and halogens.

Reactions : An organic compound containing C, H, N, S, halogens when fused with sodium metal gives the following reactions.

C + N + Na →fusion NaCN in organic compound sodium cyanide

X(Cl, Br, I) + Na →fusion NaX(X=Cl,Br, I) from organic compound sodium halide

S + 2Na →fusion Na2S from organic compound sodium sulphide If nitrogen and sulphur both are present in any

organic compound, sodium thiocyanate (NaSCN) is formed during fusion which in the presence of excess sodium, forms sodium cyanide and sodium sulphide.

Na + C + N + S →fusion NaCNS in organic compound sodium thiocyanate Detection of Nitrogen : Take a small quantity of the sodium extract in a test

tube. If not alkaline, make it alkaline by adding 2–3 drops of sodium hydroxide (NaOH) solution. To this solution, add 1 mL of freshly prepared solution of ferrous sulphate. Heat the mixture of the two solutions to boiling and then acidify it with dilute sulphuric acid. The appearance of prussion blue or green colouration or precipitate confirms the presence of nitrogen in the given organic compound.

Chemistry of the test : The following reactions describe the chemistry of the tests of nitrogen. The carbon and nitrogen present in the organic compound on fusion with sodium metal give sodium cyanide (NaCN). NaCN being ionic salt dissolves in water. So, the sodium extract contains sodium cyanide.

Sodium cyanide on reaction with ferrous sulphate gives sodium ferrocyanide. On heating, some of the ferrous salt is oxidised to the ferric salt and this reacts with sodium ferrocyanide to form ferric-ferrocyanide.

6 NaCN + FeSO4 → Na4[Fe(CN)6] + Na2SO4 sodium ferrocyanide 3Na4[Fe(CN)6] + 2Fe2(SO4)3 formed during boiling of the solution → Fe4[Fe(CN)6]3 + 6Na2SO4 prussian blue When nitrogen and sulphur both are present in any

organic compound, sodium thiocyanate is formed during fusion. When extracted with water sodium thiocynate goes into the sodium extract and gives blood red colouration with ferric ions due to the formation of ferric thiocyanate

Organic Chemistry

Fundamentals

PURIFICATION OF ORGANIC COMPOUNDS

KEY CONCEPT


Na + C + N + S NaCNSfrom organic Sod. thiocyanate

3NaCNS + Fe3+ → Fe(CNS)3 + 3Na+ ferric thiocyanate (blood red) Note : (i) Some compounds like hydrazine

(NH2NH2) although contain nitrogen, they do not respond Lassaigne's test because they do not have any carbon and hence NaCN in not formed.

(ii) Diazonium salts do not show Lassaigne's test because they are unstable and lose nitrogen as N2 gas on heating. Hence during fusion, no NaCN is formed in Lassaigne's extract due the loss of nitrogen.

Detection of Sulphur : The presence of sulphur in any organic compound is

detected by using sodium extract as follows: (a) Lead acetate test : Acidify a small portion of

sodium extract with acetic acid and add lead acetate solution to it. A black precipitate of lead sulphide indicates the presence of sulphur.

(CH3COO)2Pb + Na2S →+H PbS + 2CH3COONa

lead acetate black ppt (b) Sodium nitroprusside test : To a small quantity

of sodium extract taken in a test tube, add 2-3 drops of sodium nitroprusside solution. A violet colour indicates the presence of sulphur. This colour fades away slowly on standing.

Na2S + Na2[Fe(CN)5NO] → Na4[Fe(CN)5NOS] sodium nitroprusside violet or purple colour Detection of Halogens : The presence of halogens in any organic compound is

detected by using sodium extract (Lassaigne's extract) by silver nitrate test.

(a) Silver nitrate test: Sodium extract (or Lassaigne's extract) is boiled with dilute nitric acid to decompose sodium cyanide or sodium sulphide (if present) to hydrogen cyanide and hydrogen sulphide gases, respectively. This solution is cooled and silver nitrate solution added. A white precipitate soluble in ammonia shows chlorine, a yellowish precipitate sparingly soluble in ammonia indicates bromine, and a yellow precipitate insoluble in ammonia shows the presence of iodine in the given organic compound.

NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq) white precipitate (soluble in ammonia) NaBr(aq) + AgNO3(aq) → AgBr(s) + NaNO3(aq) light yellow ppt. (sparingly soluble in ammonia) NaI(aq) + AgNO3(aq) → AgI(s) + NaNO3(aq) yellow precipitate (insoluble in ammonia)

(b)CS2 layer test for detecting bromine and iodine: Boil a small quantity of sodium extract with dilute HNO3 for 1–2 min and cool the solution. To this solution, add a few drops of carbon disulphide (CS2) and 1–2 mL fresh chlorine water, and shake. Appearance of orange colour in the CS2 layer confirms the presence of bromine, whereas that of a violet/purple colouration confirms the presence of iodine in the compound.

2NaBr(aq) + Cl2 → 2CS Br2 + NaCl(aq) in sodium extract dissolves in CS2 to give orange colour.

2NaI(aq) + Cl2 → 2CS I2 + 2NaCl(aq) in sodium extract dissolves in CS2 to give purple/violet colour Detection of Phosphorus : In order to detect phosphorus, the organic compound

is fused with sodium peroxide, when phosphorus is converted into sodium phosphate.

5Na2O2 + )Compound(

P2 →Fuse

phosphate.Sod43PONa2 + 2Na2O

The fused mass is extracted with water and the water is extract is boiled with conc. HNO3. Upon cooling a few drops of ammonium molybdate solution are added. A yellow ppt. of ammonium phosphomolybdate indicates the presence of phosphorus in the organic compound.

Na3PO4 + 3HNO3 → H3PO4 + 3NaNO3 H3PO4 + 12(NH4)2MoO4 + 21 HNO3 →

.)pptyellow(434 PO)NH( . 12 MoO3 + 21 NH4NO3 + 12H2O

Quantitative Analysis : The quantitative analysis of an organic compound

means the estimation of percentage composition of each element present in the organic compound.

Estimation of Nitrogen : Duma's Method : Principle : A known mass of the organic substance is

heated with excess of copper oxide in an atmosphere of CO2. Carbon, hydrogen and sulphur (if present) are oxidised to CO2, H2O and SO2 while nitrogen is set free. A small amount of nitrogen may be oxidised to oxides but they are reduced back to free nitrogen by passing over a hot reduced copper gauze.

Oxides of nitrogen + Cu →∆ CuO + N2 The nitrogen thus formed is collected over conc.

KOH solution taken in Schiff's nitrometer tube which absorbs all other gases i.e., CO2, H2O vapours, SO2 etc. The volume of nitrogen collected is converted to STP and from this the precentage of nitrogen can be calculated.

% age of Nitrogen

= 22400

28 × takenSubstance of Mass

STPat N of Vol. 2 × 100


Boron Trihalides :

The trihalides of boron are electron deficient compounds having a planar structure as shown. They act as Lewis acids because of incomplete octet.

B

X

X X Planar structure of boron trihalides

120º

acidLewis3BF +

baseLewis3NH: →

productAddition3BF ← NH3

acidLewis3BF +

baseLewisF: − →

ionteFluorobora4BF−

The acid strength of trihalides decreases as :

BF3 < BCl3 < BBr3 < BI3

Explanation :

This order of acid strength is reverse of what may normally be expected on the basis of electronegativity of halogens. Since F is most electronegative, hence BF3 should be most electron deficient and thus should be strongest acid. The anomalous behaviour is explained on the basis of tendency of halogen atom to back-donate its electrons to boron atom. For example, in BF3 one of the 2p-orbital of F atom having lone pair overlaps sidewise with the empty 2p-orbital of boron atom to form pπ-pπ back bonding. This is also known as back donation. Further, due to back-π donation of three surrounding fluorine atoms. BF3 can be represented as a resonance hybrid of following three structures.

B– = F+ B– – F B– — F

F

F

F

F

F

Resonating forms of BF3

+

F + B– — FF

F≡

Probable hybridstructure

As a result of this back donation, the electron deficiency of boron gets compensated and its Lewis acid character decreases.

Now, the tendency for back donation is maximum in the case of fluorine due to its small size and more interelectronic repulsions, therefore, it is the least acidic. The tendency of back bonding falls as we move from BF3 to BCl3 and BCl3 to BBr3 due to increase in the size of halogen atoms consequently, the acidic character increase accordingly.

Empty 2p-orbital

F

π

π

2p-orbital with lone pair

B

F

F

pπ-pπ back bonding

Acidic nature of H3BO3 or B(OH)3 :

Since B(OH)3 only partially reacts with water to form H3O+ and [B(OH)4]–

, it behaves as a weak acid. Thus H3BO3 or (B(OH)3) cannot be titrated satisfactorily with NaOH, as a sharp end point is not obtained. If certain organic polyhydroxy compounds such as glycerol, mannitol or sugars are added to the titration mixture, then B(OH)3 behaves as a strong monobasic acid. It can now be titrated with NaOH, and the end point is detected using phenolphthalein as indicator (indicator changes at pH 8.3 – 10.0).

2B(OH)3 + 2NaOH

Na[B(OH)4] + metaboratesodium

22 OH2NaBO +

The added compound must be a cis-diol, to enhance the acidic properties in this way. (This means that it has OH groups on adjacent carbon atoms in the cis configuration.) The cis-diol forms

Inorganic Chemistry

Fundamentals

BORON & CARBON FAMILY

KEY CONCEPT


very stable complexes with the [B(OH)4]– formed by the forward reaction above, thus effectively removing it from solution. The reaction is reversible. Thus removal of one of the products at the right hand side of the equation upsets the balance, and the reaction proceeds completely to the right. Thus all the B(OH)3 reacts with NaOH : in effect it acts as a strong acid in the presence of the cis-diol.

– C – OH

– C – OH

HO

BOH

OH

HO +

–

–2H2O– C – O

– C – O

BOH

OH

+

HO – C –

HO – C –

–

–2H2O– C – O

– C – O

B

–O – C –

O – C –

Borax :

The most common metaborate is borax Na2[B4O5(OH)4] . 8H2O. It is a useful primary standard for titrating against acids.

(Na2[B4O5(OH)4] . 8H2O) + 2HCl →

2NaCl + 4H3BO3 + 5H2O

One of the products H3BO3 is itself a weak acid. Thus the indicator used to detect the end point of this reaction must be one that is unaffected by H3BO3. Methyl orange is normally used, which changes in the pH range 3.1 – 4.4.

One mole of borax reacts with two moles of acid. This is because when borax is dissolved in water both B(OH)3 and [B(OH)4]– are formed, but only the [B(OH)4]– reacts with HCl.

[B4O5(OH)4]2– + 5H2O 2B(OH)3 + 2[B(OH)4]–

2[B(OH)4]– + 2H3O+ → 2B(OH)3 + 4H2O

The last reaction will titrate at pH 9.2, so the indicator must have pKa < 8. Borax is also used as a buffer since its aqueous solution contains equal amounts of weak acid and its salt.

Diborane, B2H6 :

The simplest boron hydride known, is diborane. It is prepared by treating boron trifluoride with LiAlH4 in diethyl ether.

4BF3 + 3 LiAlH4 → 2B2H6 + 3LiF + 3AlF3

A convenient laboratory method for the preparation of diborane involves the oxidation of sodium borohydride with iodine.

2NaBH4 + I2 → B2H6 + 2NaI + H2

Diborane is produced on an industrial scale by the reaction of BF3 with sodium hydride.

2BF3 + 6NaH → K450 B2H6 + 6NaF

Diborane is a colourless, highly toxic gas with a b.p. of 180 K. Diborane catches fire spontaneously upon exposure to air. It burns in oxygen releasing an enormous amount of energy.

B2H6 + 3O2 → B2O3 + 3H2O;

∆cHΘ = – 1976 kJ mol–1

Most of the higher boranes are also spontaneously flammable in air. Boranes are readily hydrolysed by water to give boric acid.

B2H6 (g) + 6H2O(l) → 2B(OH)3 (aq) + 6H2 (g)

Diborane undergoes cleavage reactions with Lewis bases (L) to give borane adducts, BH3.L

B2H6 + 2 NMe3 → 2BH3.NMe3

B2H6 + 2 CO → 2BH3.CO

Reaction of ammonia with diborane gives initially B2H6. 2NH3 which is formulated as [BH2(NH3)2]+ [BH4]– ; further heating gives borazine, B3N3H6 known as "inorganic benzene" in view of its ring structure with alternate BH and NH groups.

3B2H6 + 6NH3 → 3[BH2(NH3)2]+ [BH4]–

→Heat 2B3N3H6 + 12H2

The structure of diborane is shown in Fig.(a). The four terminal hydrogen atoms and the two boron atoms lie in one plane. Above and below this plane, there are two bridging hydrogen atoms.

HHB

H

B 120º119pm

H

H

97º134pm

H

Fig. (a) The structure of diborane, B2H6

The four terminal B-H bonds are regular two centre-two electron bonds while the two bridge (B-H-B) bonds are different and can be described in terms of three centre-two electron bonds shown in Fig. (b).


Boron also forms a series of hydridoborates; the most important one is the tetrahedral [BH4]– ion. Tetrahydridoborates of several metals are known. Lithium and sodium tetrahydridoborates, also known as borohydrides, are prepared by the reaction of metal hydrides with B2H6 in diethyl ether.

2MH + B2H6 → 2M+ [BH4]– (M = Li or Na)

H H H

B B

H H H

H

H B

H

BH

H

H

Fig. (b) Bonding in diborane. Each B atom uses

sp3 hybrids for bonding. Out of the four sp3 hybrids on each B atom, one is without an electron shown in broken lines. The terminal B-H bonds are normal 2-centre-2- electron bonds but the two bridge bonds are 3-centre-2-electron bonds. The 3-centre-2-electron bridge bonds are also referred to as banana bonds.

Both LiBH4 and NaBH4 are used as reducing agents in organic synthesis.

Silicones :

These are organosilicon polymers containing Si – O – Si linkages. These are formed by the hydrolysis of alkyl or aryl substituded chlorosilanes and their subsequent polymerisation. The alkyl or aryl substitued chlorosilanes are prepared by the reaction of Grignard reagent and silicon tetrachloride.

reagentGrignard

RMgCl + SiCl4 → R – SiCl3 + MgCl2

2RMgCl + SiCl4 → R2SiCl2 + 2MgCl2

3RMgCl + SiCl4 → R3SiCl + 3MgCl2

R stands for – CH3, –C2H5 or –C6H5 groups

Hydrolysis of substituted chlorosilanes yield corresponding silanols which udergo polymerisation.

Si R R

Cl H OH

Cl H OH + Si

R R

–2HCl OHOH

Dialky silandiol Polymerisation of dialkyl silandiol yields linear

thermoplastic polymer.

HO – Si – OH + H O – Si – OH

R

R

R

R

HO – Si – O – Si – OH

R

R

R

R Polymerisation continues on both the ends and thus

chain increases in length.

RSiCl3 on hydrolysis gives a cross linked silicone. The formation can be explained in three steps :

R – Si – Cl(i)

Cl

3H2O

OH R – Si – OH

–3HCl

OH Cl

HO – Si – OH + H O – Si – OH + H O – Si – OH

R

OH

R

OH

HO – Si – O – Si – O – Si – OH

R

OH

R

OH

R

OHR

OH

(ii)


R

OH

R

OH

– O – Si – O – Si – O – Si – O –

R

O

R

O

R

OH

R

O

(iii)


H O

R

H O

R

H O

R

–3H2O

– O – Si – O – Si – O – Si – O –

R R RCross linked silicone

Cyclic (ring) silicones are formed when water is eliminated from the terminal –OH group of linear silicones.

Si O O

Si Si

R R

R R

R

R O R3SiCl on hydrolysis forms only a dimer

R3Si OH + OH Si R3 R3Si – O – Si R3


1. For SO2(g) at 273 K and 1 atm pressure, the dielectric constant (or relative permittivity) is 1.00993. This molecule has a permanent dipole moment of 1.63 D. Assuming that SO2 behaves as an ideal gas, calculate per mol of (a) total, (b) orientation, (c) induced polarizations, and (d) distortion polarizability.

Sol. We have

εr = 0εε = 1.00993

p = 1.63 D = 1.63(3.3356 × 10–30 Cm) Vm = 22414 cm3 mol–1 at 1 atm and 273 K (a) Total polarization,

Ptotal = 21

r

r

+ε−ε

ρM

= 200993.1100993.1

+− × 22.414 cm3 mol–1

= 73.95 cm3 mol–1 (b) Orientation polarization,

P0 =

ε kT3p

3N 2

0

A

=

××

−−−

−

21212

123

mNC10854.8(3mol10023.6

×××

−−

−

)K273)(KJ1038.1(3)Cm103356.363.1(

123

230

= 59.31 × 10–6 m3 mol–1 = 59.31 cm3 mol–1 (c) Induced polarization, Pind = Ptotal – P0 = 73.95 cm3 mol–1 – 59.31 cm3 mol–1 = 14.64 cm3 mol–1 (d) Distortion polarizability,

αd = A0

ind

N)3/1(Pε

= )mol10023.6)(mNC10854.83/(1

molm1064.1412321212

136–

−−−−

−

××××

= 6.46 × 10–40 C2 N–1 m 2. An electron confined to a one-dimensional box of

length 0.14 nm has a ground-state energy corresponding to the radiation of wavelength about 70 nm. Benzene, as a rough approximation, may be considered to be a two-dimensional box that encompasses the regular hexagonal shape. The C—C

bond length in benzene is 0.14 nm, so that side of the box would be about 0.28 nm. Estimate wavelength for transition from ground state to first excited state of benzene, assuming that it is π-bonding electrons that are involved.

Sol. For the one-dimensional box,

E = 2

2

m8h

ln2

Thus, the ground state energy E1 in a one-dimensional box of length 0.14 nm is

E1 = 2

2

)nm14.0(m8h

For the two-dimensional square box,

E2 = )nn(m8h 2

2212

2+

l

Now since l = 0.28 nm, we have

E2 = )nn()nm14.02(m8

h 22

212

2+

× = )nn(

4E 2

221

1 +

The various energy levels are as follows.

n1 n2 E2 1 1 E1/2

2 1 (5/4)E1

1 2 (5/4)E1

2 2 2E1

degenerate

The first three energy levels will be doubly occupied

in the ground state and hence the first excited state is obtained when the electron is promoted from n1 = 1, n2 = 2 state to n1 = 2, n2 = 2 state. Thus

∆E = 2E1 – 45 E1 =

43 E1

Since the wavelength is inversely proportional to energy, the corresponding wavelength would be (4/3)λ, i.e.

34 × 70 nm = 93 nm

3. A metal (A) gives the following observations : (i) It gives golden yellow flame. (ii) It is highly reactive and used in photoelectric cells

as well as used in the preparation of Lassaigane solution.

UNDERSTANDINGInorganic Chemistry


(iii) (A) on fusion with NaN3 and NaNO3 separately, yields an alkaline oxide (B) and an inert gas (C). The gas (C) when mixed with H2 in Haber's process gives another gas (D). (D) turns red litmus blue and gives white dense fumes with HCl.

(iv) Compound (B) react with water forming on alkaline solution (E). (E) is used for the saponification of oils and fats to give glycerol and a hard soap.

(v) (B) on heating at 670 K give (F) and (A). The compound (F) liberates H2O2 on reaction with dil. mineral acids. It is an oxidising agent and oxidises Cr(OH)3 to chromate, manganous salt to manganate, sulphides to sulphates.

(vi) (B) reacts with liquid ammonia to give (G) and (E). (G) is used for the conversion of 1, 2 dihaloalkanes into alkynes.

What are (A) to (G)? Explain the reactions involved. Sol. (i) (A) appears to be Na as it gives the golden yellow

flame. It is also used in the preparation of Lassaigane solution which is sodium extract of organic compounds.

Na + C + N → NaCN Na + Cl → NaCl 2Na + S → Na2S (ii) Compound (B) is Na2O and (C) is N2 while (D) is

NH3, as (D) is alkaline and turns red litmus blue and gives white fumes with HCl

(C) + H2 → NH3 N2 + 3H2

)D(3NH2

NH3 + HCl → NH4Cl White fumes

(iii) is prepared from Na as follows. 2NaNO3 + 10 Na →

)B(2ONa6 +

)C(2N

3NaN3 + NaNO2 → )B(2ONa2 +

)C(2N

(iv) Compound (E) is NaOH as it is used in the preparation of soaps.

)B(2ONa + H2O →

)E(NaOH2

CH2OOCC17H35

CHOOCC17H35 + 3NaOH

CH2OOCC17H35

∆

CH2OH

CH2OH + 3C17H35COONa

CH2OH (soap)

(v) (F) is sodium peroxide as only peroxides gives H2O2 on reaction with dil. acids.

)B(2ONa2

∆ → K670

)F(22ONa +

)A(Na2

)F(

22ONa + .dil

42SOH → H2O2 + Na2SO4

(F) gives the following oxidations : Cr(OH)3 + 5OH– → CrO4

2– + 4H2O + 3e–

Mn2+ + 8OH– → MnO4– + 4H2O + 5e–

S2– + 8OH– → SO42– + 4H2O + 8e–

The reduction equation of (F) is O2

2– + 2H2O + 2e– → 4OH– (vi) (G) is sodamide because it is used in the

dehydrohalogenation reactions.

)B(2ONa + NH3(l) →

)G(2NaNH +

)E(NaOH

CH3 – CH – CH2 + 2NaNH2

Br Br

∆ CH3 – C ≡ CHPropyne

+ 2NaBr + 2NH3

4. A green coloured compound (A) gave the following reactions :

(i) (A) dissolves in water to give a green solution. The solution on reaction with AgNO3 gives a white ppt. (B) which dissolves in NH4OH solution and reappears on addition of dil. HNO3. It on heating with K2Cr2O7 and conc. H2SO4 produced a red gas which dissolves in NaOH to give yellow solution (C). Addition of lead acetate solution to (C) gives a yellow ppt. which is used as a paint.

(ii) The hydroxide of cation of (A) in borax bead test gives brown colour in oxidising flame and grey colour in reducing flame.

(iii) Aqueous solution of (A) gives a black ppt. on passing H2S gas. The black ppt. dissolves in aquaregia and gives back (A).

(iv) (A) on boiling with NaHCO3 and Br2 water gives a black ppt. (D)

(v) (A) on treatment with KCN gives a light green ppt. (E) which dissolves in excess of KCN to give (F). (F) on heating with alkaline bromine water gives the same black ppt. as (D).

Identify compounds (A) to (F) and give balanced equations of the reactions.

Sol. Reaction (i) indicates that (A) contains Cl– ions because, it gives white ppt. soluble in NH4OH. It is again confirmed because it gives chromyl chloride test. The colour of oxidising and reducing flames indicate that (A) also contains Ni2+ ions. Hence, (A) is NiCl2. The different reactions are :

(i) NiCl2 + 2AgNO3 → 2AgCl + Ni(NO3)2 AgCl + 2NH3 →

lelubSo23 Cl])NH(Ag[

Ag(NH3)2Cl + 2HNO3 → )B(.pptwhite

AgCl ↓ + 2NH4NO3

The equations of chromyl chloride tests are : NiCl2 + Na2CO3 → 2NaCl + NiCO3 4NaCl + K2Cr2O7 + 6H2SO4 → 4NaHSO4 + 2KHSO4 + 3H2O +

gasdRe22ClCrO2

CrO2Cl2 + 4NaOH →)C(solutionYellow

42CrONa + 2NaCl + 2H2O

Na2CrO4 + (CH3COO)2Pb → .pptYellow

4PbCrO + 2CH3COONa

(ii) Na2B4O7 . 10H2O ∆ Na2B4O7 + 10H2O


Na2B4O7 ∆ 444 3444 21

beadtTransparen322 OBNaBO2 +

NiO + B2O3 ∆

)Brown(boratemetaNickel

22 )BO(Ni [Oxidising flame]

Ni(BO2)2 + C ∆ GreyNi + B2O3 + CO

[Reducing flame] (iii) NiCl2 + H2S → 2HCl +

.pptBlackNiS ↓

NiS + 2HCl + [O] → )A(

2NiCl + H2S ↑

(iv) )A(

2NiCl + 2NaHCO3 → NiCO3 + 2NaCl

+ CO2 + H2O

2NiCO3 + 4NaOH + [O] ∆

)D(.pptBlack

32ONi ↓

+ 2Na2CO3 + H2O (v)

)A(2NiCl + 2KCN →

)E(.pptGreen2)CN(Ni + 2KCl

Ni(CN)2 + 2KCN → )F(

42 ])CN(Ni[K

NaOH + Br2 → NaOBr + HBr

2K2[Ni(CN)4] + 4NaOH + 9NaOBr ∆

)D(32ONi ↓ + 4KCNO + 9NaBr + 4NaCNO

5. Compound (X) on reduction with LiAlH4 gives a

hydride (Y) containing 21.72% hydrogen along with other products. The compound (Y) reacts with air explosively resulting in boron trioxide. Identify (X) and (Y). Give balanced reactions involved in the formation of (Y) and its reaction with air. Draw the structure of (Y).

Sol. Since B2O3 is formed by reaction of (Y) with air, (Y) therefore should be B2H6 in which % of hydrogen is 21.72. The compound (X) on reduction with LiAlH4 gives B2H6. Thus it is boron trihalide. The reactions are shown as:

)X(

3BX4 + 3LiAlH4 → )Y(

62HB2 + 3LiX + 3AlX3

(X = Cl or Br)

)Y(62HB + 3O2 → B2O3 + 3H2O + heat

Structure of B2H6 is as follows:

B

Ht

BHt

Ht

Ht

Hb

Hb or

B

Ht

Ht Hb

Hb

B

Ht

Ht

121.5º1.33Å

97º

1.19Å1.77Å

Thus, the diborane molecule has four two-centre-two electron bonds (2c-2e– bonds) also called usual bonds and two three-centre-two-electron bonds (3c-2e– bonds) also called banana bonds. Hydrogen attached to usual and banana bonds are called Ht (terminal H) and Hb (bridged H) respectively.

TRUE OR FALSE

1. Two simple harmonic motion are represented by

the equation x1 = 5 sin

π

+π4

t2 and x2 =

5 2 (sin 2πt + cos 2πt). Their amplitudes are in the ratio 1 : 2.

2. When a dielectric is introduced between the plates of a capacitor at a constant potential difference, the charge on the plates remains unchanged.

3. Heat can never be converted completely into work.

4. The workings of a triode as an amplifier and a step-up transformer are same.

5. A 60 dB sound has twice the intensity of a 30 dB sound.

6. Two identical spherical air bubbles, one formed, inside water of a tank and the other outside the water, have equal pressure inside them.

Sol. 1. [True] 2. [False]

On introduction of dielectric the capacitance increases. Potential difference remaining constant, charge increases, because Q = CV.

3. [False] In an isothermal expansion of an ideal gas, heat can be completely converted into work.

4. [False] A step-up transformer increases the voltage but not the power. A triode increases both the voltage and the power.

5. [False] 6. [False]


1. For complex numbers z1 = x1 + iy1 and z2 = x2 + iy2 we write z1 ∩ z2, if x1 ≤ x2 and y1 ≤ y2. The for all complex numbers z with

1 ∩ z, we have z1z1

+− ∩ 0, Justify the result.

2. AP and BQ are fixed parallel tangents to a circle, and a tangent at any point c cuts them at P and Q respectively. Show that CP.CQ is independent of the position of c on the circle.

3. Let f(x) = ax2 + bx + c & g(x) = cx2 + bx + a, such that |f(0)| ≤ 1, |f(1)| ≤ 1 and |f(–1)| ≤ 1. prove that |f(x)| ≤ 5/4 and |g(x)| ≤ 2

4. A straight line is drawn throguh the origin and

parallel to the tangent to the curve a

yax 22 −+ =

ln

−+y

yaa 22

at an arbitrary point M. Show that

the locus of the points P of intersection of this straight line and the straight line parallel to the x-axis and passing through the point M is a circle.

5. Show that ∑=

+−n

0r r2r

rn

r

CC)2(

=

+

+

odd isn If,2n

1

even isn If,1n

1

6. Let In = ∫ −1

0

1n dxxtanx , then expression In in terms

of In–2.

7. If f

+

3yx =

3)y(f)x(f2 ++

for all real x and y. If f ´(2) = 2, then f(2) is - Passage : Let Z denotes the set of integers. Let p be a prime

number and let z1 ≡ 0, 1. Let f : z → z and g : z → z1 are two functions defined as follows :

f(n) = pn; if n ∈ z and g(n) = 1; if n is a perfect square = 0, otherwise.

8. g(f(x)) is - (A) Many one into (B) Many one onto (C) One one on to (D) One one into

9. f(g(x)) = p has - (A) no real root (B) at least one real root (C) infinity many roots (D) exactly one real root

10. g(f(x)) is – (A) non periodic function (B) odd function (C) even function (D) None of these

`tàxÅtà|vtÄ VtÄÄxÇzxá This section is designed to give IIT JEE aspirants a thorough grinding & exposure to variety of possible twists and turns of problems in mathematics that would be very helpful in facing IIT JEE. Each and every problem is well thought of in order to strengthen the concepts and we hope that this section would prove a rich resource for practicing challenging problems and enhancing the preparation level of IIT JEE aspirants.

By : Shailendra Maheshwari Joint Director Academics, Career Point, KotaSolut ions wil l be published in next issue

11Set


1. g(x) = sin x ; 0 ≤ x < π/2 1 ; π/2 ≤ x ≤ π sin2 x/2 ; π < x

−π→ 2xlim g(x) =

+π→ 2xlim g(x) = g(π/2) = 1

−π→x

lim g(x) = +π→x

lim g(x) = g(π) = 1

g´(π–) = g´(π +) = 0 and g´(π/2–) = g´(π/2+) = 0 Hence g(x) is continuous and differentiable in (0,∞)

2. x

xsin < xsin

)xsin(sin

Let f(θ) = θ

θsin ; 0 < θ < π/2

f ´(θ) = 2

sincosθ

θ−θθ

= 2

)tan(.cosθ

θ−θθ < 0 as tan θ > θ

so f(θ) ↓ so f(x) < f(sinx) as sin x < x

3. (i) 6C4 = 25.6 = 15

(ii) coeff. of x4 in (1 – x)–6

= 4 + 6 – 1C6 – 1 = 9C5 = 2.3.46.7.8.9 = 126

(iii) select 3 different flavours : 6C3 ways choose (at least one from each) 4 cones : 4 – 1C3 – 1 = 3C2 = 3 ways

so required ways = 6C3 × 3 = 2.34.5.6 × 3 = 60

(iv) Select 2 different flavours : 6C2 ways choose (at least one from each) 4 cones ; 4 – 1C2 – 1 = 3C1 = 3 so required ways (either 2 or 3 different flavours)

= 60 + 6C2 3 = 60 + 25.6 × 3 = 105

4. Let A at origin & P.V. of B & C are b & c .

So line AD ⇒ r = t

+

|c|c

|b|b

& line BC ⇒ r = b + ∆ ( b – c ) solve them together to find pt. D

t

+

|c|c

|b|b = b + s ( b – c )

A

D CBE

|b|

t = 1 + s ...(1)

|c|

t = –s ...(2)

so |b|

t = 1 – |c|

t ⇒ t = |c||b|

|c||b|+

use it in line AD .

pt D : |c||b|

|c||b|+

.

+

|c|c

|b|b =

|c||b||b|c|c|b

++

which divides BC in ratio of |c| : |b| similary use eq. of external angle bisector line AE

⇒ r = p

−

|c|c

|b|b

solve it with BC to find pt. E.

