XtraEdge for IIT-JEE JULY 2010 - Career Point · XtraEdge for IIT-JEE 3 JULY 2010 Volume-6 Issue-1...

XtraEdge for IIT-JEE 1 JULY 2010


Dear Students,

Is examination a common cause of stress? In most Asian cultures, the great emphasis on academic achievement and high expectations of success make it especially stressful for students. The strong negative stigma attached to failure also adds to the pressure. Like it or not, we have to accept that examinations are necessary in any educational system. Even though it is debatable whether they are accurate measures of actual ability, no better alternatives have been proposed. Examinations remain necessary to motivate students’ learning, measure their progress and ultimately, serve as evidence of attainment of certain skills, standards or qualifications. Success at examinations provides opportunities to proceed with higher education and improves employment prospects, underlining their importance. No matter how well prepared, many factors may influence one’s performance at the time of the examination and there is seemingly, no definite guarantee of success. Essentially, it is this vital importance attached to success at examinations coupled with the element of uncertainty that makes them so stressful. As with other sources of stress, the stress of examinations is not all bad. It is a strong incentive for students to study and poses a challenge for individual achievement. However, when stress becomes excessive, performance begins to suffer. There is thus a need to control levels of stress before it becomes overwhelming and detrimental. Reliase of stress is necessary for optimum performance, the means of which is relasing. Learn to relax The stress responses produces muscle tension, which you would commonly experience as backache, neck ache or tension headache at the end of the day. Often this is unconscious. So to relax these muscles, you need to consciously practice relaxation exercises. These could involve muscle relaxation, deep breathing exercises, body massage or guided imagery. Like any particular skill, you need to practice them regularly in order to reap the benefits. Another way to relax is to maintain a quiet time as part of the daily routine. Quiet time refers to a time for you with no interruption from external sources or distractions. This is a time where you may choose to just think of nothing and relax. Finally, you can always take up a hobby to help you relax. Do something you enjoy, be it listening to music. Ideally, the drive to study should be internally driven by a desire to achieve one’s own personal goals. Instead, many are driven more by the fear of failure, which is more stress-provoking and leads easily to discouragement. Attending school should not merely revolve around preparation for examinations. Interacting with teachers, socializing with friends, participating in sports or other extra-curricular activities are all valuable aspects of a ‘well-rounded’ education. Instead of wishing things would get easier, start looking at how you can get better...

Yours truly

Pramod Maheshwari, B.Tech., IIT Delhi

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Volume-6 Issue-1 July, 2010 (Monthly Magazine)

NEXT MONTHS ATTRACTIONS

Much more IIT-JEE News.

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

Challenging Problems in Physics,, Chemistry & Maths

Key Concepts & Problem Solving strategy for IIT-JEE.

Xtra Edge Test Series for JEE- 2011 & 2012

S

Success Tips for the Months

• If you don’t notice when you win, you will only notice when you lose.

• It’s not bragging if you can do it.

• Feel the power of yet. As in “I don't know how to do this yet.”

• The difficult we do immediately. The impossible takes a bit longer.

• Some look down the rapids and see the rocks. Hunters look down the rapids and see the flow around the rocks.

• To know what you are doing is an advantage. To look like you know what you are doing is essential.

• First law of expertise: Never ask a barber if you need a haircut.

• If you think you can, you are probably right. If you think you can't, you are certainly right.

• Don't do modesty unless you have earned it.

CONTENTS

INDEX PAGE

NEWS ARTICLE 4 Are Nanoparticles health hazard Nano is the new black

IITian ON THE PATH OF SUCCESS 6 Ms. Padmasree Warrior & Dr. Krishan K. Sabnani

KNOW IIT-JEE 7 Previous IIT-JEE Question

XTRAEDGE TEST SERIES 52

Class XII – IIT-JEE 2011 Paper

Class XI – IIT-JEE 2012 Paper

Regulars ..........

DYNAMIC PHYSICS 14

8-Challenging Problems [Set# 3] Students’ Forum Physics Fundamentals Capacitor - 1 Friction

CATALYSE CHEMISTRY 29

Key Concept Reaction Mechanism Energetics Understanding : Organic Chemistry

DICEY MATHS 42

Mathematical Challenges Students’ Forum Key Concept 3-Dimensional Geometry Progression & Mathematical Induction

Study Time........

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


Are Nanoparticles a Health Hazard?

Nanoparticles are a mega business opportunity for multinationals but they may pose a health hazard to users

There is a new industrial revolution taking place all around us. The only problem is we can’t see it. The building blocks, being developed at the cost of billions of dollars by scientists, governments and multinational corporations, are just a few atoms or molecules thick — nanoparticles. Many are less than 100 nanometres (nm) — one-billionth of a metre — thick. A single human red blood cell in comparison is around 500 nm in diametre. It’s a pity though that our eyesight isn’t good enough at nanometre level, for if it were, we would see that nanoparticles of precious metals like gold, silver and titanium have already made the jump from research labs to our homes. Manufactured nanoparticles are today present in thousands of consumer products around the world — silver in washing machines and water purifiers to kill bacteria, zinc in cosmetics to protect against ultraviolet rays, carbon nano-tubes in tennis rackets to make them stronger and lighter, titanium in household paints to decompose dust and grime without human intervention.

Nano is the New Black “There’s Plenty of Room at the Bottom” So went the classic lecture by the Nobel prize-winning physicist Richard Feynman in 1959 that many nano-ficionados now consider the conceptual sun of the nanotechnology universe. “Why

cannot we write the entire 24 volumes of the Encyclopedia Britannica on the head of a pin?” he asked.

“Because there isn’t much of a point, or money, in doing so!” is the answer he would have got today from nanotechnology researchers. Instead their time is mostly spent figuring out newer properties for nanoparticles which can then be embedded into commercial applications.

Nanoparticles are highly reactive and prone to unusual properties. Describing gold, a metal that is normally inert to all other chemicals, Prof. C.N.R Rao, Honorary President and Linus Pauling Professor at the Jawaharlal Nehru Centre for Advanced Scientific Research (JNCASR) and the head of its Nanoscience centre, says “At 200-300 nm thickness, gold is not metallic, it does not shine — in fact it is not gold. And at 1.5-2 nm, it reacts like mad!”

Gold that is not gold when shrunk to nanometer size might sound like an absurdity to many, but it’s exactly this change in physical properties that make nanoparticles popular.

For example zinc oxide (ZnO) and titanium dioxide (TiO2) have been used as active ingredients in sunscreens for decades because of

their ability to absorb ultraviolet rays and reflect back much of the other remaining sunlight. But they are both white — the reason many sunscreens leave a white residue on the face. When shrunk to nanometre size however, they become transparent without losing their light reflecting or absorbing abilities.

Silver, an ornamental metal and a powerful bactericide, can be reduced to nanoparticle form to destroy disease-causing bacteria from all kinds of places — kitchen counters, contaminated water, dirty clothes and stinky underarms.

Samsung claims its Silver Nano range of washing machines release hundreds of billions of silver nano-ions with each wash to kill over 99 percent of the bacteria found in dirty clothes, while the same technology when lined on the doors of their refrigerators kill bacteria that could spoil stored food. Eureka Forbes’ water purifiers use nanosilver-coated filters, developed by Prof. Pradeep, head of IIT-Madras’ Nanoscience department, to destroy harmful bacteria from drinking water. Swach, the mass-market water filter introduced by the Tata Group, also uses nanosilver (coated on rice husk particles) to purify drinking water. During the last flu pandemic threat authorities in Hong-Kong sprayed subways with nanosilver to disinfect them.

L’Oreal, the world’s largest cosmetics company, reportedly spends over $600 million each year researching and patenting nanoparticles. The head of its nanotechnology unit also sits on the management board.


Therefore a metal that is a poor second cousin to gold in a world where we value yellow over white, is the undisputed metal of choice in the nanoparticle world. In fact, silver is more popular than any other material, according to the database of consumer products using nanoparticles maintained by the Woodrow Wilson International Center for Scholars.

Obama admn nominates IIT alumnus for post of NSF Director IIT Madras alumnus Subra Suresh, popularly known as 'Bakthi Suresh' during his student days, has been nominated for the post of director of the National Science Foundation (NSF) by the Barrack Obama administration. An official relese from IIT Madras Alumni Association here said ''when confirmed by the Senate, Mr Suresh will become one of the highest ranking Indian-Americans ever to serve in an administration.'' An Indian-American technocrat, 53-year-old Subra Suresh completed his B.Tech Mechanical Engineering in 1977. Currently the dean of the MIT engineering school, he received the distinguished alumnus award in 1997. In a statement, President Obama said ''I am proud that such experienced and committed individuals have agreed to take on these important roles in my administration. I look forward to working with them in the coming months and years.'' The National Science Foundation is the funding source for nearly 20 per cent of all federally supported basic research and was an independent federal agency created by US Congress in 1950. Subra Suresh has been elected to the US National Academy of Engineering, the Indian National Academy of Engineering,

the American Academy of Arts and Sciences, the Indian Academy of Sciences in Bangalore, the German National Academy of Sciences, the Royal Spanish Academy of Sciences and the Academy of Sciences of the Developing World based in Trieste, Italy.

Major decision regarding Ganga at IIT-K KANPUR: A collaboration between the consortium of seven IITs (IIT-Kanpur, Madras, Bombay, Delhi, Kharagpur, Guwahati and Roorkee) and Ministry of Environment and Forest, Government of India, is being worked out for the purpose of cleaning the national river Ganga. The two are expected to sign an important memorandum of understanding (MoU) in this regard during Prime Minister Manmohan Singh's visit to the Indian Institute of Technology-Kanpur. Singh is scheduled to visit IIT-K on July 3 to take part in its convocation ceremony.

It will be worth mentioning here that the Central government aims at meeting the formidable challenge of cleaning the Ganga. With this goal in mind, it had launched a new initiative and established the National Ganga River Basin Authority (NGRBA) last year. The Prime Minister, who is the Chairman of the NGRBA, asked the Ministry of Environment and Forest to involve IITs in the mega project. A joint meeting of all the seven IITs was convened on March 12, 2010 in which IIT-Kanpur was represented by Prof Vinod Tare, also the convener of the mission. It is for the very first time that the Central government has involved the seven IITs together in one single project of such a large magnitude

Prof Tare further informed TOI that the 'zero discharge' of both treated and untreated sewage waste into the river had been proposed to the government under which the waste would not be allowed into the Ganga.

"We will be doing this mega project in phases. The first phase will come to an end in 18 months wherein the concept of 'zero discharge' will be put into application. We also plan to apply the 'zero discharge' formula in four cities initially, viz. Hardwar, Rishikesh, Kanpur and Allahabad. If we are able to do so in these four cities, water of the river will become clean up to Allahabad and a major work will come to an end in the first phase."

Meanwhile, Prof SG Dhande, director, IIT-Kanpur, and Prof Vinod Tare from IIT-Kanpur, Prof Devang Khakhar, director, IIT-Bombay, took part in a meeting with the officials of the Ministry of Environment and Forest in New Delhi on May 19 and discussed all important aspects of the mega project. On the occasion, the team also handed over a set of proposals to the officials of the ministry. "A detailed project report will be given later as several social, legal aspects will have to be examined," said Prof Tare.

The consortium of the seven IITs have been sanctioned Rs 16 crore by the government for formulating a proper plan of action for the purpose of cleaning the Ganga. The authority formed by the Central government has both regulatory and developmental functions. The authority will take measures for effective abatement of pollution and conservation of the Ganga in keeping with sustainable develop-ment needs.

Science Research : Conventional solar cell efficiency could be increased from the current limit of 30 percent to more than 60 percent, suggests new research on semiconductor nanocrystals, or quantum dots, led by chemist Xiaoyang Zhu at The University of Texas at Austin. The scientists have discovered a method to capture the higher energy sunlight that is lost as heat in conventional solar cells.


Padmasree Warrior is senior vice president and chief technology officer for Motorola, with responsibility for Motorola Labs, the global software group and emerging early-stage businesses. Warrior's operational responsibilities include leading a global team of 4,600 technologists, prioritizing technology programs, creating value from intellectual property, guiding creative research from innovation through early-stage commercialization, and influencing standards and roadmaps. She also serves as a technology advisor to the office of the chairman and to the board's technology and design steering committee. Before assuming her current position in January 2003, Warrior was corporate vice president and general manager of Motorola's energy systems group, where she was responsible for profit and loss, sales, marketing, engineering and manufacturing. She also was general manager of Thought beam, Inc., a wholly owned subsidiary of Motorola, where she led the commercialization evaluation team related to compound semiconductor materials research. Prior to these assignments, Warrior was corporate vice president and chief technology officer for Motorola's Semiconductor Products Sector (SPS). A Motorola since 1984, she has been instrumental in driving innovative methods for technology commercialization realizing early "time to revenue" for the corporation. She has held many leadership positions within Motorola, was appointed vice president in 1999 and was elected a corporate officer in 2000. Warrior received a M.S. degree in chemical engineering from Cornell University, and a B.S. degree in chemical engineering from the Indian Institute of Technology (IIT) in New Delhi, India. Warrior served on the Texas Governor's Council for Digital Economy, and is a member of the Texas Higher Education Board review panel. She was one of six women nationwide selected to receive the "Women Elevating Science and Technology" award from Working Woman magazine in 2001. She also is a director of Ferro Corporation.

Krishan Sabnani is Senior Vice President of the Networking Research Laboratory at Bell Labs in New Jersey. For the past 23 years Krishan has been a member of Bell Labs Research. Krishan has conceived and launched several systems projects in the areas of Internetworking and wireless networking, led successful transfers of research ideas to products in Lucent and AT&T business units and conducted extensive personal research in data and wireless networking. He has built organizations known for technical excellence by recruiting and coaching the best people in the industry. Krishan has received the 2005 IEEE Eric E. Sumner Award and the 2005 IEEE W. Wallace McDowell Award - the only person ever to receive both awards. Krishan is a Bell Labs Fellow. He is also a fellow of the Institute of Electrical and Electronic Engineers (IEEE) and the Association of Computing Machinery (ACM). He received the Leonard G. Abraham Prize Paper Award from the IEEE Communications Society in 1991. Krishan will receive the 2005 Distinguished Alumni Award from Indian Institute of Technology (IIT), New Delhi, India. He has also won the 2005 Thomas Alva Edison Patent Award from the R&D Council of New Jersey. He holds 37 patents and has published more than 70 papers. In his personal research, Krishan has made major contributions to the communications protocols area. He has designed several protocols such as SNR, RMTP, and Airmail. He has also made significant contributions to conformance test generation, protocol validation, automated converter generation, and reverse engineering. Krishan received his Ph.D. in electrical engineering from Columbia University, New York, in 1981. He joined Bell Labs in 1981. Key Awards and Honors 1. 2005 IEEE Eric E. Sumner Award, received for

seminal contributions to networking protocols 2. 2005 IEEE Computer Society W. Wallace McDowell

Award 3. 2005 IIT Delhi Outstanding Alumni Award 4. 2005 Thomas Alva Edison Patent Award 5. 1991 Leonard G. Abraham Prize Paper Award 6. 1997 Bell Labs Fellow.

Success Story Success Story This articles contains stories of person who have succeed after graduation from different IIT's

Ms. Padmasree Warrior B.Tech, IIT Madras

Dr. Krishan K. Sabnani B.Tech, IIT – Kanpur


PHYSICS

1. Masses M1, M2 and M3 are connected by strings of negligible mass which pass over massless and friction less pulleys P1 and P2 as shown in fig. The masses move such the portion of the string between P1 and P2 in parallel to the inclined plane and the portion of the string between P2 and M3 is horizontal. The masses M2 and M3 are 4.0 kg each and the coefficient of kinetic friction between the masses and the surfaces is 0.25. The inclined plane makes an angle of 37º with the horizontal. [IIT-1981]

P1

M1

M2 P2 M3

37º

If the mass M1 moves downwards with a uniform velocity, find

(a) the mass of M1 (b) The tension in the horizontal portion of the string (g = 9.8 m/sec2, sin 37º ≈ 3/5) Sol. (a) Applying Fnet = ma on M1 we get T – m1 . g = M1 × 0 = 0 ⇒ T = M1g ...(i) Applying Fnet = Ma on M2 we get T – (T´ + M2g sin θ – f) = M2 × a T = T´ + M2g sin θ + f = T´ + M2g sin θ + µN [Q f = µN = µM2 g cos θ] ∴ T = T´ + M2g sin θ + µM2g cos θ ...(ii)

P1

M1

M2

T´ P2

M2g M1g

T

T

V

θ

M2gcosθ

V

f

M2gsinθ

θ

T´

N

M3g Applying Fnet = Ma for M3 we get T´ – f ´ = M3 × 0 ⇒ T´ = f ´ = µN´ = µM3g ...(iii) Putting the value of T and T´ from (i) and (iii) in (ii)

we get M1g = µM3g + M2g sin θ – µ M2g cos θ M1 = 0.25 × 4 + 4 × sin 37º + 0.25 × 4 × cos 37º = 4.2 kg (b) The tension in the horizontal string will be T ´ = µM3g = 0.25 × 4 × 9.8 = 9.8 N

2. A 0.5 kg block slides from the point A (see fig.) on a horizontal track with an initial speed of 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 Newton/m. The part AB of the track is frictionless and the part BC has the coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2m and 2.14 m respectively, find the total distance through which the block moves before it comes to rest completely. (Take g = 10 m/s2) [IIT-1983]

A B D C Sol. K.E. of block = work against friction + P.E. of spring

21 mv2 = µk mg (2.14 + x) +

21 kx2

21 × 0.5 × 32 = 0.2 × 0.5 × 9.8(2.14 + x) +

21 2 × x2

2.14+ x + x2 = 2.25 ∴ x2 + x – 0.11 = 0

On solving we get x = – 1011

or x = 101 = 0.1 (valid answer)

Here the body stops momentarily. Restoring force at y = kx = 2 × 0.1 = 0.2 N Frictional force at y = µs mg × x = 0.22 × 0.5 × 9.8 = 1.078 N Since friction force > Restoring force the body will

stop here. ∴ The total distance travelled = AB + BD + DY = 2 + 2.14 + 0.1 = 4.24 m.