5. Consider eix(1 + eix)n = eix [1 + nC1eix + nC2ei2x + .... + nCneinx]

+ x

22ni

e . 2cosn

2x = eix + nC1ei2x + nC2ei3x +... +

nCnei(n+ 1)x

Compare real parts & get (a) Compare imag. parts & get (b)

6. Let Ei = the event that originator will not receive a letter in the ith stage.

Originator sands letters to two persons so in 1st stage he will not get letter.

Prob. that letter sent by 1st received is not received

by originator is 1

1n2

2n

CC

−

−

= )2n)(1n()3n)(2n(

−−−− =

1n3n

−−

similarly prob. that letter sent by 2nd receipiant is not

received by originator is 1n3n

−−

so P(E2) = prob. that originator not received letter in

2nd stage is = 2

1n3n

−− .

MATHEMATICAL CHALLENGES SOLUTION FOR FEBRUARY ISSUE (SET # 10)


similarly P(E3) = prob. that originator not receive letter sent by the four person getting letters from two recipients is

−−

1n3n .

−−

1n3n .

−−

1n3n .

−−

1n3n =

4

1n3n

−− =

22

1n3n

−−

Similarly, P(E4) = 8

1n3n

−− =

32

1n3n

−−

Similarly, P(Ek) = 1–k2

1n3n

−−

So the required prob. is P(E) = prob. the originator not receive letter in 1st k

stages = P(E1) . P(E2) . ........ P(Ek)

= 1k32 2....222

1n3n

−+++

−−

= 12

12.21k

1n3n −

−−

−− =

)22( k

1n3n −

−−

7. y = f(x) = ∫ −x

0

zzx 2e dz = ∫ −

x

0

zzx 2e.e dz

y´ = ∫ −x

0

zzx 2e.e dz + 1 = – ∫ −−

x

0

zzx )ze2(e21 2

dz + 1

= –

− ∫ −

x

0

zxzx0

zxz dze.xe)e.e(21 22

+ 1 = 21 xy + 1

dxdy –

21 xy = 1

I.F. = dx2/x

e ∫−= 4/x 2

e−

Sol is y . 4/x2e− = ∫ − 4/x2

e dx = ∫ −x

0

4/z2e dz

y = ∫ −x

0

4/z4/x 22ee dz.

8. ∫ sin n θ sec θ dθ = ∫ sin (n –1 + 1) θ sec θ dθ = ∫ sin (n – 1)θ + cos (n – 1)θ sin θ sec θ ) dθ = ∫ sin (n – 1)θ + [ sin (n – 1)θ cos θ – sin (n – 2)θ sec θ ] dθ = ∫ (2 sin (n – 1)θ – sin (n – 2)θ sec θ ) dθ

= – 1n

)1ncos(2−

θ− – ∫ sin (n – 2)θ secθ dθ

= ∫π

θθ−θ2

0 cos2sin8sin

21 dθ

=

θθθ−

θ− ∫

ππ 2

0

2/

0

dsec6sin7cos72

21

θθθ− ∫

π 2

0dsec2sin

θθθ−θ−− ∫

ππ

2/

0

2/0 dsec2sin)3(cos

32

52

72

21

θθθ− ∫

π 2

0dsec2sin

71 –

51 +

31 – ∫

πθθθ

2/

0dsec2sin

= 10529 + 2/

0)(cos2 πθ – ∫ θd0 = – 105181

9. 9x2 – 24xy + 16y2 – 18x – 101y + 19 = 0] (3x – 4y)2 = 18x + 101y – 19. Let the vertex of the parabola be A(α, β). Shift origin

to A and y-axis along the tangent at vertex (3x – 4y + l) . So the axis of parabola be 4x + 3y + m = 0 (along x axis) If L.R. of parabola be a then it’s equation is

2

5y4x3

+− l = a

++

5my3x4

(3x – 4y + l)2 – 5a(4x + 3y + m) = 0 9x2 – 24xy + 16y2 + (6l – 20a)x + (–8l – 15a)y + (l2 – 5am) = 0 comp. it with given equation. 6l – 20a = –18 ⇒ 24l – 80a = –72 ...(1) –8l –15a = –101 ⇒ –24l – 45a = –303 ...(2) From (1) & (2) ⇒ 125a = –375 ⇒ a = 3

10. circle : (x – 1)2 + (y – 1)2 = 1 ⇒ x2 + y2 – 2x – 2y + 1 = 0

E

D

(0,1)B

A(1,0)

Let the line be y = mx

Altitude of ∆ = 2m1

1

+

For DE length : solve line with circle. x2 + m2x2 – 2x – 2mx + 1 = 0 (1 + m2)x2 – 2(1 + m)x + 1 = 0 |x1 – x2| = 21

221 xx4)xx( −+

= 222

2

m114

)m1()m1(4

+−

++ = m2

m12

2+

|DE| = 1x 21 + |x1 – x2| = 2 2m1

m2+


1. A ray of light is coming along the line L = 0 and

strikes the plane mirror kept along the plane p = 0 at

B. It is given that L = 0 is 3

2x − = 4

1y − = 5

6z − and

p = 0 is x + y – 2z = 3, then find the co-ordinates of B and the equation of line along reflected ray.

Sol. let 3

2x − = 4

1y − = 5

6z − = λ

⇒ x = 2 + 3λ, y = 1 + 4λ, z = 6 + 5λ lies on plane x + y – 2z = 3 (2 + 3λ) + (1 + 4λ) – 2(6 + 5λ) = 3 ⇒ λ = –4 point B ≡ (–10, –15, –14) Let equation of reflected ray L1 = 0 is line joining

Q(x2, y2, z2) and B(–10, –15, –14)

i.e. 16

10x + = 20

15y + = 12

14z +

2. Six points (xi. yi), i = 1, 2, ...6 on the circle x2 + y2 = 4

such that ∑=

6

1iix = 8 and ∑

=

6

1iiy = 4. The line segment

joining orthocentre of a ∆ made by these points and the centroid of the ∆ made by other three points passes through a fixed point, find that point.

Sol. let ∑=

6

1iix = α and ∑

=

6

1iiy = β

let θ be the ortho centre of the ∆ made by (xi, yi), i = 1, 2, 3 ⇒ 0 is (x1 + x2 + x3, y1 + y2 + y3) and G be the centroid of the ∆ made by (xi, yi), i = 4, 5, 6

⇒ G is

++++3

yyy,

3xxx 654654

⇒ G is

β−βα−α

3,

311

Here you can say the point

βα

4,

4 divides to the OG

in 3 : 1.

3. Let f be a polynomial function such that f(x) . f(y) + 2 = f(x) + f(y) + f(xy) ∀ x ∈ R+,

y ∈ R+ ∪ 0 and f(x) is one one ∀ x ∈ R+ with f(0) = 1, f´(1) = 2, then find the area bounded between the curve y = x2 and y = g(x) where g(x) =

)x(f2 and x-axis and also find the no. of points of

nondifferentiability of h(x) = min g(x), x2, |1 – |x|| Sol. Let f(x) . f(y) + 2 = f(x) + f(y) + f(xy) ....(1) putting x = y = 1 f(1) . f(1) + 2 = 3f(1) ⇒ f(1) = 2, 1 f is given one-one and f(0) = 1 ⇒ f(1) = 2

replacing y by x1 in (1) than

f(x) . f

x1 + 2 = f(x) + f

x1 + 2

⇒ f(x) = 1 + xn also f´(1) = 2 ⇒ n = 2 ⇒ f(x) = 1 + x2 Now to find the area,

Area = 2 ∫

−

+

1

0

22 x

x12 dx = 2

−

π31

42 = π –

32

clearly by graph you can find there is 6 points of non differentiability.

4. If f(x) = x3 + ax2 + bx + c = 0 has three distinct integral roots and (x2 + 2x + 2)3 + a(x2 + 2x + 2)2 + b(x2 + 2x + 2) + c = 0 has no real roots then find the values of a, b and c

Sol. x3 + ax2 + bx + c = 0 has three distinct integral roots and (f(x))3 + a(f(x))2 + b(f(x)) + c = 0 has no real roots, where f(x) = x2 + 2x + 2

Let the roots of x3 + ax2 + bx + c = 0 be x1 > x2 > x3 respectively. Since f(x) can take all values from [1, ∞].

⇒ x1 ≤ 0 x2 ≤ –1 x3 ≤ –2 ⇒ a = –(x1 + x2 + x3) ≥ 3 b = x1x2 + x2x3 + x3x4 ⇒ b ≥ 2 and c = –(x1x2x3) ⇒ c ≥ 0

Expert’s Solution for Question asked by IIT-JEE Aspirants

Students' Forum

MATHS


5. Let A, B, and C are points represented by complex No. Z1, Z2, Z3. If the circum centre of the ∆ABC is at the origin and the altitude AD of the triangle meets the circumcircle again at P, then prove that P

represents the complex number 1

32

ZZZ−

.

Sol. Let circumcentre is point O. A Z1

C(Z3)B(Z2)

P (Z)

2C O

D π–2C

∠POC = π – 2C and ∠BOA = 2C Now applying coni's method Z3 = Zei(π –2C) ...(1) Z2 = Z1 ei 2 C ....(2) Multiplying (1) and (2) Z3Z2 = ZZ1 eiπ = –ZZ1

⇒ Z = –1

32

ZZZ−

6. Evaluate : ∑=

n

0rr

n C (r – nx)2 . xr(1 – x)n – r;

x ≠ 0, 1, n ∈ N > 2

Sol. ∑=

n

0rr

n C (r – nx)2 . xr(1 – x)n – r

= ∑=

n

0rr

n C (r2 + n2x2 – 2nxr) r

x1x

−(1 – x)n

= (1 – x)n

−+

− ∑∑==

n

0r

r

rn22

rn

0rr

n2

x1xCxn

x1xCr

−− ∑

=

rn

0rr

n

x1xC.rnx2

we know that rn

0rr

n yC.r∑=

= (1 + y)r ...(1)

⇒ rn

0rr

n

x1xC

−∑=

= n

x1x1

−+ = (1 – x)–n

Differentiating (1) w.r.t. y we get

⇒ 1rn

0rr

n yC.r −

=∑ = n(1 + y)n – 1

∴ rn

0rr

n yC.r∑=

= ny (1 + y)n–1 ....(2)

⇒ rn

0rr

n

x1xC.r

−∑=

= n1n

x1x1

x1x −

−+

−

= nx(1 – x)–n Differentiating (2) w.r.t. y we get

⇒ 1rn

0rr

n2 yC.r −

=∑ = n(1 + y)n–2y(n – 1) + (1 + y)

= ny(1 + y)n–2(ny + 1)

⇒ rn

0rr

n2

x1xC.r

−∑=

= n . 2n

x1x1

x1x −

−+

−

+

−1

x1nx

= nx(nx + 1 – x) (1 – x)–n given sum is equal to (1 – x)n nx(nx + 1 – x) (1 – x)–n + n2x2(1 – x)–n – 2nx . nx (1 – x)–n = nx(nx + 1 – x) + n2x2 – 2n2x2 = nx(1 – x)

Do you know

• The largest meteorite crater in the world is in Winslow, Arizona. It is 4,150 feet across and 150 feet deep.

• The human eye blinks an average of 4,200,000 times a year.

• Skylab, the first American space station, fell to the earth in thousands of pieces in 1979. Thankfully most over the ocean.

• It takes approximately 12 hours for food to entirely digest.

• Human jaw muscles can generate a force of 200 pounds (90.8 kilograms) on the molars.

• The Skylab astronauts grew 1.5 - 2.25 inches (3.8 - 5.7 centimeters) due to spinal lengthening and straightening as a result of zero gravity.

• An inch (2.5 centimeters) of rain water is equivalent to 15 inches (38.1 centimeters) of dry, powdery snow.

• Tremendous erosion at the base of Niagara Falls (USA) undermines the shale cliffs and as a result the falls have receded approximately 7 miles over the last 10,000 years.

• 40 to 50 percent of body heat can be lost through the head (no hat) as a result of its extensive circulatory network.

• A large swarm of desert locusts (Schistocerca gregaria) can consume 20,000 tons (18,160,000 kilograms) of vegetation a day.


Properties 1 :

If ∫ )x(f dx = F(x), then

∫b

a)x(f dx = F(b) – F(a), b ≥ a

Where F(x) is one of the antiderivatives of the function f(x), i.e. F´(x) = f(x) (a ≤ x ≤ b).

Remark : When evaluating integrals with the help of the above formula, the students should keep in mind the condition for its legitimate use. This formula is used to compute the definite integral of a function continuous on the interval [a, b] only when the equality F´(x) = f(x) is fulfilled in the whole interval [a, b], where F(x) is antiderivative of the function f(x). In particular, the antiderivative must be a function continuous on the whole interval [a, b]. A discontinuous function used as an antiderivative will lead to wrong result.

If F(x) = ∫x

a)t(f dt, t ≥ a, then F´(x) = f(x)

Properties of Definite Integrals :

If f(x) ≥ 0 on the interval [a, b], then ∫b

a)x(f dx ≥ 0

∫b

a)x(f dx = ∫

b

a)t(f dt

∫a

b)x(f dx = – ∫

b

a)x(f dx

∫b

a)x(f dx = ∫

c

a)x(f dx + ∫

b

c)x(f dx, a < c < b

∫a

0)x(f dx = ∫ −

a

0)xa(f dx

or ∫b

a)x(f dx = ∫ −+

b

a)xba(f dx

∫−

a

a)x(f dx =

=

=∫)x(f–f(–x)if0

f(x)f(–x) ifdx)x(f2b

a

∫a2

0)x(f dx =

=

=∫)x(f–x)–f(2aif0

f(x)x)–f(2a ifdx)x(f2b

a

Every continuous function defined on [a, b] is integrable over [a, b].

Every monotonic function defined on [a, b] is integrable over [a, b]

If f(x) is a continuous function defined on [a, b], then there exists c ∈ (a, b)such that

∫b

a)x(f dx = f(c) . (b – a)

The number f(c) = )ab(

1− ∫

b

a)x(f dx is called the

mean value of the function f(x) on the interval [a, b]. If f is continous on [a, b], then the integral function g

defined by g(x) = ∫x

adt)t(f for x ∈ [a, b] is

derivable on [a, b] and g´(x) = f(x) for all x ∈ [a, b]. If m and M are the smallest and greatest values of a

function f(x) on an interval [a, b], then

m(b – a) ≤ ∫b

a)x(f dx ≤ M(b – a)

If the function φ(x) and ψ(x) are defined on [a, b] and differentiable at a point x ∈ (a, b) and f(t) is continuous for φ(a) ≤ t ≤ ψ(b), then

dxd

∫

ψ

φ

)x(

)x(dt)t(f = f(ψ(x)) ψ´(x) – f(φ(x)) φ´(x)

∫b

adx)x(f ≤ ∫

b

a|)x(f| dx

If f2(x) and g2(x) are integrable on [a, b], then

∫b

adx)x(g)x(f ≤

2/1b

a

2 dx)x(f

∫

2/1b

a

2 dx)x(g

∫

Change of variables : If the function f(x) is continuous on [a, b] and the function x = φ(t) is continuously differentiable on the interval [t1, t2] and a = φ(t1), b = φ(t2), then

∫b

a)x(f dx = ∫ φ

2

1

t

t))t((f φ´(t) dt

Let a function f(x, α) be continuous for a ≤ x ≤ b and c ≤ α ≤ d. Then for any α ∈ [c, d], if

I(α) = ∫ αb

a),x(f dx, then I´(α) = ∫ α

b

a),x´(f dx,

DEFINITE INTEGRALS & AREA UNDER CURVES

Mathematics Fundamentals MATHS


Where I´(α) is the derivative of I(α) w.r.t. α and f´(x, α) is the derivative of f(x, α) w.r.t. α, kepping x constant.

Integrals with Infinite Limits : If a function f(x) is continuous for a ≤ x < ∞, then by

definition

∫∞

a)x(f dx = ∫∞→

b

ab)x(flim dx ....(i)

If there exists a finite limit on the right hand side of (i), then the improper integrals is said to be convergent; otherwise it is divergent.

Geometrically, the improper integral (i) for f(x) > 0, is the area of the figure bounded by the graph of the function y = f(x), the straight line x = a and the x-axis. Similarly,

∫ ∞−

b)x(f dx = ∫−∞→

b

aa)x(flim dx and

∫∞

∞−)x(f dx = ∫ ∞−

a)x(f dx + ∫

∞

a)x(f dx

properties :

∫a

0)x(fx dx =

21 a ∫

a

0)x(fx if f(a – x) = f(x)

and ∫ −+

a

0 )xa(f)x(f)x(f dx =

2a

∫π 2/

0dxxsinlog = ∫

π 2/

0dxxcoslog

= –2π log 2 =

2π log

21

Γ(n + 1) = n Γ (n), Γ(1) = 1, Γ

21 = π

If m and n are non-negative integers, then

∫π 2/

0

nm xcosxsin dx =

++

Γ

+

Γ

+

Γ

22nm2

21n

21m

Reduction Formulae of some Define Integrals :

∫∞ −

0

axe cos bx dx = 22 baa+

∫∞ −

0

axe sin bx dx = 22 bab+

∫∞ −

0

axe xndx = 1na!n+

If In = ∫π 2/

0

n dxxsin , then

In =

π−−

−−−

−−

−−−

)evenisnwhen(2

.21.....

4n5n.

2n3n.

n1n

)oddisnwhen(32.....

4n5n.

2n3n.

n1n

If In = ∫π 2/

0

n dxxcos , then

Im =

π−−

−−−

−−

−−−

)evenisnwhen(2

.21.....

4n5n.

2n3n.

n1n

)oddisnwhen(32.....

4n5n.

2n3n.

n1n

Leibnitz's Rule : If f(x) is continuous and u(x), v(x) are differentiable

functions in the interval [a, b], then

∫)x(v

)x(udt)t(f

dxd = fv(x)

dxd v(x) – fu(x)

dxd u(x)

Summation of Series by Integration :

∑−

=∞→

1n

0rn n

rflim .n1 = ∫

1

0dx)x(f

Some Important Results :

∑−

=

β+α1n

0r

)rsin( =

β

β

β−+α

21sin

n21sin)1n(

21sin

∑−

=

β+α1n

0r

)rcos( =

β

β

β−+α

21sin

n21sin)1n(

21cos

211 – 22

1 + 231 – .... =

12

2π

211 + 22

1 + 231 + .... =

6

2π

Area under Curves : Area bounded by the curve y = f(x), the x-axis and

the ordinates x = a, x = b

= ∫b

ay dx = ∫

b

a)x(f dx

y = f(x)

y x = b

O δx X

Y

Area bounded by the curve x = f(y), the y-axis and

the abscissae y = a, y = b


= ∫b

ax dy = ∫

b

a)y(f dy

y = b

x x = f(y)

O

δy

X

Y

y = a

The area of the region bounded by y1 = f1(x), y2 = f2(x)

and the ordinates x = a and x = b is given by

= ∫b

a2 )x(f dx – ∫

b

a1 )x(f dx

x =

a

B

O X

Y

A

x =

b

where f2(x) is y2 of the upper curve and f1(x) is y1 of

the lower curve, i.e. the required area

= ∫ −b

a12 )]x(f)x(f[ dx = ∫ −

b

a12 )yy( dx

f(x) ≤ 0 for all x in a ≤ x ≤ b, then area bounded by x-axis, the curve y = f(x) and the ordinates x = a, x = b is given by

= – ∫b

a)x(f dx

D

O X Y

C

A

B

If f(x) ≥ 0 for a ≤ x ≤ c and f(x) ≤ 0 for c ≤ x ≤ b, then

area bounded by y = f(x), x-axis and the ordinates x = a, x = b is given by

= ∫c

a)x(f dx + ∫ −

b

c)x(f dx = ∫

c

a)x(f dx – ∫

b

cdx)x(f

A

O N x

= a

C M

B

x =

b

f(x)≥0

f(x)≤0

ENERGY

• Mechanical energy is the sum of the potential and kinetic energy.

• Units: a = [m/sec2], F = [kg•m/sec2] (newton), work = pe= ke = [kg•m2/sec2] (joule)

• An ev is an energy unit equal to 1.6 × 10–19 joules

• Gravitational potential energy increases as height increases.

• Kinetic energy changes only if velocity changes.

• Mechanical energy (pe + ke) does not change for a free falling mass or a swinging pendulum. (when ignoring air friction)

• The units for power are [joules/sec] or the rate of change of energy.

ELECTRICITY • A coulomb is charge, an amp is current

[coulomb/sec] and a volt is potential difference [joule/coulomb].

• Short fat cold wires make the best conductors.

• Electrons and protons have equal amounts of charge (1.6 x 10-19 coulombs each).

• Adding a resistor in parallel decreases the total resistance of a circuit.

• Adding a resistor in series increases the total resistance of a circuit.

• All resistors in series have equal current (I).

• All resistors in parallel have equal voltage (V).

• If two charged spheres touch each other add the charges and divide by two to find the final charge on each sphere.

• Insulators contain no free electrons.

• Ionized gases conduct electric current using positive ions, negative ions and electrons.

• Electric fields all point in the direction of the force on a positive test charge.

• Electric fields between two parallel plates are uniform in strength except at the edges.

• Millikan determined the charge on a single electron using his famous oil-drop experiment.

• All charge changes result from the movement of electrons not protons (an object becomes positive by losing electrons).


Some Definitions : Experiment : A operation which can produce some

well defined outcomes is known as an experiment. Random experiment : If in each trail of an

experiment conducted under identical conditions, the outcome is not unique, then such an experiment is called a random experiment.

Sample space : The set of all possible outcomes in an experiment is called a sample space. For example, in a throw of dice, the sample space is 1, 2, 3, 4, 5, 6. Each element of a sample space is called a sample point.

Event : An event is a subset of a sample space. Simple event : An event containing only a single

sample point is called an elementary or simple event. Events other than elementary are called composite or compound or mixed events.

For example, in a single toss of coin, the event of getting a head is a simple event.

Here S = H, T and E = H In a simultaneous toss of two coins, the event of

getting at least one head is a compound event. Here S = HH, HT, TH, TT and E = HH, HT, TH Equally likely events : The given events are said to

be equally likely, if none of them is expected to occur in preference to the other.

Mutually exclusive events : If two or more events have no point in common, the events are said to be mutually exclusive. Thus E1 and E2 are mutually exclusive in E1 ∩ E2 = φ.

The events which are not mutually exclusive are known as compatible events.

Exhaustive events : A set of events is said to be totally exhaustive (simply exhaustive), if no event out side this set occurs and at least one of these event must happen as a result of an experiment.

Independent and dependent events : If there are events in which the occurrence of one does not depend upon the occurrence of the other, such events are known as independent events. On the other hand, if occurrence of one depend upon other, such events are known as dependent events.

Probability : In a random experiment, let S be the sample space

and E ⊆ S, then E is an event. The probability of occurrence of event E is defined as

P(E) = Sin element distinct ofnumber Ein elementsdistinct ofnumber =

n(S)n(E)

= outcomes possible all ofnumber

E of occurrence tofavourable outocomes ofnumber

Notations : Let A and B be two events, then A ∪ B or A + B stands for the occurrence of at

least one of A and B. A ∩ B or AB stands for the simultaneous

occurrence of A and B. A´ ∩ B´ stands for the non-occurrence of both A

and B. A ⊆ B stands for "the occurrence of A implies

occurrence of B". Random variable : A random variable is a real valued function whose

domain is the sample space of a random experiment. Bay’s rule : Let (Hj) be mutually exclusive events such that

P(Hj) > 0 for j = 1, 2, ..... n and S = Un

1jjH

=. Let A be

an events with P(A) > 0, then for j = 1, 2, .... , n

P

AH j =

∑=

n

1kkk

jj

)H/A(P)H(P

)H/A(P)H(P

Binomial Distribution : If the probability of happening of an event in a single

trial of an experiment be p, then the probability of happening of that event r times in n trials will be nCr pr (1 – p)n – r.

Some important results :

(A) P(A) = cases ofnumber Total

Aevent tofavourable cases ofNumber

= n(S)n(A)

PROBABILITY Mathematics Fundamentals M

ATHS

XtraEdge for IIT-JEE MARCH 2010 50

)AP( = cases ofnumber Total

Aevent tofavourablenot cases ofNumber

= n(S)

)An(

(B) Odd in favour and odds against an event : As a result of an experiment if “a” of the outcomes are favourable to an event E and b of the outcomes are against it, then we say that odds are a to b in favour of E or odds are b to a against E.

Thus odds in favour of an event E

= cases leunfavourab ofNumber

cases favourable ofNumber = ba

Similarly, odds against an event E

= cases favorable ofNumber

cases leunfavourab ofNumber = ab

Note : If odds in favour of an event are a : b, then the

probability of the occurrence of that event is

baa+

and the probability of non-occurrence of

that event is ba

b+

.

If odds against an event are a : b, then the probability of the occurrence of that event is

bab+

and the probability of non-occurrence of

that event is ba

a+

.

(C) P(A) + P( A ) = 1 0 ≤ P(A) ≤ 1 P(φ) = 0 P(S) = 1 If S = A1, A2, ..... An, then P(A1) + P(A2) + .... + P(An) = 1 If the probability of happening of an event in one

trial be p, then the probability of successive happening of that event in r trials is pr.

(D) If A and B are mutually exclusive events, then P(A ∪ B) = P(A) + P(B) or

P(A + B) = P(A) + P(B) If A and B are any two events, then P(A ∪ B) = P(A) + P(B) – P(A ∩ B) or P(A + B) = P(A) + P(B) – P(AB) If A and B are two independent events, then P(A ∩ B) = P(A) . P(B) or P(AB) = P(A) . P(B) If the probabilities of happening of n independent

events be p1, p2, ...... , pn respectively, then

(i) Probability of happening none of them = (1 – p1) (1 – p2) ........ (1 – pn) (ii) Probability of happening at least one of them = 1 – (1 – p1) (1 – p2) ....... (1 – pn) (iii) Probability of happening of first event and not

happening of the remaining = p1(1 – p2) (1 – p3) ....... (1 – pn) If A and B are any two events, then

P(A ∩ B) = P(A) . P

AB or

P(AB) = P(A) . P

AB

Where P

AB is known as conditional probability

means probability of B when A has occurred. Difference between mutually exclusiveness and

independence : Mutually exclusiveness is used when the events are taken from the same experiment and independence is used when the events are taken from the same experiments.

(E) P(A A ) = 0

P(AB) + P( AB ) = 1

P( A B) = P(B) – P(AB)

P(A B ) = P(A) – P(AB)

P(A + B) = P(A B ) + P( A B) + P(AB) Some important remark about coins, dice and playing cards : Coins : A coin has a head side and a tail side. If

an experiment consists of more than a coin, then coins are considered to be distinct if not otherwise stated.

Dice : A die (cubical) has six faces marked 1, 2, 3, 4, 5, 6. We may have tetrahedral (having four faces 1, 2, 3, 4,) or pentagonal (having five faces 1, 2, 3, 4, 5) die. As in the case of coins, If we have more than one die, then all dice are considered to be distinct if not otherwise stated.

Playing cards : A pack of playing cards usually has 52 cards. There are 4 suits (Spade, Heart, Diamond and Club) each having 13 cards. There are two colours red (Heart and Diamond) and black (Spade and Club) each having 26 cards.

In thirteen cards of each suit, there are 3 face cards or coart card namely king, queen and jack. So there are in all 12 face cards (4 kings, 4 queens and 4 jacks). Also there are 16 honour cards, 4 of each suit namely ace, king, queen and jack.


CHEMISTRY

SECTION – I Straight Objective Type

This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

1. If the wavelength of series limit of Lyman series for He+ ion is x Å then what will be the wavelength of series limit of Balmer series for Li+2 ion ?

(A) 4x9 Å (B)

7x4 Å

(C) 4x5 Å (D)

9x16 Å

2. Select the correct order of decreasing boiling point of the following compounds

N N | H

O | OH

| Me

(I) (II) (III) (IV)

(A) I > II > III > IV (B) I > III > IV > II (C) IV > III > II > I (D) IV > III > I > II 3. D-glucose and D-fructose can be distinguished by (A) Fehling's solution (B) Tollene's reagent (C) Br2/H2O (D) Benedict's test 4. Which of the following is an organo silicon polymer? (A) Silica (B) Silicon (C) Silicic Acid (D) Silicon carbide

5. Calculate the pH of a solution of 0.1 M Fe(NO3)3 if

acid dissociation constant for the given reaction is 1.0 × 10–3

[Fe(H2O)6] 3+ + H2O(l) H3O+ (aq) + [Fe(H2O)5OH]2+

(A) 1.5 (B) 2.02 (C) 2.64 (D) 3

6. A real gas of molar mass 60 g mol–1 has density at critical point equal to 0.80 g/cm3 and its critical

temperature is given by Tc = 821

104 5× K. Then the

van der Waal's constant 'a' (atm L2 mol–2) will be (A) 0.025 (B) 0.325 (C) 3.375 (D) 33.750

7. The molecular size of ICl and Br2 is approximately same, but boiling point of ICl is about 40° higher than that of Br2 this difference in boiling point is observed because-

(A) ICl bond is stronger than Br–Br bond (B) I.E. of iodine < I.E. of bromine (C) Iodine is larger than bromine (D) ICl is polar while Br2 is non polar

8. Which of the following reaction leads to formation of pair of enantiomers ?

(A) I H

→ OHCH3

(B) I H

→ OH2

(C) I

HDMSO

CN– →

MOCK TEST FOR IIT-JEE

PAPER - I

Time : 3 Hours Total Marks : 240

Instructions : • This question paper contains 60 questions in Chemistry (20), Mathematics (20) & Physics (20). • In section -I (8 Ques) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer. • In section -II (4 Ques) of each paper +4 marks will be given for correct answer –1 mark for wrong answer. • In section -III contains 2 groups of questions (2 × 3 = 6 Ques.) of each paper +4 marks will be given for each

correct answer & –1 mark for wrong answer. • In section -IV (2 Ques.) of each paper +8(2×4) marks will be given for correct answer & No Negative marking for

wrong answer.


(D) I

H

Me

DMF

OH→

SECTION – II Multiple Correct Answers Type

This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 9. Which of the following reaction occur in Bessemer's

converter ? (A) 2Cu2S + 5O2 → 2CuSO4 + 2CuO (B) 2Cu2S + 3O2 → 2Cu2O + 2SO2 ↑ (C) 2CuFeS2 + O2 → Cu2S + 2FeS + SO2 (D) FeO + SiO2 → FeSiO3

10. Identify the compounds which do not give positive

iodoform test in the following sequence of the reaction

(i) Hydrolysis (ii) Heating (iii) I2 + NaOH

(A) CH3 – CH – C – Et

O

CO2Et

(B)

CO2Et

O

(C)

O

O–Et

O

(D) Me – CH – CO2Et

CO2Et

11. A sample of H2O2 solution labelled as "28V" has

density of 265 gL–1. Identify the correct statement (s)

(A) 5.2M22OH = (B) % 5.8

vw

=

(C) 88.13m22OH =

(D) Mole fraction of H2O2 = 0.2 12. Select the correctly presented graph if v = velocity of e in Bohr's orbit r = radius of Bohr's orbit U = potential energy of e– in Bohr's orbit T = kinetic energy of e– in Bohr's orbit

(A)

n2

r

(B)

1/n2

U

(C)

n2

T

(D)

1/n

v

SECTION – III

Comprehension Type

This section contains 2 groups of questions. Each group has 3 multiple choice questions based on a paragraph. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which ONLY ONE is correct.

Paragraph # 1 (Ques. 13 to 15) Generally indicators used in acid-base titration

reactions are either weak acid or weak bases. Their equilibria may be presented as

HIn H+ + In– Where HIn is its acidic form and In– is its basic form.