A B D CRough L 2m

2.14mx

Y 3. A small sphere rolls down without slipping from the

top of a track in a vertical plane. The track in a vertical plane. The track has an elevated section and a horizontal part, The horizontal part is 1.0 meter above the ground level and the top of the track is 2.4 metres above the ground. Find the distance on the ground with respect to the point B(which is vertically

KNOW IIT-JEE By Previous Exam Questions


below the end of the track as shown in fig.) where the sphere lands. During its flight as a projectile, does the sphere continue to rotate about its centers of mass ? Explain. [IIT-1987]

A

B

1.0m

2.4

m

Sol. Applying law of conservation of energy at point D

and point A P.E. at D = P.E. at A + (K.E.)T + (K.E.)R (K.E.)T = Translational K.E.

mg (2.4) = mg (1) + 21 mv2 +

21 Iω2

(K.E.)R = Rotational K.E. Since the case is of rolling without slipping

D

2.4m

1mA

B C ∴ v = rω

∴ ω = rv where r is the radius of the sphere Also

I = 52 mr2

∴ mg(2.4) = mg(1) + 21 mv2 +

21 ×

52 mr2 × 2

2

rv

⇒ v = 4.43 m/s After point A, the body takes a parabolic path. The

vertical motion parameters of parabolic motion will be

uy = 0 S = ut + 21 at2

Sy = 1m 1 = 4.9 ty2

ay = 9.8 m/s2

∴ ty = ? ty = 9.4

1 = 0.45 sec

Applying this time in horizontal motion of parabolic path, BC = 4.43 × 0.45 = 2m

During his flight as projectile, the sphere continues to rotate because of conservation of angular momentum.

4. Two square metal plates of side 1 m are kept 0.01 m

apart like a parallel plate capacitor in air in such a way that one of their edges is perpendicular to an oil surface in a tank filled with an insulating oil. The plates are connected to a battery of emf 500 V. The plates are then lowered vertically into the oil at a

speed of 0.001 ms–1. Calculate the current drawn from the battery during the process. (Dielectric constant of oil = 11, ε0 = 8.85 × 10–12C2N–1m–1)

[IIT-1994] Sol. The adjacent figure is a case of parallel plate

capacitor. The combined capacitance will be

v

+

1–x

x1m

d C = C1 + C2

= d

)1x(k 0 ×ε +

d]1)x1[(0 ×−ε

C = d0ε

[kx + 1 – x]

After time dt, the dielectric rises by dx. The new equivalent capacitance will be

C + dC = C1´ + C2´

= d

k 0ε[(x + dx) × 1] +

d]1)dxx1[0 ×−−ε

dC = Change of capacitance in time dt

= d0ε

[kx + kdx + 1 – x – dx – kx – 1 + x]

= d0ε

(k – 1)dx

dtdC =

d0ε

(k – 1)dtdx =

d0ε

(k – 1)v ...(i)

where v = dtdx

We know that q = CV

dtdq = V

dtdC ...(ii)

⇒ I = Vd0ε

(k – 1)v

From (i) and (ii)

I = 01.0

1085.8500 12−×× (11 – 1) × 0.001

= 4.425 × 10–9 Amp.

5. Two resistors, 400 ohms, and 800 ohms are connected in series with a 6-volt battery. It is desired to measure the current in the circuit. An ammeter of a 10 ohms resistance is used for this purpose. What will be the reading in the ammeter? Similarly, If a voltmeter of 10,000 ohms resistance is used to measure the potential difference across the 400-ohms resistor, What will be the reading in the voltmeter.

[IIT-1982]


Sol. Applying Kirchoff's law moving in clockwise direction starting from battery we get

A 10Ω

400Ω 800Ω

6 volt + 6 – 10I – 400 I – 800 I = 0 ∴ 6 = 1210 I

∴ I = 1210

6 = 4.96 × 10–3 A

The voltmeter and 400 Ω resistor are in parallel and hence p.d. will be same

∴ 10,000 I1 = 400 I2 ...(i) Applying Kircoff's law in loop ABCDEA starting

from A in clockwise direction. – 400 I2 – 800 I + 6 = 0 ∴ 6 = 400 I2 + 800 (I1 + I2) ∴ 6 = 400 I2 + 800(0.04 I2 + I2) From (i) putting the value of I1 ∴ 6 = 1232 I2

V

I400Ω 800Ω

6 volt

10,000Ω

B

F G

C

I D

A E

∴ I2 = 4.87 × 10–3 Amp. Potential drop across 400 Ω resistor = I2 × 400 = 4.87 × 10–3 × 400 = 1.948 volt ≈ 1.95 volt ∴ The reading measured by voltmeter = 1.95 volt

CHEMISTRY

6. At constant temperature and volume, X decomposes as 2X(g) → 3Y(g) + 2Z(g); Px is the partial pressure of X.

Observation No.

Time (in minute) Rx (in mm of Hg)

1 0 800 2 100 400 3 200 200

(i) What is the order of reaction with respect to X ? (ii) Find the rate constant. (iii) Find the time for 75% completion of the reaction. (iv) Find the total pressure when pressure of X is

700 mm of Hg. [IIT-2005]

Sol. (i) From the given data, it is evident that the t1/2 (half-life

period)for the decomposition of X (g) is constant (100 minutes) therefore the order of reaction is one.

(ii) Rate constant, K = 2/1t

693.0

= 100693.0 = 6.93 × 10–3 min–1

(iii) Time taken for 75% completion of reaction = 2t1/2 = 2 × 100 = 200 minutes

(iv) 2x → 2Y + 2Z Initial pressure 800 0 0 Ater time t (800 – 2p) 3P 2p when the pressure of X is 700 mm of Hg the, 800 –

2P = 700 2P = 100; P = 50 mm of Hg Total pressure = 800 – 2P + 3P + 2P = 800 + 150 = 950 mm of Hg. 7. A acid solution of Cu2+ salt containing 0.4 g of Cu2+

is electrolysed until all the copper is deposited. The electrolysis is continued for seven more minutes with the volume of solution kept at 100 ml. and the current at 1.2 amp. Calculate the volume of gases evolved at NTP during the entire electrolysis. [IIT-89]

Sol. The chemical reactions taking place at the two electrodes are

At cathode : Cu2+ + 2e– → Cu H2O H+ + OH– However, note that only Cu2+ ions will be discharged

so as these are present in solution and H+ ions will be discharged only when all the cu2+ ions have been deposited.

Atcathode : 2OH– → H2O + O + 2e– O + O → O2 Thus in first case, Cu2+ ion will be discharged at the

cathode and O2 gas at the anode. Let us calculate the volume of gas (O2) discharged during electrolysis.

According to Faraday's cecond law 31.75 g Cu ≡ 8 g of oxygen ≡ 5.6 litres of O2 at NTP

0.4 g Cu = 75.316.5 × 0.4 litres of O2 at NTP

= 0.07055 litres = 70.55 ml As explained earlier, when all the Cu2+ ion will be

deposited at cathode, H+ ions will start going to cathode liberating hydrogen (H2) gas i.e.

H+ + e– H H + H → H2 However, the anode reaction remains same as

previous. Thus in the second (latter) case, amount of H2 collected at cathode should be calculated.

8 g of O2 = 1 g of H2 5.6 litres of O2 at NTP = 11.2 litres of hydrogen Quantity of electricity passed after 1st electrolysis, i.e. Q = i × t = 1.2 × 7 × 60 = 504 coulombs


504 coulombs will liberate = 96500

5046.5 × = 29.24 ml

of O2. Similarly, H2 liberated by 504 coulombs

= 11.2 × 96500

504 = 58.45 ml

Total volume of O2 liberated = 70.55 + 29.24 = 99.79 ml vol. of H2 liberated = 58.48 ml.

8. Cyclobutyl bromide on treatment with magnesium in dry ether forms an organometallic (A). The organometallic reacts with ethanal to give an alcohol (B) after mild acidification Prolonged treatment of alcohol (B) with an equivalent amount of HBr gives 1-bromo 1-methylcyclopentane (C). Write the structures of (A), (B) and explain how (C) is obtained from (B). [IIT-2001]

Sol.

Br

ether

dry→

MgBr

Cyclobutymagnesium Bromide

(A)

→+OH,CHOCH 33

CH–OH

1-Cyclobutylethanol (B)

CH3

CH–CH3

(B)

OH

→+H

CH–CH3

Oxonium ion

OH2 ⊕ →+OH– 2

CH–CH3

(2º carbocation 4-membered ring)

+

alkylshift2,1through

ansionexpring

− → H

(ring expansion) (2º carbocation in 5 membered ring

⊕ H

CH3shiftHydride

–2,1 →

CH3

(3º carbocation)

⊕ H H

→–Br

Br

1-bromo-1-methylCyclopentane

(C)

⊕ H H

CH3

9. A basic, volatile nitrogen compound gave a foul smelling gas when treated with chloroform and alcoholic potash. A 0.295 g sample of the substance. Dissolved in aq. HCl and treated with NaNO2 solution at 0ºC, liberated a colorless, odourless gas whose volume corresponded to 112 ml at STP, After the evolution of the gas was complete, the aqueous solution was distilled to give an organic liquid which did not contain nitrogen and which on warming with alkali and iodine gave a yellow precipitate. Identify the original substance. Assume that it contains one N atom per molecule. [IIT-93]

Sol. Let us summarise the given facts.

112 ml of colourless,odourless

gas at S.T.P+ Residue

(i) aq. HCl(ii) NaNO2 0ºC

Basic Nitrogen

Compound (0.295 g)

CHCl3

KOH

Foul smelling

gas

Distilled aq. Sol.

Organic liquid(no N)

OH–/I2 Yellow ppt.

Reaction of the original compound with alcoholic potash and chloroform to give foul smelling gas indicates that it contains a primary –NH2 group.

R–NH2 + CHCl3 + KOH → R–NC↑ (Basic compound) Carbylamine (foul smelling) Determination of mol. Weight of the amine. 112 ml. of gas is evolved at S.T.P. by 0.295 g of

amine

22400 ml. of gas is evolved by = 112295.0 × 22400 = 59

Hence the mol. wt. of the amine = 59 ∴ Mol. wt. of the alkyl group = 59 – 16 = 43 Nature of alkyl gp. of mol. wt.= 43 = C3H7

– Thus the amine may be either

CH3CH2CH2NH2 or

CHNH2

CH3

CH3

The reaction of amine with NaNO2 at 0ºC and all other reactions may thus be written as below.

CH3CH2CH2NH2 Cº0/NaNO)ii(

HCl)i(

2 → CH3CH2CH2OH +

N2↑ n-Propylamine → distill.sol.aq CH3CH2CH2OH

→ 2– I,OH No yellow ppt.

(CH3)CHNH2 → (CH3)2CHOH + N2↑ Isopropylamine

→ (CH3)2CHOH )reactionHaloform(

I,OH 2– → CHI3↓

(yellow) Since the given reactions correspond to

isopropylamine, the original compound is isopropylamine, (CH3)2CHNH2

10. Interpret the non-linear shape of H2S molecule and

non-planar shape of PCl3 using valence shell electron pair repulsion (VSEPR) theory. (Atomic numbers : H = 1, P = 15, S = 16, Cl = 17.) [IIT-98]

Sol. In H2S, no. of hybrid orbitals = 21 (6 + 2 – 0 + 0) = 4

Hence here sulphur is sp3 hybridised, so 16S = 1s2, 2s22p6,

44 344 21ionhybridisatsp

1z

1y

2x

2

3

p3p3p3s3


or

S

H HH H

S

Due to repulsion between lp - lp; the geometry of

H2S is distorted from tetrahedral to V-shape.

In PCl3, no. of hybrid orbitals = 21 [5 + 3 – 0 + 0] = 4

Hence, here P shows sp3-hybridisation 15P = 1s2, 2s22p6,

44 344 21ionhybridisatsp

1z

1y

1x

2

3

p3p3p3s3

P

or

P

Cl Cl

Cl Cl Cl

Cl

Thus due to repulsion between lp – bp, geometry is

distorted from tetrahedral to pyramidal.

MATHEMATICS

11. The circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the coordinate axes. If the locus of the circumcentre of the triangle is

x + y – xy + k 22 yx + = 0, find the value of k. [IIT-1987] Sol. Let OAB be the triangle in which the circle x2 + y2 – 4x – 4y + 4 = 0 is inscribed. Let the

equation of AB be by

ax

+ = 1

y B(0,b)

y´ x´ O

2 C

(a, 0)A x

1by

ax =+

1by

ax

=+

Since AB touches the circle x2 + y2 – 4x – 4y + 4 = 0. There fore,

22 b1

a1

1b2

a2

+

−+ = 2 ⇒ –

22 b1

a1

1b2

a2

+

−+

= 2

[Q O(0, 0) and C(2, 2) lie on the same side of AB

Therefore, a2 +

b2 – 1 < 0]

⇒ –22 ba

)ab–a2b2(

+

+ = 2

⇒ 2a + 2b – ab + 2 22 ba + = 0 ...(i) Let P(h, k) be the circumcentre of ∆OAB. Since

∆ OAB is a right angled triangle. So its circumcentre is the mid-point of AB.

∴ h = 2a and k =

2b

⇒ a = 2h and b = 2k ...(ii) From (i) and (ii), we get

4h + 4k – 4hk + 2 22 k4h4 + = 0

⇒ h + k – hk + 22 kh + = 0 So, the locus of P(h, k) is

x + y – xy + 22 yx + = 0 But, the locus of the circumcentre is given to be

x + y – xy + k 22 yx + = 0 Thus, the value of k is 1

12. If A, B, C are the angles of a triangle ABC and the

system of linear equations x sin A + y sin B + z sin C = 0 x sin B + y sin C + z sin A = 0 x sin C + y sin A + z sin B = 0 has a non trivial solution, prove that sin2A + sin2B + sin2C – (cos A + cos B + cos C + cos A cos B + cos B cos C + cos C cos A) = 0 [IIT-2002] Sol. The given system of linear equations has a non-trivial

solution. Therefore,

BsinAsinCsinAsinCsinBsinCsinBsinAsin

= 0

⇒ BsinAsinBsinAsinCsinAsinCsinAsinCsinBsinCsinBsinCsinBsinAsin

++++++

= 0

Applying C1 → C1 + C2 + C3

⇒ (sin A + sin B + sin C) BsinAsin1AsinCsin1CsinBsin1

= 0

⇒ BsinAsin1AsinCsin1CsinBsin1

= 0

≠=

++

02Ccos

2Bcos

2Acos4

CsinBsinAsinQ


⇒ CsinBsinBsinAsin0CsinAsinBsinCsin0

CsinBsin1

−−−− = 0

Applying R2 → R2 – R1, R3 → R3 – R1 ⇒ –(sin B – sin C)2 – (sin A – sin C) (sin A – sin B) = 0 ⇒ sin2B + sin2C – 2 sin B sin C + sin2A – sin A sin B – sin C sin A + sin B sin C = 0 ⇒ sin2A + sin2B + sin2C – sin A sin B – sin B sin C – sin C sin A = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B – cos B cos C – cos C cos A + cos (A + B) + cos (B + C) + cos (C + A) = 0 ⇒ sin2A + sin2B + sin2C – cos A cos B – cos B cos C – cos C cos A – cos A – cos B – cos C = 0 13. Determine the name of the name of the curve

described parametrically by the equations x = t2 + t + 1, y = t2 – t + 1 [IIT-1998] Sol. We have, x = t2 + t + 1 and, y = t2 – t + 1 ⇒ x + y = 2(t2 + 1) and, x – y = 2t

⇒ x + y = 2

+

− 1

2yx 2

⇒ 2(x + y) = (x – y)2 + 4 ⇒ x2 + y2 – 2xy – 2x – 2y + 4 Comparing this equation with the equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, we get a = 1, b = 1, c = 4, h = –1, g = –1 and f = –1 ∴ abc + 2fgh – af2 – bg2 – ch2 = 4 – 2 – 1 – 1 – 4 ≠ 0 and , h2 – ab = 1 – 1 = 0 Thus, we have ∆ ≠ 0 and h2 = ab So, the given equations represent a parabola.

14. Let C be any circle with centre (0, 2 ). Prove that at most two rational points can be there on C.

(A rational points is a point both of whose coordinates are rational numbers) [IIT-1997]

Sol. The equation of any circle C with centre (0, 2 ) is given by

(x – 0)2 + (y – 2 )2 = r2, where r is any positive real number.

or, x2 + y2 – 22 y = r2 – 2 If possible, let P(x1, y1), Q(x2, y2) and R(x3, y3) be

three distinct rational points on circle C. Then, 22yx 2

121 −+ y1 = r2 – 2 ...(ii)

22yx 22

22 −+ y2 = r2 – 2 ...(iii)

22yx 23

23 −+ y3 = r2 – 2 ...(iv)

We claim that at least two y1, y2, and y3 are distinct. For if y1 = y2 = y3, then P, Q and R lie on a line parallel to x-axis and a line parallel to x-axis does not cross the circle in more than two points. Thus, we have either y1 ≠ y2 or, y1 ≠ y3 or, y2 ≠ y3.

Subtracting (ii) from (iii) and (iv), we get

)yx( 22

22 + – )yx( 2

121 + – 22 (y2 – y1) = 0

and, )yx( 23

23 + – )yx( 2

121 + – 22 (y3 – y1) = 0

⇒ a1 – 2 b1 = 0 and a2 – 2 b2 = 0 ...(v) where, a1 = )yx( 2

222 + – )yx( 2

121 + , b1 = 2(y2 – y1)

a2 = )yx( 23

23 + – )yx( 2

121 + , b2 = 2(y3 – y1)

Clearly, a1, a2, b1, b2 are rational numbers as x1, x2, x3, y1, y2, y3 are rational numbers.

Since either y1 ≠ y2 or, y1 ≠ y3 ∴ Either b1 ≠ 0 or, b2 ≠ 0 If b1 ≠ 0, then a1 – 2 b1 = 0 [From (v)]

⇒ 1

1

ba = 2 ,

which is not possible because 1

1

ba is a rational

number and 2 is an irrational number. If b2 ≠ 0, then

a2 – 2 b2 = 0 ⇒ 2

2

ba = 2 ,

which is not possible because 2

2

ba is a rational

number and 2 is an irrational number. Thus, in both the cases we arrive at a contradiction.

This means that our supposition is wrong. Hence, there can be at most two rational points on circle C.

15. A rectangle PQRS has its side PQ parallel to the line

y = mx and vertices P, Q and S lie on the lines y = a, x = b and x = –b, respectively. Find the locus of the vertex R. [IIT-1996]

Sol. Let the coordinates of R be (h, k). It is given that P lies on y = a. So, let the coordinates of P be (x1, a). Since PQ is parallel to the line y = mx. Therefore,

Slope of PQ = (Slope of y = mx) = m

And, Slope of PS = –mx) y of Slope(

1=

= –m1 [∴ PS ⊥ PQ]

Now, equation of PQ is y – a = m(x – x1) ...(i)


y y = 0

PQ x = b

(0, b)x

y´

R

O (0, – b)S x´

x = –b

(0, a)

It is given that Q lies on x = b. So, Q is the point of intersection if (i) and x = b.

Putting x = b in (i), we get y = a + m(b – x1) So, coordinates of Q are (b, a + m(b – x1)).

Since PS passes through P(x1, a) and has slope – m1 .

So, Equation of PS is y – a = –m1 (x – x1) ...(ii)

It is given that S lies on x = – b. So, S is the point of intersection of (ii) and x = –b.

Solving (ii) and x = – b, we get y = a + m1 (b + x1)

So, coordinates of S are

++− )xb(

m1a,b 1

Now, Slope of RS = bh

)xb(m1ak 1

+

+−− = m

But RS is parallel to PQ.