If KIn is the indicator ionisation constant then [H+] of indicator can be expressed in the following way

[H+] = KIn . ]In[]HIn[

−

12

8 HA

HB

234

Vol.of NaOH (in mL)

pH

The indicator used in a particular acid base titration

depends on the nature of acid or base. One such indicator diagram is given.

13. The pH range of an indicator is 4-6. If it is 50%

ionised in a given solution then its ionisation constant would be

(A) 10–4 (B) 10–5

(C) 10–6 (D) None 14. Calculate the pH at equivalence point when 5 milli

mol of HB is titrated with 0.1 M NaOH. (A) 8.5 (B) 8.75 (C) 8.85 (D) 9.0 15. Which of the following indicator is most suitable for

titration of HB with strong base


(A) Phenolphthalein (8.3 – 10) (B) Bromothymol blue (6 – 7.6) (C) Methyl Red (4.2 - 6.3) (D) Malachite green (11.4–13) Paragraph # 2 (Ques. 16 to 18) A useful method to convert oxime to substituted

amide is Beckmann rearrangement which occurs through following steps,

Ph C = N

CH3 OH )I(

H→+

Ph C = N

Me OH2+

)II(

OH2 →− PhNCCH3 −=−⊕

)III(

OH2 →

CH3 –C = N–Ph

OH

(IV)

CH3 –C–NH–Ph

O

16. Rate determining step in Beckmann rearrangement is (A) I (B) II (C) III (D) IV

17. The compound Me

C = N Ph OH

when treated

with H2SO4 and hydrolysed the products formed are (A) CH3COOH and PhNH2 (B) CH3NH2 and PhCOOH (C) PhCH2NH2 and CH3COOH (D) PhCH2COOH and CH3NH2 18. In the following sequence of reaction

Ph – C – –CH3

O

64pH

OHNH2

−= → I

∆ → 5PCl P

the product P may be (A) PhCOOH

(B)

CH3– –C–NH2

O

(C)

Ph – C – NH – – CH3

O

(D)

Ph – C – NH2

O

SECTION – IV

Matrix – Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are

labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.

q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

19. Match the column Column I Column II (Compounds) (Isomers) (A) 2,3-Dihydroxy (p) Total stereoisomers butanoic acid are 10 (B) 1,3-Dichloro-1- (q) Total meso isomers methyl cyclobutane are 2 (C) 2,3,4,5-Tetrahydroxy (r) Total optical isomers hexane-1, 6-diol are 4 (D) 3-Chloro butane-2-ol (s) Total stereoisomers are 2

(t) Meso isomers zero 20. Match species of column-I with those species of

column-II which has same hybridization. Column I Column II (A) B3N3H6 (p) ClO–

(B) S2Cl2 (q) IF7

(C) XeF5– (r) CO3

2–

(D) ICl4– (s) S8

(t) XeO2F4

MATHEMATICS


This section contains 8 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. Given the function f(x) = 1 /(1 – x), the points of

discontinuity of the composite function y = f 3n (x), where f n(x) = fof …. of (n times) are (n ∈ N)

(A) 0, 1 (B) 2n (C) 3n (D) 2n + 1


2. The lines rr

= i – j + λ(2i + k) and r

r= (2i – j) + µ(i + j – k) intersect for

(A) λ = 1, µ = 1 (B) λ = 2, µ = 3 (C) all values of λ and µ (D) no value of λ and µ 3. The plane 2x – y + 3z + 5 = 0 is rotated through

90º about its line of intersection with the plane 5x – 4y + 2z + 1 = 0. The equation of the plane in the new position is

(A) 6x – 9y – 29z – 31 = 0 (B) 27x – 24y – 26z – 13 = 0 (C) 43x – 32y – 2z + 27 = 0 (D) 26x – 43y – 151z – 165 = 0 4. The tangent at the point P(x1, y1) to the parabola

y2 = 4ax meets the parabola y2 = 4a(x + b) at Q and R, the coordinates of the mid-point of QR are

(A) (x1 – a, y1 + b) (B) (x1, y1) (C) (x1 + b, y1 + a) (D) (x1 – b, y1 – b) 5. Equation of the line which is parallel to the line

common to the pair of lines given by 6x2 – xy – 12y2 = 0 and 15x2 + 14xy – 8y2 = 0 and the sum of whose intercepts on the axes is 7, is

(A) 2x – 3y = 42 (B) 3x + 4y = 12 (C) 5x – 2y = 10 (D) None of these 6. If 2 sin2 ((π/2) cos2x) = 1 – cos (π sin 2x), x ≠ (2n + 1) π/2, n ∈ I, then cos 2x is equal to (A) 1/5 (B) 3/5 (C) 4/5 (D) 1 7. The value of tan 3 α cot α cannot lie in (A) ] 0, 2/3 [ (B) ] 1/3, 3 [ (C) ] 4/3, 4 [ (D) ] 2, 10/3 [ 8. A box contains tickets numbered 1 to N. n tickets are

drawn from the box with replacement. The probability that the largest number on the tickets is k is

(A) n

Nk

(B)

n

N1k

−

(C) 0 (D) None of these


This section contains 4 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct.

9. The solution of dxdy =

xy21yx 22 ++ satisfying y(1) = 1

is given by (A) a system of hyperbola (B) a system of circles (C) y2 = x(1 + x) – 1 (D) (x – 2)2 + (y – 3)2 = 5

10. If ∫ +=−4

1

BA2dx|3x| then

(A) A = 3/2, B = 4 (B) A = 1, B = 1/2 (C) A = 2, B = –3/2 (D) A = 1/2, B = 3/2

11. If f(x) =

≤<−≤≤−−+3x2,x37

2x1,1x12x3 2. Then

(A) f(x) is increasing on [–1, 2] (B) f(x) is continuous on [–1, 3] (C) f ′(2) doesn't exist (D) f(x) has the maximum value at x = 2 12. If the line ax + by + c = 0 is a normal to the curve

xy = 1. Then (A) a > 0, b > 0 (B) a > 0, b < 0 (C) a < 0, b > 0 (D) a < 0, b < 0

SECTION – III Comprehension Type


Paragraph # 1 (Ques. 13 to 15)

The tournament for ABC Cup is arranged as per the following rules; in the beginning 16 teams are entered and divided in 2 groups of 8 teams each where the team in any group plays exactly once with all the teams in the same group.

At the end of this round top four teams from each group advance to the next round in which two teams play each other and the losing team goes out of the tournament. Then four winning teams play for semi finals and finally there is one final. The rules of the tournament are such that every match can result only in a win or a loss and not in a tie.

13. The maximum number of matches that a team going out of the tournament in the first round itself can win is -

(A) 1 (B) 2 (C) 3 (D) 4

14. The minimum number of matches that a team must win in order to qualify for the second round is -


(A) 4 (B) 5 (C) 6 (D) 7

15. Which of the following statements about a team which has already qualified for the second round is true?

(A) To win the cup it has to win exactly 14 matches (B) To win the cup it has to win exactly 3 matches (C) To win the cup it has to win exactly 4 matches (D) To win the cup it has to win exactly 5 matches

Paragraph # 2 (Ques. 16 to 18) In any ∆ABC, we known the relationship between the

sides (a, b, c), angles (A, B, C), circum-radius (R), area (∆) and other parameters.

aAsin =

bBsin =

cCsin =

R21 =

abc2∆

Using the above information, answer the following questions -

16. If two sides of a triangle are 5 and 8, and its circum

radius is 625 , then the third side can be -

(A) only 5 (B) only 7.8 (C) 5 and 7.8 (D) 6 and 7.2

17. In any triangle, a cos A + b cos B + c cos C is equal to -

(A) 4R sinA sinB sinC (B) 2R sin A sin B sin C (C) 4R cosA cosB cosC (D) 2R cos A cos B cos C

18. If 8R2 = a2 + b2 + c2, then the triangle is - (A) Right-angled (B) Isosceles (C) Equilateral (D) None of these

SECTION – IV Matrix – Match Type

This section contains 2 questions. Each question contains statements given in two columns, which have to be matched. The statements in Column I are labeled A, B, C and D, while the statements in Column II are labeled p, q, r, s and t. any given statement in Column I can have correct matching with ONE OR MORE statements (s) in column II. The appropriate bubbled corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A – p, s and t; B – q and r; C – p and q; and D – s and t; then the correct darkening of bubbles will look like the following.

q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

19. Column –I Column –II (A) If straight lines y = a1x + b and (p) a1a2 > 0

y = a2x + b meet the coordinate axes in con-cyclic points then (B) If the circles (q) b2 = a1a2 x2 + y2 + 2a1x + b = 0 and x2 + y2 + 2a2x + b = 0 touch each other then (C) If in (B) the first circle (r) b > 0 lies completely inside the second circle then (D) If the chord of contact of the (s) b = 0 tangents drawn to x2 + y2 = b2 from any point on x2 + y2 = a2

2 touches the circle x2 + y2 = a2

1, then (In all the cases a1 ≠ a2) (t) b2 – a,a2 = 0

20. Observe the following lists Column- I Column- II (A) If a, b and c are positive (p) Exactly one real numbers, then root ax3 + bx + c = 0 has (B) If c > 0 and the quadratic (q) 4a2 < (2b – c) a equation 3ax2 + 4bx + c = 0 has no real root, then (a,b,c ∈ R) (C) If the quadratic equation (r) 3a + c > 4b ax2 + bx + c = 0 has real roots and –2 lies between the roots, then (a, b, c ∈ R) (D) If the quadratic equation (s) a2 < 2a (b – 2c) ax2 + bx + c = 0 has roots α, β such that α < –2 and β > 2 then (t) 4b – c < 3a

PHYSICS



1. Three blocks are arranged as shown in which ABCD is a horizontal plane. Strings are massless and both pulley stands vertical while the strings connecting blocks m1 and m2 are also vertical and are perpendicular to faces AB and BC which are mutually perpendicular to each other. If m1 and m2 are 3 kg and 4 kg respectively. Coefficient of friction between the block m3 = 10 kg and surface is µ = 0.6 then, frictional force on m3 is -


D

A B

C

m1

m2

m3

(A) 30 N (B) 40 N (C) 50 N (D) 60 N 2. A particle of mass m is allowed to oscillate near the

minimum point of a vertical parabolic path having the equation x2 = 4ay, then the angular frequency of small oscillations of particle is –

y

x

g

x2 = 4ay m

(A) ga (B) ga2 (C) ag (D)

a2g

3. A parallel plate capacitor of plate area A and separation d is provided with thin insulating spacers to keep its plates aligned in an environment of fluctuating temperature. If the coefficient of thermal expansion of material of plate is α then the coefficient of thermal expansion (αS) of the spacers in order that the capacitance does not vary with temperature (ignore effect of spacers on capacitance)

(A) 2Sα

=α (B) αS = 3α

(C) αS = 2α (D) αS = α

4. When hydrogen like atom in excited state make a transition from excited state to ground state, most energetic photons have energy Emax = 52.224 eV and least energetic photons have energy Emin = 1.224 eV. Find the atomic number -

(A) 4 (B) 6 (C) 2 (D) 8

5. A diverging lens, focal length f1 = 20 cm is separated by 5 cm from a converging mirror, focal length f2 = 10 cm. Where should an object be placed from the lens so that a real image is formed at the object itself?

(A) 30 cm (B) 60 cm (C) 10 cm (D) 40 cm

6. If ammeter has zero resistance then –

2R

R

R

R A

ε

(A) Reading of ammeter is R6ε

(B) Reading of ammeter is R7ε

(C) Reading of ammeter is R8ε

(D) Reading of ammeter is R9ε

7. A circular disc of radius r = 5 m is rotating in horizontal plane about y-axis. y-axis a vertical axis passing through the centre of disc and x-z is the horizontal plane at ground. The height of disc above ground is h = 5 m. Small particles are ejecting from disc in horizontal direction with speed 12 m/s from the circumference of disc then the distance of these particles from origin when they hits x – z plane is -

(A) 5 m (B) 12 m (C) 13 m (D) None of these

8. The magnetic field shown in the figure consist of the two magnetic fields.

× × × × × × × × × × × × × × × ×

B

. . . . . . . . . 2B

W W

v×

If v is the velocity just required for a charge particle of mass m and charge q to pass through the magnetic field. Particle is projected with velocity 'v' then how much time does such a charge spend in the magnetic field –

(A) qB2mπ (B)

qBmπ (C)

qB4mπ (D)

qB2m3π

SECTION – II

Multiple Correct Answers Type


9. We have an infinite non-conducting sheet of negligible thickness carrying a uniform surface charge density –σ and next to it , an infinite parallel slab of thickness D with uniform volume charge density +ρ. All charges are fixed.

+ρ D

– σ

(A) Magnitude of electric field at a distance h above

the negatively charged sheet is 02

Dε

σ−ρ

(B) Magnitude of electric field inside the slab at a distance h below the negatively charged sheet

(h < D) is 02

)h2D(ε

−ρ+σ


(C) Magnitude of electric field at a distance h below

the bottom of the slab is 04

Dε

σ−ρ

(D) Magnitude of electric field at a distance h below

the bottom of the slab is 02

Dε

σ−ρ

10. A ball of mass m hits a wedge of mass '2m' with velocity 'v0' in horizontal direction and moves in vertically upward direction with velocity 'v0/2'. There is no friction between wedge and the surface –

2m 45°

m v0

(A) Coefficient of restitution between ball and wedge

is 1 (B) Coefficient of restitution between ball and wedge

is 43

(C) Impulse on wedge due to ball is 0mv25

(D) Impulse on wedge due to surface is2

mv0

11. In a modified YDSE experiment if point source of monochromatic light O is placed in such a manner

that OS1– OS2 = 4λ where λ is wavelength of light

and S1, S2 are slits separated by distance 2λ. Then value/s of θ for which a maxima is obtained will be –

θ O

S1

S2 (A) sin–1 (1/8) (B) sin–1 (–1/4) (C) sin–1 (5/6) (D) sin–1 (–7/8)

12. A source of sound producing sound of frequency 100 Hz is moving towards a wall with velocity 20 m/s as shown. If O1 and O2 are two stationary observers and beat frequencies heard by them are f1 and f2 respectively between sound directly from source and sound reflected from wall. Then: (Assume velocity of sound to be 330 m/s)

wall

O1 O2 S

(A) f1 = 0 Hz (B) f2 = 0 Hz (C) f1 > f2 (D) f1 < f2

SECTION – III

Comprehension Type


Paragraph # 1 (Ques. 13 to 15) The system shown in the diagram consists of two flat

conducting strips of length l, width b (perpendicular to plane of diagram) separated by a small gap 'a' (where a <<< b, l). The right ends of the strips are shorted and a battery of voltage V0 is connected across the left ends. The current is assumed to flow only parallel to the l-dimension of the strips. Neglect all resistances.

a

i

i

l

13. What is the self inductance of the circuit -

(A) b

a0lµ (B)

ab0 lµ

(C) 2

20

balµ

(D) 2

20

ba lµ

14. What is the voltage across the strips as a function of the distance 'x' from the shorted end. (L is a self inductance) :

(A) bLxaV00µ

(B) xbL

aV00µ

(C) bLVax

0

0µ (D)

bxLaV00µ

15. What is the rate of flow of energy down the system as a function of distance from shorted end. Where t-time and x-distance from shorted end. (L is a self inductance) :

(A) L

tVbxa 00µ (B) 2

200

LtV

bxaµ

(C) bL

txaV200µ

(D) None of these

Paragraph # 2 (Ques. 16 to 18) Nuclear reactions are performed for artificial

transmutation of elements for there are two types of nuclear reactions, exoergic and endoergic. In exoergic reactions energy is released. In endoergic reactions energy has to supplied for the reaction proceed. In exoergic reactions nuclear energy is converted into kinetic energy. In endoergic reactions, energy input is required in the form of kinetic energy to be converted into nuclear binding energy. The


minimum energy required for the reaction to take place is called threshold energy consider the reaction-

p + H31 → H2

1 + H21

Atomic masses are :- m( H1

1 –atom) = 1.007825 amu (u)

m( H31 –atom) = 3.016049 amu (u)

m( H21 –atom) = 2.014102 amu (u)

16. Protons are incident on H31 at rest. The threshold

energy for the reaction is - (A) 5.4 MeV (B) 10 MeV (C) 2 MeV (D) 8 MeV

17. When H31 are incident on protons. Threshold energy

is - (A) 10 MeV (B) 8 MeV (C) 16 MeV (D) 20 MeV 18. Which of the following is true -

(A) less energy is required for nuclear reaction if light part is at rest and heavy particle is incident

(B) More energy is required for nuclear reaction when heavy particle is incident and light particle is at rest

(C) Threshold energy does not depend on which particle is at rest

(D) Threshold energy does not depend on Q-value of reaction

SECTION – IV

Matrix – Match Type


q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

19. A bird in air is diving vertically over a tank with speed 6 cm/s. Base of the tank is silvered. A fish in

the tank is rising upward along the same line with speed 8 cm/s. (Take µwater = 4/3)

Column I Column II (in cm/s) (A) Speed of the image of (p) 16 fish as seen by the bird directly (B) Speed of the image of (q) 0 fish formed after reflection from the mirror as seen by the bird (C) Speed of image bird (r) 12 relative to the fish looking upwards (D) Speed of image bird (s) 8 relative to the fish looking downwards in the mirror

(t) 10 20. Column I shows various mass distribution and

Column II contains magnitude of gravitation field vs 'r' and magnitude of gravitation potential vs 'r' graphs. Where 'r' is distance from centre 'C'.

Column I Column II

(A)

r

P C

R (p)

E r

Solid sphere of radius 'R' from which sphere of radius 'R/2' is removed

(B)

r

P C

R (q)

E r

Cross section of long solid cylinder from which a cylinder of radius 'R/2' is removed

(C)

r

PC

R

(r)

v r

Ring of radius 'R', point 'P' is at the axis of ring perpendicular to ring and passing through its centre


(D)

R C

P

r

(s)

r

v

There identical masses placed at the corner of equilateral triangle, Point 'P' lies of a line perpendicular to plane

of triangle and passing through its centroid

(t)

v r

CHEMISTRY



1. Copper reduces NO3– to NO and NO2. The formation

of product depends on the concentration HNO3 in solution. Assuming [Cu +2] = 0.1 M,

PNO = 2NOP = 10–3 bar. At what approximate

concentration of HNO3 thermodynamic tendency for reduction of NO3

– into NO and NO2 by copper is same. Cu/Cu 2E +° = 0.34 V, NO/NO3

E −° = 0.96 V,

23 NO/NOE −° = 0.79 V]

(A) 100.12 (B) 100.45 (C) 100.66 (D) 101.23 2. Identify the final product P in following sequence of

reaction –

Br OEt

Mg

2

→ A +

O

CH3 + →OH

OEt

3

2 P

(A)

OH

CH3

(B)

OH

CH3

(C)

O

CH3

OH (D)

OH

CH3

3. A bromide ion does not interfere with the chromyl chloride test because when a bromide ion is present

(A) CrO2Br2 formed is less volatile than CrO2Cl2 (B) CrO2Br2 does not react with NaOH (C) Br2 is liberated which leaves NaOH solution

colourless (D) no gaseous substance containing bromine is

produced 4. The correct order of reactivity towards diazo

coupling with phenol in basic medium would be-

+N2Me2N

(I),

+N2O2N

(II)

MOCK TEST FOR IIT-JEE

PAPER - II


Instructions : • This question paper contains 57 questions in Chemistry (19,) Mathematics (19) & Physics (19). • In section -I (4 Ques) of each paper +3 marks will be given for correct answer & –1 mark for wrong answer. • In section -II (5 Ques) of each paper +4 marks will be given for correct answer & –1 mark for wrong answer • In section -III (2 Ques.) of each paper +8(2×4) marks will be given for correct answer. No Negative marking for

wrong answer. • In section -IV (8 Ques.) of each paper +4 marks will be given for correct answer & –1 mark for wrong answer.


+ N2 CH3O

(III) ,

+ N2 CH3

(IV)

(A) I > II > III > IV (B) IV > III > II > I (C) II > IV > III > I (D) III > I > IV > II


This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 5. For the reaction A → B, the rate law expression is

–dt

]A[d = k[A]1/2. If initial concentration of [A] is [A0]

then which of the following is/are correct ? (A) The integrated rate expression is

k =t2 ([A0]1/2 – [A]1/2)

(B) The graph of ]A[ v/s t will be

]A[

t (C) The time taken for 75% completion of reaction

t3/4 = k]A[ 0

(D) The half life period t1/2 = k]A[)12(2 0−

6.

Ph

Ph

H Me HO2C CO2H

H Me →∆ A & B, which of the

following statements is/are incorrect ? (A) A & B are structural isomers (B) A & B are enantiomers (C) A & B are geometrical isomers (D) A & B are diastereomers

7. Which of the following is/are correct ? (A) N(CH3)3 and N(SiMe3)3 are not isostructural (B) Methylisocyanate is bent but silyl isocyanate is

linear (C) In trisilyl amine all N–Si bond lengths are equal

but shorter than the expected N–Si bond length (D) Trisilyl amine is weaker base than trimethyl

amine

8. 100 ppm hardness of water equal to -

(A) 100 mg of CaCO3 in one litre of water (B) 120 mg of MgSO4 in one litre of water (C) 84 mg of MgCO3 in one litre of water (D) 111g of CaCl2 in one litre of water

9. Which of the following reactions will proceed with retention of configuration only at the chiral carbon ?

(A) H3C C C

Br O

OΘ H

⊕ →

Ag

OH2

(B) H3C

C

I

H — CH2 — S — CH3 ⊕

→Ag

OH2

(C)

CH3 Br

C2H5⊕

→Ag

OH2

(D) All of these

SECTION – III Matrix - Match Type


q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

10. Column – I Column – II (A) N2(g)+3H2(g) 2NH3(g), (p) K increases ∆H = –ve with increase in T (B) N2(g)+O2(g) 2NO(g), (q) K decreases with ∆H = +ve increase in T (C) A(g)+B(g) 2C(g), (r) Pressure has no


∆H = +ve effect (D) PCl5(g) PCl3(g) (s) Product moles + Cl2(g), ∆H = +ve increases when

He is added at constant P. (t) Products moles

increases when He is added at

constant V

11. Column – I Column – II

(A) CH3–C–CH3

OH

CN

→–EtO (p) E1

(B) Ph–CH–CH3

Br∆

→–EtO (q) E2

(C) CH–Br

Me

Εt

Me

ΕtΕtOH

∆

(r) E1CB


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(D) Ph–N–CH2CH3

O–

CH3

+ (s) Ei (elimination

intramolecular) (t) SNi (substitution

nucleophilic internal)

SECTION – IV Integer answer type

This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9. The appropriate bubbles below the respective question numbers in the ORS have to be darkened. For example, if the correct answers to question numbers X, Y, Z and W (say) are 6, 0, 9 and 2, respectively, then the correct darkening of bubbles will look like the following :

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. Equal volumes of 0.02 M AgNO3 & 0.02 M HCN

were mixed. If the [Ag+] at equilibrium was 10–n. Find n. Given Ka(HCN) = 4 × 10–10, Ksp(AgCN) = 4 × 10–16.

13. CH3CH2– −CH|CH3

CH2CH3 → νh,Cl2 Products. Total number of isomers in the product is.

14. Haemoglobin contains 0.25% iron by weight. The molecular weight of haemoglobin is 89600. Calculate the number of iron atoms per molecule of haemoglobin.

15. Two liquids A and B form an ideal solution at temperature T. When the total vapour pressure above the solution is 600 torr, the amount of A in the vapour phase is 0.35 and in the liquid phase is 0.70. What is the vapour pressure of pure A ? Express your answer after divide actual answer by 100.

16. The value of x in the complex Hx[Co(CO)4] is

17. After losing a number of α and β particles, Th23290 is

changed into .Pb20882 The number of β particles lost

are.

18. Calculate enthalpy change (in calories) adiabatic compression of one mole of an ideal monoatomic gas against constant external pressure of 2 atm starting from initial pressure of 1 atm and initial temperature of 300 K. (R = 2 cal/mol degree) Give your answer after divide actual answer by 100.

19. Calculate the emf of the cell Cd|Cd2+ (0.10M)1|| H+(0.20M)| Pt, H2(0.5 atm) [Given: EºCd2+/Cd = – 0.403 V,

F

RT303.2 = 0.0591]

Round off your answer after multiplying actual answer by 10.

MATHEMATICS



1. The points of contact of the vertical tangents to x = 2 – 3 sin θ, y = 3 + 2 cos θ are

(A) (2, 5), (2, 1) (B) (–1, 3), (5, 3) (C) (2, 5), (5, 3) (D) (–1, 3), (2, 1)

2. Let f(x) = sin x; g(x) = x2 and h(x) = log x.

If u (x) = h (f (g (x))), then 2

2

dxud is -

(A) 2 cos3 x (B) 2 cot x2 – 4x2 cosec2 x2 (C) 2x cot x2 (D) –2 cosec2 x

3. If p1and p2 are the lengths of the perpendiculars from the points (2, 3, 4) and (1, 1, 4) respectively on the plane 3x – 6y + 2z + 11 = 0, then p1, p2 are the roots of the equation

(A) p2 – 23p + 7 = 0 (B) 7p2 – 23p + 16 = 0 (C) p2 – 17p + 16 = 0 (D) p2 – 16p + 7 = 0


4. Equation of a common tangents to the curves y2 = 8x and xy = –1 is

(A) 3y = 9x + 2 (B) y = 2x + 1 (C) 2y = x + 8 (D) y = x + 2



5. If A (α, β) =

αα−αα

βe000cossin0sincos

, then -

(A) A (α, β)′ = A (–α, β) (B) A (α, β)–1 = A (–α, –β) (C) Adj (A (α, β)) = eβ A(–α, –β) (D) A (α, β)′ = A (α, –β) 6. A man observes that the angle of elevation of the top

of a tower from a point P on the ground is θ. He moves a certain distance towards the foot of the tower and finds that the angle of elevation of the top has doubled. He further moves a distance 3/4 of the previous and finds that the angle of elevation is three times that at P, then -

(A) sin θ = 12/5

(B) sin 2θ = 1/6

(C) sin 3θ = 33/52

(D) sin θ = 12/7

7. If α, β are the roots of the equation ax2 + bx + c = 0

and α′, β′ those of a′ x2 + b′ x + c′ = 0 and the circle having A(α, α′) and B(β, β′) as diameter passes

through the origin, and the point

'a'b,

ab then -

(A) a′c + ac′ = 0 (B) a′b ± ab′ = 0 (C) b′c + bc′ = 0 (D) a2b′ 2 + a′ 2b2 = 0

8. If f(x) =

=

≠+

0xfor0

0xfor)x1log(

xcoslogx2

then - (A) f(x) is continuous at x = 0

(B) f(x) is continuous at x = 0 but not differentiable at x = 0

(C) f(x) is derivable at x = 0 (D) f(x) is not continuous at x = 0

9. The value ∫ −1

0

1cot (1 + x2 – x)dx is -

(A) π/2 – log 2 (B) π – log 2

(C) π/4 – log 2 (D) 2 ∫ −1

0

1tan x dx

SECTION – III Matrix - Match Type


q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

10.

Column -I Column -II

(A) Two distinct chords passing through (a, 2a) of parabola y2 = 4ax are bisected by the line x + y = 1, then the length of latus rectum can be

(p) –1


(B) A circle drawn through the POI of the parabola y = x2 – 5x + 4 and x axis such that origin lies outside it. The length of a tangent to the circle from the origin is equal to

(q) 0

(C) If y + b = m1 (x + a) and y + b = m2 (x + a) are tangents to y2 = 4ax then m1m2 equal to

(r) 1

(D) If the point (h, –1) is outside to both the parabolas y2 = |x|, then the integral part of h can be equal to

(s) 2

(t) 3

11. A man takes a step forward with probability 0.4 and backward with probability 0.6. Suppose the man takes 11 steps and p4 denotes the probability that the man is r steps away from his initial position, then value of

Column-I Column-II (A) p1 (p) 0 (B) p3 (q) 11C5 (0.24)5 (C) p0 (r) (0.4)11 + (0.6)11 (D) p11 (s) 11C4 (0.24)4 (0.28) (t) 11 C6 (0.24)5



0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. If z ≠ 0 and 2 + cos θ + i sin θ = 3/z, then find the value of 2(z + z ) – |z|2.

13. Find the number of rational roots of p(x) = 2x98 + 3x97 + 2x96 + … + 2x + 3 = 0.

14. An infinite G.P. is selected from

1,21 ,

41 ,

81 , …….

to coverage to 1/7. If 1/2a is the first term of such a G.P., find a.

15. Let A =

α00

0 and

(A + I)50 – 50A =

dcba

, find a + b + c + d.

16. If a = (0, 1, –1) and c = (1, 1, 1) are given vectors, then |b|2 where b satisfies a × b + c = 0 and a . b = 3 is _____.

17. ABC is an isosceles triangle inscribed in a circle of radius r. if AB = AC and h is the altitude from A to BC. If the triangle ABC has perimeter P and area ∆

then 512lim0h→

r 3P∆ is equal to.

18. A function f(x) is defined for x > 0 and satisfies f(x2) = x3 for all x > 0. Then the value of f ′(4) is ___.

19. The value of y ( )318 − if (1 + x2)

dxdy = x(1 – y),

y(0) = 4/3 is _____.

PHYSICS


This section contains 4 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct. 1. A pendulum of length 10 cm is hanged by wall

making an angle 3° with vertical. It is swinged to position B. Time period of pendulum will be –


6° 3°

A

B

(A) π/5 sec (B) 152π sec

(C) π/6 sec (D) Subsequent motion will not be periodic

2. A rope of mass M is hanged from two support 'A' & 'B' as shown in figure. Maximum and minimum tension in the rope is –

θ1

θ2 B

A

(A) )sin(

cosMg

21

2θ+θθ ,

)(sincosMg

21

1θ+θθ

(B) Mg, Mg cos θ1

(C) )(sin

cosMg

21

2θ+θθ ,

)(sincoscosMg

21

21

θ+θθθ

(D) )(cos

cosMg

12

2

θ−θθ ,

)(sincosMg

21

1θ+θθ

3. Two ring of mass m and 2 m are connected with a mass less spring and can slips over two frictionless parallel horizontal rails as shown in figure. Ring of mass m is given velocity 'v0' in the direction shown. Maximum stretch in spring will be –

m v0

k

2m

(A) 0vkm (B) 0v

km3

(C) 0vk3m2 (D) 0v

km2

4. A block 'B' is just fitting between two plane inclined at an angle 'θ'. The combination of plane is inclined at angle 'α' with horizontal. If coefficient of friction

between block and the plane 'µ' is insufficient to stop slipping, then acceleration of block is –

α

B

θ

(A) gsin α – µ cos α

(B)

θα

µ−α)2/sin(

cossing

(C) g sin α – 2 µ cos α cos θ/2 (D) g sin α – µ cos α. sin (θ/2)


This section contains 5 multiple correct answer(s) type questions. Each question has 4 choices (A), (B), (C) and (D), out of which ONE OR MORE is/are correct. 5. In a modified YDSE experiment if point source of

monochromatic light O is placed in such a manner

that OS1– OS2 = 4λ where λ is wavelength of light

and S1, S2 are slits separated by distance 2λ. Then value/s of θ for which a maxima is obtained will be –

θ O

S1

S2 (A) sin–1 (1/8) (B) sin–1 (–1/4) (C) sin–1 (5/6) (D) sin–1 (–7/8) 6. A particle of charge q and mass m moves

rectilinearly under the action of an electric field E = α – βx. Here α and β are positive constants and x is the distance from the point where the particle was initially at rest then-

(A) motion of particle is oscillatory (B) amplitude of the particle is α/β

(C) mean position of the particle is at x = βα

(D) the maximum acceleration of the particle is mqα


7. Speed of a body moving in a circular path changes with time as v = 2t, then –

(A) Magnitude of acceleration remains constant (B) Magnitude of acceleration increases (C) Angle between velocity and acceleration remains

constant (D) Angle between velocity and acceleration increases

8. A wave disturbance in medium is given by y(x,t) = 0.04 cos (25πt + π/2) cos (5πx), where x and y are in meter and t is in second –

(A) An antinode occurs at x = 0 (B) Speed of wave is 5 m/s (C) A node occurs at x = 20 cm (D) Maximum velocity of medium particle is π m/s

9. Radius of rod changes linearly from a to b (a > b).Temperature of the two ends are maintained at θ1 and θ2 (θ1 > θ2) respectively. Let 'i' be the heat passing per unit cross sectional area of rod and 'θ' be temperature at a distance 'r' from one end (having cross section radius 'a'), then which of the following graphs is/are correct –

(A)

θ1 θ2

θ

r b

(B)

θ1

θ2 θ

r b

(C) i

r b

(D) i

r b

SECTION – III

Matrix - Match Type


q r

p s

r

p q

t

s t

r

A

B

C

D

p q r s t

10. Column-I contains the process of emission of electrons while column-II contains the method to achieve emission. Match column I and II.

Column I Column II (A) Thermionic emission (p) By irradiating with light (B) Photoelectric (q) By applied strong emission electric field (C) Field emission (r) By colliding accelerated electrons on metals (D) Secondary emission (s) By heating (t) None of these

11. Column-I contains different processes undergone by a diatomic ideal gas. Column-II change in different parameter of ideal gas.

Column I Column II (A) PV–1 = constant and (p) Heat is given to volume is increased gas twice (B) P2V = constant and (q) Heat is rejected by pressure is increased gas twice (C) PV6/5 = constant and (r) Work done by gas volume is reduced is negative to half the initial volume (D) PV2 = constant (s) Internal energy and pressure is increase increased 3 times (t) None of these




0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

12. A small body is projected with a velocity just

sufficient to make it reach from the surface of a planet (of radius 2R and mass 4 M) to the surface of another planet (of radius R = 1000 km and mass M). The distance between the centres of the two spherical planets is 6 R. Find the distance of the body from the centre of bigger planet in kilometers where the speed of the body is minimum. Assume motion of body along the line joining centres of planets. (Ans... ×103)

13. Light from sun is found to be maximum intensity

near 470 nm. Distance of sun from earth is 1.5 × 1011 and radius of sun is 7 × 108 m. Treating sun as a black body, calculate the intensity of radiation from sun at the surface of earth in watt/m2. (Ans. ... ×103)

14. 1 gm of helium gas undergoes process ABCA as

shown in figure. Calculate the maximum temperature of gas in degree centigrade. (Ans. ... ×102)

A C

B 225

200

3 4

(in kPa) P

V (in liters) 15. Consider the interference at P between waves

emanating from three coherent sources in same phase located at S1, S2 and S3. If intensity due to each

source is I0 = 12 W/m2 at P and D2

d2=

3λ then

calculate the resultant intensity at P. (in W/m2). (Ans.. ×101)

d

d

S1

S2

S3

P

Screen

D>>d 16. A cubical block of mass 6 kg and side 16.1 cm is

placed on frictionless horizontal surface. It is hit by a cue at the top as to impart-impulse in horizontal direction. Minimum impulse imparted to topple the block must be greater than.

17. An unstable element is produced in a nuclear reactor

at a constant rate. If its half life is 100 years, how much time in years is required to produce 50% of the equilibrium quantity ? (Ans.... ×102)

18. The wavelength of light incident on a metal surface is

reduced from 300 nm to 200 nm (both are less than threshold wavelength). What is the change in the stopping potential for photoelectrons emitted from the surface. (Take h = 6.6 × 10– 34 J-sec)

19. A mercury pallet is trapped in a tube as shown in

figure. The tube is slowly heated to expel all mercury inside it (Isothermal condition). Calculate the heat given to the tube in J. (ρHg = 13.6 gm/cc, Atmospheric pressure = 105 Pa, cross–section area of tube = 2 cm2)

10 cm

5 cm

10 cm


PHYSICS

1. An elevator, in which a man is standing, is moving upwards with a speed of 10 m/s. If the man drops the coin from a height of 2.45 m, it reaches the floor of the elevator after a time -

(A) 2

1 sec (B) 2 sec

(C) 2 sec (D) 21 sec

2. Graph between the mass of liquid inside the capillary

and radius of capillary is –

(A)

r

m (B)

r

m

(C)

r

m (D)

r

m

3. If the De –Broglie wavelength of an electron in first Bohr's orbit be λ then the minimum radial distance between the electrons in the first and second Bohr's orbit is –

(A) λ (B) 2λ (C) 2λ (D)

πλ2

4. Photoelectron are emitted with maximum kinetic energy E from a metal surface when light of frequency υ falls on it when light of frequency υ' falls on the same metal, the max. KE. of emitted Photoelectrons is found to be 2E then υ' is -

(A) υ´ = υ (B) υ´ = 2υ (C) υ´ > 2υ (D) υ´ < 2υ

5. If input in a full-wave rectifier is e = 50 sin 314t volt, diode resistance is 100 Ω and load resistance is 1K Ω then. (1) Pulse frequency output voltage is 100. (2) Input power is 1136 mw (3) Output power is 827 mw (4) Efficiency is 81.2 % (A) 1, 3 (B) 1, 2 (C) 1, 2, 3 (D) 1,2,3,4 6. An n-p-n transistor circuit is arranged as shown, it is

a –

R = 10 K V

V

NPN

(A) Common base amplifier circuit (B) Common-emitter amplifier circuit (C) Common-collector amplifier circuit (D) None

7. At what temperature will wood and iron appear equally hot or equally cold -

(A) 37° C

(B) 98.6° (C) Temperature of human body

(D) all of the above 8. In an Experiment to find loss of energy w.r.t time in

case of swinging simple pendulum mark graph between (amplitude)2 and time is –

SYLLABUS : Physics : Full syllabus Chemistry : Full syllabus Mathematics : Full syllabus


Instructions : • Part A – Physics (144 Marks) – Questions No. 1 to 2 and 9 to 30 consist FOUR (4) marks each and Question No.

3 to 8 consist EIGHT (8) marks each for each correct response. Part B – Chemistry (144 Marks) – Questions No. 31 to 39 and 46 to 60 consist FOUR (4) marks each and

Question No. 40 to 45 consist EIGHT (8) marks each for each correct response. Part C – Mathematics (144 Marks) – Questions No.61 to 82 and 89 to 90 consist FOUR (4) marks each and

Question No. 83 to 88 consist EIGHT (8) marks each for each correct response • For each incorrect response, ¼ (one fourth) of the weightage marks allotted of the would be deducted.

MOCK TEST - AIEEE PATTERN


(A)

a2

t

(B)

a2

t

(C)

a2

t

(D)

a2

t

9. A body weights 24.2g when placed is one pan of a balance and 20g when placed in other. What is the true mass of the body if the arms have un equal length -

(A) 24.2g (B) 20g (C) 22.1g (D) 22g

10. In the YDSE apparatus shown in the fig ∆x is the path difference between S2P and S1P. Now a glass slab is introduced in front of S2 then the number of fring between O and P will –

S1

P

O

S2

(A) Increase (B) decrease

(C) Many increase or decrease depends upon ∆x (D) remains constant 11. A capacitor is charged until its stored energy is 3 J

and the charging battery is removed. Now another uncharged capacitor is connected across it and it is found that charge distributes equally. The final value of total energy stored in the electric fields is -

(A) 1.5 J (B) 3 J (C) 2.5 J (D) 2 J

12. A potential difference of 30 V is applied between the ends of a conductor of length 100 m and resistance 0.5 Ω and uniform area of cross-section. The total linear momentum of free electrons is -

(A) 3.4 × 10–6 kg/s (B) 4.3 × 10–6 kg/s (C) 3.4 × 10–8 kg/s (D) 4.3 × 10–8 kg/s

13. A certain unknown resistance is connected in the left gap and a resistance box in the right gap of a metre bridge. By introducing a resistance of 10 Ω with the help of resistance box, the balance point is determined. If the balance point shift by 20 cm on increasing the resistance from the resistance box by 12.5 Ω ,then value of unknown resistance is -

(A) 15 Ω (B) 25 Ω (C) 10 Ω (D) 20 Ω

14. A cyclotron is accelerating proton, where the applied magnetic field is 2T and the potential gap is 100 keV. To acquire a kinetic energy of 20 MeV, the number of turns, the proton has to move between the dees is -

(A) 200 (B) 300 (C) 150 (D) 100 15. A magnet is suspended horizontally in the earth's

magnetic field. When it is displaced and released, it oscillates in a horizontal plane with a period T. If a piece of wood of same M.I as the magnet is attached to the magnet is attached to the magnet, the new period of oscillation of the system would be -

(A) 3T

(B) 2T (C)

2T (D) T2

16. A square loop of side 1m is placed in a perpendicular

magnetic field. Half of the area of the loop lies inside the magnetic field. A battery of emf 10V and negligible internal resistance is connected in the loop. The magnetic field change with time according to the relation B = (0.01 –2t) tesla. The total emf of the battery will be -

X X X X X

X X X X X

X X X X X

–

+ 10V

(A) 11V (B) 9 V (C) 12 V (D) 6 V

17. A Smooth ring of Mass 'M' is threaded on a string as shown in the figure. Various portions of strings are vertical. What is the condition if the ring alone is to remain at rest –

m m'

M

(A) M4 =

m1 + 'm

1 (B) M2 =

m1 +

'm1

(C) M1 =

m1 +

'm1 (D)

M3 =

m1 +

'm1


18. A Particle of mass 100 gm moves in a potential well given by U = 8x2 – 4x + 400 Joule. Find its acceleration at a distance of 25 cm from equilibrium in positive direction -

(A) 0 (B) 40 m/s2

(C) –40 m/s2 (D) 20 m/s2

19. Side of a cube is measured with a standard vernier

callipers. The main scale reads 10 mm and first division of vernier scale coincides with that of main scale. Measured value of side of cube is -

(A) 1.1 cm (B) 1.01 cm (C) 1.001 cm (D) 1.02 cm 20. While measuring the speed of sound by performing a

resonance column experiment a student gets the first resonance condition at a column length of 18 cm during winter. Repeating the same experiment during summer, she measures the column length to be x cm for the second resonance then -

(A) x > 54 (B) 54 > x > 36 (C) 36 > x > 18 (D) 18 > x 21. An ideal gas is taken through the cycle

A → B → C → A as shown in figure. If the net heat supplied to the gas in the cycle is 5 J. The work done by the gas in the process C → A is –

10

C

A

B

1

2

V(m3)

P (N/m3) (A) –5J (B) –10J (C) –15J (D) –20 J 22. The apparent frequency of the whistle of an engine

changes by the ratio 5/3 as the engine passes a stationary observer. If the velocity of sound is 340 m/s, then the velocity of the engine is -

(A) 340 m/s (B) 170 m/s (C) 85 m/s (D) 42.5 m/s 23. A ray incident at sphere an angle of incidence 60°

enters a glass sphere of R.I µ = 3 . This ray is reflected and refracted at the farther surface of the sphere. The angle between reflected and refracted rays in this surface is -

(A) 90° (B) 60° (C) 70° (D) 40°

Statement based questions: Each of the questions (Q.24 to 25) given below consist of Statement-I and Statement-II. Use the following Key to choose the appropriate answer. (A) If both Statement-I and Statement-II are true, and

Statement-II is the correct explanation of Statement-I. (B) If both Statement - I and Statement-II are true but

Statement-II is not the correct explanation of Statement-I.

(C) If Statement-I is true but Statement-II is false. (D) If Statement-I is false but Statement-II is true. 24. Statement-I : Heat supplied to a gas in a process is

100 J and work done by the gas in the same process is 120 J, then pressure of the gas in the process should increase.

Statement-II : Work done by the gas is greater than the heat supplied to the gas. Hence, internal energy of the gas should decrease.

25. Statement-I : In the circuit shown in figure current I

through the battery rises instantly to its steady state value V/R when the switch is closed, provided R = C/L .

I

R

C

V S

R

L

Statement-II : At R = C/L , τL = τC.

Passage Based questions (Q. 26 to 27) : When two concentric shells are connected by a thin

conducting wire, whole of the charge of inner shell transfers to the outer shell and potential difference between them becomes zero.

R 2R

+σ

–σ

Questions: Surface charge densities of two thin

concentric spherical shells are σ and – σ respectively. They radii are R and 2R. Now they are connected by a thin wire.

26. Potential on either of the shells will be -

(A) –02R3

εσ (B) –

0

R2εσ

(C) –02

Rε

σ (D) zero


27. Suppose electric field at a distance r (> 2R) was E1 before connecting the two shells and E2 after

connecting the two shells, 1

2

EE then is -

(A) zero (B) 1 (C) 2 (D) 1/2

Passage Based questions (Q. 28 to 30) : In perfectly inelastic collision between two bodies

momentum remains constant and the bodies stick together. Angular frequency of a spring block system

is ω =mK and maximum speed of particle is SHM is

ωA, where A is the amplitude. Question: Two identical blocks P and Q have mass

m each. They are attached to two identical springs initially unstretched. Now the left spring (along with P) is compressed by A/2 and the right spring (along with Q) is compressed by A. Both the blocks are released simultaneously. They collide perfectly inelastically. Initially time period of both the block was T.

P2A A

Q 28. The time period of oscillation of combined mass is -

(A) 2

T (B) T2 (C) T (D) 2T

29. The amplitude of combined mass is -

(A) 4A (B)

2A (C)

3A2 (D)

4A3

30. What is energy of oscillation of the combined mass ?

(A) 21 kA2 (B)

41 kA2

(C) 81 kA2 (D)

161 kA2

CHEMISTRY

31. Which of the following oxides of Chromium is amphoteric ?

(A) CrO (B) Cr2O3 (C) CrO3 (D) CrO5

32. Which of the following gives foul smelling gas with smell of rotten eggs with dil H2SO4?

(A) CO32– (B) SO3

2– (C) S2– (D) NO2–

33. The Nessler's reagent is used for the detection of ammonia the active species involved in this is-

(A) HgCl2 (B) HgCl42+

(C) HgI42– (D) Hg2I2

34. Which of the following has minimum flocculating power-

(A) Pb+2 (B) Pb+4 (C) Sr+2 (D) Na+

35. The conjugate base of H3BO3 is- (A) B(OH)4

– (B) H2BO3–

(C) HBO3– (D) H4BO3

+

36. Calculate equivalent weight of C6H12O6 in given redox change-

C6H12O6 →CO2 (A) M/2 (B) M/4 (C) M/24 (D) M/6

37. Give correct IUPAC name for CHO

CHO

CHO

-

(A) 3-Aldo pentane-1,5-dial (B) 3-Formyl-1,5-pentanedial (C) Propane-1,2,3-tricarbaldehyde (D) Propane-1,2,3- trial

38. Which of the following compounds would be hydrolysed most easily-

(A) C2H5Br (B) H3C-Br (C) H2C=CH-Br (D) H2C=CH-CH2Br

39. Which of the following alcohols is most soluble in H2O -

(A) n-Butyl alcohol (B) iso-Butyl alcohol (C) sec-Butyl alcohol (D) tert-Butyl alcohol

40. Which of the following reaction does not give amine- (A) R-X + NH3 →

(B) R-CH=NOH → OHHC/Na 52

(C) R-CN →+H/OH2

(D) R-CONH2 → 4LiAlH

41. Aniline reacts with conc. HNO3 to give -

(A) NH2H2N

(B)

NH2

NO2

NH2

NO2

and

(C)

O

O

(D) NO2


42. Benzaldehyde reacts with ammonia to form- (A) Benzal amine (B) Urotropine (C) Hydrobenzamide (D) Benzaldehyde ammonia

43. Several blocks of magnesium are fixed to the bottom of a ship to-

(A) Prevent puncturing by under sea rocks. (B) Keep away the sharks (C) Make a ship lighter (D) Prevent action of water & salt

44. Which of the following metal reacts with hot solution of NaOH and liberates H2 gas-

(A) Tin (B) Lead (C) Zinc (D) None of These

45. Which of the following compound is aspirin- (A) Methyl salicylate (B) Acetyl salicylic Acid (C) Phenyl salicylate (D) Salicylic acid

46. A coloured ppt. is obtained when H2S gas is passed through an aqueous solution of the salt in presence of NH4OH. The ppt. dissolves in dil. HCl and reacts with NaOH to give white ppt which on standing turns into brown/black mass. The cation present in the salt is -

(A) Cu2+ (B) Mg2+ (C) Ni2+ (D) Mn2+ 47. The pair in which both species have the same

magnetic moment is - (A) [Cr(H2O)6]2+ [CoCl4]2– (B) [Cr(H2O)6]2+ [Fe(H2O)6]2+ (C) [Mn(H2O)6]2+ [Cr(H2O)6]2+ (D) [CoCl4]2– [Fe(H2O)6]2+ 48. A solid has a bcc structure. If the distance of closest

approach between the two atoms is 1.73Å. The edge length of cell is-

(A) 200pm (B) 400pm (C) 100pm (D) 150pm 49. Equilibrium constant for the reaction A2 + B2 2AB is 100 at 25°C. What will be the

rate constants for the reactions (i) 2AB A2 + B2 (ii) 1/2 A2 + 1/2 B2 AB at 25°C. (A) 10–2, 10 (B) 10–4, 100 (C) 10–2, 100 (D) 10–2, 200

50. Find the t2/3of the first order reaction in which K1 = 5.48×10–14sec–1.

(A) 2.01×1013sec (B) 2.01×1010sec (C) 2.01×1016sec (D) 2.01×105sec 51. The electrode potential of Mg+2/Mg electrode in

which concentration of Mg+2 is 0.01 M is, Given that

ºMg/Mg2E + = –2.36 volt.

(A) 2.36 volt (B) –2.36 volt (C) 2.42 volt (D) –2.42 volt 52. 50ml of 2N CH3COOH mixed with 10ml of 1N

CH3COONa solution will have pH of.......(Ka= 10–5) (A) 4 (B) 5 (C) 6 (D) 7

53. An aqueous solution containing 2gm of solute dissolved in 100 gm of water freezes at –0.5°C. What is the molecular wt. of solute ? Molar heat of fusion of ice at 0°C is 1.44 KCals. & R = 2cals-

(A) 74.4 (B) 84.6 (C) 48.6 (D) 90.2

54. Calculate the maximum no. of possible e– for which 4 < n + l ≤ 6 -

(A) 18 (B) 36 (C) 72 (D) 4

55. Point out wrong statement about Resonance:- (A) Resonance structures should have equal energy. (B) In resonance structures, the constituent atoms

should in the same position. (C) In resonance structures, there should not be same

no. of e– pairs. (D) Resonance structures should differ only in the

location of electrons around the constituent atom 56. Which is not true about the following reaction ?

R–COCl + H2 4BaSO

Pd→ R–CHO + HCl

(A) It is Rosenmund reduction reaction (B) BaSO4 increases the efficiency of the catalyst (C) BaSO4 decreases the efficiency of catalyst (D) Pd-is heterogeneous catalyst

57. C2H5MgBr OH)ii(

CO)i(

2

2 → CnH2n+1COOH

The value of n is- (A) 1 (B) 2 (C) 3 (D) 0

58. The intermediate stages during the reduction of nitrobenzene to Aniline are-


(A) N=O (B)

NHOH

(C) N–N

H H

(D) (A) and (B) both

59. The volume of 10 vol. of H2O2 required to liberate 500 cm3 of O2 at STP is-

(A) 50ml (B) 5.0ml (C) 15ml (D) 100ml

60. Silver ore dissolved in dilute soln of NaCN in the presence of air to form-

(A) AgCN (B) [Ag(CN)2]– (C) AgCNO (D) [Ag(CN)4]3–

MATHEMATICS

61. If sinx + cos x = then tan2x is -

(A)1725 (B)

257 (C)

725 (D)

724

62. If tan(x + y) = 33 and x = tan–13 then y will be - (A) 0.3 (B) tan–1(1.3)

(C) tan–1 (0.3) (D) tan–1

181

63. The variance of the first n natural numbers is -

(A) 12

1n 2 − (B) ( )12

1nn 2 −

(C) 12

1n 2 + (D) ( )12

1nn 2 +

64. The negation of statement (p ∧ q ) → (q ∨ ~ r) will be. (A) (~ p ∨ ~ q) → (~ q ∧ r) (B) (p ∧ q) ∨ (~ q ∧ r) (C) (p ∧ q) ∧ (~ q ∧ r) (D) (p ∧ q) ∧ (q ∨ ~ r) 65. The angle between the lines given by the equation

ay2 – (1 + λ2)xy – ax2 = 0 is same as the angle between the lines -

(A) 5x2 +2xy–3y2 = 0 (B) x2 – y2 = 100 (C) xy = 0 (D) B & C both

66. The vector ((i – j) × (j – k)) × (i + 5k) is equal to - (A) 5i – 4j – k (B) 3i – 2j + 5k (C) 4i – 5j – k (D) 5i + 4j – k

67. If A = x; x2 – 5x + 6 = 0 B = 2, 4, C = 4, 5 then A × (B ∩ C) is - (A) (2, 4), (3, 4) (B) (4, 2), (4, 3) (C) (2, 4),(3, 4), (4, 4) (D) (2, 2),(3, 3),(4, 4), (5, 5)

68. If a > 0, b > 0, c > 0 then (a + b) (b + c) (c + a) is greater than -

(A) 2(a + b + c) (B) 6abc (C) 3(a + b + c) (D) 8abc

69. The order of the differential eqn of all conics whose axes coincide with the axes of coordinates is -

(A) 2 (B) 3 (C) 4 (D) 1

70. The area bounded by y = [x] and the two ordinates x = 1 and x = 1.7 is -

(A) 1017 (B) b = 1 (C)

517 (D)

107

71. The range of the function

f(x) = 1x

])1x[sin(4

2

++π is -

(A) [0, 1] (B) [–1, 1] (C) 0 (D) None of these

72. If A and B are square matrices of the same order and AB = 3I then A–1 is equal to -

(A) 3B (B) 31 B

(C) 3B–1 (D) None of these

73. The value of k for which points A(1, 0, 3), B(–1, 3, 4), C(1, 2, 1) & D(k, 2, 5) are coplaner, is - (A) 1 (B) 2 (C) 0 (D) –1

74. Area of triangle formed by the positive x-axis, the normal & the tangent to the circle x2 + y2 = 4 at point (1, 3 ) is -

(A) 23 sq. unit (B) 3 sq. unit

(C) 2 3 sq. unit (D) 6 sq. unit

75. A & B are two candidates seeking admission in AIEEE. The probability that A is selected is 0.5 and probability that both A & B are selected is at most 0.3. The probability of B getting selected can not exceed

(A) 0.6 (B) 0.7 (C) 0.8 (D) 0.9


76. If a plane meets the co-ordinate axes in A, B & C such that the centroid of triangle ABC is a point (1, 2r, 3r2) then equation of plane is -

(A) x +r2

y + 2r3z = 9 (B) x +

ry + 2r

z = 3

(C) 6x +ry3 + 2r

z2 =18 (D) None of these

77. If the equation of an ellipse whose focus is (–1, 1) & eccentricity is 1/2 and directrix is x – y + 3 = 0 is

ax2 + by2 + 2hxy + 2gx + 2fy + c = 0 then - (A) h2 – abc = 0 (B) a + b = c (C) a + b = h (D) a + b = 2ch

78. If shortest distance of a line x – y = 3a from a point (a, b) is |a| then b/a must be root of equation (a, b, ∈ Ro)

(A) x2 + 4x – 2 = 0 (B) x2 + 4x + 2 = 0 (C) x2 – x – 1 = 0 (D) None of these

79. If both the roots of equation 2x2 + 3 2 x + 6 = 0 are real and equal then both the roots of equation

x2 – bx + 1 = 0 are (A) Imaginary (B) both are –ve (C) One is +ve other is –ve (D) both are +ve

80. If pth, qth and rth term of a A.P. are three consecutive terms of G.P. find common ratio of the G.P.

(A) 1 (B) rpqp

−−

(C) pqqr

−− (D) None of these

81. There are four balls of different colours and four boxes of coluers same as those of the balls. The number of ways in which the balls, one in each box, could be placed such that a ball does not go to box of its own colour is :

(A) 8 (B) 7 (C) 9 (D) None of these

82. If T0, T1, T2......are the terms in expansion of (x + a)n then value of

(T0 –T2 + T4 – ...........)2 + (T1 – T3 + T5–........)2 is - (A) (x2 + a2) (B) (x2 + a2)n (C) (x2 + a2)1/n (D) (x2 + a2)–1/n

83. If the capital latters denote the cofactors of the

corresponding small letters in the determinant

∆ =

333

222

111

cbacbacba

then the value of the determinant

333

222

111

CBACBACBA

is -

(A) ∆ (B) ∆2 (C) 2∆ (D) 0

84. The value of ∑=

π

+π16

1r 17r2cosi

17r2sin is -

(A) 1 (B) i (C) –i (D) – 1

85. +→ 0xlim

2

31

2

21

21

x31xx3tan

x1x1cos

x1x2sinx

−−

+−

+

−−

−

=

(A) 1/2 (B) 1/3 (C) 1/4 (D) 1/6

86. f(x) = xcosdxsincxcosbxsina

++ decreases for all x if -

(A) ad – bc < 0 (B) ad – bc > 0 (C) ab – cd > 0 (D) ab – cd < 0 87. The minimum value of 27cos2x.81sin2x is -

(A) 2431 (B) – 5

(C) 1/5 (D) None of these

88. The value of ∫π+

+2/a

a

44 )xcosx(sin dx is -

(A) 83 a2π (B) a(π/2)2

(C) 3π/8 (D) None of these 89. The number of products that can be formed with

10 prime number taken two or more at a time is - (A) 210 (B) 210–1 (C) 210–11 (D) 210–10 90. If A ≡ (3, 4) & B is a variable point on line |x| = 6 if

AB ≤ 4 then no of position of point B with integral co-ordinates is -

(A) 5 (B) 6 (C) 10 (D) 12


PHYSICS 1. A particle is given an initial speed u inside a smooth

spherical shell of radius R = 1 m that it is just able to complete the circle. Acceleration of the particle when its velocity is vertical is -

u R

(A) g 10 (B) g

(C) g 2 (D) 3g 2. A rigid rod leans against a vertical wall (y-axis) as

shown in figure. The other end of the rod is on the horizontal floor. Point A is pushed downwards with constant velocity. Path of the centre of the rod is –

A

x B

y

(A) a straight line passing through origin (B) a straight line not passing through origin (C) a circle of radius l/2 and centre at origin (D) a circle of radius l/2 but centre not at origin 3. The height at which the acceleration due to gravity

becomes 9g (where g = the acceleration due to

gravity on the surface of the earth) in terms of R, the radius of the earth, is –

(A) 2R (B) 2

R

(C) R / 2 (D) R2

4. A system is shown in the figure. The time period for small oscillations of the two blocks will be -

m m 2k k

(A) 2π km3 (B) 2π

k4m3

(C) 2π k8m3 (D) 2π

k2m3

5. The displacement of two identical particles executing

SHM are represented by equations

x1 = 4 sin

π

+6

t10 and x2 = 5 cos ωt

For what value of ω energy of both the particles is same ?

(A) 16 unit (B) 6 unit (C) 4 unit (D) 8 unit 6. A solid sphere of mass M and radius R is placed on a

smooth horizontal surface. It is given a horizontal impulse J at a height h above the centre of mass and sphere starts rolling then, the value of h and speed of centre of mass are –

J

M C R

µ = 0

h

(A) h = 52 R and v =

MJ

(B) h = 52 R and v =

52

MJ

(C) h = 57 R and v =

57

MJ

(D) h = R57 and v =

MJ

MOCK TEST – BIT-SAT

Time : 3 Hours Total Marks : 450 Instructions :

• This question paper contains 150 questions in Physics (40) Chemistry (40), Mathematics (45), Logical Reasoning (10) & English (15). There is Negative Marking

• Each question has four option & out of them, ONLY ONE is the correct answer. There is – ve marking.

• +3 Marks for each correct & – 1 Mark for the incorrect answer.


7. As shown in figure, wheel A of radius rA = 10 cm is coupled by belt B to wheel C of radius rC = 25 cm. The angular speed of wheel A is increased from rest at a constant rate of 1.6 rad/s2. Time after which wheel C reaches a rotational speed of 100 rpm, assuming the belt does not slip, is nearly-

B

A C

(A) 4 sec (B) 8 sec (C) 12 sec (D) 16 sec

8. Let rRQ)r(P

4π= be the charge density distribution

for a solid sphere of radius R and total charge Q. For a point ‘p’ inside the sphere at distance r1 from the centre of the sphere, the magnitude of electric field is-

(A) 0 (B) 2

10 r4Q∈π

(C) 4

0

21

R4rQ

∈π (D)

40

21

R3rQ

∈π

9. An isolated and charged spherical soap bubble has a

radius 'r' and the pressure inside is atmospheric. If 'T' is the surface tension of soap solution, then charge on drop is -

(A) 0

rT22∈

(B) 8 π r 0rT2 ∈

(C) 8 π r 0rT ∈ (D) 8 π r 0

rT2∈

10. Current versus time and voltage versus time graphs

of a circuit element are shown in figure.

4.0 sec

1.0 amp

t(s)

I(A)

4.0 Volt

t(s)

V(Volt)

4.0 sec

The type of the circuit element is : (A) capacitance of 2 F (B) resistance of 2Ω (C) capacitance of 1 F (D) a voltage source of e.m.f 1 V 11. Three identical metal plates of area 'A' are at distance

d1 & d2 from each other. Metal plate A is uncharged, while plate B & C have respective charges +q & – q. If metal plates A &C are connected by switch K

through a consumer of unknown resistance, what energy does the consumer give out to its surrounding?

Assume d1 = d2 = d A

B

C

+q

–q

K

(A) A4dq

0

2

ε (B)

Adq

0

2

ε (C)

A2dq

0

2

ε (D)

Adq2

0

2

ε

12. Consider the network of equal resistances (each R)

shown in Figure. Then the effective resistance between points A an B is –

A

B

(A) (5/3) R (B) (5/6) R (C) (5/12) R (D) None of these 13. A current of 2 ampere flows in a system of

conductors as shown in the following figure. The potential difference (VA – VB) will be - (in volt)

3Ω2Ω

3Ω 2Ω

2 amp

A

B

C D

(A) +2 (B) +1 (C) –1 (D) –2 14. Consider a toroid of circular cross-section of radius b,

major radius R much greater than minor radius b, (see diagram) find the total energy stored in magnetic field of toroid –


(A) 0

222

2RbB

µπ (B)

0

222

4RbB

µπ

(C) 0

222

8RbB

µπ (D)

0

222 RbBµπ

15. AB and CD are smooth parallel rails, separated by a distance L and inclined to the horizontal at an angle θ. A uniform magnetic field of magnitude B, directed vertically upwards, exists in the region. EF is a conductor of mass m, carrying a current I. For EF to be in equilibrium:

A θ

θ C

L

D

B

E

F

(A) I must flow from E to F (B) BIL = mg cos θ (C) BIL = mg sin θ (D) BIL = mg 16. In the circuit shown the cell is ideal. The coil has an

inductance of 4H and zero resistance. F is a fuse of zero resistance and will blow when the current through it reaches 5A.The switch is closed at t = 0. The fuse will blow -

+ – 2V S

F L=4H

(A) after 5 sec (B) after 2 sec (C) after 10 sec (D) almost at once 17. In the circuit shown X is joined to Y for a long time

and then X is joined to Z. The total heat produced in R2 is –

Z

R2

X Y

R1 E

L

Fig.