∴ bh

)xb(m1ak 1

+

+−− = m

⇒ b + x1 = m(k – a) – m2(h + b) ...(iii) Similarly,

Slope of RQ = bh

)xb(mak 1

−−−−

But, RQ is perpendicular to PQ whose slope is m.

∴ bh

)xb(mak 1

−−−− = –

m1

⇒ b – x1 = m1 (k – a) + 2m

1 (h – a) ...(iv)

We have only one variable x1. To eliminate x1, add (iii) and (iv) to obtain

2b = (k – a)

+

m1m – m2(h + b) + 2m

1 (h – b)

⇒ 2b = (k – a)

+m

1m2– h

+2

4

m1m – b

+2

4

m1m

⇒ (k–a)

+m

1m2– 2

22

m)1m)(1m(h +− – 2

22

m)1m(b + =0

⇒ (k – a) – m

)1m(h 2 − –m

)1m(b 2 + = 0

⇒ m(k – a) – h(m2 – 1) – b(m2 + 1) = 0 Hence, the locus of R(h, k) is m(y – a) – x(m2 – 1) – b(m2 + 1) = 0

BEWARE OF THE BATTERIES THAT YOU USE!

Have you ever noticed that the batteries are becoming smaller and smaller day after day? Many scientists and researchers have been finding the effective way to shrink the batteries into the smallest size as possible!

In this case, Jae Kwon, an assistant professor of Electrical and computer engineering has recently developed a nuclear energy source, which is smaller, lighter and more efficient than the common batteries.

Kwon’s described that the new discovered radioisotope battery can provide power density as much as six orders of magnitude higher than chemical batteries.

Kwon and his research team members have been cooperated and working on building a small nuclear battery. According to the information, the radioisotope batteries are having the size and thickness of a penny, but it’s powerful enough to power various micro or nanoelectromechanical systems.

Even though the nuclear power sources have always been a safety concern, they’ve claimed to be safe, as the nuclear power sources have been used for powering many types of devices, including the pace-makers, space satellites and underwater systems.

Kwon’s battery is in a liquid semiconductor rather than a solid semiconductor, as he believed that the liquid semiconductor can overcome the problem, where the lattice structure of the semiconductor being damaged, if it’s in the solid semiconductor form!


1. A circuit consisting of a constant e.m.f. ‘E’, a self induction ‘L’ and a resistance ‘R’ is closed at t = 0. The relation between the current I in the circuit and time t is as shown by curve ‘a’ in the figure. When one or more of parameters E, R and L are changed, the curve ‘b’ is obtained. The steady state current is same in both the cases. Then it is possible that

(a)(b)

t

I

(A) E and R are kept constant and L is increased

(B) E and R are kept constant and L is decreased (C) E and R are both halved and L is kept constant (D) E and L are kept constant and R is decreased

2. Consider a resistor of uniform cross section area connected to a battery of internal resistance zero. If the length of the resistor is doubled by stretching it then (A) current will become four times (B) the electric field in the wire will become half (C) the thermal power produced by the resistor

will become one fourth (D) the product of the current density and

conductance will become half

3. In front of an earthed conductor a point charge +q is placed as shown in figure

+q

(A) On the surface of conductor the net charge is

always negative (B) On the surface of conductor at the same points

charges are negative and at some points charges may be positive distributed non uniformly

(C) Inside the conductor electric field due to point charge is non-zero

(D) None of these

Passage # (Q. No. 4 to Q. No. 6) Resistance value of an unknown resistor is calculated using the formula R= V/I where V and I be the readings of the voltmeter and the ammeter respectively. Consider the circuits below. The internal resistances of the voltmeter and the ammeter (RV and RG respectively) are finite and non zero.

AR

V

E r

AR

V

E r

Fig. (A) Fig. (B) Let RA and RB the calculated values in the two

cases A and B respectively. 4. The relation between RA and the actual value R is (A) R > RA (B) R < RA (C) R = RA (D) dependent upon E and r 5. The relation between RB and the actual value R is (A) R< RB (B) R > RB (C) R = RB (D) dependent upon E and R 6. If the resistance of voltmeter is RV = 1 KΩ and

that of ammeter is RG = 1Ω, the magnitude of the percentage error in the measurement of R (the value of R is nearly 10 Ω) is

(A) zero in both cases (B) non-zero but equal in both cases (C) more in circuit A (D) more in circuit B

Passage # (Q. No. 7 to Q. No. 8) The figure shows the interference pattern obtained in

a double-slit experiment using light of wavelength 600 nm. 1,2,3,4 and 5 are marked on five fringes.

7. The third order bright fringe is (A) 2 (B) 3 (C) 4 (D) 5

8. Which fringe results from a phase difference of π4 between the light waves incidenting from two slits

(A) 2 (B) 3 (C) 4 (D) 5

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 # 3


1.[C] Initially the potential at centre of sphere is

xQ

41

xQ2

41

xQ

41V

000C

3πε

=πε

+πε

=

After the sphere grounded, potential at centre becomes zero. Let the net charge on sphere finally be q.

xQr3qor0

xQ3

41

rq

41

00==

πε+

πε∴

∴The charge flowing out of sphere is xQr3

2. [D] The velocity is maximum at mean position. Hence the magnetic force on block is maximum, at its mean position. The magnetic force on the block while it crosses the mean position towards right and left is as shown

N1 vmax

Mg + qvmaxB Case-1

N2

vmax

mg Case-2

qvmaxB

Hence normal reaction is maximum in case-1 and minimum in case-2. Hence correct option is D.

3. (A) → Q,R, (B) → P,S, (C) → P,R, (D) → Q,S

4. [A,B,C] Total charge = ∫ == C10curveunderareaIdt

Average current = A5dt

Idt=

∫∫

Total heat produced = ∫ RdtI2

∫ =+−=2

0

2 J3

200dt.1.)10t5(

Maximum power = I2R, when I is maximum current = 100 × 1 =100W.

5. [B,D] Equivalent circuit

x x

x

r/2 r/2 εε

r

Induced emf 2aB

2rBe

22 ω=

ω= (∴ Radius = a)

By nodal equation, nodal

0r

0xr

ex4 =

−

+

−

5x = 4e x = 4e/5

and 5

aB2rxi

2ω==

=

+=

r5e4

4/rrei

6. (A) → R , (B) → Q,S, (C) → P, (D) → Q,R

(A) =→F constant and 0Fu =×

→→

Therefore initial velocity is either in direction of constant force or opposite to it. Hence the particle will move in straight line and speed may increase or decrease.

(B) 0F.u =→→

and =→F constant

Initial velocity is perpendicular to constant force, hence the path will be parabolic with speed of particle increasing.

(C) 0F.v =→→

means instantaneous velocity is always perpendicular to force. Hence the speed

will remain constant. And also =→

|F| constant. Since the particle moves in one plane, the resulting motion has to be circular.

(D) ∧∧→

−= j3i2u and ∧∧→

−= j9i6a . Hence initial velocity is in same direction of constant acceleration, therefore particle moves in straight line with increasing speed.

7. [B]

OrCθ

v0qr R

qB

mvr 0= ; θ= sinr2R

θ=

sinm2qBRv0

8. [B,C] ××

××

× × h

drr

r2NIhdr

hdrBd 0

πµ

=×=φ

tsinIR

bRlog2

hNmax

0total ω××

+π×µ

=φ

dtd

e totalφ=

Solution Physics Challenging Problems

Set # 2

8 Questions were Published in June Issue


1. A particle of mass m moves along a horizontal circle of radius R such that normal acceleration of particle varies with time as an = Kt2 , where K is a constant. Calculate

(i) tangential force on particle at time t, (ii) total force on particle at time t, (iii) Power developed by total force at time t, and (iv) average power developed by total force over

first t second. Sol. Since, the particle is moving along a circle,

therefore, its normal acceleration is centripetal acceleration i.e. v2/R, where v is velocity of particle at time t.

∴ Rv2

= Kt2 or v = t. KR

…(1) Due to centripetal acceleration, particle follows a

circular path but due to it velocity its magnitude does not change. Velocity magnitude increases due to tangential acceleration alone.

∴ Tangential acceleration, at = dtd . v = KR

∴ Tangential force, Ft = mat = m KR Ans. (i) Normal force, Fn = man = mKt2 ∴ Resultant force on particle,

F = 2n

2t FF +

= m )KR(K 4t+

Ans. (ii)

Since, power developed by force →F is given by

P = →→vF , therefore, power developed by normal

force Fn is always zero because its direction is always perpendicular to the instantaneous direction of motion of the particle. Hence, power is developed by tangential force alone. Figure

Ft

Fnv

i.e. P = →

tF →v = mkRt Ans.(iii)

Since, resultant force equals (mass × acceleration), therefore, resultant force is used to accelerate the

body. It means that velocity of the body increases due to resultant force. Hence power developed by resultant force is used to increase kinetic energy of the body.

∴ Average power developed by resultant force = Average rate of increase of KE

Initial kinetic energy (at t = 0), E0 = 0

Kinetic energy at time t, E = 21 mv2 =

21 mKRt2

…(2)

∴ Average power = tE–E 0 =

21 mKRt` Ans. (iv)

2. Figure Shows a particle of mass m = 100 gm, attached with four identical springs, each of length l = 10 cm. Initial tension in each spring is F0 = 25 newton. Neglecting gravity, calculate period of small oscillations of the particle along a line perpendicular to the plane of the figure

B

C PA

D

m

Sol. Let the particle be displaced slightly through x along a line normal to plane of the figure. Then each spring is further elongated. Since, springs are identical, therefore, increase in tension of each spring will be the same. Let this increase be dE0.

P(F0 + dF0)(F0 + dF0)

CAC

ll

θθ

First considering forces exerted by spring AP and

CP only as shown in Figure. Restoring force produced by these two springs = (F0

+ dF0) 2 sin θ

Since x is very small, therefore, sin l

x≈θ

Neglecting product of very small quantities, restoring force produced by these two springs

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

Students' ForumPHYSICS


= 2F0l

x

Similarly, restoring force produced by two remaining springs BP and DP will also be equal to

l

xF2 0

∴ Resultant restoring force, F = 2 ×

l

xF2 0 = l

0F4.x

∴ Restoring acceleration is directly proportional to displacement x, therefore, the particle executes SHM,

∴ Its period T = onacceleratintdisplaceme2π

or T = 0F4

m2 lπ =

0Fml

π = 0.02π sec

Ans.

3. Find the electric potential, at any point on the axis of a uniformly charged circular disc, whose surface charge density is σ, radius, a.

Sol. Let us consider a small elemental thin ring of width dy.

Area of the ring = 2πy dy Charge on this elemental ring = (2πy dy)σ Again, we can consider that this ring is divided into

a large number of small elements. Each such element e is at the same distance from P. Hence, potential produced by this ring of width dy at the point P is given by dV, where

dV = 220 yr

dyy24

1

+

σππε

e

y

O

a r P

22 yr +

Again, each ring as we go from centre to rim,

produces different contributions. Since the distance of each ring from P changes as y increases from 0 to a, hence, total potential produced by the whole ring,

V = ∫ =

a

0ydV

= ∫ πεπσa

0 042 .

22 yr

ydy

+

= ∫+ε

σ a

0 220 yr

ydy2

Put )yr( 22 + = P

∴ r2 + y2 = p2 or 2ydy = 2pdp

∴ ∫+ 22 yr

ydy = ∫ ppdp

= ∫dp = p = 22 yr +

∴ ∫+ε

σ a

0 220 yr

ydy2

= a

0

22

0yr

+

εσ

=

−+

εσ rar

222

0

∴ V =

−+

εσ rar

222

0

As a special case, if r >> a

22 ar + = r2/12

ra1

+ ≈ r

+ 2

2

ra

211

or, V = 02ε

σ .r2

a 2 ×

ππ =

r4q

0πε

[Q πa2 = A and Aσ = q] i.e., the result is the same as if all the charge is

concentrated at the centre of the ring. 4. Three concentric, conducting spherical shells A,B

and C have radii a = 10 cm, b = 20 cm and c = 30 cm respectively. The innermost shell A is earthed and charges q2 = 4µC and q3 = 3 µC are given to shells B and C respectively. Calculate charge q1 induced on shell A and energy stored in the system.

Sol. System of three concentric shells is as shown in Figure(A) Since, Shell A is earthed, therefore its potential is zero. But its potential is

a

b

A

q1

q2

q3

CB

c

(A)

a

b

A (–3µC)

C B c

(B)

(+3µC)

(–1µC)(+1µC)

(+4µC)

V =

++πε c

qb

qaq

41 321

0

∴

++c

qb

qaq 321 = 0

or q1 = – 3µC Ans. Now, charges on different surface will be as shown

in Figure(B) to calculate energy stored in the system, it can be as considered in three parts : (i) a spherical capacitor having radii a and b and

having charge |q1| 3µC.


Its capacitance, C1 = )a–b(

ab4 0πε

(ii) a spherical capacitor having radii b and c and having charge (q2 + q1) = 1 µC.

Its capacitance. C2 = )b–c(

bc4 0πε

(iii) an isolated sphere of radius c and having charge (q3 + q2 + q1) = 4µC

∴ Energy stored in the system

= 1

21

C2q +

2

221

C2)qq( + +

3

2321

C2)qqq( ++

= 0.45 joule Ans. 5. In the circuit shown in Figure emf of each battery is

E = 20 volts and capacitance is C1 = 5 µF, C2 = 3 µF and C3 = 6 µF. Calculate charge on capacitor C3 when switch S is closed and steady state is reached. Calculate also, heat generated in the circuit.

E +–

C1

C2

C3 C1

C2

S

+–E

Sol. When Switch S is closed and steady state is reached, the circuit becomes symmetric about the dotted line shown in Figure(A).

E +–

C1

C2

C3 C1

C2+–E

(A) Right part of the circuit is exactly mirror image of

the left part. Hence, charges on both plates of capacitor C3 should be identical. But charges on plates of a capacitor are always opposite to each other. It means one of the plates is always positively charged and the other is negatively charged. Both these conditions can be satisfied only if charge on capacitor C3 is zero.

To calculate heat generated in the circuit initial and final charges on all the capacitors must be known.

First analyse the circuit when switch S was open Charges on capacitors will be as shown in

Figure (B)

E

+

–

C1

C2

C3 C1

C2

S

+– E

– – –+ ++ +

–+–(q1 –q2) q2

(B)

Applying Kirchhoff's voltage law on left mesh, of Figure (B)

1

1

Cq +

2

21

Cq–q – E = 0 …(1)

For middle mesh,

3

2

Cq +

2

2

Cq –

2

21

Cq–q = 0 …(2)

From equation (1) and (2), q1 = 50 µC and q2 = 20 µC

Now consider the circuit when switch S is closed and steady state is reached. The circuit will be as shown in Figure (C)

E+

–

C1

C2

C3 C1

C2+–E

– – ++ + +

– –q qq q

(C) Applying Kirchhoff's voltage law on left mesh of

Figure(C)

1C

q + 2C

q – E = 0 or q = 37.5 µC.

After shorting of switch S, increase in charge on capacitor C1 of

Left mesh, ∆q1 = (q – q1) = – 12.5 µC That for capacitor C2 of left mesh, ∆q2 = q – (q1 –q2) = 7.5 µC That for capacitor C3 , ∆q3 = 0 – q2 = – 20µC That for capacitor C2 of right mesh, ∆q4 = (q –q2) = 17.5 µC That for capacitor C1 or right mesh, ∆q5 = q – 0 = 37.5 µC Since, heat generated in the circuit is given by

H = ∑ ∆C2)q( 2

H = 1

21

C2)q(∆ +

2

22

C2)q(∆ +

3

23

C2)q(∆

+ 2

24

C2)q(∆ +

1

25

C2)q(∆

= 250 × 10–6 joule Ans.


Capacitance : Whenever charge is given to a conductor of any

shape its potential increases. The more the charge (Q) given to the conductor the more is its potential (V) i.e. Q ∝ V

⇒ Q = CV where C is constant of proportionality called

capacitance of the conductor C = Q/V, C = Q SI unit of capacitance is farad (F) and 1 F = 1

coulomb/volt (1CV–1) Energy stored in a charged capacitor :

W = 20CV

21 =

C2Q2

= 21 QV0

Capacitance of an isolated sphere : Let a conducting sphere of radius a acquire a

potential V when a charge Q is given to it. The potential acquired by the sphere is

V = a4

Q

0πε ⇒ C =

VQ = 4πε0a

Charge sharing Between two charged conductors :

C1

V1

C2

V2

q1 = C1V1 q2 = C2V2

(Initially)

C1

V

C2

V

q´1 = C1V q´2 = C2V

(Finally)

V = 21

2211

CCVCVC

++

There is always a loss in energy during the sharing process as some energy gets converted to heat.

Loss = – ∆U =

+ 21

21

CCCC

21 (V1 – V2)2

Capacitor or Condenser : An arrangement which has capability of collecting

(and storing) charge and whose capacitance can be varied is called a capacitor (or condenser)

The capacitance of a capacitor depends. (a) directly on the size of the conductors of the

capacitor.

(b) directly on the dielectric constant K of the medium between the conductors.

(c) inversely on the distance of separation between the conductor.

Principle of a condenser : Consider a conducting plate A which is given a

charge Q such that its potential rises to V. Then C = Q/V Let us place another identical conducting plate B

parallel to it such that charge is induced on plate B (as shown in figure).

++++++++

++++++++

A

Q

If V– is the potential at A due to induced negative charge on B and V+ is the potential at A due to induced positive charge on B, then

++++++++

++++++++

A++++++++

– – – – – – – –

B

C´ = ´V

Q = −+ −+ VVV

Q

Since V´ < V (as the induced negative charge lies closer to the plate A in comparison to induced positive charge).

⇒ C´ > C Further, if B is earthed from the outer side

(see figure) then Vn = V – V– as the entire positive charge flows to the earth. So

C" = nVQ =

−− VVQ ⇒ Cn >> C

So, if an identical earthed conductor is placed in the viscinty of a charged conductor then the capacitance of the charged conductor increases appreciably. This is the principle of a parallel plate capacitor.

Capacitor-1 PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Parallel Plate Capacitor :

–σ

A B

+ σ

d

+ + + + + + + +

– – – – – – – –

+ + + + + + + +

A B

A = Area of plate d = Separation between the plates

E =εσ =

0kεσ

It consists of two metallic plates A and B each of area

A at separation d. Plate A is positively charged and plate B is earthed. If K is the dielectric constant of the material medium and E is the field that exists between the two plates, then

E = εσ =

0Kεσ

=σ=

Aqand

dVEQ

⇒ dV =

AKq

0ε

⇒ C = Vq =

dAK 0ε

If medium between the plates is air or vacuum, then K = 1

⇒ C0 = dA0ε

Special Case I : When the space between the parallel plate capacitor

is partly filled with a dielectric of thickness t(<d) If no slab is introduced between the plates of the

capacitor, then a field E0 given by E0 = 0ε

σ , exists in

a space d. E=

KE0

+σ

dt

On inserting the slab of thickness t, a field E = KE0

exists inside the slab of thickness t and a field E0 exists in remaining space (d – t). If V is total potential then

V = E0(d – t) + Et

⇒ C = Vq =

−−

ε

K11td

A0

Special Case II : When the space between the parallel plate capacitor

is partly filled by a conducting slab of thickness t(<d).