(A) 21

2

R2LE (B) 2

2

2

R2LE

(C) 21

2

RR2LE (D) 3

1

22

R2RLE

18. A step down transformer reduces 220 V to 110 V. The primary draws 5 ampere of current and secondary supplies 9 ampere. The efficiency of transformer is -

(A) 20% (B) 44% (C) 90% (D)100% 19. Of the following transitions in hydrogen atom, the

one which gives emission line of minimum frequency is -

(A) n = 1 to n = 2 (B) n = 3 to n = 10 (C) n = 10 to n = 3 (D) n = 2 to n = 1 20. In uranium (Z = 92) the K absorption edge is

0.107 Å and the Kα line is 0.126 Å the, wavelength of the L absorption edge is -

(A) 0.7 Å (B) 1 Å (C) 2 Å (D) 3.2 Å 21. A material whose K absorption edge is 0.15 Å is

irradiated with 0.1 Å X-rays. The maximum kinetic energy of photoelectrons that are emitted from K-shell is-

(A) 41 KeV (B) 51 KeV (C) 61 KeV (D) 71 KeV 22. The element which has Kα X-ray line whose

wavelength is 0.18 nm is – (A) Iron (B) Cobalt (C) Nickel (D) Copper 23. The momentum of a photon having energy equal to

the rest energy of an electron is: (A) zero (B) 2.73 × 10–22 kg ms–1 (C) 1.99 × 10–24 kg ms–1 (D) infinite 24. A parallel beam of uniform, monochromatic light of

wavelength 2640 Å has an intensity of 100 W/m2. The number of photons in 1 mm3 of this radiation are –

(A) 222 (B) 335 (C) 442 (D) 555 25. The figure shows the variation of photo current with

anode potential for a photo-sensitive surface for three different radiations. Let Ia, Ib and Ic be the intensities and fa, fb and fc be the frequencies for the curves a, b and c respectively -

b a c

O Anode potential

Photo current


(A) fa = fb and Ia ≠ Ib

(B) fa = fc and Ia = Ic (C) fa = fb and Ia = Ib

(D) fb = fc and Ib = Ic 26. The internal resistance of a cell is determined by

using a potentiometer. In an experiment, an internal resistance of 100 Ω is used across the given cell. When the key K2 is closed, the balance length on the potentiometer decreases from 90 cm to 72 cm. Calculate the internal resistance of the cell -

(A) 100Ω (B) 75Ω (C) 50Ω (D) 25Ω 27. In the potentiometer arrangement shown, the driving

cell D has e.m.f. E and internal resistance r. The cell C whose e.m.f. is to be measured has e.m.f. E/2 and internal resistance 2r. The potentiometer wire is 100 cm long. If the balance is obtained the length AP = l, then-

D(E,r)

P B A

C (E/2, 2r) G

(A) l = 50 cm (B) l > 50 cm (C) l < 50 cm (D) Balance will not obtained 28. The figure shows a metre-bridge circuit, with AB =

100 cm, X = 12 Ω and R = 18 Ω, and the jockey J in the position of balance.

R

+ –

X A J B

If R is now made 8 Ω, through what distance will J have to be moved to obtain balance?

(A) 10 cm (B) 20 cm (C) 30 cm (D) 40 cm 29. The pitch of a screw gauge is 0.1 cm. The number of

divisions on its circular scale is 100. In the measurement of diameter of a wire with this screw gauge the linear scale reading is 'N' cm and the number of division on the reference line is n. Then the radius of the wire in cm will be -

(A) N + 0.01 n (B) N + 0.001 n (C) 0.5 N + 0.001 n (D) 5(0.1 N+0.0001 n) 30. When Li6

3 is bombarded with 4 MeV deutrons, one reaction that is observed is the formation of two α-particles, each with 13.2 MeV of energy. The Q-value for this reaction is -

(A) 13.2 MeV (B) 26.4 MeV (C) 22.4 MeV (D) 4 MeV 31. In a radioactive decay, let N represent the number of

residual active nuclei, D the number of daughter nuclei, and R the rate of decay at any time t. Three curves are shown in Fig. The correct ones are –

t(1)

NNR

t(2)

t(3)

D

(A) 1 and 3 (B) 2 and 3 (C) 1 and 2 (D) all three 32. Young's double slit experiment is made in a liquid. The

10th bright fringe in liquid lies where 6th dark fringe lies in vacuum. The refractive index of the liquid is approximately-

(A) 1.8 (B) 1.54 (C) 1.67 (D) 1.2 33. A particle moves in a circle of diameter 1 cm with a

constant angular velocity. A concave mirror of focal length 10 cm is placed with its principal axis passing through the centre of the circle and perpendicular to its plane. The distance between the pole of the mirror and the centre of the circle is 30 cm. The ratio of acceleration of image to that of object is -

(A) 21 (B)

41 (C) 2 (D) 4

34. A concave mirror of focal length 15 cm forms an

image having twice the linear dimensions of the object. The position of the object when the image is virtual will be-

(A) 22.5 cm (B) 7.5 cm (C) 30 cm (D) 45 cm 35. A telescope has focal length of objective and

eye-piece as 200 cm and 5 cm. What is the magnification of telescope ?

(A) 40 (B) 80 (C) 50 (D) 101 36. A compound microscope has magnifying power as 32

and magnifying power of eye-piece is 4, then the magnifying power of objective is -

(A) 8 (B) 10 (C) 6 (D) 12


37. Two blocks are connected by a massless string through an ideal pulley as shown. A force of 22N is applied on block B when initially the blocks are at rest. Then speed of centre of mass of block A and block B, 2 sec, after the application of force is (masses of A and B are 4 kg and 6 kg respectively and surfaces are smooth) –

A 4kg

B 6kg

F = 22 N

(A) 1.4 m/s2 (B) 1 m/s2 (C) 2 m/s2 (D) None of these 38. A chain of length 1.5 πR and mass ‘m’ is put on a

mounted half cylinder as shown in figure. Chain is pulled by vertically downward force 2 mg. Assuming surfaces to be friction less, acceleration of chain is –

R

F = 2mg

(A) 2g (B) 3g2 (C)

2g (D)

3g5

39. In hydraulic press radii of connecting pipes r1 and r2

are in ratio 1 : 2. In order to lift a heavy mass M on larger piston, the small piston must be pressed through a minimum force f equal to -

f M

(A) Mg (B) Mg/2 (C) Mg/4 (D) Mg/8 40. A uniform rod of length 2.0 m specific gravity

0.5 and mass 2 kg is hinged atone end to the bottom of a tank of water (specific gravity = 1.0) filled upto a height of 1.0 m as shown in figure. Taking the case θ ≠ 0º the force exerted by the hinge on the rod is : (g = 10 m/s2) –

θ 1.0 m

O

(A) 10.2 N upwards (B) 4.2 N downwards (C) 8.3 N downwards (D) 6.2 N upwards

CHEMISTRY

1. According to Bohr’s theory, angular momentum of an electron in fourth orbit is -

(A) π2

h (B) π4

h (C) πh2 (D)

πh4

2. 1.25g of a solid dibasic acid is completely neutralized by 25 ml. of 0.25 molar Ba(OH)2 solution. Molecular mass of the acid is -

(A) 100 (B) 150 (C) 120 (D) 200 3. Rates of effusion of hydrogen and deuterium under

similar conditions are in the ratio - (A) 1 : 1 (B) 2 : 1 (C) 2 : 1 (D) 1 : 4 4. For equilibrium NH4HS(s) NH3(g) + H2S(g)

KC = 1.8 × 10–4 at 298 K. The value of Kp at 298 K is-

(A) 0.108 (B) 4.4 × 10–3

(C) 1.8 × 10–4 (D) 4.4 × 10–4

5. Given that H2O (l) → H2O(g) ; ∆H = + 43.7 kJ H2O (s) → H2O (l) ; ∆H = + 6.05 kJ ∆Hsublimation of ice is - (A) 49.75 kJ mol–1 (B) 37.65 kJ mol–1 (C) 43.7 kJ mol–1 (D) – 43.67 kJ mol–1

6. Which of the following is a Lewis base ? (A) CO2 (B) BF3 (C) Al3+ (D) CH3NH2

7. The solubility product Ksp of sparingly soluble salt Ag2CrO4 is 4 × 10–12. The solubility of the salt is -

(A) 1 × 10–12 M (B) 2 × 10–6 M (C) 1 × 10–6 M (D) 1 × 10–4 M 8. Which of the following chemical reactions depicts

the oxidising behaviour of H2SO4 ? (A) 2HI + H2SO4 → I2 + SO2 + 2H2O (B) Ca(OH)2 + H2SO4 → CaSO4 + 2H2O (C) NaCl + H2SO4 → NaHSO4 + HCl (D) 2PCl5 + H2SO4 → 2POCl3 + 2HCl + SO2Cl2


9. Potassium has a bcc structure with nearest neighbour distance of 4.52 Å. If atomic mass of potassium is 3a, its density is -

(A) 454 kg m–3 (B) 804 kg m–3

(C) 852 kg m–3 (D) 900 kg m–3 10. If 0

Zn/Zn2E + = – 0.763 V and 0Cd/Cd2E + = – 0.403 V,

the emf of the cell Zn | Zn2+ ||Cd2+|Cd (a = 0.004), (a = 0.2) will be

given by -

(A) E = – 0.36 + 2059.0 log

2004.0

(B) E = + 0.36 +2059.0 log

204.0

(C) E = – 0.36 + 2059.0 log

004.02.0

(D) E = + 0.36 + 2059.0 log

004.02.0

11. The value of P° for benzene of certain temperature is 640 mm of Hg. The vapour pressure of solution containing 2.5 g of a certain substance ‘A’ in 39.0 g of benzene is 600 mm of Hg. The molecular mass of A is -

(A) 65.25 (B) 130 (C) 40 (D) 80

12. For adsorption, ∆H is - (A) + ve (B) – ve (C) zero (D) may + ve or –ve 13. A reaction which is of first order w.r.t. reactant A,

has a rate constant 6 min–1. If we start with [A] = 0.5 mol L–1, when would [A] reach the value of 0.05 mol L–1 ?

(A) 0.384 min (B) 0.15 min (C) 3 min (D) 3.84 min

14. The number of molecules present in 1 cm3 of water is (density of H2O = 1 g cm–3)

(A) 2.7 × 1018 (B) 3.3 × 1022

(C) 6.02 × 1020 (D) 1000

15. CH3NH2 + CHCl3 + KOH → Nitrogen containing compound + KCl + H2O

Nitrogen containing compound is – (A) CH3 – C ≡ N (B) CH3 – NH – CH3 (C) CH3 – Ν ≡ C+ (D) CH3 – N+ ≡ C–

16. 4-methyl benzene sulphonic acid react with sodium acetate to give –

(A)

CH3

SO3Na

; CH3COOH

(B)

COONa

CH3

; SO3

(C)

Br

; SO3

(D)

SO2 – O – C – CH3

CH3

O; NaOH

17. The product(s) obtained via oxymercuration (HgSO4 + H2SO4) of 1-butyne would be –

(A) CH3CH2COCH3 (B) CH3CH2CH2CHO (C) CH3CH2CHO + HCHO (D) CH3CH2COOH + HCOOH

18. Acetophenone is prepared by the reaction of which of the following in the presence of AlCl3 catalyst –

(A) Phenol and acetic acid (B) Benzene and acetone (C) Benzene and acetyl chloride (D) Phenol and acetone

19.

OCH3

CH3

heat2NaOH/Br.1 2

− →

(A)

OCH3

CH3

Br (B)

OCH3

CH3Br

(C)

OCH3

CH3Br

(D)

OCH3

CH3

Br

20. Phenol → 422 SOH/NaNO B → OH2 C →NaOH D Name of the above reaction is – (A) Libermann's reaction (B) Phthalein fusion test (C) Reimer-Tiemann reaction (D) Schotten-Baumann reaction

21.

CCl3

→ Fe/Brof.eqv1 2 A. Compound A is -


(A)

CCl3

Br

(B)

CCl3

Br

(C)

CCl3

Br Br

(D)

CCl3

Br

22. In a reaction

CH2 = CH2 acidusHypochloro → M →R

CH2 – OHCH2 – OH

where M = molecule R = Reagent M and R are (A) CH3CH2Cl and NaOH (B) CH2Cl – CH2OH and aq. NaHCO3 (C) CH3CH2OH and HCl (D) CH2 = CH2 and heat 23. Which of the following will have least hindered

rotation about carbon-carbon bond – (A) Ethane (B) Ethylene (C) Acetylene (D) Hexachloroethane 24. Which is least reactive towards nucleophilic

substitution (SN2) (A) CH2 = CH2 – CH2 – Cl

(B)

CH3 CH3 – C – Cl

CH3

(C) Cl

(D) Cl

CH3 – CH – CH3

25. Among the following the least stable reasonance

structure is –

(A)

N - - -

O

O -

- -

(B)

N -

- O

O -

-

(C)

N - - O

O -

-

(D) -

-O

O -

- N- -

26. Homolytic fission of C–C bond in ethane gives an intermediate in which carbon is -

(A) sp3 hybridised (B) sp2 hybridised (C) sp hybridised (D) sp2d hybridised

27. The IUPAC name of the compound is –

(A) (2E, 4E)-2, 4-hexadiene (B) (2Z, 4Z)-2, 4-hexadiene (C) (2Z, 4E)-2, 4-hexadiene (D) (2E, 4Z)-4, 2-hexadiene

28. The brown ring test for −2NO and −

3NO is due to the formation of complex ion with the formula –

(A) [Fe(H2O)6]2+ (B) [Fe(NO)(CN)5]2+ (C) [Fe(H2O)5NO]2+ (D) [Fe(H2O) (NO)5]2+

29. The correct order for the wavelength of absorption in the visible region is –

(A) [Ni (NO2)6]4– < [Ni(NH3)6]2+ < [Ni(H2O)6]2+ (B) [Ni (NO2)6]4– < [Ni(H2O)6]2+ < [Ni(NH3)6]2+ (C) [Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni (NO2)6]4– (D) [Ni(NH3)6]2+ < [Ni(H2O)6]2+ < [Ni (NO2)6]4– 30. In nitroprusside ion,, the iron and NO exists as Fe (II)

and NO+ rather than Fe(III) and NO these forms can be differentiated by –

(A) Estimating the concentration of iron (B) Measuring the concentration of CN– (C) Measuring the solid state magnetic moment (D) Thermally decomposing the compound 31. Four reactions are given below I 2Li + 2H2O → 2LiOH + H2 II 2Na + 2H2O → 2NaOH + H2 III 2LiNO3 →heat 2LiNO2 + O2

IV 2NaNO3 →heat 2NaNO2 + O2 Which of the above if any is wrong (A) IV (B) III (C) I (D) None of these 32. Name of the structure of silicates in which three

oxygen atoms of [SiO4]4– are shared is – (A) Pyrosilicate (B) Sheet silicate (C) Linear chain silicate (D) Three dimensional silicate 33. The metallic lusture exhibited by sodium is explained

by – (A) Diffusion of sodium ions (B) Oscillation of loose electron (C) Excitation of free protons (D) Existence of body centred cubic lattice 34. Hydrogen is evolved by the action of cold dil. HNO3

on – (A) Fe (B) Mn (C) Cu (D) Al


35. 'Lapis-Lazuli' is a blue coloured precious stone. It is mineral of the class –

(A) Sodium alumino silicate (B) Zinc-cobaltate (C) Basic copper carbonate (D) Prussian blue 36. In which of the following arrangements the order is

not according to the property indicating against it – (A) Al3+ < Mg2+ < Na+ < F– (increasing ionic size) (B) B < C < N < O (increasing first I.E.) (C) I < Br < F < Cl (increasing electron gain enthalpy (–ve)) (D) Li < Na < K < Rb (increasing metallic radius) 37. Which set of hybridisation is correct for the

following compound NO2, SF4, −

6PF (A) sp, sp2, sp3 (B) sp, sp3d, sp3d2 (C) sp2, sp3, d2sp3 (D) sp3, sp3d2, sp3d2 38. The increasing order of atomic radius for the

elements Na, Rb, K and Mg is – (A) Mg < Na < K < Rb (B) K < Na < Mg < Rb (C) Na < Mg < K < Rb (D) Rb < K < Mg < Na 39. Which of the following ion forms a hydroxide highly

soluble in water – (A) Ni2+ (B) K+ (C) Zn2+ (D) Al3+

40. When CO2 is bubbled into an aqueous solution of Na2CO3 the following is formed –

(A) NaOH (B) NaHCO3

(C) H2O (D) OH–

MATHEMATICS

1. y = 2x2 – log | x | passes - (A) two minima & one maxima (B) Two maxima and one minima (C) Only two minima (D) Only two maxima

2. The function f(x) = 1 + x sin x [cos x],

0 < x ≤ 2π [ . ] = G.I.F.

(A) is continuous on (0, π/2) (B) is strictly decreasing in (0, π/2) (C) is strictly increasing in (0, π/2) (D) has global maximum value 2

3. If the radius of a spherical balloon is measured with in 1 % the error (in percent) in the volume is –

(A) 4 π r2 % (B) 3 %

(C) %788

(D) None

4. The relation R defined on the set A = 1, 2, 3 is given by R = (1, 1) (2, 2) then number of correct choices from the following is -

(i) reflexive (ii) symmetric (iii) Transitive (iv) anti symmetric (A) 1 (B) 2 (C) 3 (D) 4 5. Let U be the universal set and A ∪ B ∪ C = U then

(A – B) ∪ (B – C) ∪ (C – A)c = (A) A ∩ (B ∩ C) (B) A ∩ (B ∪ C) (C) (A ∩ B ∩ C) (D) None of these 6. If A and B are square matrices of same size and

| B | ≠ 0 then (B–1 AB)4 = (A) (B4)–1 AB4 (B) BA4B–1 (C) B–1A4B (D) None of these

7. Let g(x) = )3('f)2('f)('f)3(f)2(f)(f

)3x(f)2x(f)x(f

αααααα

α+α+α+

where α is a constant then 0x

lim→ x

)x(g =

(A) 0 (B) 1 (C) –1 (D) None

8. If ∆1 = ba2ey2x4z2ed2f

and ∆2 = z2y2x4

fed2eba2

then

∆1/∆2 =

(A) 1 (B) 2 (C) 21 (D) None

9. The number of ways in which 20 one rupee coin can be distributed among 5 people such that each person gets at least 3 rupee is –

(A) 26 (B) 63 (C) 125 (D) None

10. The total number of six digit number x1 x2 x3 x4 x5 x6 have the property that

x1 < x2 ≤ x3 < x4 < x5 ≤ x6 is equal to – (A) 10C6 (B) 12C6 (C) 11C6 (D) None

11. 2

+++ ....4

a2

a142

; a = logen is equal to –

(A) n

)1n( − (B) n

1n2 − (C) n

1n + (D) n

1n2 +


12. The term independent of x in 18

x2x

− is –

(A) 18C626 (B) 18C12212 (C) 18C828 (D) None of these

13. 1 + 32 .

21 +

5.35.2 2

21

+

9.6.38.5.2 3

21

+ . . . . =

(A) 21/3 (B) 31/4 (C) 41/3 (D) 31/3 14. If ω is imaginary cube root of unity then arg(iω) + arg (iω2) (A) 0 (B) π/2 (C) π (D) None

15. ∑=

n

1r 2 4log1

r

is equal to –

(A) 4

)1n(n + (B) 2

)1n(n +

(C) n(n + 1) (D) None of these 16. If a1, a2, . . . . a15 are in A.P. and a1 + a8 + a15 = 15

then a2 + a3 + a8 + a13 + a14 = (A) 15 (B) 10 (C) 25 (D) None 17. Let a, b, c be positive real numbers, such that

bx2 + 22 b4)ca( ++ x + (a + c) ≥ 0 ∀ x ∈ R then a, b, c are in –

(A) G.P. (B) A.P. (C) H.P. (D) None 18. If a1 < a2 < a3 < a4 < a5 < a6 then the equation (x – a1)

(x – a3) (x – a5) + 2(x – a2) (x – a4) (x – a6) = 0 has – (A) Four real roots (B) One real root (C) One real root in each interval (a1, a2), (a3, a4) and

(a5, a6) (D) None of these

19. Solution of the differential equation xdx + zdy + (y + 2z)dz = 0 is – (A) x2 + 2yz + 2z2 = c (B) x2 + yz + z2 = c (c) x2 + 2yz + z2 = c (D) None of these 20. The slope of the tangent to the curve y = f(x) at (x,

f(x)) is (2x + 1). If the curve passes through the point (1, 2), then the area bounded by the curve, x-axis and the lines x = 1, x = 0 is –

(A) 5/6 (B) 6/5 (C) 6 (D) 1

21. The maximum area of a rectangle whose two consecutive vertices lie on the x-axis and another two lie on the curve

y = e–|x| is equal to – (A) 2e (B) 2/e (C) e (D) 1/e

22. ∫ xsin 2 dx =

(A) – cos x + C (B) cos x + C (C) – cos x sgn sin x+C (D) None of these

23. ∫ φφ−φ

)x()x(f)x()x('f)x(')x(f log

)x(f)x(φ dx =

(A) log )x(f)x(φ + C (B)

2

)x(f)x(log

21

φ + C

(C) )x(f)x(φ log

)x(f)x(φ + C (D) None of these

24. Segment of the tangent to the curve xy = c2 at the point (x′, y′) which is contained between the co-ordinate axes is bisected at the point –

(A) (–x′, y′) (B) (y′, x′)

(C)

2'y,

2'x (D) None of these

25. There is a point P(a, a, a) on the line passing through the origin and equally inclined with axes the equation of the plane perpendicular to OP and passing through P cuts the intercepts on axes the sum of whose reciprocals is –

(A) a (B) 3/2a (C) 3a/2 (D) 1/a

26. If ar

= k17j5ip ++ and kj13iq2b ++=r

have equal magnitude and p, q are positive integer ∈ [1, 1000] then the total number of ordered pair (p, q) is –

(A) 33 (B) 32 (C) 31 (D) None

27. If ar

, br

, cr

be such that | cbarrr

++ | = 1, bacrrr

×λ=

and | ar

| = 2

1 , | br

| =3

1 , | cr

| = 6

1 then the angle

between ar

and br

is- (A) π/6 (B) π/4 (C) π/3 (D) π/2

28. The equation a8

x 2

− +

2ay2

−= 1 will represent an

ellipse if – (A) a ∈ (1, 4) (B) a ∈ (– ∞, 2) ∪ (8, ∞) (C) a ∈ (2, 8) (D) None of these

29. Angle between the tangent drawn to y2 = 4x at the point where it is intersected by line y = x – 1 is –

(A) π/6 (B) π/3 (C) π/4 (D) π/2

30. Consider four circles (x ± 1)2 + (y ± 1)2 = 1 equation of smaller circle touching these four circles is –

(A) x2 + y2 = 3 – 2 (B) x2 + y2 = 6 – 23

(C) x2 + y2 = 5 – 22 (D) x2 + y2 = 3 – 22


31. If the point P(a, a2) lies completely inside the triangle formed by the lines x = 0, y = 0 and x + y = 2 then exhaustive range of 'a' is –

(A) a ∈ (0, 1) (B) a ∈ (1, 2 )

(C) a ∈ ( 2 – 1, 2 ) (D) a ∈ ( 2 – 1, 1) 32. The distance between the orthocentre and the

circumcentre of the triangle with vertices (0, 0) (0, a) and (b, 0) is –

(A) )ba(21 22 + (B) a + b

(C) a – b (D) 2

ba 22 +

33. If the sides of a ∆ are 3 : 7 : 8 then R : r is equal to – (A) 2 : 7 (B) 7 : 2 (C) 3 : 7 (D) None 34. The equation sin x (sin x + cos x) = K has real

solution then K belongs to –

(A)

+2

21,0 (B) (2 – 3 , 2 + 3 )

(C) (0, 32 ) (D)

+−2

21,2

21

35. The function f(x) = xtanx1

x+

, (0, 2π ) has –

(A) One point of minimum (B) One point of maximum (C) No extreme point (D) Two point of maximum 36. If solution of the equation

3cos2θ – 32 sinθ cosθ – 3 sin2θ = 0 are nπ + rπ and

sn π

+π then | r – s | =

(A) 3 (B) 9 (C) 7 (D) 1

37. If cot–1 πn >

6π , n ∈ N then maximum value of n =

(A) 6 (B) 5 (C) 4 (D) 3

38. Period of the function f(x) = sin 3πx + tan π [x] where [.] and . denote the integral part and fractional part respectively, is given by –

(A) 1 (B) 2 (C) 3 (D) π

39. The domain and range of f(x) = cos–1

x|x|log ]x[ .

Where [.] denotes the greatest integer function respectively –

(A) [ 1, ∞), [0, 2π ] (B) [2, ∞), [0,

2π ]

(C) [2, ∞), 2π (D) [1, ∞), 0

40. The graph of the function y = f(x) has a unique

tangent not parallel to x-axis at the point (a, 0) through which the graph passes, then

axlim→ )x(f3

)x(f61loge + is –

(A) 1 (B) 0 (C) 2 (D) None

41. If P = +→5x

lim]x[x20x9x2

−+− –

–4xlim→ ]x[x

20x9x2

−+− and

Q = +→4x

lim]x[x20x9x2

−+− –

−→5xlim

]x[x20x9x2

−+−

[.] = G.I.F. then QP =

(A) 1 (B) 2 (C) 3 (D) None

42. Let f(x) =

=

≠−−

1x;0

1x;1x1]x[

2

22

2

then at x = 1, f(x) is –

(A) Differentiable (B) Discontinuous (C) Continuous not differentiable (D) None of these

43. If (a + bx)ey/x = x then 2y

1 (xy1 – y)2 =

(Α) x3 (B) 3x2 (C) 1/x3 (D) None 44. If f(x) is continuous function such that

∫+

=1n

n

3ndx)x(f n ∈ Z then ∫−

3

2

dx)x(f =

(A) 16 (B) 0 (C) 2 (D) None

45. If x2f(x) +

x1f = 2 for all x except at x = 0 then

∫3

3/1

dx)x(f =

(A) 4/3 (B) 8/3 (C) 1/3 (D) None


LOGICAL REASONING

1. Fill in the blank spaces 6, 13, 28, ? . . . . (A) 56 (B) 57 (C) 58 (D) 59 2. Choose the best alternative Car : Petrol : : T.V. : ? (A) Electricity (B) Transmission (C) Entertainment (D) Antenna 3. Pick the odd one out – (A) Titan (B) Mercury (C) Earth (D) Jupiter 4. Direction : In questions, find out which of the

figures (A), (B), (C) and (D) can be formed from the pieces given in (x).

(x)

(A)

(B)

(C)

(D)

5. Directions : In question, choose the set of figures

which follows the given rule. Rule : Closed figures become more and more open

and open figures more and more closed.

(A)

(B)

(C)

(D)

6. Directions : In question below, you are given a

figure (X) followed by four figures (A), (B), (C) and (D) such that (X) is embedded in one of them. Trace out the correct alternative.

(x)

(A)

(B)

(C)

(D)

7. Directions : In following question, find out which of

the answer figures (A), (B), (C) and (D) completes the figure – matrix ?

?

(A)

(B)

(C)

(D)

8. Directions : The questions that follow contain a set

of three figure X, Y and Z showing a sequence of folding of piece of paper. Fig. (Z) shows the manner in which the folded paper has been cut. These three figure are followed by four answer figure from which you have to choose a figure which would most closely resemble the unfolded form of figure. (Z)

X Y Z

(A)

A

(B)

B

(C)

C

(D)

D

9. Direction : In following questions, complete the

missing portion of the given pattern by selecting from the given alternatives (A), (B), (C) and (D).

(X)

?

(A)

(B)

(C) (D)

10. Directions : In question below, you are given a

figure (x) followed by four figures (A), (B), (C) and (D) such that (X) is embedded in one of them. Trace out the correct alternative.

(X)

(A)

(B)

(C) (D)


ENGLISH

1. Find the correctly spelt word – (A) Geraff (B) Giraffe (C) Giraf (D) Gerraffe 2. Find out that word where the spelling is wrong – (A) Puncture (B) Puntuation (C) Pudding (D) Pungent 3. Pick up the correct synonym for the following words Plush : (A) Luxurious (B) Delicious (C) Comforting (D) Tasty 4. Choose the alternative which can replace the word

printed in underline without changing the meaning of the sentence.

When he returned, he was accompanied by 'sprightly' young girl.

(A) Lively (B) Beautiful (C) Sportive (D) Intelligent 5. Choose one alternative which is opposite in meaning

to the given word : Astute : (A) Wicked (B) Impolite (C) Cowardly (D) Foolish 6. Choose the word which is closest to the 'opposite' in

meaning of the underlined word Many snakes are 'innocuous' : (A) Deadly (B) Ferocious (C) Poisonous (D) Harmful 7. Choose the one which can be substituted for the

given words/sentences : Giving undue favours to one's kith and kin' (A) Corruption (B) Worldliness (C) Favouritism (D) Nepotism 8. Find out which one of the words given below the

sentence can most appropriately replace the group of words underlined in the sentence :

The bus has to "go back and forth" every six hours. (A) Cross (B) Shuttle (C) Travel (D) Run 9. Read both the sentences carefully and decide on their

correctness on the basis of the underlined words : 1. I am out of practise these days 2. I practice law

(A) Only 1 is correct (B) Only 2 is correct (C) Both the sentences 1 & 2 are correct (D) Both the sentences 1 & 2 are incorrect 10. Which one of the two sentences given below is

wrong on the basis of the underlined words : 1. He is a very "ingenuous" businessman. 2. I like him for his "Ingenious" nature. (A) Sentence 1 is correct (B) Sentence 2 is correct (C) Both the sentences can be made correct by

interchanging the underlined words. (D) Both the sentences can not be interchanged hence,

both are wrong. 11. Choose from the given words below the two

sentences, that word which has the same meaning and can be used in the same context as the part given underlined in both the sentences :

1. His "aloof" behaviour is an indication of his arrogance.

2. During our field visits we visited "remote" parts of Rajasthan.

(A) Far-off (B) Introvert (C) Distant (D) Depressed 12. Find out which part of the sentence has an error. If

there is no mistake, the answer is 'No error'.

)a("daysMeatless" /

)b(madebeenhave /

)c(filmaoint /

)d(ErrorNo

(A) Meatless days (B) have been made (C) into a film (D) No Error 13. Which part of the following sentence has an error ? If

the sentence is correct, the answer will be 'No Error".

)a(forwardLooking /

)b(to /

)c(hereyoumeet /

)d(ErrorNo

(A) looking forward (B) to (C) meet you here (D) No error 14. Choose the one which best expresses the meaning of

the given Idiom/Proverb : The 'pros and cons' (A) Good and Evil (B) Former and Latter (C) For and Against a thing (D) Foul and Fair 15. Replace the underlined word with one of the given

options : The Second World War started in 1939. (A) Broke out (B) Set out (C) Took out (D) Went out


CHEMISTRY

1.[D] For He+, 1

1λ

= RH × Z2

∞− 2

111 = 4RH

or, HR4

1 = x = λ1

∴ For Li+2 2

1λ

= RH × z2

∞− 22 )(

1)2(

1

4R91 H

2=

λ; λ2 =

9x16

R94

H= Å

2.[B] I – maximum H-bonding III – H-bonding IV – More electronegative oxygen II – Spherical Hydrocarbon part

3.[D]

4.[B] Silicon contains Si – C bond.

5.[B] K = )1(C

C.Cα−αα , Q Fe+3 = 0.1

Solving the quadratic equation for α = a = 0.095 ∴ H+ = Cα = 0.1 × 0.095 = 0.0095 ∴ pH = 2.02

6.[C] VC = 80.0

60 = 75 cm3 mol–1

Q b = 3

753

VC = cm3mol–1 = 0.025L mol–1

∴ TC = Rb27a8 ⇒ a =

8Rb27.TC = 3.375

7.[D] Due to dipole-dipole interaction ICl has stronger intermolecular force of attraction.

8.[B] Since the nucleophile is weaker and solvent is highly polar therefore reaction proceed through carbocation intermediate

9.[A,B,D] Except CuFeS2 + O2 → Cu2S + 2FeS + SO2 all the reactions are occurring in Bessemer's converter

10.[A,C,D]

CO2Et

OC–CH3

→

+OH3

CO2H

O

CH3 →∆

O

NaOH

I2→ COONa

+ CHI3

B will give +ve iodoform test.

11.[A,B,C,D]

M = 2.11

282.11'v'

= = 2.5

∴ 1 L contain 2.5 moles of H2O2 Mass of H2O2 = 2.5 × 34 = 85g wt. of 1L solution = 265 gL–1 ∴ OH2

W = 265 – 85 = 180 g

∴ moles of H2O = 18

180 = 10

∴ molality 'm' = 2.5 × 180

1000 = 13.88

mole fraction, 22OHX =

5.125.2

nn

T

OH 22 = = 0.2

% 1000

345.2vw ×

= × 100 = 8.5

12.[A,B,C,D]

(A) z

nr2

α (B) P.E α – 2

2

nz

(C) K.E α 2

2

nz (D) v α

nz

13.[B] Also, pH = pKIn ± 1 or, pKIn + 1 = 6 and pKIn– 1 = 4 ∴ pKIn = 5 ⇒ KIn = 10–5

14.[C] At VNaOH = 0, pH = 3 ⇒ H+ = 0.001

∴ ka = α−α

1001.0 = 10–5

⇒ α = 0.01 ⇒ c = 0.1

Now, C = v

nHB or 0.1 = v5 Q v = 50 mL

∴ [NaB] = 100

5volumeTotal

HBofmolm= = 0.05

∴ pH = 21 [pKw + pKa + log C] = 8.85

SOLUTION FOR MOCK TEST IIT-JEE (PAPER - I)


15.[A] Since pH corresponding to steepest point is > 8 ∴ most suitable indicator would be phenolphthalein

16.[B] R D S is the step in which intermediate carbocation intermediate is formed

17.[B] Me

pH C = N

OH

→+H

Ph– C – NHCH3

O

→

+OH3 PhCOOH + CH3NH2

18.[C]

Ph – C – – CH3

O

→ OHNH2

CH3

C = N or OH

Ph

CH3

C = N

OH

Ph

PCl5 ∆ PCl5 ∆

Ph – C – NH – – CH3

O

CH3 – –C–NH – Ph

O

Column Matching 19. [A] → r,t; [B] → s; [C] → p; [D] → r,t

Stereoisomers Optical Meso (A) 4 4 0 (B) 2 0 0 (C) 10 8 2 (D) 4 4 0 20. [A] → r; [B] → p,s; [C] → q; [D] → t

A → r — sp2 B → p,s — sp3 C → q — sp3d3 D → t — sp3d2

MATHEMATICS

1.[A] The point x = 1 is a discontinuity of the function

f(x) = 1/(1 – x). If x ≠ 1, then

u(x) = f(f(x)) = x

1x − . Hence x = 0 is a point of

discontinuity of the function u. If x ≠ 0, and x ≠ 1, then fofof(x) = x. Hence y = f 3n(x) = (f

3(x)n) = x is continuous everywhere. Therefore, 0 and 1 are the only points of discontinuities of y.

2.[D] The given lines intersect, if the shortest distance between the lines is zero.

We know that the shortest distance between the lines r = a1 + (λ 1b

r) and r = a2 + λb2 is

|bb|

|bb).aa(|

21

2121

××−

So the shortest distance between the given lines is zero if

(i – j – (2i – j) . (2i + k) × (i + j – k) = 0

L.H.S. = 111

102001

−

− = 1 ≠ 0

Hence the given lines do not intersect.

3.[B] Equation of a plane passing through the line of intersection of the given planes is

2x – y + 3z + 5 + λ (5x – 4y – 2z + 1) = 0 or (2 + 5λ)x – (1 + 4λ)y + (3 – 2λ)z + 5 + λ = 0 This will be perpendicular to the plane

2x – y + 3z + 5 = 0 if 2(2 + 5λ) + (1 + 4λ) + 3(3 – 2λ) = 0 ⇒ λ = –7/4 and the required equation of the plane is 4(2x – y + 3z + 5) – 7(5x – 4y – 2z + 1) = 0 ⇒ 27x – 24y – 26z – 13 = 0

4.[B] Equation of the tangents at P(x1, y1) to the parabola y2 = 4ax is yy1 = 2a(x + x1)

or 2ax – y1y + 2ax1 = 0 ….(i) If M(h, k) is the mid-point of QR, then equation

of QR a chord of the parabola y2 = 4a(x + b) in term of its mid-point is ky – 2a (x + h) – 4ab = k2 – 4a (h + b) (using T = S ′ ) or 2ax – ky + k2 – 2ah = 0 …...(ii) Since (i) and (ii) represent the same line, we have

a2a2 =

ky1 =

ah2kax2

21

−

⇒ k = y1 and k2 – 2ah = 2ax1 ⇒ 2

1y – 2ah = 2ax1 ⇒ 4ax1 – 2ax1 = 2ah ⇒ h = x1

5.[B] 6x2 – xy – 12y2 = 0 ⇒ (2x – 3y) (3x + 4y) = 0 ….(i) and 15x2 + 14xy – 8y2 = 0 ⇒ (5x – 2y) (3x + 4y) = 0 ….(ii) Equation of the line common to (i) and (ii) is 3x + 4y = 0 Equation of any line parallel to (ii) is

3x + 4y = k or 3/k

x + 4/k

y = 1

If 3k +

4k = 7, then k = 12 and the equation of

the required line is 3x + 4y = 12

6.[B] The given equation is equivalent to 2 sin2 ((π/2) cos2 x ) = 2 sin2 ((π/2) sin 2x) ⇒ cos2 x = sin 2x ⇒ cos x (cos x – 2 sin x) = 0


⇒ 1 – 2 tan x = 0 as cos x ≠ 0, x ≠ (2n + 1)π/2 ⇒ tan x = 1/2

⇒ cos 2x = 53

xtan1xtan1

2

2=

+−

7.[B] tan 3 α cot α = )tan31(tan

tantan32

3

α−αα−α

= α−α−

2

2

tan31tan3 = x (say)

⇒ tan2 α = 1x33x

−− = 2)1x3(

)3x)(1x3(−

−−

Since tan2 α is non-negative, either x < 1/3 or x ≥ 3, so x cannot lie between 1/3 and 3.

8.[D] Let X denote the largest number on the n tickets drawn. We have

P(X ≤ k) = n

Nk

and P(X ≤ k – 1) =

n

N1k

−

Thus P(X = k) = n

Nk

–

n

N1k

−

9.[A,C] Rewriting the given equation as

2xydxdy – y2 = 1 + x2 ⇒ 2y

dxdy –

x1 y2 =

x1 + x.

Putting y2 = u. We have dxdu –

x1 u =

x1 + x.

The I.F. of this equation is 1/x, so

u . x1 = ∫

+1

x12 dx = –

x1 + x + C

⇒ y2 = (x2 – 1) + Cx. Since y(1) = 1 so C = 1, hence y2 = x(1 + x) –1 which represents a system of hyperbola.

10.[B,C,D] ∫ −4

1

dx|3x| = ∫ −3

1

dx|3x| + ∫ −4

3

dx|3x|

= ∫∫ −+−−4

3

3

1

)3x()3x( dx

=

−− x3

2x 2 4

3

23

1x3

2x

−+

= 2 + 1/2 = 5/2 Thus 2A + B = 5/2. The values in (B), (C), (D)

satisfy this equation.

11.[A, B, C, D] For x ∈ [–1, 2), f ′(x) = 6x + 12 > 0 on (–1, 2), so

f increases on [–1, 2]. f is trivally continuous on [–1, 3] except possibly x = 2. At point x = 2. f(2) = 35 and

)x(flim2x +→

= )x37(lim2x

−+→

= 35,

)x(flim2x −→

= −→2x

lim (3x2 + 12x – 1)

= 3 × 4 + 12 × 2 – 1 = 35. Thus, f is continuous at x = 2 as well. Now

f ′( 2+) = h

)2(f)h2(flim0h

−++→

= 1h

35)h2(37lim0h

−=−+−

−→

Similarly f ′(2–) = 24 ≠ f ′(2 +) As f increases on [–1, 2] and decreases on [2, 3]

so f has a maximum at x = 2. 12.[B,C] Differentiating the equation of curve xy = 1,

we have x dxdy + y = 0 ⇒

dxdy = – y/x

Hence the slope of normal = x/y. Morever the slope of the line ax + by + c = 0 is –a/b. So we have x/y = –a/b, i.e., bx + ay = 0 solving this with xy = 1, we have x2 = –a/b. So we must have a/b < 0, i.e., a > 0, b < 0 or a < 0, b > 0.

13. [C] Each team will play 7 matches and so any team can win any no. of matches between 0 to 7.

0, 1, 2, 3, 4, 5, 6, 7 Four team will be selected (7, 6, 5, 4).

Thus, team which win only 3 matches will be out of the first round.

14. [A] And from the above question minimum number of matches that a team must win in order to qualify for second round is 4 matches.

15. [B] In second round it has to win one match. Then one in semi final and one final.

16.[C] R = Asin2

a

⇒ sin A =

6/2525

×=

53 ,

a = 5, b = 8, R = 25/6 ∴ cos A = 4/5 or – 4/5

∴ cos A = bc2

acb 222 −+ ; ∴ 54 =

c8225c64 2

××−+

−= solutionsanygivenotdoes

54Acos

⇒ c2 – 64c + 195 = 0 ⇒ c = 5, 7.8.

17.[A] a cos A + b cos B + c cos C = 2R sin A cos A + 2R sin B cos B + 2R sin C cos C = R sin 2A + sin 2B + sin 2C= 4R sin A sin B sin C

18. [A] 8R2 = a2 + b2 + c2 ⇒ 8R2 = 4R2 sin2A + 4R2sin2B + 4 sin2C

⇒ 2 = 2

A2cos1− + 2

B2cos1− + 2

C2cos1−

⇒ cos 2A + cos 2B + cos 2C + 1 = 0 ⇒ –1 – 4 cos A cos B cos C + 1 = 0 ⇒ cos A. cos B. cos C = 0

⇒ A = 2π

or B =

2π or C =

2π

19. (A) → p,q,r,t; (B) → s; (C) → q,t; (D) → p,q,r,t


20. (A) → p,q,r,s,t; (B) → r,t; (C) → q; (D) → q,s (A) Let f(x) = ax3 + bx + c ⇒ f ′(x) = 3ax2 + b > 0 ∀ x ∴ f(x) increases for all x. (B) f(x) = 3ax2 + 4bx + c then f(x) has same sign for

all x. Q f(0) = c > 0 ⇒ f(x) > 0; ∀ x ∈ R, thus f(–1) > 0 (C) If f(x) = ax2 + bx + c, then af(–2) < 0 (D) If f(x) = ax2 + bx + c then af(–2) < 0, and

af(–1/2) < 0 and af (–1) < 0

PHYSICS

1.[C] Net force on m3 = 22 )40()30( + = 50 N and limiting friction on m3 = µm3g = 60 N ∴ System remain in equilibrium and friction on

m3 = 50 N 2.[D]

y

x θ

N

mg sin θ mg cos θ

ma = – mg sin θ a = – g sin θ or a = – g tan θ … (1) (as θ is small) Now, x2 = 4ay

∴ a2

xdxdy

=

∴ a = – ga2

x

– ω2x = – a2

gx

ω = a2

g

3.[C] C = xA0ε , where x is separation between plates

dTdC

C1 =

dTdA

A1 –

dTdx

x1

for dTdC = 0,

dTdx

x1 =

dTdA

A1

⇒ αS = 2α

4.[C] Maximum energy is liberated for transition

En → 1 and minimum energy for En → En–1

Hence , 121 E

nE

− = 52.224 eV … (1)

21

21

)1n(E

nE

−− = 1.224 eV … (2)

Solving (1) and (2) E1 = – 54.4 eV

E1 = 2

2

1Z6.13

−

Z = 2 5.[B]

f =10 f =20

2f =20

O I1C

5 For lens f = –20, u = ?, v = – (20 – 5) = – 15

201

− =

151

−–

u1

u1 =

151

201

−

u = – 60 6.[B]

A

Equivalent circuit

2R

R

R

R

ε

⇓

i Req ε

Req = 2R

3R2

+

= 6R7

i = eqRε


R7

6i ε=

Req

⇓

A R

R

A 2R

B ε

R72ε

R73ε

R76ε

Using Kirchoff's first law current through

AB = R7R7

2R7

3 ε=

ε−

ε

7.[C] R = gh2u =

105212 × = 12 m

∴ Distance from origin = 22 )12(5 + = 13 m

8.[B]

π/2

2B B π

×

Time = qB2m

qB2m π

+π =

qBmπ

9.[A,B,D] At a distance h above the sheet

E = Esheet + Eslab =00 2

D2 ε

ρ+

εσ− =

02D

εσ−ρ

At a distance h below the top surface of slab Eslab

= 02

)h2D(ε−ρ

E = Esheet + Eslab=00 2

)h2D(2 ε

−ρ+

εσ =

02)h2D(

ε−ρ+σ

At a distance h below the bottom surface of the

slab = 00 2

D2 ε

ρ+

εσ− =

02D

εσ−ρ

10.[A,C,D]

2m 45°

m v0/2

v

45°

45° v v0

v0/2

Momentum of ball + wedge system is conserved

in horizontal direction

⇒ v = v0/2 v = velocity of wedge after collision.

∴ e = °

°+°45sinv

45sin2/v45cos.2/v

0

00 = 1

Impulse on wedge due to ball = Impulse on ball due to wedge = change in momentum of ball.

= 4vmvm

20

220

2 +

= 25 mv0

0i mvp =

2mvp 0

f =

ip

fp

Impulse on wedge due to surface = Change in momentum of ball + wedge system

= 2

mv0

11.[A,C]

θ S1

S2

2λ

P

δ = 2π –

λπ2 (2λ sin θ)

⇒ δ = 2π – 4π sin θ

For maxima, δ = nπ where n = 0, ± 1, ± 2 . . .

2π – 4π sin θ = nπ

sin θ = 4

n21

−

n = 0, sin θ = 81

n = ± 1, sin θ = 83,

81

+−

n = ± 2, sin θ = 83

− , 85

n = ± 3, sin θ = 85

− , 87

n = 4, sin θ = 87

−

12.[B,C] fap. = realS

fvv

v

−

, when source is moving

towards observer


and fap = realS

fvv

v

+

, when source is moving

away from observer and fbeat = f1 ~ f2 Passage # 1 (Q. 13 to 15) 13.[A]

l b

a

B = µ0J

× × × × ×

. . . . .

Flux = B × Area = µ0J × l × a

φ = b

ai0 ××µ l

φ = Li

L = iφ

= b

a0lµ

14.[A] Vx = dtdiLiLi =

Using kirchoff 's law

V0 – dtdiL = 0

i = L

tV0

L1 = bxa0µ

Vx = bxa0µ ×

dtdi =

LV

bxa 00 ×

µ

15.[B] Energy flow rate = Vx × i

= L

tVbLxaV 000 ×

µ

= 2

200

bLtxaVµ

Passage # 2 (Q. 16 to 18) 16.[A] When protons are incident on H3

1

X = H11 ; X = H3

1

Kth = 4.033 MeV

+

016049.3007825.11

= 5.381 MeV

17.[C] When H31 is incident on protons ;

x = H31 ; X = H1

1

Kth = (4.033 MeV)

+

007825.1016049.31

= 16.10 MeV 18.[C] Above calculations shown that, less energy is

required for a nuclear reaction if a light particle is incident on a heavy target than if a heavy particle is incident on a light target.

Column Matching :

19. [A] → r,; [B] → q; [C] → p; [D] → p

(A) Velocity of fish in air = 8 × 43 = 6 ↑

Velocity of fish w.r.t. bird = 6 + 6 = 12 ↑ (B) Velocity of image of fish after reflection from

mirror in air = 438× = 6 ↓

w.r.t. bird = – 6 + 6 = 0

(C) Velocity of bird as seen from water = 6 × 34 = 8 ↓

w.r.t. fish = 8 + 8 = 16 ↓ (D) Velocity of bird in water after reflection from

mirror = 8 ↓ w.r.t. fish = 8 – 8 = 0

20. [A] → p,s; [B] → p; [C] → q,r; [D] → q,r

(A) E = 3

G4πρ (R/2) : r ≤ R

[ρ = density , R = Radius of bigger sphere]

= 2rGM : r > R

|V| = 3G2 ρπ

−− 22

)2/Rr(4R3 : r ≤ R

= r

GM : r > R

(B) E = 2πGρ . (R/2) : r ≤ R

= r

RG2π 2ρ : r > R

(C) E = GM 3/222 )r(Rr

+

V = 1/222 )r(RMG

+

(D) E = G.3m 3/22

2)r

3R(

r

+

V = 1/22

2)r

3R(

mG.3

+


CHEMISTRY

1.[C] 3Cu(s) + 8H+ +2 −3NO → 3Cu+2 + 2NO + 4H2O

Cu(s) + 4H+ + −3NO2 → Cu+2 + 2NO2 + 2H2O

Let concentration of HNO3 is x, then [H+] = [ −

3NO ] = x

The tendency for the reduction of −3NO to NO

and NO2 will be same and at that stage, Ecell must have same value

∴ NO/NO3E − – cu/Cu 2E + = −−

23 NO/NOE – Cu/Cu 2E +

or NO/NO3E − =

23 NO/NOE −

or 0.96 – 3

0591.0 log 5

3–

x10

= 0.79 – 1

0591.0 log 3

3–

x10

∴ log x = 0.656 ≈ 0.66 ; ∴ x = 100.66

2.[B]

Br OEt

Mg

2

→

Oδ–

MgBr CH3δ+ +

→

OH

CH3

3.[C] Br– Changes to Br2

4.[C] Diazo coupling reaction is an ES reaction in which electrophile is diazo component. Greater electro philicity of diazo group makes it more reactive

EWG ERG NO2– , CH3–, CH3O–, Me2N–

5.[A,B,C,D]

– 2/1]A[]A[d = kdt

Integrating both sides, we get –2[A]1/2 = kt + C

When t = 0 the [A] = [A0] So C = – 2[A0]1/2

∴ k =t2 ([A]1/2 – [A]1/2)

After rearrangement [A]1/2 = –2kt + [A0]1/2

y = mx + C So graph will be

t

[A]1/2

[A0]1/2

Slope = –k/2

t ¾ =k

]A[ 0 and t ½ =k

]A[)12(2 0−

6.[A,B,C]

On heating one molecule of carbon dioxide comes out. Ph

Ph

H MeHO2C CO2H

H Me

2CO−

∆→

Ph

Ph

H Me H CO2H H Me

Ph

Ph

H MeHO2C H

H Me

[A] [B] Both have plane of symmetry hence optically inactive

7.[A,B,C,D]

8.[A,B,C] (A) 100 mg of CaCO3 in 1000 ml = 100 g CaCO3 in

106 ml = 100 ppm (B) 120 mg of MgSO4 in 103 ml = 120 g MgSO4 in

106 ml = 100 g CaCO3 in 106ml = 100 ppm (C) 84 mg of MgCO3 in 103 ml = 84 g MgCO3 in

106 ml = 100 g CaCO3 in 106ml = 100 ppm (D) 111 g of CaCl2 in 103 ml = 100 × 103 g of CaCO3

in 106 ml = 100000 ppm

SOLUTION FOR MOCK TEST

PAPER - II IIT-JEE (PAPER - II)


9.[D] Neighbouring group participation occurs through two consecutive SN2 substitution with inversion of configuration, thus the net result is retention of configuration.

Column Matching : 10. (A) → q ; (B) → p,r ; (C) → p,s ; (D) → p,s * For exothermic reaction, ∆H = –ve, ↑ in T backward

whereas for endothermic, ↑ in T forward. * Reactions for which ∆ng = 0, pressure has no effect

11. (A) → r ; (B) → q ; (C) → p ; (D) → s

Ph–CH–CH3

Br ∆

→–EtO Ph – CH = CH2 (E2)

CH–C–Br

Εt

Me Me

Εt →EtOH

–C

Εt

MeMe

Εt ⊕

E1

Εt

Me Me

Εt Ph–N+–CH2–CH2

CH3

H O–

Ei CH2 = CH2

CH3–C–CH3

CN

OH

CB1E

–EtO → CH3–C–CH3

O

Numerical Response type questions : 12. [4]

Volume of both AgNO3 & HCN are equal so concentration is halved. [AgNO3] = [HCN] = 0.01M HCN(aq) H+(aq)+CN–(aq); Ka = 4×10–10 …(i) Ag+(aq) + CN–(aq) AgCN(s);

K =spK

1 = 161041

−× …(ii)

Adding Ag+(aq) + HCN(aq) H+(aq) + AgCN(s);

K =sp

a

KK = 106

Initial 0.01 0.01 – – At. Eq. x x 0.01 0.01

(Since K is very large so almost entirely forward shifted)

⇒ K =]HCN][Ag[

]H[+

+

= 2

2

x10−

= 106

⇒ x2 = 10–8 ⇒ x = 10–4 = 10–n ⇒ n = 4 13. [8]

CH3 – CH2– CH – CH2 – CH3

CH3

6

1 2 3 4 5

→ νh,Cl2 products.

Reaction at C–1 and C–5 are enantiotropic face give racemic products (2)

Reaction at C–2 and C–3 it self cause racemization in addition to that these are diastereotopic faces so diastereomers form (4) Reaction at C–3 (1) Reaction at C–6 (1) Total isomers are (8)

14. [4]

89600g haemoglobin =100

8960025.0 × = 224 g Fe

No. of Fe =56224 = 4

15. [3] xA = 0.70 and yA = 0.35 p = 600 Torr

ºAp =

A

A

xpy =

70.060035.0 × = 300 Torr

16. [1] 27 – (–x) + 8 = 36 x + 35 = 36 x = 36 – 35 = 1

17. [6]

18. [6] nCV (T2 – T1) = – Popp (V2 – V1) Popp = P2

n23 R (T2 – 300) = –2

−

1)300(nR

2nRT2

23 (T2 – 300) = –2

2)600T( 2 −

5T2 = 2100 T2 = 420 K ∆U = nCV (T2 – T1)

= 1×23 × 2(420 – 300)

= 3 × 120 = 360 Cal. ∆H = ∆U + nR(∆T) = 360 + 1 × 2 × 120 = 360 + 240 = 600 Cal. 19. [4] The cell reaction is Cd + 2H+ (0.2M) → Cd2+ (0.1M) + H2 (0.5 atm) º

cellE = ºH½/H 2

E + – ºCd/Cd2E +

= 0 – (–0.403) = 0.403 V

Ecell = ºcellE –

nFRT303.2 log 2

H2

]H[

P]Cd[2

+

+ ×

= 0.403 –2

0591.0 log 2)2.0(5.01.0 ×

= 0.403 – 0.003 = 0.400 V


MATHEMATICS

1.[B] For vertical tangents θd

dx = 0 so, we have

–3cos θ = 0 ⇒ = 2π or

23π . Corresponding to

these values of θ, we have

x = 2 – 3 sin 2π = –1; y = 3 + 2 cos

2π = 3;

x = 2 – 3 sin 2

3π = 2 + 3 = 5

y = 3 + 2 cos 2

3π = 3

Thus the required points are (–1, 3), (5, 3). 2.[B] u(x) = h(f (g(x))) = h(f(x2)) = h (sin x2) = log sin x2 Hence u′ (x) = 2 x cot x2 and

u′′ (x) = 2 cot x2 – 4x2 cosec2 x2. 3.[B] We have

p1 = 222 )2()6(3

11423623

+−+

+×+×−× =77 = 1

and p2 = 222 )2()6(3

11421623

+−+

+×+×−× = 7

16

so, that p1, p2 are the roots of the equation

p2 –

+

7161 p +

716 = 0 ⇒ 7p2 – 23p + 16 = 0

4.[D] Equation of a tangent at (at2, 2at) to y2 = 8x is ty = x + at2 where 4a = 8 i.e. a = 2 ⇒ ty = x + 2t2 which intersects the curve xy = –1

at the points given by t

)t2x(x 2+ = –1

clearly t ≠ 0 or x2 + 2t2x + t = 0 and will be a tangent to the curve if the roots of this quadratic equation are equal, for which 4t4 – 4t = 0 ⇒ t = 0 or t = 1 and an equation of a common tangent is y = x + 2.

5.[A,B,C] We have

A (α, β)′ =

ααα−α

βe000cossin0sincos

=

α−α−−α−α−

βe000)cos()sin(0)sin()cos(

= A(– α, β) Also, A(α, β) A(–α, –β)

=

αα−αα

βe000cossin0sincos

ααα−α

β−e000cossin0sincos

= I

⇒ A(α, β)–1 = A(–α, –β) Next, Adj A(α, β) = |A (α, β)|A(α, β)–1

= eβ A(–α, –β). 6.[A,B,C] Let AB be the tower, ∠APB = θ, ∠AQB = 2θ and ∠ARB = 3θ Then QR = (3/4) PQ and ∠PBQ = ∠QBR = θ ⇒ BQ is the bisector of ∠PBR (Fig.)

Pθ

θ

3θ 2θ

B

θ

Q R A

90º – 3θ

⇒ BRPB =

QRPQ

⇒θθ3eccosAB

eccosAB =34

⇒θθ

sin3sin

=34

⇒ 3 – 4 sin2 θ = 4/3 ⇒ 12 sin2 θ = 5 ⇒ sin θ = 12/5 . cos 2θ = 1 – 2 sin2 θ = 1/6 sin 3θ = sin θ (3 – 4 sin2 θ) = 33/52 . 7.[A,D] We have α+β = –b/a, αβ = c/a, α′+β′ = –b′/a′ and

α′β′ = c′/a′. Therefore, equation of the circle having AB as

diameter is (x – α) (x – β) + (y – α′) (y – β′) = 0 ⇒ x2 – (α+β)x + αβ + y2 – (α′ + β′)y + α′β′ = 0

⇒ x2 +ab x +

ac + y2 +

'a'b y +

'a'c = 0

⇒ aa′ (x2 + y2) + a′ bx + ab′ y + a′ c + ac′ = 0 Since it passes through the origin, a′ c + ac′ = 0

and through (b/a, b′/a′)

aa′

+ 2

2

2

2

'a'b

ab + a′ b ×

ab + ab′ ×

'a'b = 0

⇒ a′2 b2 + a2 b′2 = 0

8.[A,C] 0x

)0(f)x(flim0x −

−→

=)x1(log

xcosloglim 20x +→

=)x1log(

x.)2/x(4

)2/x(sin2.xcos1

)]xcos1(1log[lim 2

2

2

2

0x +−−−

→

=2

0x0y 2/x)2/xsin(.lim

y)y1log(lim

21

−

→→

21

)x1log(xlim. 2

2

0x−=

+→


where y = 1 – cos x. So f is derivable at x = 0 and hence also continuous.

9.[A,D] cot–1 (1 + x2 – x) = tan–1

−−−+

)x1(x1x1x

= tan–1 x + tan–1 (1 – x)

I = ∫ −+−1

0

21 dx)xx1(cot = ∫ −1

0

1 dxxtan

+ ∫ −−1

0

1 dx)x1(tan

= ∫ ∫∫ −−− =+1

0

1

0

11

0

11 xdxtan2xdxtanxdxtan

= 2x tan–1 ] ∫ +−

1

0 210 dx

x1x2x

= 2tan–1 (1) – log(1 + x2)]10

= 2(π/4) – log 2 = π/2 – log 2. 10. A → r, s; B → s; C → p;D → p, q

(A) Any point on the line (t, 1– t). The chord with this as mid point T = S1 passes through the point (a, 2a)

⇒ (1 – t)2 = 2a (1 – a) > 0 ⇒ a ∈ (0, 1) Q LR ∈ (0, 4)

(B) POI are (1, 0) and (4, 0) and circle is (x – 1) (x – 4) + y2 + λ y = 0 the length of the tangent from (0, 0) is √4 = 2

(C) The ⊥r tangents from any point to the parabola intersect on the directrix.

(D) 1 – | h | > 0 ⇒ h ∈ (–1, 1). 11. A → q,t; B → s; C → p; D → r

Let X = the number of steps taken in the forward direction, then X ~ B (n, p) with n = 11, p = 0.4.

p1 = P(X = 5) + P(X = 6) = 11C5 p5q6 + 11C6p6q5 = 11C5 (pq)5 = 11C5 (0.24)5 = 11C6 (0.24)5 p3 = P(X = 4) + P(X = 7) = 11C4 p4 q7 + 11C5 P7 p7 q4

= 11C4 (pq)4 (1 – 3pq)

= 11C4 (0.24)4 (0.28)

p0 = 0 and p11 = P(X = 0) + P(X = 11) = (0.4)11 + (0.6)11

Numerical Response type questions : 12. [3] We have

2|z|9 = (2 + cos θ)2 + sin2 θ = 5 + 4 cos θ (1)

and

z3 +

z3 = 4 + 2 cos θ (2)

Eliminating θ from (1) and (2), we get

2|z|9 – 6

+

z1

z1 = – 3

⇒ 3 = 2( z + z) – |z|2

13. [2] p(x) = (2x + 3) (x97 + x96 + … + 1). = (2x + 3) ( x + 1) (x96 + x94 + … + x2 + 1) Also, x96 + x94 + … + x2 + 1 > ∀ x ∈ R.

14. [3] Let the required G.P. be

,2

1,2

1,21

b2abaa ++……

Its sum is 71

122

2/112/1

b

ab

b

a=

−=

−

−

We can take a = 3, b = 3.

15. [2] As A2 = O, Ak = O ∀ k ≥ 2. Thus, (A + I)50 = I + 50A ⇒ (A + I)50 – 50A = I ∴ a = 1, b = 0, c = 0, d = 1

16. [6] Let b = xi + yj + zk. So a × b = (z + y) i – xj – xk a × b + c = 0 ⇒ z + y + 1 = 0, –x + 1 = 0 ⇒ x = 1 a . b = 3 ⇒ y – z = 3

Solving these equations we have y = 1, z = –2. Thus b = (1, 1, –2). i.e. |b|2 = 6.

17. [4] We have BC = 2BD, AD = h and OD = h – r.

∴ BC = 22 )rh(r2 −− = 2hhr22 −

⇒ AB = 22 hhhr2 +− = hr2 so that P = 2AB + BC

= ]hr2hhr2[2 2 +−

Also the area of ∆ABC is

∆ = BD × AD = 2hhr2h − .


A

r

B C D

O

r

∴ 3P∆ = 3

2

2

hr2hhr28

hhr2h

+−

−

=( )3r2hr28

hr2

+−

−

⇒ 30h P512lim ∆

→

= 512 r ( )3r228

r2 = 4

18. [3]

Note that it is not given that f is a differentiable function we have

f ′ (4) =h

)4(f)h4(flim0h

−+→

= ( )h

)2(f)h4(flim22

0h

−+→

=h

8)h4(lim2/3

0h

−+→

=h

]1)4/h1[(8lim2/3

0h

−+→

=h

1)h(04h

2318

lim

2

0h

−++

→=

h

)h(0h838

lim

2

0h

+

→

=0h

lim→

[3 = 0(h)] = 3

19. [3] The given equation can be written as

dxdy + 2x1

x+

y = 2x1x

+

This is a linear equation with I.F. 2x1+ . So

y 2x1+ = dxx1x 2∫ + + C.