It no conducting slab is introduced between the

plates, then a field E0 = 0ε

σ exists in a space d. If C0

be the capacitance (without the introduction of

conducting slab), then C0 = dA0ε

+σ

dt

E= 0

E0

On inserting the slab, field inside it is zero and so a

field E0 = 0ε

σ now exists in a space (d – t)

⇒ V = E0(d – t)

⇒ V = 0ε

σ (d – t)

⇒ V = 0A

qε

(d – t)

⇒ C = Vq =

tdA0

−ε

⇒ C =

−

ε

dtt1d

A0

⇒ C =

−

dt1

C0

Since d – t < d ⇒ C > C0 i.e. Capacitance increases on insertion of conducting

slab between the plates of capacitor.

Charge induced on a dielectric :

++++++++

––––––––

–qp

+ + + + + + + +

– – – – – – – –

+qp

–q+q E0

Ep

E = E0 – Ep

Resultant dielectric field within the plates is E = E0 – Ep


⇒ E = 0

1ε

(σ – σp) ...(1)

Also E = 0Kε

σ ...(2)

Compare (1) and (2), we get

0

1ε

(σ – σp) = 0Kε

σ

⇒ σp = σ

−

K11

⇒ Aqp =

−

K11

Aq

⇒ qp = q

−

K11

Spherical capacitor : B

C2

A

a C1

b

let C1 be the capacitance in between the two

conductors and C2 be capacitance out side both. To find C1 : Imagine the outer surface of B to be earthed. Then –q

is the charge induced on the inner surface of B. If V is the potential difference between the two

surfaces, then

V = Ka4

q

0πε +

Kb4q

0πε−

⇒ V = K4

q

0πε

−

b1

a1

⇒ C = Vq = 4πε0K

− abab ...(1)

To find C, Imagine A to be made open circuited (i.e. made non

conducting), then C2 = 4πε0Kb ...(2) Case I : When battery is connected to B and A is

earthed. Then C1 and C2 are in parallel ⇒ C = C1 + C2

⇒ C = 4πε0K

− abab + 4πε0Kb

⇒ C = 4πε0K

− abb2

Case II : When battery is connected to A, then C1 and C2 are in series.

⇒ C1 =

1C1 +

2C1

⇒ C1 =

abab −

K41

0πε +

Kb41

0πε

⇒ C1 =

+

−πε

1a

abKb4

1

0

⇒ C1 =

πε ab

Kb41

0

⇒ C = 4πε0Ka Case III : When battery connected to A and B is

earthed. Then C2 can be omitted as it will not receive any charge.

So, C = C1

⇒ C = 4πε0K

− abab

Case IV : When battery connected to B and A is open circuited (or made non conducted) then C1 can be omitted (as it is open circuited). So,

C = C2 ⇒ C = 4πε0Kb

Cylindrical capacitor : Let inner cylinder be given a charge per unit length

of λ

=

l

q . A charge – q is induced on length l at

inner surface of outer cylinder

ba q –q

l

E = r2 0πε

λ for a < r < b

⇒ – drdV =

Kr2 0πελ

⇒ ∫surfaceouter

surfaceinner

dV = – ∫=

=πε

λbr

ar0 rdr

K2

⇒ Vinner surface – Vouter surface = K2 0πε

λ loge

ab

Since, inner surface is at higher potential and outer at lower potential, so


–q

b a+q

Gaussian surface

⇒ Vouter surface – Vinner surface = K2 0πε

λ loge

ab

⇒ Vinner surface – Vouter surface = K2

q

0lπεloge

ab

⇒ C = surfaceoutersurfaceinner VV

q−

=

πε

ablog

K2

e

0l

⇒ C =

πε

ablog

K2

e

0l

(A) Energy stored in a capacitor E = 21 CV2 =

21 QV

= C

Q21 2

and energy stored per unit volume = 21

ε0E2

Note: The energy is stored in a capacitor is in the form of electric field between the plates.

(B) A parallel plate capacitor is charged by a battery and then the capacitor disconnected from the battery (a) If the distance between plates of the capacitor is increased then the new parameter of the capacitors as compared to the previous parameters is

q' = q; C' = 'dA0ε

, V' = 'C'q ,

E' = 'd'V (charge will not change) Energy =

21 C'V'2

If a dielectric slab (dielectric constant k) is introduced between the plates then

q' = q, C' = kC, V' = kV , E' =

kE U'(Energy) =

kU

(charge will not change)

(C) A parallel plate capacitor is charged by a battery. (a) If the distance between plates of the capacitor is

Increased (with the battery connected) then the new parameters of the capacitors as compared to the previous parameters is

V' = V, C' = 'dA0ε

, q' = C'V', E' = d

'V ,

Energy U' = 21 C'V'2

(b) If a dielectric slab (dielectric constant k) is introduced between two plates then

V' = V, C' = kC, q' = kq, E' = kE ; U' = KU

(p.d.) will not change)

1. A capacitor of 20 µF and charged to 500 volt is

connected in parallel with another capacitor of 10 µF charged to 200 volt. Find the common potential.

Sol. Charge on one capacitor q1 = C1V1 ∴ q1 = 20 × 10–6 × 500 = 0.01 coulomb Charge on second capacitor q2 = 10 × 10–6 × 200 = 0.002 coulomb The charge on the two capacitors q = q1 + q2 = 0.01 + 0.002 = 0.003 coulomb Total capacity C = C1 + C2 = 20 × 10–6 + 10 × 10–6 = 30 × 10–6 Farad. Common potential = q/C

= 61030012.0

−× = 400 Volt.

2. A battery of 10V is connected to a capacitor of

capacity of 0.1 F. The battery is now removed and this capacitor is connected to a second uncharged capacitor. If the charge distributes equally on these two capacitors, find the total energy stored in the two capacitors. Further, compare this energy with the initial energy stored in the first capacitor.

Sol. The initial energy stored in the first capacitor.

U0 = 21 CV2

= 21 × 0.1 × (10)2 = 5.0 J

When this capacitor is connected to the second uncharged capacitor, the charge distributes equally. This shows that the capacitance of the second capacitor is also C. The voltage across each capacitor will be V/2. If U be the energy stored in the two capacitors, then

U = 21 C

2

2V

+

21 C

2

2V

= 41 CV2 = 2.5 J

0U

U = 0.55.2 =

21

Solved Examples


3. Two isolated metallic solid spheres of radii R and 2R are charged such that both of these have same charge density σ. The spheres are located far away from each other, and connected by a thin conducting wire. Find the new charge density on the bigger sphere.

Sol. Charge on smaller sphere Q1 = 4πR2 . σ Charge on bigger sphere Q2 = 4π(2R)2σ = 16πR2σ ∴ Total charge Q = Q1 + Q2 = 20πR2σ ...(1) Capacitances of two spherical conductors are C1 = 4πε0R and C2 = 4πε0(2R) ∴ Total capacitance C = C1 + C2 = 12πε0R ...(2) After connection, the common potential V is given by

V = CQ =

R12R20

0

2

πεσπ =

03R5ε

σ

New charge on bigger sphere Q2

´ = C2V

= 4πε0R(2R) × (5Rσ/3ε0) = 3R40 2σπ

Surface density

σ2´ = areasurface

´Q2 = 2

2

)R2(4

3R40

π

σπ

= 65

σ.

4. A 8 µF capacitor C1 is charged to V0 = 120 volt. The

charging battery is then removed and the capacitor is connected in parallel to an uncharged 4 µF capacitor C2 (a) what is the potential difference V across the combination ? (b) What is the stored energy before and after the switch S is thrown ?

S

V0 C1 C2

Sol. (a) Let q0 be the charge on C1 initially. Then q0 = C1 V0 when C1 is connected to C2 in parallel, the charge q0

is distributed between C1 and C2. Let q1 and q2 be the charges on C1 and C2 respectively. Now let V be the potential difference across each condenser.

Now q0 = q1 + q2 or C1V0 = C1V + C2V

∴ V = 21

1

CCC+

V0 = µF4µF8

µF8+

(120 V)

= 80 volt.

(b) Initial energy stored

U0 = 21 C1V0

2

= 21 (8 × 10–6) (120)2

= 5.76 × 10–2 Joule Final energy stored

U = 21 C1V2 +

21 C2V2

= 21 (8 × 10–6)(80)2 +

21 (4 × 10–6)(80)2

= 3.84 × 10–2 joule Final energy is less than the initial energy. The loss

of energy appears as heat in connecting wires.

5. Calculate the capacitance of a parallel plate condenser, with plate area A and distance between plates d, when filled with a dielectric whose dielectric constant varies as

ε(x) = ε0 + βx 0 < x < 2d

ε(x) = ε0 + β(d – x) 2d < x < d

For what value of β would the capacity of the condenser be twice that when it is without any dielectric.

Sol. The capacitance in series is given by

´C

1 = 1C

1 + 2C

1

∴ ´C

1 = A1 ×

−β+ε

+β+ε ∫∫

d

2/d 0

2/d

0 0 )xd(dx

xdx

= βA

1 [log(ε0 + βx) 2/d0 – log(ε0 + β(d – x) d

2/d ]

=

β+ε−ε−

ε−

β+ε

β 2dlogloglog

2dlog

A1

0000

=

ε−

β+ε

β 00 log2dlog

A2

= βA

2 log

εβ+ε

0

0 2/d

The capacitance C of a condenser without dielectric is given by

C = d

A 0ε

According to the question, C´ = 2C

∴ 0A

2ε

log

εβ+ε

0

0 2/d =

A2d

0ε

β = d

4 0εlog

εβ+ε

0

0 2/d


Friction : Whenever there is a relative motion between two

surfaces in contact with each other, an opposing force comes into play which forbids the relative motion of two bodies. This opposing force is called the force of friction.

Ex. : If a book on a table slides from left to right along the surface of a table, a frictional force directed from right to left acts on the book.

Frictional force may also exist between the surfaces when there is no relative motion. Frictional forces arise due to molecular interactions.

Static and Kinetic Friction : The frictional force between two surface before the

relative motion actually starts is called static frictional force or static friction, While the frictional force between two surfaces in contact and in relative motion is called kinetic frictional force or kinetic friction.

Static friction is a self adjusting force and it adjusts both in magnitude and direction automatically. Its magnitude is always equal to external effective applied force, tending to cause the relative motion and its direction is always opposite to that of external applied force.

So, when a body is not in motion or equilibrium, then Force of static Friction = Applied External Force Limiting friction, coefficients of friction and angle of friction : Consider a block resting on a rough horizontal

surface. The forces acting on the block are its weight mg downwards and normal reaction N acting upward. Such that N = mg.

N R

f

θ

mg

P(<f))

M

Now suppose a force Fapp is applied to the block to

the right, then there will arise a frictional force f directed to the left (opposite to direction of applied

force), which prevents the motion of the block. Let the resultant of N

rand F

r be R

r which makes an

angle θ with normal reaction Nr

. Resolving Rr

along Nr

and Fr

, we get R cos θ = N and R sin θ = f For equilibrium N = mg and f = Fapp If we increase the pull Fapp continuously, the force of

friction increases and a stage comes when the body is just on the state of moving. This state is called limiting equilibrium. Under this condition the frictional force is maximum and is equal to applied force.

Limiting Friction : The maximum value of static frictional force exerted

between two surfaces in contact parallel to surfaces for a given normal force between when the body is on the verge of motion them is called limiting friction.

Angles of Friction : Angle of friction (θ) is the angle which the resultant

of force of static friction (f) and normal (N) makes with the normal reaction

The Coefficient of Friction (µ) : It is defined as the ratio of limiting friction F to the

normal reaction N between two surface in contact, i.e., µ = F/N ...(3) from figure, tan θ = F/N ...(4) Equation (3) and (4) µ = tan θ Static and Kinetic Regions : If a graph is plotted between applied force and

frictional force, the graph is obtained. In figure AC is limiting or (maximum) static friction and BD is kinetic friction. Obviously, kinetic friction is less than static friction.

If relative motion is absent and is at the verge of start µ = µs, the coefficient of static friction but if relative motion is present µ = µk, the coefficient of kinetic friction.

The coefficient of friction depends on the (a) strength of molecular interaction between the

surfaces in contact, (b) roughness of the two surface in contact. Whenever we are dealing with problem involving

friction we can follow the following analysis flow chart.

Friction

PHYSICS FUNDAMENTAL FOR IIT-JEE

KEY CONCEPTS & PROBLEM SOLVING STRATEGY


Read the problem Carefully

Find the value of applied force Fapp and limiting force

of Static Friction (fs)

If Fapp < fs then body does not move and the force of friction

f = Fapp

if Fapp > fs then body moves

If Fapp = fs the body is on the verge of motion (still in

equilibrium)

EITHER with Constant Velocity

OR with an Acceleration a

On a Level Track

On an inclined Plane

Fapp – fk = 0 or Fapp = fk

On a Level Track

On an inclined Plane

Fapp – fk = ma

Applied Pull = fk fk = mg sinθ Fapp – µkmg = ma

or Fapp = m(a + µkg) Fapp – µkmg cosθ = ma

or Fapp = m(a + µkg cosθ)

Laws of static and kinetic friction : (a) The force of limiting friction is directly

proportional to normal reaction for the same two surfaces in contact and acts opposite to direction of pull.

The kinetic friction is also proportional to normal reaction and acts opposite to direction of instantaneous relative motion. The kinetic friction is less than the static friction.

(b) The force of limiting (or static) friction is independent of area of contact of bodies as long as normal reaction remains the same.

The kinetic friction (to a good approximation) is independent of velocity, provided the velocity is neither too large nor too small.

Angle of repose (α) This is concerned with an inclined plane on which a

block rests, exerting its weight on the plane. The angle of repose α is the angle which an inclined

plane makes with the horizontal such that a body placed on it is on the verge of motion (is just about to loose the state of rest).

Under this condition the forces acting on the block are: (a) its weight mg, downward, (b) normal reaction N, normal to plane, (c) a force of friction fs, parallel and tangential to

plane upward. Taking α as angle of inclination of the plane with the

horizontal and resolving mg, parallel and

perpendicular to inclined plane, then for equilibrium, we get

N = mg cos α and fs = mg sin α

⇒ tan α = Nfs

N

mg mg cos α

fs

mg sinα

Tendency to slide

α Frictional force on a bicycle in motion : (a) When a wheel is rotated about its axle without

sliding, the frictional force acting on it is the rolling friction and it acts opposite to the direction of tendency of motion of a points of its contacts with the ground. In case the wheel rotates clockwise and frictional force (f) on wheel is forward. In case the wheel rotates anticlockwise, the frictional force (f) on wheel is backward.

(b) When the bicycle is pedalled, the force exerted on the rear wheel through the pedal-chain-axle system is in backward direction, therefore force of friction on rear wheel is forward. The front wheel of cycle moves by itself in forward direction, hence the force of friction of front-wheel is in backward direction.


(c) When the bicycle is not pedalled, no external force is being exerted, both wheels move forward by itself due to inertia and so the net frictional force on both wheels is in backward direction.

Solved Examples

1. A block of mass 5 kg is placed on a slope which makes an angle of 20º with the horizontal and is given a velocity of 10 m/sec up the slope. Assuming that the coefficient of sliding friction between the block and the slope is 0.20, find how far the block travels up the slope ? Take g = 10 m/sec2.

Sol. This situation is shown in fig. R

mg mg cos 20º

u = 10 m/s

mg sin 20º

x

20º The component of the weight perpendicular to plane = mg cos 20º = 5 × 10 × 0.9397 = 46.98 N The component of the weight parallel to the plane = mg sin 20º = 5 × 10 × 0.3420 = 17.10 N From figure R = mg cos 20º = 46.98 N Here the coefficient of kinetic friction = 0.2 Thus the frictional force X = 0.2 × 46.98 = 9.39 N The frictional force will be downward because the

motion is in the upward direction. The resultant force parallel to the plane is given by = X + mg sin 20º = 9.39 + 17.10 = 26.49 N From Newton's law F = ma, i.e., 26.49 = 5 × a

∴ a = 549.26 = 5.29 m/s2 downward

When the block is given a velocity 10 m/s in the upward direction we have

u = 10 m/s, v = 0, a = – 5.9 m/s2. (Taking the direction up the plane as positive) Let s be the distance traveled by the block. Using the formula v2 = u2 + 2a s, we have 0 = (10)2 – 2 × 5.29 × s

or s = 29.52

100×

= 9.45 m.

2. A block is projected up with 10 m/s along a fixed inclined plane of inclination 37º with the horizontal. If the time of ascend from the point of projection is half the time of descend to the same point, find the distance travelled by the block during the up and down journey.

Sol. Let µ, t1 and t2 be the coefficient of friction between the plane and the block, time of ascend and time of descend respectively.

The retardation while going up

a1 = g (sin θ + µ cos θ) = 10

+

54µ

53

The acceleration while descending

a2 = g(sin θ – µ cos θ) = 10

−

54µ

53

Now, s = distance of ascend = distance of descend. As final velocity is zero, we have 0 = u – a1t1 or u = a1t1

Now s = a1t12 –

21 a1t1

2 = 21 a1t1

2

s = a1t12 =

21 a2t2

2 and t2 = 2t1

∴

1

2

aa =

2

2

1

tt

or

+

−

54µ

53

54µ

53

= 2

21

Solving we get µ = (9/20)

Again a1 = 10

×+

54

209

53 = 9.6 m/sec2

∴ s = 1

2

a2u =

6.92)10( 2

× = 5.21 meter

So total distance = 2s = 10.42 metre 3. A block weighing 20 nt is at rest on a horizontal

table. The coefficient of static friction between block and table is 0.50. (a) What is the magnitude of the horizontal force that will just start the block moving ? (b) What is the magnitude of a force acting upward 60º from the horizontal that will just start the block moving ? (c) If the force acts down at 60º from the horizontal how large can it be without causing the block to move ?

Sol. (a) As shown in fig. the horizontal force F that will just start the block moving is equal to the maximum force of static friction. Thus,

R

W

F µR

F = µR = µW = 0.50 × 20 nt. = 10.0 nt. (b) The forces acting on the block are shown in fig.

R

W

Fcos θ µR

F sin θ F

θ

The applied force is inclined at an angle θ in the

upward direction. Its horizontal and vertical


components are F cos θ and F sin θ respectively. In equilibrium.

F cos θ = µR and F sin θ + R = W or R = (W – F sin θ) ∴ F cos θ = µ(W – F sin θ) = µ W – µF sin θ F (cos θ + µ sin θ) = µW

or F = θ+θ sinµcos

µW

Here µ = 0.50, W = 20 nt. and θ = 60º

F = º60sin5.0º60cos

2050.0+

× = 866.05.050.0

10×+

= 10.72 nt. (c) In this case,

R

W

Fcos θ

µR F sin θ

F θ

F cos θ = µR and R = W + F sin θ Solving we get,

F = θ−θ sinµcos

µW = 866.05.050.0

2050.0×−

×

= 149.2 nt. 4. Two blocks, m1 = 2kg and m2 = 4kg, are connected

with a light string that runs over a frictionless peg to a hanging block with a mass M as shown in fig. (a). The coefficient of sliding friction between block m2 and the horizontal surface at the speeds involved is µk = 0.2. The coefficient of static friction between the two blocks is µS = 0.4. What is the maximum mass M for the hanging block if the block m1 is not to slip on block m2 while m2 is sliding over the surface ?

Sol. The relevant free body diagrams are shown in fig.(b) Using two body system, we have

m1 m2

ms = 0.4 N1

m1g M

µk = 0.2 f1 f

T

N

(m1+m2)g Mg

T+

a

(a) (b) N – (m1 + m2)g = 0 ...(1) T – F = (m1 + m2)a ...(2) For hanging block Mg – T = Ma ...(3) From eqs. (2) and (3), Mg – f = (M + m1 + m2)a But f = µkN = µk(m1 + m2)g [Q using eq. (1)] ∴ Mg – µk(m1 + m2)g = (m + m1 + m2)a

or a = )mmM(

g)mm(µM

21

21k

+++− ...(4)

From free body diagram of mass m1, we have N1 – m1g = 0 and f1 = m1a It should be noticed that the force f1 accelerates m1 to

the right. Just before slipping occurs, we find

1

1

Nf = µS or µS =

gmam

1

1 = ga

∴ µS = )mmM(

)mm(µM

21

21k

+++− ....(5)

Solving eq. (5) for M, we have

M = S

21kS

µ1)mm(µµ(

−++

or M = )4.01(

)kg4kg2)(2.04.0(−

++ = 6 kg.

5. In fig.(a) the blocks A, B and C weight are 3kg, 4kg

and 8kg respectively. The coefficient of sliding friction between any two surfaces is 0.25. A is held at rest by a massless rigid rod fixed to the wall, while B and C are connected by a light flexible cord passing around a fixed frictionless pulley. Find the force P necessary to drag C along the horizontal surface to the left at a constant speed. Assume that the arrangement shown in the diagram, B on C and A on B is maintained all the throughout.

A

B

CP

Sol. When block C moves towards left, B moves towards right, while A is fixed. There would be a tension T in the string. Under this condition, let us consider the frictional forces between different surfaces.

Frictional force between A and B = µR = 0.25 × 3 Frictional force between C and B = µR = 0.25 (3 + 4) = 0.25 × 7 Frictional force between C and surface = 0.25(3 + 4 + 8) = 0.25 × 15 Considering fig. (b)

0.25(3 + 4)

C P0.25 × 15

TB0.25 × 3

T0.25(3 + 4)

Fig (b) Tension in the string = Frictional forces at upper and

lower surfaces of block B or T = 0.25 × 3 + 0.25 × 7 = 2.5 kg wt. For block C, P = T + Frictional force between C and B + Frictional

force between C and surface = 2.5 + 0.25 × (3 + 4) + 0.25 × (15) = 8 kg wt. = 8 × 9.8 = 78.4 newton


Elimination reactions : The elimination reactions are reverse of addition

reactions. In these reactions two atoms or group attached to the adjacent carbon atoms of the substrate molecule are eliminated to form a multiple bond. In these reactions a atom or group from α-carbon atom and a proton from the β-carbon are eliminated.

– C – C –

X α –HX – C = C –

β

H In eliminations reactions, the presence of one

hydrogen on the β-carbon atom is necessary. In general the elimination reactions are divided into two types, i.e., bimolecular elimination reactions (E2) and unimolecular elimination reactions (E1).

Bimolecular elimination reactions (E2) : In these elimination reactions, the rate of elimination

depends on the concentration of the substrate and the nucleophile and the reaction is of second order. It is represented as E2. Like SN2 reaction, the E2 reaction is also one step process. In these reactions abstraction of proton from the β-carbon atom and the expulsion of an atom or group from the α-carbon atom occur simultaneously. The mechanism of this reaction is represented as follows:

R – CH – CH2

B HB:

β α

Transition state

R — CH — CH2

X

δ+

δ –

RCH = CH2 + BH + X⊕ Θ

H

X

The above reaction is a one step process and passes

through a transition state. This reaction is also known as 1, 2-elimination or simply β-elimination. In these reactions, the two groups to be eliminated (i.e., H and X) are trans to each other and hence E2 reactions are generally trans elimination.

The second-order elimination reaction may also proceed in two steps (as in E1 elimination which will be discussed subsequently). In this mechanism, the base removes the hydrogen in the first step to form an intermediate carbanion. In the second step, the intermediate carbanion looses the leaving group. The second step is slow and is rate determining step.

– C – C – + OEt

HFast – C – C –

Br

Θ

Br

(First step)–

– C – C –Slow

– C = C – + Br–

Br

(Second step)Θ

The rate of this reaction is dependent on the

carbanion (conjugate base of the substrate). So this mechanism is called ElcB mechanism (Elimination, Unimolecular from conjugate base).

E1cB mechanism is not common for the E2 reactions. The carbanion mechanism occurs only where the carbanion from the substrate is stabilized and where the leaving group is a poor leaving group. A typical example, which follows E1cB mechanism is the formation of 1,1-dichloro-2,2-difluoroethene from 1,1-dichloro-2,2,2-trifluoroethane in presence of sodium ethoxide.

CHCl2 – CF3

C2H5ONaCl2C – CF3 –F–

Cl2C = CF2

1,1-Dichloro-2,2,2-trifluoroethane

Carbanion 1,1-Dichloro-2,2-difluoroethane

–

In the above case the carbanion is strongly stabilized

due to –I effect of halogens. Also F– is a poor leaving group.

A distinction between the E2 and E1cB mechanism can be made by tracer experiments. Thus, the reaction of 1-bromo-2-phenylethane (this substrate was selected as Ph group is expected to increase the acidity of β-hydrogen and also to stabilize the carbanion) with C2H5OD gives back the starting 1-bromo-2-phenylethane. If the carbanion mechanism had operated, the deuterium would have been found in the recovered 1-bromo-2-phenylethane, which is not the case.

C6H5CH2CH2Br + C2H5O–

C6H5CHCH2Br + C2H5OH1-Bromo-2-phenylethane

C6H5CHCH2Br + C2H5OD

C6H5CHCH2Br + OC2H5

D

In case the above reaction is allowed to go to

completion, the product obtained will be

Organic Chemistry

Fundamentals

REACTION MECHANISM

KEY CONCEPT


PhCH – CH2Br Ph – CH – CH2Br

C2H5OD

Fast

Ph – CH – CH2Br

D

OEt

PhCD – CH2BrPhCD = CH2 + BrStyrene

– H EtO–

The styrene obtained does not contain any deuterium

(contrary to what has been shown in the above E1cB mechanism). So in the above reaction E2 mechanism operates.

The E2 mechanism is supported by the following evidences.

(i) During elimination, there is no rearranged product obtained. This is due to the fact that E2 is a single step process and does not involve the formation of intermediate carbocation (the carbocations are known to undergo rearrangement).

(ii) The E2 mechanism finds support from isotope labeling experiments. Dehydrohalogenation of unlabelled 1-bromopropane is seven times faster than the dehydrohalogenation of CH3CD2CH2Br.

CH3CH2CH2Br → 2E CH3CH = CH2

CH3CD – CH2 E2 CH3CD = CH2

Br

D In E2 mechanism a hydrogen (from CH3CH2CH2Br)

or a deuterium (from CH3CD2CH2Br) has to be abstracted. It is known that the C – D bond is stronger than the C – H bond and requires more energy to be broken. Therefore, rate of elimination in CH3CD2CH2Br should be slower. In fact, it has been found that in the unlabelled alkyl halides the elimination rate is seven times more than in labelled alkyl halides.

Unsymmetrical substrate which has hydrogen attached to two different β-carbons can affored two alkenes. For example, 2-bromobutane on dehydrohalogenation may give 1-butene or 2-butene.

CH3 – CHCH2CH3

Br–HBr

CH2 = CHCH2CH3 + CH3CH=CHCH3

2-Bromobutane

1-Butene 2-Butene In a similar way, decomposition of sec-butyl-

trimethylammonium hydroxide may give a mixture of two alkenes. The question arises as to which alkene will be obtained in major amount in the above

dehydrohalogenation. The orientation of the reaction is determined by Hafmann and Saytzeff Rule.

Hofmann Rule : This rule is applicable for those substrates in which α-carbon atom is attached to a positively charged atom. According to this rule, in the elimination reaction of positively charged species, the major product will be the alkene which is least substituted.

CH3CH2NCH2CH2CH3

CH3

CH3

OH–

+ Heating

CH2 = CH2 + CH3CH2CH2N(CH3)2

CH3CH2S(CH3)2

+ C2H5O CH2 = CH2 + S(CH3)2

–

Saytzeff Rule : In case of unsymmetrical alkyl

halides, for example in 2-bromobutane, the course of elimination is determined by Saytzeff Rule. According to this rule, hydrogen is eliminated preferentially from the carbon atom which has less number of hydrogen atoms and so the highly substituted alkene is the major product.

CH3CH2 – CH – CH3

Br

2-Bromobutane

alk.KOH

CH3CH = CHCH3 + CH3CH2CH = CH22-Butene(80%) 1-Butene (20%)

CH3CH2 – C – CH3

CH3

2-Bromo-2-methylbutane

C2H5O–

Br

CH3CH = C – CH3 + CH3CH2C = CH2

CH3 CH3

2-methyl-2-butene (71%)

2-methyl-1-butene(29%)

The formation of highly substituted alkene can be explained as follows.

The transition states of less substituted and more substituted alkenes from an alkyl halide are represented as shown below:

CH3CH2CH — CH2

H OR

Br

δ–

δ– T.S. of less substituted alkene

CH3CH — CHCH3

RO H

Br

δ–

δ– T.S. of more substituted alkene

Both the transition states have partial double bond character. However, the transition state leading to more stable alkene is more stabilized and is of lower energy. Thus, the more stable alkene is formed as the major product.


less substitued alkene more substitued alkenepredominant product

Energy diagram for a typical E2 reaction, showing why the more substituted alkene predominates

Reaction progress

E

RX + base

Hofmann rule can be understood by considering the

mechanism of elimination reaction of quaternary ammonium hydroxide.

H – C – CH2 – N – CH2CH2CH3

Route a

H

H

CH3

CH3 B–

β´ β´´

+

CH2 = CH2 + (CH3)2N(CH2)2CH3 Another possibility is :

CH3CH2 – N – CH2 – C – CH3

Route b

H

H

CH3

CH3 B–

β´ +

(CH3)2NCH2CH3 + CH3CH = CH2

β´´

In the above reaction the strong electron-withdrawing

group makes the hydrogens of the β-carbons more acidic for facile abstraction by the base. In this compound, with alternate β-hydrogens (marked β´ and β´´), the β" hydrogen are less acidic due to +I effect of the adjacent methyl group. Hence β´- hydrogen is relatively more acidic and is removed to give the alkene (ethene) by route a.

In elimination reactions steric effect also plays an important role. Thus, dehydrohalogenation of alkyl halide using the bulky base leads to the formation of terminal alkene as the major product.

CH3CH2CHCH3

Br t-BuO

–

CH3CH2CH=CH2 + CH3CH=CHCH32-Bromobutane (73%) (27%)

Unimolecular elimination reactions (E1) : In these reactions the rate of elimination is dependent

only on the concentration of the substrate and is independent of the concentration of the nucleophile and the reaction is of first order, (E1). Like SN1

reaction the E1 reaction is also a two step process. The first step is the slow ionization of alkyl halide to give the carbocation. The second step involves the fast abstraction of a proton from the adjacent β-carbon atom giving rise to the formation of an alkene.

CH3 — C — X

CH3

CH3

Slow CH3 — C + X–CH3

CH3

+

Carbocation

CCH3

CH3

Fast CH3C = CH2 + BH

CH3

2-Methylpropene

CH2 – H:B

–

Carbocation In case the substrate is such that more than one

alkenes can be formed, that alkene will predominate which has larger number of alkyl groups on the double bonded carbon (this is as per Saytzeffs rule. This can be visualised since the substituted alkyl groups will stabilise the alkene by hyperconjugation.

CH3 — C — C — CH3

CH3

H

Br

H

–HBrCH3 — C == C — CH3

CH3

H

+ CH3 — CH — CH == CH2

CH3

2-Bromo-3-methylbutane 2-Methyl-2-butene (major)

3-Methyl-1-butene (minor) The acid catalysed dehydration of alcohols also

follows E1 mechanism.

(CH3)3COH H2SO4 (CH3)3C—OH2

⊕ –H2O (CH3)3C⊕

t-Butyl alcohol

H3C — C

CH3

CH2 – H

CH3 — C == CH2

CH3

2-Methylpropene

⊕

In the E1 mechanism the rate of reaction is

determined by the rate of formation of carbocation, which in turn depends on the stability of carbocation. Due to the formation of carbocation, these may undergo rearrangements. This has been experimentally confirmed.

(CH3)3CCHCH3 C2H5OH

(CH3)3CCH = CH2 + CH.3C = CCH3

Br

CH3

CH3


Thermodynamics deals with the transfer of heat

between a chemical system and its surroundings when a reaction or phase change takes place within the system. The entire formulation of thermodynamics is based on two fundamental laws which have been established on the basis of experimental on the basis of experimental behaviour of macroscopic aggregates of matter, collected over a long period of time.

Since First Law of thermodynamics The internal energy of a system can be changed by

transferring heat to/from the system from/to the surroundings. It can also be changed by doing the mechanical work on/by the system by/on the surroundings. These facts are represented in the form of the first law of thermodynamics as

dU = dq + dw or ∆U = q + w Since heat given to the system and work done on the

system raise the internal energy of the system, these two operations are assigned positive values. The converse of the two operations, viz., heat given out and work done by the system are assigned negative values.

The expression of work done by/on a gaseous system is given by

dw = – pext dV Where pext is the external pressure against which the

volume gaseous system is changed by an amount dV. For a constant external pressure, we have

w = – nRT ln (V2/V1) where V1 and V2 are the initial and final volumes of

the gaseous system. If pext differs from the pressure of the gas by

infinitesimal amount, the work is said to be carried out under reversible condition. In this case, the expression of work under constant temperature condition is given by

w = – nRT ln (V2/V1) ' Note that for V2 > V1, there occurs an expansion of

the gas. The work is done by the system on the surroundings and it carries a negative sign.

For V1 > V2, there occurs compression of the gaseous system. The work is done by the surroundings on the system and it carries a positive sign.

Internal energy and enthalpy From the first law of thermodynamics, it can be

shown that the heat transferred at constant volume

changes the internal energy of the system, whereas that at constant pressure changes the enthalpy of the system . ∆U = nCv,m (T2 – T1) ; and ∆H = nCp, m (T2 – T1)

where Cv,m and Cp,m are the molar heat capacities at constant volume and constant pressure, respectively.

In the laboratory, the majority of chemical reactions are carried out under the condition of constant pressure, and thus the heat transferred in such a system is equal to the enthalpy change in a chemical reaction. Since the enthalpy of a system can also change due to the variation in temperature and pressure, it is, therefore essential that the reactants and products in a chemical reaction must have the same temperature and pressure.

Enthalpy change of a chemical equation The enthalpy change of thermochemical equation is

∆H = ∑)products(

j,mjHv – ∑)tstanreac(

i,miHv

where Hm,i refers to the molar enthalpy of species i in the balanced chemical equation and vi the corresponding stoichiometric coefficient. The unit of ∆H are kJ mol–1 .

Two types of reactions may be distinguished. (a) Exothermic reactions For these ∆H is negative,

which implies negative qp and hence release of heat when reactants are converted into products. In this case

ΣH(products) < ΣH(reactants) (b) Endothermic reactions For these ∆H is is positive,

which implies positive qp and hence absorption of heat when reactants are converted into products. In this case

ΣH(products) > ΣH(reactants)

Molar enthalpies of formations It is not possible to determine the absolute value of

enthalpy of a substance. However, based on the following convention, the relative values of standard molar enthalpies of formation (the term standard indicates of pressure of 1 bar) other substances can be determined.

The enthalpy of formation of every element in its stable states of aggregation at 1 bar and 25ºC is assigned a zero value.

For example, ∆fHº (graphite) = 0 ∆fHº (Br2, 1) = 0 ∆fHº (S, rhombic) = 0 ∆fHº (H2, g) = 0 and so on.]

Physical Chemistry

Fundamentals

ENERGETICS

KEY CONCEPT


The enthalpy change of a chemical equation can be

computed by using the expression ∆rHº = ∑

)products(

)reactors(

0if

0if H–H ∑∆∆

Hess's law of constant heat summation Since the molar enthalpies of formation of reactants

and products involved in a chemical equation have definite values, the enthalpy change of (or heat involved in) the chemical equation will have a definite value, irrespective of the fact whether the reaction is carried out in one step or more than one step. This fact is known as Hess's law of constant heat summation. For example,

(i) C(graphite) + O2((g) → CO2(g) ∆H1 = – 393.51 kJ mol–1

(ii) C(graphite) + 21 O2(g) → CO(g)

∆H2 = – 110.52 kJ mol–1

CO(g) + 21 O2(g) → CO2(g)

∆H3 = – 282.99 kJ mol–1 Obviously, ∆H1 = ∆H2 + ∆H3

Types of reactions and corresponding enthalpy changes The enthalpy change in a reaction is suitable named

according to the type of reaction in question. Two types of reaction are specifically defined as follows.