= (1/3) (1 + x2)3/2 + C ⇒ y = (1/3) (1 + x2) + C(1 + x2)–1/2

34 = y(0) =

31 + C ⇒ C = 1.

Hence y( 8 ) =31 . 9 +

31 ⇒ y( 8 ) –

31 = 3.

PHYSICS

1.[B]

6° 3°

A

B C

Time taken by pendulum in going from A to B

= 4T where T = 2π

gl

Time taken by pendulum in going from B to C

= 12T

∴ Time period of pendulum

=

+

12T

4T2

= 3T2 =

32 .

5π =

152π sec

2.[C]

θ1

θ2 B

A

C Free body diagram of rope AB −

θ1 θ2

Mg

TA TB

TA cos θ1 = TB cos θ2 TA sin θ1 + TB sin θ2 = Mg

∴ TA =)sin(

cosMg

21

2

θ+θθ

TB = )(sin

cosMg

21

1

θ+θθ


Free body diagram of point AC –

θ1

M′g

TA

TC

Horizontal equilibrium : TA cos θ1 = TC

∴ TC = )(sin

coscosMg

21

21

θ+θθθ

Tension will be maximum at A and minimum at C. 3.[C] Maximum expansion in spring is given by

2maxkx

21 = 2

0v21

µ

[µ = Reduced mass]

⇒ xmax = kµ . v0 =

k3m2 v0

4.[B] Force diagram of block for the view shown

view

N N

θ

mg cos α

N N 180°– θ

mg cos α

⇒ N = )2/sin(2

cosmgθ

α

∴ Net friction up the plane = 2 µN

= µmg )2/sin(

cosθ

α

∴ a = g

θα

µα)2/sin(

cos–sin

5.[A,D]

θ S1

S2

2λ

P

δ = 2π –

λπ2 (2λ sin θ)

⇒ δ = 2π – 4π sin θ

For maxima, δ = nπ where n = 0, ± 1, ± 2 . . .

2π – 4π sin θ = nπ

sin θ = 4

n21

−

n = 0, sin θ = 81

n = ± 1, sin θ = 83,

81

+−

n = ± 2, sin θ = 83

− , 85

n = ± 3, sin θ = 85

− , 87

n = 4, sin θ = 87

−

6.[A,B,C,D]

a = mF =

mqE =

m)x(q β−α

a = 0 at x = βα

Force on the particle is zero at x = βα

So, mean position of particle is at x = βα

dxvdv = )x(

mq

β−α

Solve for v =

β

−α x2m

qx2

v = 0 at x = 0 and x = βα2

x = βα2 with mean position at

βα

=x .Therefore

amplitude of particle is βα .

Maximum acceleration of particle is at extreme

position (at x = 0 or x = βα2 ) and αmax = m

qα .


7.[B,D] As v = 2t and let radius of circular path is r then,

aT = dtdv = 2, ar =

rv2

= rt4 2

Therefore,

a = 2r

2T aa +

a = 2

4

rt164 +

∴ (B) and (D) are correct.

8.[A,B,D] A = 0.04 m ω = 25π k = 5π ∴ Position of antinodes is given by cos (5πx) = 1 ⇒ x = 0, 20 cm ....

v = kω = 5 m/s

vmax = Aω = π m/s

9.[A,C] Temperature gradient at distance 'r'

drdθ =

K1 × 2x

1π

[Where x = cross-sectional radius at distance 'r'] ⇒ Temperature increases with increase in 'r'.

10. (A) → (S) ; (B) →(P) ; (C) → (Q) ; (D) → (R) 11. (A) → (P,S) ; (B) → (Q,R) ;

(C) → (Q,R,S) ; (D) → (P,Q,R) For process PVn = constant Molar heat capacity of gas

C = R

−

−−γ 1n

11

1 ... (i)

Here, γ = 7/5

PV = nRT = 1nVttancos

−

= Constant ×

−

n11

P For n = 1 : Temperature with increase in volume work done positive Hence heat is absorbed by system.

For n =21 : Temperature and volume decrease

with increase in pressure

∴ Work done negative Hence heat is rejected

For n =

56 : Temperature increases with decrease

in volume work done negative For n = 2 : Temperature increase with increase

in pressure work done negative Hence heat is absorbed.

Numerical Response type questions : 12. [4]

M,R

4M, 2R

6R – x

V = 0 E = 0

x

2xm4.G = 2)x–R6(

GM

2 (6 R – x) = x x = 4 R = 4000 km = 4 × 103 km.

13. [1] Temperature of sun is given by

T = m

bλ

[b : wein constant]

= 6127 K

∴ Intensity at earth = 2d4u

π

= 2

4

d4ATπ

σ

[d = 1.5 × 1011 m, T = 6127 K, A = 4π r2, r = 7 × 108 m] = 1740.1 W/m2 = 1.740 × 103

14. [1] Temperature of gas will increase with increase in

volume and becomes maximum at ‘C’

∴ Tmax = nRPV =

314.841

10410200 3–3

×

×××

= 384.89 K = 112ºC = 1.12 × 102 ≈ 1


15. [3]

S1

S2

S3

P

2d3

θ1

θ2

S2P – S1P = d sin θ1 = d. D

2/d = D2

d2=

3λ

S3P – S2P = d sin θ2 = d D

2/d3 = D2d3 2

= λ

120° I0

4I0

IResultant = I0 + 4I0 + 0I2 0I4 cos120° = 3I0 16. [4] Let, a = side of cube

P = Impulse imparted P

a

A B ∴ After hitting,

v0 = mP and ω0 =

I2Pa

[I = Moment of inertial about axis passing through centre of mass]

For just toppling

20I

21

ω = mga

−

21

21

(Applying energy conservation between situation A and B)

⇒ P = aI2 0ω

= 4 kg m/s

17. [1] Let rate of production = R

∴ dtdN = NR λ−

dtdN + λN = R

eλt dtdN + λNeλt = Reλt

dt

)Ne(d tλ

= Reλt

Neλt = λ

λtRe + C

At t = 0, N = 0 ⇒ C = λ

−R

∴ N = λR (1– e–λt)

At equilibrium quantity N = λR for t → ∞

∴ λ2

R = λR (1 – e–λt)

⇒ e–λt = 21

t = λ

2nl = T1/2 = 100 years = 1 × 102

18. [2]

eV1 = 1

hcλ

– φ

eV2 = 2

hcλ

– φ

e(V2 – V1) = hc

λλλ−λ

21

21

V2 – V1 = ehc

λλλ−λ

21

21

= 19–

834–

106.1103106.6

×××× × 5–106

100×

= 3266 × 10–34 + 8 + 2 + 19 + 5

= 1633

= 2.0625 volt ≈ 2 volt

19. [2]

A

l

Wgas = pressureexternalgravity WW +−

= mgl + P0lA [m = mass of Hg pallet] = 2.136 J ∴ ∆Q = ∆W = 2.136 J ≈ 2


PHYSICS

1.[A] Initially relative velocity between coin and lift is zero.

h = ut + 21 gt2 here u = 0

t = gh2 =

8.945.22× =

21 sec.

2.[C] Mass of liquid inside the capillary = πr2 h d = (πr h d). r since, hr = constant ∴ mass of liquid inside α r 3.[D] 2π r = nλ

or r = πλ

2n

⇒ r2 – r 1 = πλ

22 –

πλ2

= π

λ2

4.[C] KE = hυ + φ …..(i) 2KE = hυ' + φ ……(ii) or 2 (hυ + φ) = hυ' + φ

or υ' = 2υ + hφ ⇒ υ' > 2υ

5.[C] 6.[B]

7.[D]

8.[A] E = 21 Ka2

E = 21 mw2a2

So a2 t2 = hence (A) constant 9.[C] M = 21MM

M = 202.24 × = 484 = 22g

10.[D] Due to introduction of glass slab, number of fring between two points remains constant because if n number of fring crosses point P, then same number of fring also crosses point O in same sence.

11.[A] Energy stored in capacitor is, 21 CV2 = 3 J

On connecting this capacitor to an uncharged capacitor since charge distributes equally, hence both capacitors are of same capacity

Now common potential, V' = CC

)0(CCV++ =

2V

Total energy stored in two capacitors is –

U' = 21 CV'2 +

21 CV'2 = C

2

2V

= 41 CV2 =

23 = 1.5 J

12.[C] Current, I = RV =

5.030 = 60 A

Total no. of free e– s, N = nAl

and linear momentum of each e¯s, P = mvα ∴ Total momentum of all free e¯s, P = (nAl) (mυα)

But I = neAυα , so nAυα = eI

∴ P = emIl = 19–

31

106.1

101.910060

×

××× −

3.4 ×10–8 kg/s 13.[A] Let unknown resistance be X. In first case, if

l is the balancing length then, RX = ( )l

l

−100

∴ X = ( )l

l

−10010 …………..(i)

In second case, R = 10 + 12.5 = 22.5 Ω and l ' = l – 20

∴ RX =

)'100('l

l

− or

5.22X = ( )20100

20−−

−l

l


PAPER - II AIEEE


or X = ( )l

l

−−

120205.22 ……….(ii)

Solving eqs (i) and (ii) we get l = 60 cm and X = 15Ω

14.[D] Energy gained in one movement across the gap = 100 KeV

The energy gained in one turn = 200 KeV

N = 3

6

102001020

×

× = 100

15.[D] T = 2π HMB

I

T' = 2π HMB

I2 = 2 T

16.[B] emf induced is,

e = – dtdφ =

dtAdB

− = 2

2l− × (–2)

= l2 = 1 Volt ∴ Resultant emf = 10 – 1 = 9V

17.[A] Since 'M' is at rest the tension in the string

= 2

Mg Let acceleration of m and m' is 'f' one

will move downward and other will move upward

mg – 2

Mg = mf ……..(i)

2

Mg – m'g = m'f …….(ii)

Solving equation (i) and (ii)

M4 =

m1 +

1m1

18.[C] F = dxdu

− = –(16x – 4) = 4 – 16x

At equilibirum 4 – 16x0 = 0

and x0 = 41 m = 25 cm.

At the given position x = (25 + 25) = 50 cm. ⇒ F = 4 – 16 × .5 = –4 N.

or a = 1.04− = – 40 m/s2

19.[B] LC = 101 mm =

1001 cm = 0.01 cm

Side of cube = 1010 + 1×.01 = 1.01 cm

20.[A] As temperature increases, frequency increases so it will be more than thrice the length, hence (A)

x > 54

21.[A]

B C2

V

1 A p

P 10

dwABCA = dQABCA

21 (2–1) (10–P) = 5

5 – 2P = 5 ⇒ P = 0

dwCA = 21

− [2 – 1] [0 + 10] = –5J

22.[C] "n'n =

s

sVVVV

−+ =

35

3V + 3Vs = 5V – 5Vs

Vs = 4V =

4340 = 85 m/s

23.[A]

θ θ θ φ

60°60°

µ = 3

θ

°sin

60sin = 13

23 = 3 sinθ

sinθ = ½ θ = 30° ∴ φ = 90°

Statement based : (Sol. 24 to 25)

24.[D]

25.[B]

Passage based : (Sol. 26 to 30)

26.[A] Inner charge also flows to outer shell. Hence, net charge on it q = q1 + q0

= (σ)(4πR2) + (–σ) (4π) (2R)2 = – 12σπR2

∴ V = 04

1πε

.)R2(

q =R8R12

0

2

πεσπ−

= – 02R3

εσ


27.[B] Using the Gauss theorem we can see that net charge inside the Gaussian sphere passing from the point under consideration has no changed.

Therefore, E1 = E2 or 2

1

EE = 1.

28.[C] They will collide at their mean positions because time period of both are same and that is

2πKm . After collision combined mass is 2m

and Keff = 2K. Hence, time period remains unchanged.

29.[A] From conservation of linear momentum we can see that velocity of combined mass just after

collision is v = 4Aω .

m

A

m2A

← → ωω

⇒ mm2Av

←ω

=

Before collision Just after collision Since, this is the velocity at mean position. Hence, v = ω′ A′

or 4Aω = ω′ A′

or A′ =4A because ω′ = ω =

mK

or m2K2

30.[D] E =21 (2m)v2 =

21 (2m)

2

4A

ω

=16

Am 22ω =16

kA2

=ω

m2k2or

mk2

or E =21 (2m)

2

4A

=

16kA2

CHEMISTRY

31.[B] 32.[C] Sulphide ions produces rotten egg smell on

treating with dil. H2SO4

33.[C] HgI42–

34.[D] Smaller the valency, smaller the flocculating

power.

35.[A] H3BO3 + OH– → H4 BO4– or B(OH)4

–

36.[C] C6H12O6 → CO2

O.N. = C ⇒ 0 C = +4

Change in O.N. per atom = 4 Change in O.N. for 6 C atom = 24 So equivalent weight = M/24

37.[C]

38.[D] Allylic halides undergo substitution very easily.

39.[D] Solubility of alcohols in water increases with increase in branching of alkyl group.

40.[C] R–CN on hydrolysis gives RCOOH.

41.[C] Conc. HNO3 oxidises aniline to p–Benzoquinone

42.[C]

43.[D] Mg prevent corrosive action of water & salt by providing cathodic protection.

44.[D] Sn + 2NaOH + H2O → Na2 SnO3+2H2 Pb + 2NaOH → Na2PbO2+H2

Zn + 2NaOH → Na2 ZnO2 + H2

45.[B]

OCOCH3

COOHAcetyl salicylic acid (Aspirin)

46.[D]

47.[B] The magnetic moments depends upon the no. of unpaired electrons i.e. )2n(n +

no. of unpaired electrons in Cr2+ → Four Co2+ → three Fe2+ → Four Mn2+ → Five Both Cr2+and Fe2+ have four electrons

48.[A] for bcc lattice, l = 0.433a lattice is half of the body diagonal that is

2

a3 = 1.73 Å ; So, a = 200 pm

49.[A] K = 100 = ]B][A[

]AB[

22

2

K1 = 222

]AB[]B][A[ ,

Thus K1 = 1/K = 10–2

K2 = 2/12

2/12 ]B[]A[

]AB[ = K = 10


50.[A] a = initial conc.

(a–x) = a – 3a2 = a/3

t2/3 = K303.2 log

xaa−

= 14–1048.5303.2×

log

3/aa

= 2.01×1013 sec.

51.[D] Mg/Mg 2E + = º

Mg/2MgE

+ +

20591.0 log

]Mg[]Mg[ 2+

= –2.36 + 2

0591.0 log

101.0

= – 2.36 + 2

0591.0 (–2)

= – 2.42 volt

52.[A] Milli equivalents of CH3COOH = 50×2 = 100 Milli equivalents of CH3COONa = 10×1 = 10

pH = pKa + log

acidsalt

= –log (10–5) + log

10010 = 5–1 = 4

53.[A] Lf = 1.44 cals heat of fusion of Ice per gram

Lf = 18

100044.1 × = 80 cals/gm.

Now: Kf = f

2f

L.1000T.R

Kf = 801000)273(2 2

×× = 1.86°C

So: ∆Tf = Kf × m = Kf × WM

1000w×

×

0.5 = 100M

1000286.1×

××

⇒ M = 74.4 gm. 54.[B] → 4 < n + l ≤ 6 n + l = 5 & n + l = 6 5 + 0 5s2 6 + 0 = 6s2 4 + 1 4p6 5 + 1 = 5p6 3 + 2 3d10 4 + 2 = 4d10 Total e– = 18 Total e– = 18 So total e– = 18+18 = 36

55.[A] Resonance structure may or may not have equal energy.

56.[B] BaSO4, poisons the catalyst and decreases the efficiency. Hence, to check further reduction of aldehydes to alcohols.

57.[B] C2H5 MgBr OH)ii(

CO)i(

2

2 → C2H5COOH

Hence the value of n = 2

58.[D]

NO2 2[H] N=O 2[H]

NHOH 2[H] NH2

59.[A] Q10 ml of oxygen is obtained at STP from H2O2 = 1ml

∴ 500 ml of O2 is obtained at STP= = 50ml 60.[B] Ag2S + 4 NaCN → Na [Ag (CN)2] + Na2S Na2S + 2O2 + H2O → Na2S2O3 + 2NaOH The complex Na [Ag (CN)2] contains [Ag (CN)2]– ions

MATHEMATICS

61.[D] sinx + cosx = 51

⇒ sin2x + cos2x + 2sinx cosx = 251

⇒ 1 + sin2x = 251 ⇒ sin2x = –

2524

⇒ cos2x = – 257 ⇒ tan2x =

724

62.[C] tan (x + y) = 33 ⇒ x + y = tan–133 ⇒ y = tan–133 – x ⇒ y = tan–133 – tan–13

⇒ y = tan–1

+

−)3(331

333 ⇒ y = tan–1(0.3)

63.[A] x = n

n.........21 +++ = 2

1n +

variance σ2 =

n

x2i∑ – ( )x 2

σ2 = n

n2∑ – 2

21n

+

σ2 = n6

)1n2)(1n(n ++ – 2

21n

+ =

121n 2 −


64.[C] ~ (a → b) = a ∧ ~ b ∴ (p ∧ q) ∧ ~ (q ∨ ~ r) = (p ∧ q) ∧ (~ q ∧ r)

65.[C] Lines ay2 – (1 + λ2)xy – ax2 = 0 are perpendicular because, coeff x2+coeff y2 = 0 option (B) is not pair of lines, option (C)

represent two lines x = 0 & y = 0 which are perpendicular hence option(C)

66.[A] ( ar

× br

) × cr

= ( ar

. cr

) br

– ( br

. cr

) ar

here a

r = i – j, b

r = j – k, c

r= i + 5k

⇒ [(i – j). (i + 5k)] (j – k) – [(j – k).(i + 5k)](i – j) ⇒ 1(j – k) – (–5)(i – j) ⇒ 5i – 4j – k

67.[A] A = 3, 2; B = 2, 4; C = 4, 5 ⇒ A ×(B ∩ C) = 3, 2 × 4 ⇒ A× (B ∩ C) = (3, 4),(2, 4)

68.[D] for two +ive numbers A.M. ≥ G.M.

∴ 2

ba + ≥ ab ; 2

cb + ≥ bc ; 2

ac + ≥ ca

multiplying all three

8

)ac)(cb)(ba( +++ ≥ abc

(a + b) (b + c) (c + a) ≥ 8abc

69.[A] since the general equation of all conics whose axes coincide with the axes of co-ordinates

if ax2 + by2 = 1 Q it has two arbitrary constant a, b ∴ its differential equation will be of order 2

70.[D] Required area = ∫7.1

1]x[ = ∫

7.1

1xd1

= 7.11)x( = 1.7 – 1 = 0.7 = 7/10

71.[C] [x2 + 1] → is always integer ∴ π [x2 + 1] → will always be multiple of π ∴ sin(π [x2 + 1]) = 0 Q sin nπ = 0 ∴ f(x) = 0 ∀ x ∈ R

72.[B] AB = 3I ⇒ 31 (AB) = I

⇒ det

AB

31 = 1 ⇒

n

31

det (AB) = 1

⇒ n

31

(detA) (detB) = 1

⇒ det A ≠ 0 ⇒ A–1 exist

∴ 31 AB = I ⇒ A

B

31 = I

⇒ A–1 = 31 B Q AA–1 = I

73.[D] If points are coplaner

⇒ [ ]AD,AC,AB = 0

⇒ 221k220

132

−−

− = 0 ⇒ k = – 1

74.[C]

A M

O

P(1, √3)

From figure we have to find out Area of ∆ OPA tangent at P(1, 3 ) given by x + 3 y = 4 cut x axis at A (4, 0)

⇒ OA = 4 , PM = 3

Area (∆ OPA) = 21 . 4 3 = 2 3

75.[C] given P(A) = 0.5 & P(A ∩ B) ≤ 0.3 we have P(A ∪ B) = P(A) + P(B) – P(A ∩ B) P(A ∪ B) ≤ 1 P(A ∪ B) + P(A ∩ B) ≤ 1+ .3 P(B) ≤ 1 + 0.3 – 0.5 ≤ 0 .8

76.[C] Let equation of plane is

ax +

by +

cz = 1 meets axes in

A(a, 0, 0), B(0, b, 0) & C(0, 0, c)

centroid

3c,

3b,

3a ≡ (1, 2r, 3r2)

⇒ a = 3, b = 6r, c = 9r2

plane 3x +

r6y + 2r9

z = 1

6r2x + 3ry + 2z = 18r2

77.[D] equation of ellipse is given by

(x + 1)2 + (y – 1)2 = 41

2

23yx

+−

or 8(x + 1)2 + 8(y –1)2 = (x – y + 3)2 7x2 + 7y2 +2xy + 10x – 10y + 7 = 0 compare with given eqn a = 7, b = 7, c = 7, h = 1 a + b = 14 = 2 c.h

78.[B] shortest distance must he perpendicular distance

⇒ 2

30ba −− = |a| ⇒ |2a + b| = 2 |a|


⇒ (2a + b)2 = 2a2 ⇒ 2a2 + b2 + 4ab = 0

⇒ 2

2

ab + 4

ab + 2 = 0

clearly ab is root of eqn x2 + 4x + 2 = 0

79.[D] Q root of eqn 2x2 + 3 2 x + b = 0 are real and equal ∴ D = 0

( )223 – 4.2.b = 0

b = 8

18 = 9/4 ∴ b is +ve

Now in equation x2 – bx + 1 = 0 coff. of x2 & constant term are +ve and coff. of x is –ve ∴ both root of equation will be +ve 80.[C] Let first term of A.P. is A and common

difference is d ∴ Tp = A + (p – 1)d Tq = A + (q – 1)d Tr = A + (r – 1)d These three are consecutive terms of G.P. Let

common ratio of G.P. is R

∴ R = p

q

TT

= q

r

TT =

qp

rq

TTTT

−

−

(by property of ratio & proportion)

R = qp

rq

TTTT

−

− =

d)1q(Ad)1p(Ad)1r(A]d)1q(A[

−+−−+−+−−+

= qprq

−− =

pqqr

−−

81.[C] If n distinct objects are arranged in a row then

No. of arrangement in which none of them occupies its original place is

n

−+−+−+−

n1)1(.......

41

31

21

111 n

here in Question n = 4

∴ ways are 4

+−+−

41

31

21

111

= 24

+−

241

61

21 = 24

+−

241412 = 9

82.[B] (x + a)n = nC0xn + nC1xn–1 + nC2xn–2a2 + nC3xn–3a3 + nC4xn–4a4.......+ nCnan = (nC0xn

+ nC2xn–2a2 + nC4xn–4a4 +....)

+ (nC1xn–1 + nC3xn–1a3 + ....) = (T0 + T2 + T4 +....) + (T1 + T3 + .....) replace a by ia & –ia (x + ia)n = nC0xn + i nC1xn–1a – nC2xn–2a2 –

i nC3xn–3a3 +nC4xn–4a4 +.........+ in nCnan

= (nC0xn – nC2xn–2a2 + nC4xn–4a4........) + i([nC1xn–1a – nC3xn–3a3 +.........) = (T0 – T2 + T4–....) + i(T1 – T3 + T5.....) (i) and(x – ia)n = (T0 – T2 + T4..) – i(T1–T3 + T5....)

(ii) multiplying (i) & (ii) (x + ia)n(x – ia)n = (T0 – T2 + T4.....)2 +(T1– T3 + T5....)2

⇒ (x2 + a2)n

83.[B] Let A =

333

222

111

cbacbacba

then

∆ = det A =

333

222

111

cbacbacba

and

333

222

111

CBACBACBA

= |(adj A)´| = | adj A|

= |A|3–1 = |A|2 = ∆2 Q |adj A| = |A|n–1

84.[C] sin17r2 π + i cos

17r2 π

= i

ππ

17r2sini–

17r2cos

= i

π−

17r2i

e

∴ ∑=

π

+π16

1r 17r2cosi

17r2sin = ∑

=

π−16

1r

17r2i

ei

= i

++++

π−

ππ−

π1732i

176i–

174i

172i–

e.......eee G.P.

= i ( )

−

−

π−

ππ−

17/2i

1617/2i)17/2i(

e1

e1e

= i ( )

−−

π−

π−π−

17/2i

17/32i17/2i

e1e1e

= i

π−π−π− 17/2ie–1

17/34ie–17/2ie = i

π

ππ−

/172i–

2i17/2i

e–1e–e

π

π−

/17i2–

/17i2

e–11–e = i = – i Q ei2π = 1

85.[B] put x = tan θ as x → 0+

∴ 0lim→θ

θθθθ

3.22.tan =

31


86.[A] f(x) = xcosdxsincxcosbxsina

++

f '(x) =

( )2xcosdxsinc]xsindxcosc][xcosbxsina[]xsinbxcosa][xcosdxsinc[

+

−+−−+

= 2

2222

)xcosdxsinc()xsinx(cosbc)xsinx(cosad

++−+

f '(x) = 2)xcosdxsinc(bcad

+−

Q f(x) is decreasing for all x if and only if f '(x) < 0 for all x ∴ ad – bc < 0 ⇒ ad < bc 87.[A] Let y = 27cos2x. 81sin2x = 33cos2x. 34sin2x = 33cos2x + 4sin2x ∴ y will be minimum when 3cos2x +4sin2x is minimum

+≤θ+θ≤+− 2222 bacosbsinaba

∴ min. of (3cos2x + asin2x) = 22 43 +− = –5

∴ y min. = 3–5 = 2431

88.[C] Let f(x) = sin4x + cos4x is periodic with period π/2

∴ ∫π+

+2/a

a

44 )xcosx(sin dx

= ∫π

+2/

0

44 )xcosx(sin dx

= 2 ∫π 2/

0

4 xsin = 2.2

.21.

43 π =

83π

89.[C] Number of product of numbers taken two or

more at a time Q all 10 nos. are prime ∴ all are different ∴ Nos of product of nos. taken two or more at a

time is 10C2 + 10C3 +10C4 ...... + 10C10 Q 10C0 + 10C1 + 10C2 + 10C3 + ......... 10C10 = 210 ∴ 1+10 + 10C2 + 10C3 + 10C4 +.........+10C10 = 210 10C2 + 10C3 + ............... 10C10 = 210 – 11 90.[A] Any point on line |x| = 6 has co-ordinates

(± 6, λ) if A(3, 4) & B(– 6, λ); AB ≤ 4 (3 + 6)2 + ( λ – 4)2 ≤ 16 ( λ – 4)2 + 81 –16 ≤ 0 No real λ

For A(3, 4) & B (6, λ); AB ≤ 4 32 + (λ – 4)2 ≤ 16 (λ – 4)2 ≤ 7 4 – 7 ≤ λ ≤ 4 + 7 ≈ 1.3 ≤ λ ≤ 6.3 integral λ is 2, 3, 4, 5, 6 i.e. 5 values

Interesting Science Facts

• Alfred Nobel invented dynamite in 1866.

• Wilhelm Rontgen won the first Nobel Prize for physics for discovering X-rays in 1895.

• The tallest tree ever was an Australian eucalyptus - In 1872 it was measured at 435 feet tall.

• Christian Barnard performed the first heart transplant in 1967 - the patient lived for 18 days.

• The wingspan of a Boeing 747 is longer than the Wright brother’s first flight.

• An electric eel can produce a shock of up to 650 00.

• There are 60,000 miles of blood vessels in the human body.

• An individual blood cell takes about 60 seconds to make a complete circuit of the body.

• On the day that Alexander Graham Bell was buried the entire US telephone system was shut down for 1 minute in tribute.

• The low frequency call of the humpback whale is the loudest noise made by a living creature.

• The call of the humpback whale is louder than Concorde and can be heard from 500 miles away.

• A quarter of the world’s plants are threatened with extinction by the year 2010.

• Each person sheds 40lbs of skin in his or her lifetime.

• At 15 inches the eyes of giant squids are the largest on the planet.

• The largest galaxies contain a million, million stars.


PHYSICS

1.[A] u2 = 5gR ∴ v2 = u2 – 2gR = 5gR – 2gR = 3gR

u

v B

Tangential acceleration at B is at = g (downwards) Centripetal acceleration at B is

aC = Rv2

= 3g

∴ Total acceleration will be

a = 2t

2C aa + = g 10

2.[C] Let l be the length of the rod and θ the angle of

rod with x-axis (horizontal) at some instant of time. Co-ordinates of the centre of rod at this instant of time are

A

x B

y

l

θ

x =2l cos θ

and y =2l sin θ

Squaring and adding Eqs. (1) and (2), we get:

x2 + y2 =4

2l

Which is an equation of a circle of radius 2l

and centre at origin.

3.[A] Q g = 2)hR(GH+

∴ 2R9GM = 2)hR(

GM+

⇒ 3R = R + h ⇒ h = 2R So option (1) is correct. 4.[B] Both the spring are in series

Keq = k2k)k2(k

+ =

3k2

Time period T = 2π eqK

µ

where µ = 21

21

mmm.m

+

Here µ = 2m

∴ T = 2π k23.

2m = 2π

k4m3

Alternative method :

m m 2k/3

m m 2k/3

x2 x1 ∴ mx1 = mx2 ⇒ x1 = x2 force equation for first block;

3k2 (x1+x2) = m 2

12

dtxd

Put x1 = x2 ⇒ 21

2

dtxd +

m3k4 x1 = 0

⇒ ω2 = m3k4

∴ T = 2π k4m3

5.[D] E1 = E2

∴ 21

1m 21ω 2

1A = 21

2m 22ω 2

2A


PAPER - II

BIT-SAT


but m1 = m2 ∴ 2

1ω × 16 = 22ω × 25

∴ 100 × 16 = ω2 × 25 ω = 8 units 6.[A] Let the force producing impulse J is F then

F × h = 52 mR2 × α

and F = ma (where a = Rα)

∴ mah = 52 mRa ⇒ h =

52 R

Also impulse = change in momentum or J = Mv 7.[D] at = αA rA = αCrC

αC = αA

C

Arr = 1.6 ×

2510 = 0.64 rad/s2

t = C

Cαω =

64.060

2100 π×

= 16.35 sec.

8.[C] P(r) = rRQ

4π

r1 + + +

++ +

++ + +++

+ + +

+

+ +++ ++

+ +

+

From Gauss law

∫ ds.E = 0

enqε

= 0

VdV

ε

ρ∫

= ∫ επ

π1r

O04

2/

Rdrr4Qr

E.4πr12 =

4r4

RQ 4

14 π

π

ε0

E = 40

21

R4Qr

πε

9.[B] Inside pressure must be rT4 greater than outside

pressure in bubble. This excess pressure is provided by charge on bubble.

rT4 =

0

2

2 ∈σ

rT4

= 0

42

2

2r16Q

∈×π…

π=σ 2r4

Q

Q = 8πr 0rT2 ∈ 10.[C] In case of a capacitor q = CV

∴ i = dtdq = C

dtdV

dtdV =

0.40.4 V/s = 1.0 V/s

Therefore, if C = 1 F then i = 1×1 = 1A (constant)

11.[A]

+q –q

Energy loss = CC

C21 2

+×V2 =

2C

21 × 2

2

Cq

= C4

q2 =

A4dq

0

2

ε

12.[C] RAB = R

65||R

65 =

125 R

13.[B]

3Ω

2Ω

3Ω 2Ω

2A

A

B

C D

1A

1A

VA – VC = i R = 1 × 3 = 3V .......(i) VB – VC = 1 × 2 = 2V ........(ii) VA – VB = 3 – 2 = 1V

14.[D] B = R2Ni0µ

φ = πb2 × B × N φ = Li

L = iφ =

R2bN 22

0µ , with b <<< R

Energy = 21 Li2 =

R4iN 22

0µ b2

Pa

Pa


15.[A] N cos θ

N sin θ

mg

F = BIL

N Bθ

θ

N cos θ = mg …..(1) [⊗ indicates current I is flowing into the paper] N sin θ = BIL …..(2)

∴ tan θ = mgBIL

16.[C] E = dtdiL

di = dtLE

i = tLE

i = t42

×

i = 0.5t 5 = 0.5t t = 10 sec

17.[A] Steady state current in L = i0 =1R

E Energy

stored in L =21 L

2

1RE

= heat produced in R2

during discharge = 21

2

R2LE .