Enthalpy of formation: the enthalpy of combustion of a given substance is defined as the enthalpy change when 1 mole of a given substance is formed, starting from the elements in their stable states of aggregation. A few examples are

H2(g) + 21 O2(g) → H2O(1)

∆fHº(H2O, 1) = – 285.77 kJ mol–1

12C(graphite) + 11H2(g) + 2

11 O2(g) → C12H22O11(s)

∆fHº(C12H22O11,s) = –2218 kJ mol–1 Enthalpy of Combustion: The enthalpy of

combustion of a given substance is defined as the enthalpy change when one mole of this substance combines with requisite amount of oxygen to form products in their stable states of aggregation. A few examples are

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆cH(CH4, g) = –74.85 kJ mol –1 C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l) ∆cH(C12H22O11, s) = – 5644 kJ mol–1 Similarly, one can mane the enthalpy change based

on the type of reaction. A few examples are Enthalpy of fusion: H2O(s) → K15.273 H2O(l)

∆fusH = 6 kJ mol–1 Enthalpy of vaporization :

H2O(1 ) → K15.373 H2O(g)

∆vapH = 40.6 kJ mol–1 Enthalpy of sublimation : I2(s) → I2(g) ∆subH = 63.4 kJ mol–1 Enthalpy of transition : C(graphite) → C(diamond) ∆trsH = 1.90 kJ mol–1 Enthalpy of neutralization : H+(aq)+OH–(aq)→H2O(l) ∆neutH = – 57.3 kJ mol–1 Enthalpy of ionization : HCN(aq) → H+(aq)+CN–(aq) ∆ionizH = 45.17 kJ mol–1

Relation between ∆H and ∆U of a chemical equation Since, H = U + pV, we have

∆H = ∆U + ∆(pV) = ∆U + (∆vg) RT where ∆vg is the change in the stoichiometric number

of gaseous molecules in converting reactants to products and is given as

∆vg = ∑)products(

i,gv – ∑)tstanreac(

i,gv

rFor a reaction involving condensed phases

∆H ~– ∆U Bond Enthalpies: Bond enthalpy of a given bond is

defined as the average enthalpies required to dissociate the said bond present in different gaseous compounds into free atoms in the gaseous phase. The bond enthalpy may be distinguished from bond dissociation enthalpy which is enthalpy required to dissociate a given bond of some specific molecule. It is possible to construct a table listing the average bond enthalpies of different types of bonds and with the help of this, one can estimate the enthalpy change of a chemical equation involving gaseous species. For example, for a reaction

H2(g) + Cl2(g) → 2HCl(g) we can write ∆H = + ε(H–H) + ε(Cl–Cl) –2ε(H–Cl) Second Law of thermodynamics The second law of thermodynamics identifies a state

function, called the entropy, which provides a criterion for identifying reversible or irreversible nature of the given process undergone by a system. The entropy of the universe (system + surroundings) increases for irreversible processes whereas it remains constant for reversible processes.

The entropy function has been identified with the disorderliness of the system–larger the disorderliness, larger the entropy of the system. Foe example, for a substance in three states of matter we have

S(gaseous state) > > S(liquid state) > (solid state) Expression of Entropy Function For a system which involves transferring

infinitesimal heat at constant temperature, the entropy change of the system is given by

dS = T

dqrev


For finite heat transferred at constant temperature, we have

∆S = T

qrev

For example, for a pure substance we have

∆vapS = b

vap

TH∆

and ∆fusS = m

fus

TH∆

where the subscripts vap and fus represent vaporization and fusion, respectively.

Gibbs Function Gibbs function (or energy) or simply free energy is

defined as G = H – TS

For a process occurring at constant T and P, the change in Gibbs function is given by

∆G = ∆H – T ∆S For a process to be spontaneous, the value of ∆G is

negative. For a nonspontaneous reaction, ∆G is positive. For a reaction at equilibrium, ∆G = 0 and temperature at which the system occurs at equilibrium is given by

Teq = ∆H/∆S Pressure-Volume Work An ideal gas can undergo expansion of compression

under isothermal or adiabatic conditions. The expansion ant compression may be carried out under reversible or irreversible conditions. We give below the expressions of p-V work under different conditions.

Isothermal p–V Work In this case, temperature of the system remains

constant, ie. ∆T = 0 For irreversible condition: w = – Pext (V2 – V1) For reversible condition: w = – nRT In (V2/V1) Adiabatic p–V Work In this case, heat can neither enter to or leave from

the system, i.e. q = 0. From first law of thermodynamics, it follows that

∆U = w where ∆U is given by

∆U = Cv(T2 – T1) For a gas undergoing adiabatic irreversible volume

change, the expression of work is given by w = – Pext (V2 –V1)

For an ideal gas undergoing adiabatic reversible expansion/compression, we also have

pVγ = constant pTγ(1–γ) = constant and TVγ–1 = constant

here γ = Cp,m/Cv,m The symbols Cp,m and Cv,m represent molar heat capacities at constant pressure and volume conditions, respectively.

For a monatomic ideal gas: Cv,m = (3/2)R; Cp,m = (5/2)R; and γ = 5/3 For a diatomic ideal gas: Cv,m = (5/2)R; Cp,m = (7/2)R; and γ = 7/5

CHEMISTRY JOKES If you didn't get the joke, you probably didn't understand the science behind it. If this is the case, it's a chance for you to learn a little chemistry.

Chemistry Joke 1:

Outside his buckyball home, one molecule overheard another molecule saying, "I'm positive that a free electron once stripped me of an electron after he lepton me. You gotta keep your ion them."

Chemistry Joke 2: A chemistry professor couldn't resist interjecting a little philosophy into a class lecture. He interrupted his discussion on balancing chemical equations, saying, "Remember, if you're not part of the solution, you're part of the precipitate!"

Chemistry Joke 3: One day on the Tonight Show, Jay Leno showed a classified add that read: "Do you have mole problems? If so, call Avogadro at 602-1023."

Chemistry Joke 4: A student comes into his lab class right at the end of the hour. Fearing he'll get an "F", he asks a fellow student what she's been doing. "We've been observing water under the microscope. We're suppose to write up what we see." The page of her notebook is filled with little figures resembling circles and ellipses with hair on them. The panic-stricken student hears the bell go off, opens his notebook and writes, "During this laboratory, I examined water under the microscope and I saw twice as many H's as O's."


1. An organic compound (A), C10H15N, undergoes carbylamine reaction but no diazotization. It reacts with HNO2 giving off N2 and a compound (B), C10H14O. (B) reacts with Lucas reagent immediately, but no colour in Victor meyer's test. (B) on heating with conc. H2SO4 eliminates water to give (C), C10H12, which decolourises Br2/CCl4 and cold dilute neutral KMnO4 solution. (C) on ozonolysis gives (D), C7H6O and (E), C3H6O. Compound (E) on heating with I2 and NaOH produced yellow precipitate and sodium acetate. Compound (D) reacts with conc. NaOH to give (F) and (G). Compound (G) on heating with sodalime gives benzene. Compound (F) gives a red colour with ceric ammonium nitrate, and on oxidation and heating the product with sodalime produced benzene. What are (A) to (G) ?

Sol. The given data are :

C10H15N HNO2 –N2;–H2O (A)

C10H14O Conc. H2SO4

∆;–H2O(B) C10H12

(C)(I) O3

(II) H2/Pd C7H6O + C3H6O

(D) (E)

C7H6O Conc. NaOH

(D) (F) + (G) Sodalime C6H6

[O]Product ∆ C6H6

Sodalime

)E(63 OHC → +NaOHI2 2 CHI3 ↓ + CH3COONa

+ 3NaI + 3H2O Since (C) decolourise Br2/CCl4 and KMnO4 colour,

hence it has C=C bond. Its ozonolysis gives (D) and (E). Among these (D) undergoes Cannizaro's reaction, while (E) gives iodoform test, hence (D) is benzaldehyde and (E) acetone. Now joining (D) and (E), the structure of (C) can be determined.

–2[O]

C=O + O =C

H CH3

CH3

C=C

HCH3

CH3

(D) (E)

(C) Since (C) is produced from (B), which is a t-alcohol,

as it gives Lucas test immediately, hence (B) is.

C–CH

H CH3

CH3

(B) OH

As (B) is obtained by the action of HNO2 on (A),

hence (A) would be

C–CH

H CH3

CH3

(A) NH2

2. Two isomeric compounds (A) and (B) have the

molecular formula C7H9N. (A) being soluble in water, the solution being alkaline to litmus It does not undergoes diazotization, but show carbylamine reaction and mustard oil reaction, it reacts with acetyl chloride and acetic anhydride. Its product with benzene sulphonyl chloride dissolves in KOH. (B) on the other hand, does not dissolve in water, but undergoes diazotization. Its product with C6H5SO2Cl dissolves in KOH. Its salt undergo hydrolysis in aqueous solution showing an acidic test. What are (A) and (B) ?

Sol. As both (A) and (B) give products with C6H5SO2Cl, which are soluble in KOH, they contain –NH2 group. (B) can be diazotized so contains – NH2 in the nucleus. (A) cannot be diazotized, hence contains –NH2 in the side chain. The number of carbon and hydrogen atoms also indicates aromatic character.

On the basis of above considerations we may show that (A) is benzylamine and (B) o–, m– or p-toludine.

CH2NH2

Benzylamine(A)

CHCl3 + 3KOH

CH2OH

CH2NC

CS2 + HgCl2 CH2NCS

CH2NHSO2C6H5C6H5SO2Cl

– HCl

NaNO2 + HCl

KOH

–H2OCH2 N–SO2C6H5

K Soluble

C6H5CH2NH2 + HOH → C6H5CH2N+H3OH–

UNDERSTANDINGOrganic Chemistry

U n d e r s t a n d i n g


C6H4

CH3 NH2

NaNO2 + HCl C6H4 CH3 N2Cl (B)

C6H5SO2Cl–HCl

C6H4 CH3 NHSO2C6H5

KOH C6H4 CH3

NKSO2C6H5

Soluble 3. An organic compound (A) of molecular weight

140.5, has 68.32% C, 6.4% H and 25.26% Cl. Hydrolysis of (A) with dilute acid gives compound (B), C8H10O. (B) can be oxidised under milder conditions to (C), C8H8O. (C) forms a phenyl hydrazone (D) with C6H5NHNH2 and gives positive iodoform test. What are (A) to (D) ?

Sol. (i) Calculation of empirical formula of (A)

Element % Relative no. of atoms

Simplest ratio

C 68.32 12

32.68 = 5.59 71.059.5 = 8

H 6.4 14.6 = 6.40

71.040.6 = 9

Cl 25.26 5.3526.25 = 0.71

71.071.0 = 1

Empirical formula = C8H9Cl Empirical formula wt. = 140.5 Molecular weight = Emp. formula weight Hence, Molecular formula = Empirical formula = C8H9Cl (ii) Given that

)A(98 ClHC →HOH

)B(98 OHHC

OH

]O[

2−→

)C(88 OHC

(iii) (C) reacts with C6H5NHNH2 to give C8H8=N.NHC6H5, hence (C) contains a C = O group.

(iv) (C) on heating with I2 + NaOH gives CHI3, hence (C) contains –COCH3 group. Thus (C) is

– C – CH3

O

(v) Oxidation of (B) gives (C), hence (B) is a

secondary alcohol, i.e.,

– CH – CH3

OH (vi) (B) is obtained by the hydrolysis of (A), hence it

is :

– CH.CH3

Cl

1-Chloro-1-phenyl ethane Now, different reactions are as follows :

– CH.CH3

Cl(A)

– CH–CH3

OH (B)

HOH/H+

–HCl

– COCH3

(C)

[O]–H2O

– C = O + H2 N . NH . C6H5

CH3(C) – C=N.NH.C6H5

–H2O CH3

– C – CH3 + 3I2 + 4NaOH → CHI3 + 3NaI

O(C) – COONa

+ 3H2O +

∆

Yellow ppt

4. An organic compound (A), C4H9Cl, on reacting with

aqueous KOH gives (B) and on reaction with alcoholic KOH gives (C) which is also formed on passing vapours of (B) over heated copper. The compound (C) readily decolourises bromine water. Ozonolysis of (C) gives two compounds (D) and (E). Compound (D) reacts with NH2OH to gives (F) and the compound (E) reacts with NaOH to give an alcohol (G) and sodium salt (H) of an acid. (D) can also be prepared from propyne on treatment with water in presence of Hg++ and H2SO4. Identify (A) to (H) with proper reasoning.

Sol. C4H9Cl (A)

(Alkyl halide)

Alc. KOH ∆; –KCl

C4H8 (C)

(Alkene) Aq.KOH ∆; –KCl ∆; –H2O

Cu C4H9OH (B)

(Alcohol)

We know that p-alcohol on heating with Cu gives aldehyde while s-alcohol under similar conditions gives ketone. Thus, (B) is a t-alcohol because it, on heating with Cu gives an alkene (C). Since a t-alcohol is obtained by the hydrolysis of a t-alkyl halide, hence (A) is t-butyl chloride.

(A) =

3

33

CH|

CHCCH|Cl

−− and (B) =

3

33

CH|

CHCCH|

OH

−−

The alkene (C) on ozonolysis gives (D) and (E), hence (C) is not symmetrical alkene. In these compound (E) gives Cannizaro's reaction with NaOH. So, (E) is an aldehyde which does not contain α–H atom. Hence it is HCHO. Compound (D) can also be prepared by the hydration of propyne in the presence of acidic solution and Hg++.


CH3 – C ≡ CH + H2O +

++

→H

Hg

OH|

CHCCH 23 =−

→

)D(O||

CHCCH 33 −−

Hence (D) is acetone and (E) is formaldehyde. Therefore, alkene (C) is 2-methyl propene. (CH3)2–C=CH2

(D) reacts with hydroxyl amine (NH2OH) to form oxime (F).

CH3 CH3

C = O + H2 NOH –H2O CH3

CH3 C = NOH

(D) (F)

(B) =

3

33

CH|

CHCCH|

OH

−− and (A) =

3

33

CH|

CHCCH|Cl

−−

Reactions :

)A(3

33

CH|

CHCCH|Cl

−−KCl;–

KOH.Aq

∆ →

)B(3

33

CH|

CHCCH|OH

−−

OH

Cº300/Cu

2− →

)C(3

223

CH|

OHCHCCH +=−

OH;KCl

/KOH.Alc

2−−

∆ →

)C(3

23

CH|

CHCCH =−

CH3 – C = CH2 (I) O3

(II) H2O/Zn C = O + H – C – H

CH3

CH3

CH3

O

(C) (D)

(E)

C = O + H2NOH ∆ –H2O

C = NOHCH3

CH3 (D) (F)

CH3

CH3

)E(HCHO2 + NaOH →

)G(3OHCH +

)H(HCOONa

CH3 – C ≡ CH + H2O +

++

→H

Hg

CH3 – C – CH3

O

(D)

5. A hydrocarbon (A) of the formula C8H10, on ozonolysis gives compound (B), C4H6O2, only. The compound (B) can also be obtained from the alkyl bromide, (C) (C3H4Br) upon treatment with Mg in dry ether, followed by treatment with CO2 and acidification. Identify (A), (B) and (C) and also equations for the reactions.

Sol. A(C8H10) OH)ii(

O)i(

2

3 →)B(

264 OHC

Since compound (A) adds one mol of O3, hence it

should have either C = C or a – C ≡ C – bond. If it was alkene its formula should be C8H16 (CnH2n), and if it was alkyne it should have the formula C8H14; it means it is neither a simple alkenen or simple alkyne. However it is definite that the compound has an unsaturated group, it appears that it is a cyclosubstituted ethyne.

H – C ≡ C – H 106HC

H2

+

− → C3H5 – C ≡ C – C3H5

the C3H5 – corresponds to cyclopropyl (∆) radical, hence compund (A) is

CH – C≡C – CH

CH2

CH2

CH2

CH2 1,2-dicyclopropyl ethyne

The ozonolysis of above compound would give two moles of cyclopropane carboxylic acid (C4H6O2).

CH – C≡C – CH

CH2

CH2

CH2

CH2 (A)

(i) O3

CH – C — C – CH CH2

CH2

CH2

CH2 (A)

H2O

O

O O

Warm

CH – C – C – CH

CH2

CH2

CH2

CH2 O O

(B)

+ H2O2

CH2

CH2

CH – COOH 2

Compound (B) is prepared from cyclopropyl bromide

as follows :

CH – Br CH2

CH2

CH2

CH2

Cyclopropyl magnesium bromide

CH . MgBr Mg

ether

C=O∆

O

CH .COOMgBr

CH2

CH2

CH2

CH2

Addition compound

CH–COOH HOH

dil. HCl; –MgBrOH


1. Let y = f(x) be a curve satisfying

dxdy – y ln 2 = 2sin x (cos x – 1). ln2, then

(A) y is bounded when x → ∞ (B) f(x) = 2sin x + c . 2x, where c is an arbitrary

constant (C) y = 2sinx, y is bounded when x → ∞ (D) f(x) = 2sinx does not have any solution if y is not

bounded. 2. In a right angled triangle the length of its hypotenuse

is four times the length of the perpendicular drawn from its orthocentre on the hypotenuse. The acute angles of the triangle can be

(A) 6π ,

3π (B)

8π ,

83π

(C) 6π ,

4π (D)

12π ,

125π

3. Let a, b ∈ R such that 0 < a < 1 and 0 < b < 1. The

values of a and b such that the complex number z1 = –a + i, z2 = –1 + bi and z3 = 0 form an equilateral triangle are

(A) 3,32 − (B) 32,32 −−

(C) 32,3 − (D) None of these 4. If c1 is a fixed circle and c2 is a variable circle with

fixed radius. The common transverse tangents to c1 and c2 are perpendicular to each other. The locus of the centre of variable circle is :

(A) circle (B) ellipse (C) hyperbola (D) parabola 5. The length of the latus rectum of the parabola 169 (x – 1)2 + (y – 3)2 = (5x – 12y + 17)2 is –

(A) 1314 (B)

1356 (C)

1328 (D) None

6. Evaluate : ∫ −+

x3cos21x4cosx5cos dx

7. Find all the real values of a, for which the roots of the equation x2 – 2x – a2 + 1 = 0 lie between the roots of equation x2 – 2(a + 1) x + a(a – 1) = 0

8. Given the base of a triangle and the sum of its sides

prove that the locus of the centre of its incircle is an ellipse.

9. A bag contains 7 tickets marked with the number 0,

1, 2, 3, 4, 5, 6 respectively. A ticket is drawn and replaced. Then the chance that after 4 drawings the sum of the numbers drawn is 8, is –

10. A polynomial in x of degree greater than 3 leaves

remainders 2, 1 and – 1 when divided by (x – 1), (x + 2) and (x + 1) respectively. What would be the remainder if the polynomial is divided by (x2 – 1) (x + 2) ?

`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

3Set

• William Bottke at Cornell University in the US has calculated that at least 900 asteroids of a kilometre or more across regularly sweep across Earth's path.

• The Dutch astronomer Christiaan Huygens (1629 - 1695) drew Mars using an advanced telescope of his own design. He recorded a large, dark spot on Mars, probably Syrtis Major. He noticed that the spot returned to the same position at the same time the next day, and calculated that Mars has a 24 hour period. (It is actually 24 hours and 37 minutes)

• Space debris travels through space at over 18,000 mph.

• The nucleus of Comet Halley is approximately 16x8x8 kilometers. Contrary to prior expectations, Halley's nucleus is very dark: its albedo is only about 0.03 making it darker than coal and one of the darkest objects in the solar system.


1. 1st box can be filled in 4 ways. Next each box can be filled in 3 ways (except the ball

of colour in previous box). Hence the required no. of ways = 4 × 35 = 972 2. Given |A| ≠ 0; AA–1 = I ⇒ (AA–1)T = IT (A–1)TAT = I (as A is symmetric) (A–1)T A = I so by the definition of inverse A–1 = (A–1)T Hence A–1 is also symmetric. 3. The normal to hyperbola at the point P(a sec θ, b tan θ) is ax cos θ + by cot θ = a2 + b2 If it passes through (h, k) then a h cos θ + b k cot θ = a2 + b2 ...(1)

Let z = eiθ = cos θ + i sin θ then put cos θ = z2

1z2 +

and cot θ = i

−+

1z1z

2

2 in the equation (1).

ahz4 + 2(i bk – (a2 + b2)) z3 + 2(i bk + (a2 + b2))z – ah = 0 z1, z2, z3, z4 are its four solutions so Σ z1z2 = 0 = Σ )(i 21e θ+θ = 0 Σ (cos (θ1 + θ2) + i sin (θ1 + θ2)) = 0 Hence Σ cos(θ1 + θ2) = Σ sin(θ1 + θ2) = 0 4. Planes are – x – 2y – 2z + 9 = 0 ....(1) and 4x – 3y + 12z + 13 = 0 ...(2) The plane bisecting the angle b/w these planes

containing origin is

3

9z2y2x +−−− = + 13

13z12y3x4 ++−

i.e. 25x + 17y + 62z – 78 = 0 ...(3) If θ be the angle between (1) & (3) then

cos θ = 475861

⇒ tan θ = 61

1037 < 1

Hence plane given by (3) is bisecting the acute angle between given two planes also. Hence the conclusion holds true.

5. Let I2 = ∫ −−)b(f

)a(f

221 )a))y(f(( dy

Let f–1(y) = x ⇒ f(x) = y

I2 = ∫ −b

a

22 )ax( f´(x) dx

Using by parts

I2 = ( )ba22 )x(f)ax( − – ∫b

ax2 f(x) dx

= (b2 – a2) f(b) – ∫b

ax2 f(x) dx

= ∫b

ax2 f(b) dx – ∫

b

ax2 f(x) dx

= ∫b

ax2 (f(b) – f(x)) dx

Hence 2

1

II =

21

6. y + y1 = 2

⇒ y = 1

x + x1 = 25 +

⇒ x2 + 2x1

= ( 5 + 2) – 2 = 5

x4 + 4x1 = 5 – 2

⇒ x8 + 8x1 = 9 – 2

⇒ x16 + 16x1 = 49 – 2

⇒ 47 + 1 + 1 = 49

MATHEMATICAL CHALLENGES SOLUTION FOR JUNE ISSUE (SET # 2)


7. Let the radius of S2 is r

6

6

6

6rr

2 r + r + 6 = 2 6

r = 6

+

−

1212

= 6(3 – 22 )

= 18 – 212 8. S1 = 2 + 4 + 6 + .... + 120

= 2

60 (2 + 120)

= 30 × 122 = 3660 S2 = 7 + 14 + 21 + ..... + 119

= 2

17 (7 + 119)

= 17 × 63 = 1071 S3 = 14 + 28 + ..... + 102

= 28 (14 + 112)

= 4 × 126 = 504

Ans. = 2

121120× – 3660 – 1071 + 504

= 7260 – 4731 + 504 = 2529 + 504 = 3033 9. Here F(x) is even function so f(x) = F(–x) = F(x) ⇒ f(–x) = f(x) g(x) = –F(x) = – f(x) = –f(–x) h(x) = –F(–x) = – F(x) = –f(x) Ans. (C) 10. f(x) + h(x) = f(x) – f(x) = 0 g(x) – h(x) = – f(x) + f(x) = 0 F(x) + f(x) ≠ 0 f(x) – g(x) = f(x) + f(x) ≠ 0 Ans. (B)

Brief Description : zirconium is a grayish-white lustrous metal. The finely divided metal can ignite spontaneously in air, especially at elevated temperatures. The solid metal is much more difficult to ignite. The inherent toxicity of zirconium compounds is low. Hafnium is invariably found in zirconium ores, and the separation is difficult. Commercial grade zirconium contains from 1 to 3% hafnium. The hafnium is removed from the zirconium used in the nuclear power industry.

Zirconium is found in S-type stars, and has been identified in the sun and meteorites. Analyses of lunar rock samples show a surprisingly high zirconium oxide content as compared with terrestrial rocks. Some forms of zircon (ZrSiO4) have excellent gemstone qualities. Table: basic information about and classifications of zirconium. • Name : Zirconium • Symbol : Zr • Atomic number : 40 • Atomic weight : 91.224 (2) • Standard state : solid at 298 K • CAS Registry ID : 7440-67-7 • Group in periodic table : 4 • Group name : (none) • Period in periodic table : 5 • Block in periodic table : d-block • Colour : silvery white • Classification : Metallic ISOLATION Isolation : zirconium is available from commercial sources so preparation in the laboratory is not normally required. In industry, reduction of ores with carbon is not a useful option as intractable carbides are produced. As for titanium, the Kroll method is used for zirconium and involves the action of chlorine and carbon upon baddeleyite (ZrO2). The resultant zirconium tetrachloride, ZrCl4, is separated from the iron trichloride, FeCl3, by fractional distillation. Finally ZrCl4 is reduced to metallic zirconium by reduction with magnesium, Mg. Air is excluded so as to prevent contamination of the product with oxygen or nitrogen. ZrO2 + 2Cl2 + 2C (900°C) → ZrCl4 + 2CO ZrCl4 + 2Mg (1100°C) → 2MgCl2 + Zr Excess magnesium and magnesium dichloride is removed from the product by treatment with water and hydrochloric acid to leave a zirconium "sponge". This can be melted under helium by electrical heating.


1. Let f(x) be a function which satisfy the equation

f(xy) = f(x) + f(y) for all x > 0, y > 0 such that f ´(1) = 2. Find the area of region bounded by the curves y = f(x), y = |x3 – 6x2 + 11x – 6| and x = 0

Sol. Take x = y = 1 ⇒ f(1) = 0

Now f

x1.x = f(x) + f

x1

⇒ f

x1 = – f(x)

f

yx = f(x) + f

y1 = f(x) – f(y)

f ´(x) = h

)x(f)hx(flim0h

−+→

=

+

→ hhxf

h1lim

0h =

x.xh

)1(fxh1f

lim0h

−

+

→

= x

)1´(f = x2

⇒ f(x) = 2 log |x| + c ⇒ c = 0 when x = 1; as f(1) = 0 ⇒ f(x) = 2logx ∴ Required area

= ∫ −+−1

0

23 dx)6x11x6x( + dye0

–

2/y∫ ∞

= 47 sq units

2. Let a1, a2, ......, an be real constant, x be a real variable

and f(x) = cos(a1 + x) + 21 cos(a2 + x) +

41 cos(a3 + x)

+...... + 1n21

− cos(an + x). Given that f(x1) = f(x2) = 0,

prove that (x2 – x1) = mπ for integer m. Sol. f(x) may be written as,

f(x) = ∑=

−

n

1k1k2

1 cos(ak + x)

= ∑=

−

n

1k1k2

1 cosak. cos x – sin ak . sin x

= cos x .

∑

=−

n

1k1kk

2acos – sin x

∑

=−

n

1k1kk

2asin

= A cos x – B sin x, where A = ∑=

−

n

1k1kk

2acos and

B = ∑=

−

n

1k1kk

2asin

since f(x1) = f(x2) = 0 ⇒ A cos x1 – B sin x1 = 0 and A cos x2 – B sin x2 = 0

⇒ tan x1 = BA

⇒ tan x2 = BA

⇒ tan x1 = tan x2 ⇒ (x2 – x1) = mπ 3. Let a variable chord from (–1, 0) point to the circle

(x – 2)2 + y2 = 1, makes a intercept of length 'l' on the circle and length of perpendicular from centre of the circle to chord is 'p'. find the range of 'λ' such that l2 + 3λ p2 + 5 = 0.

Sol. We have OB2 = OD2 + BD2

(–1, 0) 0

AD B

p

1 = p2 + 4

2l ⇒ p2 = 4

4 2l−

we have been given, l2 + 3λp2 + 5 = 0

l2 + 4

3λ

−4

4 2l + 5 = 0

l2 = 432012

−λ+λ

clearly 0 ≤ l2 < 4

⇒ 0 ≤ )3/4()3/5(4

−λ+λ < 4

⇒ λ ∈ (– ∞, –5/3) [as λ can not be + ve]

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

Students' ForumMATHS


4. Find all possible negative real values of 'a' such that

∫ −− −0

a

tt2 )9.29( dt ≥ 0

Sol. Here, ∫ −−0

a

tt2 )9.29( dt ≥ 0

⇒ 0

a

tt2

9log9.2

9log29

−−

−

−−

≥ 0

⇒ ( )0att2 )49(9 −− +− ≥ 0

⇒ 9–2a – 4.9–a + 3 ≥ 0 t2 – 4t + 3 ≥ 0 where t = 9–a and t ∈(1, ∞) (t – 1) (t – 3) ≥ 0 ⇒ t ≤ 1 or t ≥ 3 ⇒ t ≥ 3 is possible as t > 1

9–a ≥ 3 ⇒ a ≤ – 21

5. Let a0, a1, .... an – 1 be real numbers where n ≥ 1 and het f(x) = xn + an – 1 xn –1 + ..... + a0 be such that : |f(0)| = f(1) and each root of f(x) = 0 is real and lies between 0 and 1. Prove that the product of the roots

does not exceed n21 .

Sol. Let, f(x) = (x – α1) (x – α2) ..... (x – αn) where α1, α2 ........, αn are the roots of f(x) = 0 since |f(0)| = f(1) ∴ α1 . α2 ...... αn = (1 – α1) (1 – α2) ...... (1 – αn) ⇒ (α1 . α2 ....... αn)2 ⇒ α1(1 – α1) α2(1 – α2) ...... αn(1 – αn) ⇒ ( Π αi)2 = II αi(1 – αi) . i = 1, 2, ..... n

Now, ( Π αi)2 = Π αi(1 – αi) ≤ Π2

i

2)1(

α−+α

= n221

Since GM ≤ AM

⇒ ( Π αi) ≤ n21

6. If the equation az2 + z + 1 = 0 has a purely imaginary

root where a = cos θ + i sin θ, i = 1− . Then find the interval in which the function, f(x) = x3 – 3x2 + 3(1 + cos θ)x + 5 is increasing.

Sol. We have, az2 + z + 1 = 0 ...(i) ⇒ az2 + z + 1 = 0 Taking conjugate on both sides ⇒ a z 2 + z + 1 = 0 ...(ii) Eliminating z from eq. (i) and (ii) by cross

multiplication rule, ( a – a)2 + 2(a + a ) = 0

⇒ 2

2aa

− +

2

2aa

+ = 0

⇒ – 2

i2aa

− +

2

2aa

+ = 0

⇒ – sin2θ + cos θ = 0 ⇒ cos θ = sin2θ ...(iii)

Now, f(x) = x3 – 3x2 + 3(1 + cos θ)x + 5 f ´(x) = 3x2 – 6x + 3 (1 + cos θ) ∴ Discriminate (D) = 36 – 36(1 + cos θ) = – 36 cos θ = – 36 sin2θ < 0 ⇒ f(x) is increasing ∀ x ∈ R

Regents Physics You Should Know Electricity

1. A coulomb is charge, an amp is current [coulomb/sec] and a volt is potential difference [joule/coulomb].

2. Short fat cold wires make the best conductors. 3. Electrons and protons have equal amounts of

charge (1.6 x 10-19 coulombs each). 4. Adding a resistor in parallel decreases the total

resistance of a circuit. 5. Adding a resistor in series increases the total

resistance of a circuit. 6. All resistors in series have equal current (I). 7. All resistors in parallel have equal voltage (V). 8. If two charged spheres touch each other add

the charges and divide by two to find the final charge on each sphere.

9. Insulators contain no free electrons. 10. Ionized gases conduct electric current using

positive ions, negative ions and electrons. 11. Electric fields all point in the direction of the

force on a positive test charge. 12. Electric fields between two parallel plates are

uniform in strength except at the edges. 13. Millikan determined the charge on a single

electron using his famous oil-drop experiment. 14. All charge changes result from the movement

of electrons not protons (an object becomes positive by losing electrons)


Coordinates of a point :

x y

P (x,y,z)

z

N

O

Y

Z

L

M

X

x-coordinate = perpendicular distance of P from yz-plane y-coordinate = perpendicular distance of P from zx-plane z-coordinate = perpendicular distance of P from xy-plane Coordinates of a point on the coordinate planes and axes: yz-plane : x = 0 zx-plane : y = 0 xy-plane : z = 0 x-axis : y = 0, z = 0 y-axis : y = 0, x = 0 z-axis : x = 0, y = 0 Distance between two points : If P(x1, y1, z1) and Q(x2, y2, z2) are two points, then

distance between them

PQ = 221

221

221 )zz()yy()xx( −+−+−

Coordinates of division point : Coordinates of the point dividing the line joining two

points P(x1, y1, z1) and Q(x2, y2, z2) in the ratio m1 : m2 are

(i) in case of internal division

++

++

++

21

1221

21

1221

21

1221mm

zmzm,mm

ymym,mm

xmxm

(ii) in case of external division

−−

−−

−−

21

1221

21

1221

21

1221

mmzmzm

,mm

ymym,

mmxmxm

Note: When m1, m2 are in opposite sign, then division will be external.

Coordinates of the midpoint: When division point is the mid-point of PQ, then

ration will be 1 : 1; hence coordinates of the mid-point of PQ are

+++

2zz,

2yy,

2xx 212121

Coordinates of the general point : The coordinates of any point lying on the line joining

points P(x1, y1, z1) and Q(x2, y2, z2) may be taken as

++

++

++

1kzkz,

1kyky,

1kxkx 121212

which divides PQ in the ratio k : 1. This is called general point on the line PQ.

Division by coordinate planes : The ratios in which the line segment PQ joining

P(x1, y1, z1) and Q(x2, y2, z2) is divided by coordinate planes are as follows : (i) by yz-plane : –x1/x2 ratio (ii) by zx-plane : – y1/y2 ratio (iii) by xy-plane : –z1/z2 ratio Coordinates of the centroid : (i) If (x1, y1, z1); (x2, y2, z2) and (x3, y3, z3) are

vertices of a triangle then coordinates of its centroid are

++++++3

zzz,

3yyy

,3

xxx 321321321

(ii) If (xr, yr, zr); r = 1, 2, 3, 4 are vertices of a tetrahedron, then coordinates of its centroid are

+++++++++4

zzzz,

4yyyy

,4

xxxx 432143214321

Direction cosines of a line [Dc's] : The cosines of the angles made by a line with

positive direction of coordinate axes are called the direction cosines of that line.

Let α, β, γ be the angles made by a line AB with positive direction of coordinate axes then cos α, cos β, cos γ are the direction cosines of AB which are generally denoted by l, m, n. Hence

l = cos α, m = cos β, n = cos γ

3-DIMENSIONAL GEOMETRY

Mathematics Fundamentals MATHS


x-axis makes 0º, 90º and 90º angles with three coordinate axes, so its direction cosines are cos 0º, cos 90º, cos 90º i.e. 1, 0, 0. Similarly direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively. Hence

dc's of x-axis = 1, 0, 0 dc's of y-axis = 0, 1, 0 dc's of z-axis = 0, 0, 1 Relation between dc's ∴ l2

+ m2 + n2 = 1 Direction ratios of a line [DR's] : Three numbers which are proportional to the

direction cosines of a line are called the direction ratios of that line. If a, b, c are such numbers which are proportional to the direction cosines l, m, n of a line then a, b, c are direction ratios of the line. Hence

⇒ l = ± 222 cba

a

++,

m = ± 222 cba

b

++, n = ±

222 cba

c

++

Direction cosines of a line joining two points : Let ≡ (x1, y1, z1) and Q ≡ (x2, y2, z2); then (i) dr's of PQ : (x2 – x1), (y2 – y1), (z2 – z1)

(ii)dc's of PQ : PQ

zz,PQ

yy,PQ

xx 121212 −−−

i.e., 2

12

122

12

122

12

12

)xx(

zz,)xx(

yy,)xx(

xx

−Σ

−

−Σ

−

−Σ

−

Angle between two lines : Case I. When dc's of the lines are given If l1, m1, and l2,m2 n2 are dc's of given two lines, then

the angle θ between them is given by cos θ = l1l2 + m1m2 + n1n2

sin θ = 21221

21221

21221 )nn()nmnm()mm( llll −+−+−

The value of sin θ can easily be obtained by the following form :

sin θ = 2

22

112

22

112

22

11nn

nmnm

mm

l

l

l

l++

Case II. When dr's of the lines are given If a1, b1, c1 and a2, b2, c2 are dr's of given two lines,

then the angle θ between them is given by

cos θ = 22

22

22

21

21

21

212121

cbacba

ccbbaa

++++

++

sin θ = 22

22

22

21

21

21

21221

cbacba

)baba(

++++

−Σ

Conditions of parallelism and perpendicularity of two lines : Case I. When dc's of two lines AB and CD, say l1,

m1,n1 and l 2, m2, n2 are known AB || CD ⇔ l 1 = l 2, m1 = m2, n1 = n2 AB ⊥ CD ⇔ l 1 l 2 + m1m2 + n1n2 = 0. Case II. When dr's of two lines AB and CD, say : a1,

b1, c1 and a2, b2, c2 are known

AB || CD ⇔ 2

1

2

1

2

1cc

bb

aa

==

AB ⊥ CD ⇔ a1a2 + b1b2 + c1c2 = 0. Area of a triangle : Let A(x1, y1, z1); B(x2, y2, z2) and C(x3, y3, z3) are

vertices of a triangle. Then dr's of AB = x2 – x1, y2 – y1, z2 – z1 = a1, b1, c1 (say)

and AB = 21

21

21 cba ++

dr's of BC = x3 – x2, y3 – y2, z3 – z2 = a2, b2, c2(say)

and BC = 22

22

22 cba ++

Now sin B = 22

21

21221

aa

)cbcb(

ΣΣ

−Σ

= BC.AB

)cbcb( 21221 −Σ

∴ Area of ∆ABC = 21 AB. BC sin B

= 21221 )cbcb(

21

−Σ

Projection of a line segment joining two points on a line : Let PQ be a line segment where P ≡ (x1, y1, z1) and

Q ≡ (x2, y2, z2); and AB be a given line with dc's as l, m, n. If P'Q' be the projection of PQ on AB, then

P'Q' = PQ cos θ where θ is the angle between PQ and AB. On

replacing the value of cos θ in this, we shall get the following value of P'Q'.