18.[C] % efficiency = in

out

PP × 100 =

52209110

×× × 100

= 90%

19.[C] ν = RcZ2

− 2

221 n

1n1

20.[A] n = ∞

M N

L

K E(K) E(Kα)

E(L)

E(K) = K

hcλ

= 107.0

4.12 = 115.9 KeV

E(Kα) = E(K) – E(L) = αλ

hc

= 98.4 KeV EL = E(K) – E(Kα) = 115.4 – 98.4 EL = 17.5 KeV

λL = ELhc =

KeV5.17KeVÅ4.12 = 0.709Å

21.[A] | EK | = K

hcλ

= Å15.0

Å.KeV4.12 = 82.7 KeV

The energy of incident photon

Eν = λhc =

1.04.12 = 124 KeV

The maximum kinetic energy is Kmax = Eν – |EK| = 41.3 KeV ~– 41 KeV 22.[B] Frequency corresponding to wavelength of

0.180 nm is ν = c/λ = 1.67 × 1018 Hz From Mosley's law

ν = 43 cR (Z – 1)2

(Z – 1) = cR34 = 26

27Z = Hence element is cobalt.

23.[B] p = λh =

cE =

ccm 2

e = mec

24.[C] I = A.t

Nh∆

ν = 2mm1

cmm1

ch.N

×

λ

N = 34–28

10–9–

1063.6)103(10264010100

××××××

~− 442 photons/mm3 25.[A] Stopping potential of a & b same so frequency fa = fb

* Saturated photoelectric current of b & c same so intensity Ib = Ic > Ia

26.[D] r = R

−1

2

1

l

l = 100

−1

7290

r = 25 Ω 27.[B] VA – VB < Ε VA – VP = Ε/2


PA

BA

VVVV

−− =

l

l 0

VA – VB = l

l 0

2E < E

l > 20l

⇒ l > 50 cm

28.[B] RX =

ll−100

for balance

Initially, 1812 =

ll−100

, finally 8

12 =l'

l'−100

or JJ′ = l′ – l = 20 cm

29.[D] Least count = 100

1.0

2r = N + n (100

1.0 )

r = 2N +

2n (0.001)

r = 0.5 N + n (0.0005) r = 5 (0.1 N + 0.0001 n) 30.[C] HLi 2

163 + → HeHe 4

242 +

Q = (K )KK(21 αα + – Kd

= (13.2 + 13.2) MeV – 4MeV = 22.4 MeV 31.[D] N = N0e–λt, D = N0 (1 – e–λt)

R = R0e–λt, NR

λ===λ

λ

0

0t–

0

t–0

NR

eNeR

= const. 32.[C]

10 β1 = 10 × dD

µλ

in liquid

β2 = dDλ

6β2 = 10β1

dD6λ =

dD10

µλ

µ = 6

10 = 1.67

33.[A] f = 10 cm

1 cm

30 cm

uf

f|m|−

= =3010

10+−

− = 21

o

Irr = m =

21

ωo = ωI

∴ o

Iaa =

o

Irr =

21

34.[B] f = – 15cm

for virtual & 2 times large image m = + 2

m = uf

f−

or + 2 = u–15

15−

−

– 30 – 2u = – 15 – 2u = 15 u = – 7.5 cm

35.[A] MP = 405

200ff

e

0 ==

36.[A] M.P. = m0 × me

32 = m0 × 4 ∴ m0 = 8

37.[A] F – 2T = 6a and T = 4 × 2a ∴ F – 16 a = 6a

⇒ a = 22F ⇒ a = 1m/s2

∴ aCM = 10

2416 ×+× = 1.4 m/s2

38.[D] Acceleration of chain is given by

a = m

3/mgmg2 − = 3

g5

39.[C] According to Pascal’s principle

2

1ff =

2

1AA = 2

2

21

rr =

41

f1 = 41 Mg

40.[C] Length of rod inside the water = 1.0 secθ = secθ

θ 1.0 m

O

F

W

Upthrust F =

22 (sec θ)

500

1 (1000) (10)


or F = 20 sec θ Weight of rod W = 2 × 10 = 20 N

For rotational equilibrium of rod net torque about O should be zero.

∴ F

2

secθ (sin θ) = W = (1.0 sin θ)

or 220 sec2θ = 20

or θ = 45º ∴ F = 20 sec 45º = 20 2 N

CHEMISTRY

1.[C] mvr = π2

nh = 4 × π2

h = πh2

2.[D] Meq. of Acid = Meq. of Ba(OH)2

⇒ 2/M

25.1 × 1000 = (0.25 × 2) × 25

⇒ M = 200

3.[B] )D(

)H(

2

2

rr

= )H(

)D(

2

2

MM

= 24 =

12

4.[A] From Kp = Kc (RT)∆ng = 1.8 × 10–4 × (0.082 × 298)2 = 0.108 5.[A] ∆Hsublimation = ∆Hfusion + ∆Hvap 6.[D] In CH3NH2, N has one lone pair of electrons. 7.[D] Ksp = 4s3 = 4 × 10–12 ⇒ s = 10–4 M 8.[A] Oxidant is the one whose O.N. decreases during

the reaction. H2SO4 (O.N. of S = + 6) changes to SO2 (O.N. of S = + 4)

9.[D] d = A

3 NaMZ

×× =

2333–

3–

1002.632

1052.44

)10a3(2

××

××

××

= 900 kg m–3

10.[B] Ecell = n059.0E0

cell + log ]Anode[]Cathode[

= [– 0.0403 – (– 0.763)] + 2059.0 log

2.0004.0

= + 0.36 +2059.0 log

204.0

11.[D] °

∆PP =

AB

AB

WMMW

or MB = )P/P(W

MW

A

AB

°∆

⇒ MB = 4039

640785.2×

×× = 80

12.[B] Adsorption is exothermic process due to

attraction between adsorbate and adsorbent.

13.[A] t = k303.2 log

xaa−

= 6303.2 log

05.05.0

= 0.384 min 14.[B] 1 cm3 H2O = 1 g H2O

No. of molecules in 1 g H2O = 18

023.61× × 1023

= 3.3 × 1022

15.[D] Isocyanide test also known as carbylamine test. 16.[A] 4-methyl benzene sulphonic acid is stronger

than acetic acid thus it will release acetic acid from sodium acetate.

17.[A] CH3 – CH2 – C ≡ CH + H2O

But–1–yne 42

4

SOH

HgSO →

CH3 – CH2 – C = CH2 OH

→ − mtautomerisenolKeto CH3 – CH2 – C – CH3

O

18.[C]

+ CH3COCl → 3AlCl.Anhyd

COCH3

+ HCl

19.[B]

OCH3

CH3 Br

20.[A] Libermann's reaction


21.[A]

CCl3

→ Fe/Brat.eqv1 2

CCl3 Br

22.[B] CH2 = CH2 →HOCl

Cl CH2 – CH2

OH → 3NaHCO.aq

CH2 – OH CH2 – OH (glycol)

23.[A] Free rotation around carbon-carbon bond takes

place easily in alkanes. Now ethane and hexachloroethane both are alkanes, but in hexachloroethane bulky chlorine atom is present while ethane is least hindered.

24.[C] Due to the presence of –Cl group which is a +M

group. 25.[A] Due to similar charges on adjacent atom the

structure is least stable.

N

- -

O

O -

-

- -

26.[B] CH3 – CH3 fissionbondHomolytic →

radicalfreemethyl33 HCHC

••+

Free radical is formed which is sp2 hybridised

C — H

H H

27.[A] If atom or group of higher priority are on

opposite direction at the double bond of each carbon atom then the configuration is known as E and if they are in same direction then the configuration is known as Z-configuration.

(2E, 4E) – 2, 4-hexadiene 28.[C] The brown ring test for −

2NO and −3NO is due

to formation of [Fe(H2O)5NO]2+

29.[A] The absorption of energy or observation of

color in a complex transition compounds depend upon the charge of the metal ion and the nature of the ligand attached. The same metal ion with different ligands shows different absorption depending upon the type of ligand, the presence of weak field ligand make the central metal ion to absorb low energies i.e. of higher wavelength.

30.[C] The existance of Fe2+ and NO+ in nitroprusside ion [Fe(CN)5NO]2– can be established by measuring the magnetic moment of the solid compound which should correspond to Fe2+ = 3d6 four unpaired electron.

31.[B] LiNO3 on heating gives 4LiNO3 →∆ 2Li2O(s) + 4NO2 + O2 32.[D] Three dimensional sheet structure are formed

when three oxygen atoms of each [SiO4]4–

tetrahedral are shared. 33.[B] Due to oscillation of free electron Na metal

shows metallic lusture. 34.[B] Mn + 2HNO3 → Mn (NO3)2 + H2 35.[A] 'Lapis Lazuli' is the aluminium silicate present

in the earth rocks as blue stone. 36.[B] B < C < N < O when we move from B to O in a

periodic table the first ionization enthalpy increase due to the attraction of nucleus towards the outer most of electron and IE of N > O.

37.[B] NO2 → sp SF4 → sp3d

−6PF → sp3d2

38.[A] Mg belongs to group 2. Therefore its size is less

than that of Na. 39.[B] Alkali metal hydroxide KOH is highly soluble

in water. 40.[B] Na2CO3 + H2O + CO2 → 2NaHCO3

MATHEMATICS

1.[C] y = 2x2 – log | x |

dxdy = 4x –

|x|1 ×

x|x| = 4x –

x1

dxdy =

x

)21x)(

21x(4 −+

– + – +–1/2 1/2 0

∴ y has minima at x = –21 and x =

21 but x = 0

is not point of maxima as x = 0 is not in the domain.


2.[A] f(x) = 1 + x sin x [cos x]

Q 0 < x ≤ 2π ⇒ 0 ≤ cos x < 1

⇒ [cos x] = 0 ∴ f(x) = 1 ∴ f(x) is a constant function and hence

continuous. It neither strictly increasing nor decreasing.

3.[B] It is given that

rSr × 100 = 1

v = 34

πr3 ⇒ log v = log 3π4 + 3 log r

v1

δv = r3

δr

vvδ × 100 =

rr3δ × 100

= 3 × 1 = 3 Hence error in volume is with in 3 % 4.[C] Set A = 1, 2, 3 and R = (1, 1), (2, 2) Since (3, 3) ∈/ R it is not reflexive Since R–1 = (1, 1) (2, 2) = R, R is symmetric Since the situation in (a, b), (b, c) ∈ R does not

arise in R, R is also transitive. Also R ∩ R–1 = (1, 1), (2, 2) ⊂ DA = (1, 1)

(2, 2) (3, 3) ⇒ R is anti symmetric Hence (ii) (iii) and (iv) are correct. 5.[C] (A – B) ∪ (B – C) ∪ (C – A)C = (A ∪ B ∪ C)

– (A – B) ∪ (B – C) ∪ (C – A) = A ∩ B ∩ C Q A ∪ B ∪ C = universal set 6.[C] (B–1 AB)2 = (B–1AB) (B–1 AB) = (B–1ABB–1

AB) = (B–1 AIAB) = (B–1 A2B) (B–1AB)3 = (B–1AB)2 (B–1AB) = (B–1A2B) (B–

1AB) = (B–1A2BB–1AB) = (B–1A2IAB) = (B–1 A3B) Now (B–1 AB)4 = (B–1AB)3 (B–1AB) = (B–1A3B) (B–1AB) = B–1A4B

7.[A] Q g(0) = 0 ∴ 0x

lim→ x

)x(g

00 form

∴ 0x

lim→

g′(x) = g′(0) ... (1)

Q g(x) = )3('f)2('f)('f)3(f)2(f)(f

)3x(f)2x(f)x(f

αααααα

α+α+α+

∴ g′ (x) = )3('f)2('f)('f)3(f)2(f)(f

)3x('f)2x('f)x('f

αααααα

α+α+α+

∴ g′ (0) = 0

0x

lim→ x

)x(g = g′(0) = 0

8.[A] ∆1 =ba2ey2x4z2

ed2f=

bea2y2z2x4efd2

– (C1 ↔ C2)

= eba2z2y2x4

fed2 (C2 ↔ C3)

= –fed2z2y2x4

eba2 (R1 ↔ R3)

=z2y2x4

fed2eba2

= ∆2 (R2 ↔ R3) ∴ ∆1/∆2 = 1

9.[D] Q All coins are identical ∴ First we will give 3 coin to each person so

that every one has at least 3 rupee, now rest 5 coin we have to distribute among 5 person in such a way that any one can get any no. of coin.

∴ Total no. of ways 5 + 5 – 1C5 – 1 = 9C4 = 126 Q No. of ways of distributing n identical thing

among r person when any one can get any no. of thing is n + r –1Cr–1

10.[C] x1 < x2 ≤ x3 < x4 < x5 ≤ x6 gives rise to the

following four cases x1 < x2 < x3 < x4 < x5 < x6

x1 < x2 = x3 < x4 < x5 < x6 x1 < x2 < x3 < x4 < x5 = x6 x1 < x2 = x3 < x4 < x5 = x6 ∴ Total ways 9C6 + 9C5 + 9C5 + 9C4 = 10C6 + 10C5 = 11C6

11.[D] 2

+++ ....4

a2

a142

= ea + e–a

= eln n + e– ln n = n + n1 =

n1n2 +


12.[A] 18

x2x

−

Let (r +1 )th term is independent of x

∴ r = 1

21

02118

+

−× = 6 Q r =

β+α−α mn

∴ (r + 1) = 7th term is independ of x

∴ 7th term is 18C6 ( x )18–6 6

x2

−

= 18C6 26

13.[C] S = 1 + 32 .

21 +

6.35.2 2

21

+

9.6.38.5.2 3

21

+ . .

. .

= 1 + 1

3/2

21 +

2)3/5)(3/2( 2

21

+

3)3/8)(3/5)(3/2( 3

21

+ . . .

= 3/2

211

−

− =

32

21 −

= 3/22 = 41/3

14.[D] Q ω = 2

3i1+− ⇒ iω = 2

i3 −−

∴ arg (iω) = π + 6π

Q ω2 = 2

3i1−− ⇒ iω2 = 2

i3 −

∴ arg (iω2) = 2π – 6π

∴ arg iω + arg iω2 = 3π

15.[A] Q 4log

1r2

= 4log

r1

1

2

= 2r

∴ ∑=

n

1r 2r =

+

2)1n(n

21 =

4)1n(n +

16.[C] a1 + a8 + a15 = 3a1 + 21d = 15 ⇒ a1 + 7d = 5 a2 + a3 + a8 + a13 + a14 = 5 a1 + 35d = 5(a1 + 7d) = 5 × 5 = 25 17.[B] Qb > 0 ∴ D ≤ 0

(a + c)2 + 4b2 – 4b (a + c) ≤ 0 ⇒ a2 + c2 + 2ac + 4b2 – 4ab – 4bc ≤ 0 ⇒ (a + c – 2b)2 ≤ 0 ⇒ 2b = a + c i.e. a, b, c are in A.P. 18.[C] f(x) = (x – a1) (x – a3) (x – a5) + 2 (x – a2) (x –

a4) (x – a6) = 0 a1 < a2 < a3 < a4 < a5 < a6 f(a1) = 2 (a1 – a2) (a1 – a4) (a1 – a6) < 0 f(a2) = (a2 – a1) (a2 – a3) (a2 – a5) > 0 ∴ At least one real root lies in (a1, a2) Similarly, at least one real roots lies in each

interval (a3, a4) and (a5, a6) But f(x) is cubic, therefore there are only three

roots. Hence the equation f(x) = 0 has one real roots in

each interval (a1,a2) (a3,a4) and (a5,a6) 19.[A] xdx + zdy + (y + 2z)dz = 0 ⇒ xdx + 2zdz + zdy + ydz = 0 xdx + 2zdz + d(yz) = 0

2

x2 + z2 + yz = c

20.[A] slope of tangent

dxdy = 2x + 1

⇒ y = x2 + x + C when x = 1, y = 2 ∴ 2 = 1 + 1 + C ⇒ C = 0 ∴ y = x2 + x

∴ Required area = dx)xx(1

0

2∫ +

= 1

0

23

2x

3x

+ =

65

21.[B]

y = ex

R

y = e–x

Q

P S O

y = e–|x| =

<

≥

0x:e

0x:ex

x–

By symmetry Let P = (t , 0) then Q( t, e–t), R = (–t, e–t) and S = (– t, 0) ∴ Area of rectangle = 2te–t = f(t) say


then dtdf = 2–te–t + e–t = 0 ⇒ t = 1

2

2

dtfd = 2 – (1 – t)e–t – e–t < 0 for t = 1

Hence, maximum area 2/e

22.[C] I = ∫ dx|xsin| =

<−

≥

∫∫

0xsinifxdxsin

0xsinifxdxsin

= – cos x + C if sin x ≥ 0 cos x + C if sin x < 0 = cos x . sgn (sinx) + C

Qsgn (sinx) = xsin

|xsin| =

<−=>

0xsin;10xsin;00xsin;1

23.[B] From the option dxd

2

)x(f)x(log

21

φ

= log

φ)x(f)x(

φ−φ

2)x(f)x('f)x()x(')x(f ×

)x()x(f

φ

= )x()x(f

)x('f)x()x(')x(fφφ−φ × log

φ)x(f)x(

∴ ∫ φφ−φ

)x()x(f)x('f)x(x')x(f log

)x(f)x(φ dx

= 21

2

)x(f)x(log

φ + C

24.[D] dxdy = – 2

2

xc

⇒ )'y,'x(dx

dy

= 2

2

)'x(c− = 2)'x(

'y'x− = –

'x'y

∴ Equation of tangent at (x′,y′) is

y – y′ = – 'x'y (x – x′)

Which meets the co-ordinate axes at A and B (say) then A = (2x′, 0), B = (0, 2y′)

Mid point of AB is (x′, y′) 25.[D] D.R's of OP = a, a, ,a ∴ Equation of plane ⊥ to OP and passing

through P is a(x – a) + a(y – a) + a (z – a) = 0 ⇒ x + y + z = 3a Intercepts on axes made by the planes are 3a, 3a, 3a ∴ Sum of reciprocal of the intercepts

= a31

a31

a31

++ = a1

26.[C] 2|a|

r= p2 + 25 + 289 = p2 + 314

| br

|2 = 4q + 169 + 1 = 4q + 170 According to question | a

r |2 = | br

|2 ⇒ p2 + 314 = 4q + 170 ⇒ p2 = 4q – 144 = 4(q – 36) p, q are +ve integer 1 ≤ p, q ≤ 1000 p is even integer let p = 2 K then 4K2 = 4(q – 36) ⇒ K2 = q – 36 Q 1 ≤ K2 ≤ 964 ⇒ 1 ≤ K ≤ 31 ∴ Number of ordered pairs (p, q) = 31

27.[D] ∴ | cbarrr

++ | = 1 ⇒ | a

r|2 + | b

r|2 + | c

r|2 + 2( b.a

rr+ c.b

rr+ a.c

rr) = 1

⇒ 61

31

21

++ + 2( a.brr

) = 1

⊥⊥

×λ=

bcandac

bacrrrr

rrrQ

⇒ a.brr

= 0 ⇒ br

⊥ ar

∴ Angle between them is 2π

28.[D] a8

x 2

−+

2ay2

− = 1 will represents an ellipse is

8 – a > 0, a – 2 > 0 and 8 – a ≠ a – 2 ⇒ a < 8, a > 2 and a ≠ 5 ∴ a∈ (2, 8) – 5 29.[D] y = x – 1 is a focal chord of the parabola y2 = 4x. Therefore tangent at its extremities are

perpendiculars.

30.[D]

(1, 1)

(1, –1)

(–1, 1) A1

B1

B

A O

(–1, –1)

A1B1 = 22

AB = 22 – 2 = 2( 2 – 1)

OA = 2 – 1 taking origin as centre and OA as radius circle

will touches all four circle ∴ equation of circle is


x2 + y2 = ( 2 – 1)2

x2 + y2 = 3 – 2 2 31.[A]

P(a, a2) (2, 0)

(0, 2)

Clearly a > 0 Also P lies on that side of line x + y = 2 Where origin lies ∴ a + a2 – 2 < 0 ⇒ (a – 1) (a + 2) < 0 ⇒ – 2 < a < 1 but a > 0 ∴ 0 < a < 1 ∴ a ∈ (0, 1) 32.[D] Triangle is right angled at O(0, 0). Therefore

orthocentre is O(0, 0) and circumcentre is mid

point of hypotense i.e.

2b,

2a

∴ Distance between orthocentre and

circumcentre = 22 ba21

+

33.[B] Let a = 3K, b = 7 K and c = 8K

∴ s = 2

cba ++ = 9K

there rR =

∆4abc .

∆s =

)cs)(bs)(as(s4abcs

−−−

= KK2K6.4

K8.K7.K3 = 27 ⇒

rR =

27

34.[D] sinx (sinx + cosx) = K ⇒ sin2 x + sinx cos x = K

⇒ 2

x2cos1− + 2

x2sin = K

⇒ 21 (sin2x – cos2x + 1) = K

Q – 2 ≤ sin 2x – cos 2x ≤ 2

⇒ 2

21− ≤ 2

1x2cosx2sin +− ≤ 2

12 +

⇒ 2

21− ≤ K ≤ 2

12 +

35.[B] y = f(x) = xtanx1

x+

⇒ dxdy = 2

2

)xtanx1()xsecxx(tanxxtanx1

++−+

= 2

22

)xtanx1(xsecx1

+−

O

y = x2 y = cos2x

π/2 π

dxdy

= 0 ⇒ x2 = cos2x

There is only one point in (0, 2π ) say x, at

which dxdy

= 0

at x1 – h ⇒ x2 < cos2x ∴ dxdy > 0

& at x1 + h ⇒ x2 > cos2 x ∴ dxdy < 0

∴ at x1 slope change from + ve to – ve ∴ There is only one critical point

in

π

2,0 at which f(x) has local maxima.

36.[B] 3cos2θ – 32 sinθ cosθ + 3sin2θ = 0

( 3 cosθ + sinθ) (cosθ – 3 sinθ) = 0

⇒ tan θ = 3

1 or tanθ = – 3

∴ θ = nπ + 6π or θ = nπ –

3π

∴ | r – s| = | – 3 – 6| = 9

37.[B] cot–1

πn >

6π ⇒

6π < cot–1

πn < π, n∈ N

Q cot–1x ∈ 0 (a, π)

⇒ – ∞ < πn < 3

– ∞ < n < 3 π – ∞ < n < 5.4 ⇒ max. n = 5 Q n ∈ N 38.[A] ∴ tan π [x] = 0 ∀ x ∈ R since [x] ∈ Z Period of x = 1 ⇒ Period of sin 3πx = 1 Hence period of f(x) = 1


39.[C] f(x) = cos–1

x|x|log ]x[

For domain x

|x| > 0

⇒ x ∈ (0, ∞) and [x] > 0 and [x] ≠ 1 ⇒ x ≥ 2 ∴ x ∈ [2, ∞)

⇒ x

|x| = 1 then log[x]

x|x| = 0

f(x) = cos–10 = 2π

40.[C] ∴ f(a) = 0

∴ ax

lim→ )x(f3

)x(f61loge +

00 form

⇒ ax

lim→ )x(f6

)x(f61log2 e +

× = 2 × 1 = 2

Q 0x

lim→ x

x1loge + = 1

41.[D] +→5x

lim]x[x20x9x2

−+−

=0h

lim→ ]h5[h5

20)h5(9)h5( 2

+−+++−+ =

0hlim→ h

hh2 + = 0

–4x

lim→ ]x[x

20x9x2

−+− =

0hlim→ ]h4[h4

20)h4(9)h4( 2

−−−+−−−

= 0h

lim→ h1

hh2

−+

= 0 ∴ P = 0

42.[B] f(1) = 0

f(1 + 0) = 0h

lim→ 1)h1(

1])h1[(2

2

−+−+ =

0hlim→ 2hh2

11+−

= 0

f(1 – h) = 0h

lim→ 1)h1(

1–])h–1[(2

2

−−=

0hlim→ 2hh2

10+−

− = ∞

⇒ f(x) is discontinuous at x = 1 43.[A] (a + bx)ey/x = x ... (1) Differentiating, w.r.t. x we get

bey/x + (a + bx)ey/x .

−

21

xyy.x = 1

⇒ bey/x + x.

−

21

xyxy = 1 Q (a + bx)ey/x = x

⇒ bxey/x + xy1 – y = x ⇒ xy1 – y = x – bxey/x ⇒ xy1 – y = aey/x ... (2) (from (1))

⇒ xy2 + y1 – y1 = aey/x

−

21

xyxy

⇒ x3y2 = aey/x (xy1 – y) = (xy1 – y)2 (from 2)

⇒ 2y

1 (xy1 – y)2 = x3

44.[B] ∫−

3

2

dx)x(f = ∫−

−

1

2

dxt)x(f ∫−

0

1

dx)x(f + ∫1

0

dx)x(f

+ ∫2

1

dx)x(f + ∫3

2

dx)x(f

= (–2)3 + (–1)3 + 03 + 13 + 23 = 0

45.[B] x2f(x) +

x1f = 2

I = ∫3

3/1

dx)x(f put x = t1 , dx = dt

t12−

⇒ I = – dtt1.

t1f 2

3/1

3∫

= dx

x1.

x1f 2

3

3/1∫

⇒ 2I = ∫

+

3

3/12 dx

x1f

x1)x(f

= ∫

+

3

3/12

2 dxx1

x1f)x(fx = ∫

3

3/12 dx

x2

= 3

3/1x12

− = – 2

− 331 =

316

⇒ I = 38

LOGICAL REASONING 1. [D] The pattern is x2 +1, x2 + 2, . . . . Missing number = 28 × 2 + 3 = 59 2. [A] A car runs on petrol and a television works by

electricity.

3.[A] All except Titans are planets of the solar system.

4. [C] 5. [B]


6. [D]

7.[B] The third figure in each row comprises of parts

which are not common to the first two figure.

8. [A]

9. [C]

10.[A]

ENGLISH 1.[B] Geraff : Incorrect spelling. • 'e' should be replaced with 'i' • The word should end with 'e' after 'ff' Giraffe : Correct spelling. Giraf : 'fe' is to be added in the end. Gerraffe : • 'Ge' is to be replaced with 'Gi' to make the

correct spelling.

2.[B] Puncture : No error. It makes the tyre flat. Puntuation : Error of spelling Correct spelling is 'Punctuation' Hence 'c' is missing. Pudding : No error It is used as 'Dessert' Pungent : No Error It is some what 'sharp' and 'shrill'.

3.[A] Luxurious : (Plush) Something full of all 'amenities' making life

'cozy' and 'snug'.

Delicious : Irrelevant as it means 'something very tasty.'

Comforting : 'Irrelevant' as it means 'giving necessary comforts', whereas 'Plush' means more than comforts.

Tasty : (Irrelevant) It means 'delicious' 4.[A] Lively : Correct synonym to 'sprightly' as both

means, 'someone dashing/energetic/enthusiastic'.

Beautiful : (Irrelevant) Sportive : (Irrelevant) Intelligent : (Irrelevant) 5.[D] Wicked : It is almost a synonym to 'Astute' Impolite : Irrelevant because it is the antonym

of 'polite'. Cowardly : Irrelevant as it is the opposite of

'bravely'. Foolish : (It's the correct antonym of 'Astute'

which itself means 'clever, shrewd'. 6.[D] Deadly : It means 'Fatal'. Hence, this is not a proper antonym to

'innocuous'. Ferocious : It means 'horrible' Hence, irrelevant to the opposite of 'innocuous'. Poisonous : It means 'venomous'. Hence, an irrelevant 'antonym'. Harmful : It is a perfect antonym of innocuous

which itself means 'harmless'. 7.[D] Corruption : Irrelevant Worldliness : Irrelevant Favouritism : Irrelevant Nepotism : (Correct Answer) because It's a kind of corruption in which the authority

in power takes the advantage of giving opportunity to their relatives in their self interest.

8.[B] Cross : (to pass by, to intersect) It means different Hence, irrelevant. Shuttle : (Proper answer)


It's a kind of "regular beats" of an air flight or bus service between the two stations.

Travel : It means to journey. Hence, irrelavent. Run : (to move regularly) Hence, irrelevant. 9.[D] Only 1 is correct : Inappropriate answer because sentence 1 can't

be correct using 'practise' as it is a verb, whereas the required word should be a noun.

Only 2 is correct : Sentence 2 is also wrong because the word

'practice' is wrongly used as a verb. It should be a verb like 'practise'. Hence, incorrect answer.

Both the sentences 1 and 2 are correct. This is not relevant. Both the sentences 1 and 2 are not correct. Correct option, if both the words, i.e. 'practice'

and 'practise' are interchanged respectively, it really makes a meaningful sentence.

10.[C] Sentence 1 is correct : This option is wrong because the word

'ingenuous' means 'frank and simple' which is inappropriate.

Sentence 2 is correct : This option is also wrong because the word

'ingenious' means 'clever or prudent' and this is inappropriate.

Both the words, i.e. 'ingenuous' and 'ingenious' if interchanged together respectively, it really makes both the sentences meaningful.

Hence, appropriate option. Both the sentences can't be interchanged.

This is an incorrect option because words have been misinterpreted together.

Incorrect option.

11.[C] Far off : It can't be used in place of 'aloof' as far off'

means long-long ago. Hence, incorrect alternative . Introvert : It means 'self-centred', Hence, It is an incorrect alternative. distance : This is an appropriate word because

one of the meaning of 'aloof' is distant also while keeping distance between two nouns.

Depressed : (it means 'hopeless') Hence, quite irrelevant.

12.[A] "Meatless days" This is the name of a novel. Hence, no error is there.

Have been made : (Erroneous) Because 'have' should be replaced with 'has'

because 'meatless days' is a singular noun. Into a film : No error in this part of the sentence. No error : Incorrect option because there is an

error in the sentence.

13.[C] Looking forward : (No error) This is a phrase. 'to' (no error) This is a preposition. 'Meet you here' (erroneous) Because 'meet will be replaced with 'meeting' Phrase 'looking forward to' is followed by

present participle (V. I + ing) form of the Verb. No error : (incorrect option) Part 'C' is erroneous.

14.[C] Good and Evil This is a wrong interpretation. Former and Latter : Wrong interpretation. For and against a thing. Appropriate option as it really suits the Idiom

ins and outs. Foul and Fair : (by hook or by crook) This is an inappropriate option.

15.[A] Broke out : (to start suddenly) 'Correct and relevant' option because it is used

for 'wars' and 'diseases' e.g. cholera broke out in Surat in 1985.

Set out : (to start) it is different because it is used when one leaves

for somewhere e.g. He set out on his long voyage to Achilese. took out : (incorrect use) Because it means differently. e.g. He took out a one rupee coin to give to the

beggar. Went out : (Incorrect use) Because meaning is

different e.g. : The light went out when I was preparing

for my Board Exams. Hence, inappropriate

Students' Forum - Career Point

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