P'Q' = l (x2 – x1) + m(y2 – y1) + n (z2 – z1) Projection of PQ on x-axis : a = |x2 – x1| Projection of PQ on y-axis : b = |y2 – y1| Projection of PQ on z-axis : c = |z2 – z1|

Length of line segment PQ = 222 cba ++

* If the given lines are l

α−x = m

y β− = n

z γ− and

´´x

l

α− = ´m

´y β− = ´n

´z γ− , then condition for

intersection is


If the given lines are l

α−x = m

y β− = n

z γ− and

´´x

l

α− = ´m

ý β− = ń

´z γ− , then condition for

intersections is

ń´m

nm´–´´–

l

l

γγβ−βαα = 0

Plane containing the above two lines is

ń´m´

nm–zy–x

l

l

γβ−α = 0

Condition of coplanarity if both the lines are in general form: Let the lines be ax + by + cz + d = 0 = a´x + bý + c´z + d´ and αx + βy + γz + δ = 0 = α´x + βý + γ´z + δ´

These are coplanar if

´´´´

´dc´bádcba

δγβαδγβα

= 0

Reduction of non-symmetrical form to symmetrical form: Let equation of the line in non-symmetrical form be' a1x + b1y + c1z + d1 = 0; a2x + b2y + c2z + d2 = 0. To find the equation of the line in symmetrical form,

we must know (i) its direction ratios (ii) coordinates of any point on it.

Direction ratios : Let l, m, n be the direction ratios of the line. Since the line lies in both the planes, it must be perpendicular to normals of both planes. So

a1l + b1m + c1n = 0; a2l + b2m + c2n = 0 From these equations, proportional values of l, m, n

can be found by cross-multiplication as

1221 cbcb −

l = 1221 acac

m−

= 1221 baba

n−

Point on the line : Note that as l, m, n cannot be zero simultaneously, so at least one must be non-zero. Let a1b2 – a2b1 ≠ 0, then the line cannot be parallel to xy-plane, so it intersect it. Let it intersect xy-plane in (x1,y1, 0). Then

a1x1 + b1y1 + d1 = 0 and a2x1 + b2y1 + d2 = 0 Solving these, we get a point on the line. Then its

equation becomes

1221

1

cbcbxx

−− =

1221

1

acacyy

−− =

1221 baba0z

−−

or 1221

1221

1221

cbcbbabadbdbx

−−−

−=

1221

1221

1221

acacbabaadady

−−−

−=

1221 baba0z

−−

Note : If l ≠ 0, take a point on yz –plane as (0, t1, z1) and if m ≠ 0, take a point on xz-plane as (x1, 0, z1)

Skew lines : The straight lines which are not parallel and non-coplanar i.e. non-intersecting are called skew lines.

If ∆ = ń´m´

nm–zy–x

l

l

γβ−α ≠ 0, the lines are skew.

Shortest distance : Suppose the equation of the lines

are l

α−x = m

y β− = n

z γ−

and ´

´xl

α− = ´m

ý β− = ń

´z γ− . Then

S.D. = 2)n´m´mn(

)m´´m)(ńń´)(()n´m´mn´)((

−Σ

−−β−β+−α−α llll

= ń´m´

nm´–´´–

l

l

γγβ−βαα

Some results for plane and straight line: (i) General equation of a plane : ax + by + cz + d = 0 where a, b, c are dr's of a normal to this plane. (ii) Equation of a straight line :

General form :

=+++=+++

0dzcybxa0dzcybxa

2222

1111

(In fact it is the straight line which is the intersection of two given planes)

Symmetric form : czz

byy

axx 111 −

=−

=−

where (x1, y1, z1) is a point on this line and a, b, c are its dr's

(iii) Angle between two planes : If θ be the angle between planes a1x + b1y c1z + d1 = 0

and a2x + b2y + c2z + d2 = 0, then

cos θ = 22

22

22

21

21

21

212121

cbacba

ccbbaa

++++

++

(In fact angle between two planes is the angle between their normals.)

Further above two planes are

parallel ⇔ 2

1

2

1

2

1

cc

bb

aa

==

perpendicular ⇔ a1a2 + b1b2 + c1c2 = 0


Arithmetic Progression (AP) AP is a progression in which the difference between

any two consecutive terms is constant. This constant difference is called common difference (c.d.) and generally it is denoted by d.

Standard form: Its standard form is a + (a + d) + (a + 2d) +.......... General term : Tn = a + (n – 1) d If Tn = l then it should be noted that

(i) d

a1n)ii(1nad −

+=−−

=ll

Note: cab2APinarec,b,a +=⇔

Sum of n terms of an AP :

)a(2nSn l+=

where l is last term (nth term). Replacing the value of l, it takes the form

]d)1n(a2[2nSn −+=

Arithmetic Mean : (i) If A be the AM between two numbers a and b,

then )ba(21A +=

(ii) The AM of n numbers a1, a2,..............,an

= n1 (a1 + a2 +........+ an)

(iii) n AM's between two numbers If A1, A2,....., An be n AM's between a and b then a A1, A2,....., An, b is an AP of (n + 2) terms. Its common

difference d is given by

Tn+2 = b = a + (n + 1)d ⇒ d = 1nab

+−

so A1 = a + d, A2 = a + 2d,....., An = a + nd. Sum of n AM's between a and b ∴ ΣAn = n(A) Assuming numbers in AP : (i) When number of terms be odd Three terms : a –d, a, a + d

Five terms : a – 2d, a–d, a, a + d, a + 2d ................ ....... ....... ....... ....... (ii) When number of terms be even Four terms: a – 3d, a – d, a + d, a + 3d Six terms : a –5d, a – 3d, a –d, a + d, a+3d, a + 5d ............... ........ ...... ...... ...... ...... Geometrical Progression (GP) : A progression is called a GP if the ratio of its each

term to its previous term is always constant. This constant ratio is called its common ratio and it is generally denoted by r.

Standard form : Its standard form is a + ar + ar2 +......... General term : Tn = arn–1

a, b, c are in GP ⇔ bc

ab

= ⇔ b2 = ac

Sum of n terms of a GP : The sum of n terms of a GP a + ar + ar2

+....... is given by

Sn =

>−−

=−

−

<−−

=−−

1rwhen,1rar

1r)1r(a

1rwhen,r1ra

r1)r1(a

n

n

l

l

when l = Tn. Sum of an infinite GP : (i) When r > 1, then rn → ∞, so Sn → ∞ Thus when

r > 1, the sum S of infinite GP = ∞ (ii) When | r | < 1, then rn → 0, so

S = r1

a−

(iii) When r = 1, then each term is a so S = ∞. Geometric Mean : (i) If G be the GM between a and b then

G = ab (ii) G.M. of n numbers a1, a2 ......, an = (a1a2a3 .....an)1/n (iii) n GM’s between two numbers ⇒ r = (b/a)1/n+1

PROGRESSION & MATHEMATICAL INDUCTION

Mathematics Fundamentals MATHS

XtraEdge for IIT-JEE 51 JUNE 2010

Product of n GM's between a and b Product of GM's = (ab)n/2 = Gn Assuming numbers in GP : (i) When number of terms be odd Three terms : a/r, a, ar Five terms : a/r2, a/r, a, ar, ar2 ............... .. ..... ..... ..... ..... ..... (ii) When number of terms be even Four terms : a/r3, a/r, ar, ar3 Six terms : a/r5, a/r3, a/r, ar, ar3, ar5

Arithmetic-Geometric Progression : If each term of a progression is the product of the

corresponding terms of an AP and a GP, then it is called arithmetic-geometric progression (AGP). For example:

a, (a + d)r, (a + 2d)r2 ....... Tn = [a + (n – 1)d] rn–1

Sn = 2

1n

)r1()r1(dr

r1a

−

−+

−

− –

r1r]d)1n(a[ n

−−+

S∞ = 2)r1(dr

r1a

−+

− | r | < 1

Harmonic Progression : A progression is called a harmonic progression (HP)

if the reciprocals of its terms are in AP.

Standard form : d2a

1da

1a1

++

++ +.............

General term : d)1n(a

1Tn −+=

∴ a, b, c are in HP ⇔ c1

a1

b2

+= ⇔ b = ca

ac2+

Harmonic Mean : (i) If H be a HM between two numbers a and b, then

ba

ab2H+

= or b1

a1

H2

+=

(ii) To find n HM's between a and b we first find n AM's between 1/a and 1/b, then their reciprocals will be the required HM's.

Relations between AM, GM and HM : G2

= AH A > G > H, when a, b > 0. If A and AM and GM respectively between two

positive numbers, then those numbers are

2222 GAA,GAA −−−+ Some Important Results : If number of terms in an AP/GP/HP is odd then

its mid term is the AM/GM/HM between the first and last term.

If number of terms in an AP/GP/HP is even the AM/GM/HM of its two middle terms is equal to the AM/GM/HM between the first and last term.

a, b, c are in AP, GP and HP ⇔ a = b = c a, b, c are in AP and HP ⇒ a, b,c are in GP. a, b, c are in AP

⇔ ab1,

ca1,

bc1 are in AP. ⇔ bc, ca, ab are in HP.

a, b, c are in GP ⇔ a2, b2, c2 are in GP. a, b, c are in GP ⇔ loga, logb, logc are in AP. a, b, c are in GP ⇔ logam logbm, logcm are in HP. a, b, c d are in GP ⇔ a + b, b + c, c + d are in GP. a, b, c are in AP ⇔ αa, αb, αc

are in GP (α ∈ R0) Principle of Mathematical Induction : It states that any statement P(n) is true for all positive

integral values of n if (i) P(1) is true i.e., it is true for n = 1. (ii) P(m) is true ⇒ P(m + 1) is also true i.e., if the statement is true for n = m then it must also

be true for n = m + 1. Some Formula based on the Principle of Induction :

Σn = 1 + 2 + 3 +....... + n = 2

)1n(n +

(Sum of first n natural numbers) Σ(2n – 1) = 1 + 3 + 5 + ... + (2n – 1) = n2 (Sum of first n odd numbers) Σ2n = 2 + 4 + 6 + ...... + 2n = n(n + 1) (Sum of first n even numbers)

Σn2 = 12 + 22 + 32 +.......+ n2 =

6)1n2()1n(n ++

(Sum of the squares of first n natural numbers)

Σn3 = 13 + 23 + 33 +.......+ n3 =

4)1n(n 22 +

(Sum of the cubes of first n natural numbers) Application in Solving Objective Question : For solving objective question related to natural

numbers we find out the correct alternative by negative examination of this principle. If the given statement is P(n), then by putting n = 1, 2, 3, ..... in P(n), we decide the correct answer.

We also use the above formulae established by this principle to find the sum of n terms of a given series. For this we first express Tn as a polynomial in n and then for finding Sn, we put Σ before each term of this polynomial and then use above results of Σn, Σn2, Σn3 etc.

XtraEdge for IIT-JEE JULY 2010 52

PHYSICS

Questions 1 to 6 are multiple choice questions. Each question has four choices (A), (B), (C) and (D), out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 5 marks will be given for each correct answer and – 2 mark for each wrong answer.

1. The figure shows two isotherms at temperatures T1 and T2. A gas is taken from one isotherm to another isotherm through different processes. Then change in internal energy ∆U has relation –

b

c d

e a

P

T1

T2

V (A) ∆Uab > ∆Uac > ∆Uad > ∆Uac (B) ∆Uab = ∆Uac > ∆Uad > ∆Uac (C) ∆Uab = ∆Uac = ∆Uad = ∆Uac

(D) ∆Uab < ∆Uac < ∆Uad < ∆Uac

2. An ideal gas whose adiabatic exponent is γ is expanded so that the amount of heat transferred to the gas is equal to the decrease of its internal energy. Molar heat capacity of the gas for this process is -

(A) γ–1

R (B) 1–

Rγ

(C) R

(D) R/2

3.

A B C x

T

(a)

A B C x

T

(b)

A B C x

T

(c)

The above graphs shows conduction of heat through materials A,B,C connected in series. Graph shows variations of temperature with distnace x-axis. Which of the above graph are not possible -

(A) a, b, c (B) a, b (C) a, c

(D) b, c

IIT-JEE 2011

XtraEdge Test Series # 3

Based on New Pattern

Time : 3 Hours Syllabus : Physics : Calorimetry, K.T.G., Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave, Sound wave, Doppler's effect, Atomic Structure, Radioactivity, X-ray, Nuclear Physics, Matter Waves, Photoelectric Effect, Practical Physics. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements, Nitrogen Family, Oxygen Family, Halogen Family & Noble Gas, Salt Analysis, Metallurgy, Co-ordination Compounds, Transitional Elements. Mathematics: Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D, Probability, Determinants, Matrices. Instructions : Section - I • Question 1 to 6 are multiple choice questions with only one correct answer. +5 marks will be awarded for correct answer and

-2 mark for wrong answer. Section - II • Question 7 to 12 are passage based single correct type questions. +3 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - III • Question 13 to 14 are Column Match type questions 8 marks will be awarded for correct answer and 0 mark for wrong answer. Section - IV • Question 15 to 19 are numerical response questions (with single digit Answer). 3 marks will be awarded for correct answer

and 0 mark for wrong answer.


4. Figure shows a rectangular pulse and a triangular pulse approaching each other along x-axis. The pulse speed is 0.5 cm/s. What is the resultant displacement of medium particles due to superposition of waves x = 0.5 cm and t = 2 sec. 0.5 cm/s 0.5 cm/s

x (cm)3 210–1–2

1

2

y (cm)

(A) 3.5 cm (B) 2.5 cm (C) 4 cm

(D) 3 cm

5. Choose the correct statement (s) related to the photocurrent and the potential difference between the plate and the collector -

(A) Photocurrent always increase with the increase in potential difference

(B) when the potential difference is zero, the photocurrent is also zero

(C) Photocurrent attain a saturation value of some positive value of the potential difference

(D) None of these

6. Binding energy per nucleon of 1H2 and 2He4 are 1.1 MeV and 7.0 MeV respectively. Energy released in the process 1H2 + 1H2 = 2He4 is -

(A) 20.8 MeV (B) 16.6 MeV (C) 25.2 MeV (D) 23.6 MeV

This section contains 2 paragraphs, each has 3 multiple choice questions. (Questions 7 to 12) Each question has 4 choices (A), (B), (C) and (D) out of which ONLY ONE is correct. Mark your response in OMR sheet against the question number of that question. + 3 marks will be given for each correct answer and – 1 mark for each wrong answer.

Passage : I (Ques. 7 to 9)

Two parallel plates in vacuum, separated by a small distance which is small compared with their linear dimensions, are at temperatures T1 & T2 respectively (T1 > T2). The plates are black bodies. Another plate (black body) at temperature T0 is a kept in between the two plates.

7. Temperature of the plate kept i.e. T0 is -

(A) 2

TT 42

41 + = T0

4 (B) 2

T–T 42

41 = T0

4

(C) 21TT = T0 (D) 2T14 – T2

4 = T04

8. Energy absorbed by the plate having temperature T0, per sec per unit area is -

(A) σ(T14 –T2

4) (B) 2

)T–T( 40

41σ

(C) 2

)T–T( 42

41σ

(D) 2

)T–T( 42

40σ

9. If 'n' black Body plates are placed in between the two plates having temperature T1 & T2, then the net rate of emission of radiant energy by first plate is -

(A) nσ (T1

4 – T24) (B)

1n +σ (T1

4 – T24)

(C) 1n

n+

σ (T14 – T2

4)

(D) 1–n

σ (T14 – T2

4)

Passage: II (Ques. 10 to 12)

Two hydrogen like atoms A and equal number of protons and neutrons. The energy difference between the radiation corresponding to first Balmer lines emitted A and B is 5.667 eV. When the atoms A and B moving with the dame velocity, strikes a heavy target they rebound back with the same velocity. In this process the atom B imparts twice the momentum to the target than the A imparts.

10. Ionization energy of Atom B is - (A) 27.2 eV (B) 13.6 eV (C) 10.2 eV (D) 54.4 eV 11. Atomic number of atom A is - (A) 1 (B) 2 (C) 3

(D) 4 12. Mass number of atom B & atom A (A) 2, 4 (B) 4, 2 (C) 2, 1

(D) 4, 1

This section contains 2 questions (Questions 13, 14). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer.


13. Match the Column-I with Column-II : Column-I Column-II (A) An electron moves (P) Total Energy

in an orbit in a = 2

EnergyPotential

Bohr atom (B) As a satellite moves (Q) Kinetic Energy = in a circular orbit Magnitude of Energy around total earth (C) In Rutherford's (R) Motion is under a α-scattering experiment central force as an α-particle moves in the electric field of a nucleus

(D) As an object, released (S) Mechanical energy from some height above is conserved ground, falls towards earth, assuming negligible air resistance (T) None of these 14. Match the following : Column-I` Column-II (A) Steady state (P) A blackened platinum wire, when gradually heated appear first red and then blue. (B) Wein's displacement (Q) Radiated power is law proportional to fourth power of absolute temperature of body (C) Stefan's law (R) Energy absorbed is equal to energy emitted (D) Black body (S) Absorptive power of body is unity (T) None of these

This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 OMR has 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

012

3

4

56

7

8

9

0 1 2

3

4

5 6

7

8

9

X Y Z W

15. Four cylindrical rods of same material with length and radius (l,r), (2l,r), (2l,2r) and (l,2r) are connected between two reservoirs at 0ºC and 100ºC. Find the ratio of the maximum to minimum rate of conduction in them.

16. Using solar constant S = 20 kilo cal/mm –m2 and Joule's constant J = 4.2 J/cal, find the pressure exerted by sunlight ? (Ans. in …… × 10–6 N/m2)

17. The minimum intensity of audible sound is 10–12

W/m2 sec and density of air is 1.3 kg/m3. If the frequency of sound is 1000 Hz, find the amplitude (Ans. in …… × 10–11 m) of vibration ? [Speed of sound = 332 ms–1]

18. The size of a nucleus is of the order of –14 m. Calculate the velocity with which protons move inside the nucleus. The mass of a proton = 1.675 × 10–27 kg. [Ans. in …… × 107 ms–1]

19. Find the change in frequency of red light whose original frequency is 7.3 × 1014 Hz when it falls through 22.5 m losing gravitational potential energy.

[Ans. in …… ×103]

CHEMISTRY


1. The optically active tartaric acid is named as D–(+)– tartaric acid because it has a positive -

(A) optical rotation and is derived from D-glucose (B) pH in organic solvent (C) optical rotation and is derived from D–(+)–

glyceraldehyde (D) optical rotation only when substituted by

deuterium

2. Which of the following compounds is not coloured ? (A) Na2[CuCl4] (B) Na2[CdCl2] (C) K4[Fe(CN)6] (D) K3[Fe(CN)6]

3. The brown ring complex compound is formulated as [Fe(H2O)5(NO)]SO4. The oxidation state of iron is -

(A) 1 (B) 0 (C) 2

(D) 3

4. The following equilibrium is established when HCl is dissolved the acetic acid,

HCl + CH3COOH Cl– + CH3COOH2+

the set that characterises the conjugate acid-base pairs is- (A) (HCl, CH3COOH) and (Cl–, CH3COOH2

+) (B) (HCl, CH3COOH2

+) and (CH3COOH, Cl–) (C) (CH3COOH2

+) HCl) and (Cl–, CH3COOH)

(D) (HCl, Cl–) and (CH3COOH2+, CH3COOH)


5. Pure ammonia is placed in a vessel at a temperature where its dissociation constant(α) is appreciable. At equilibrium -

(A) kp does not change significantly with pressure (B) does not change with pressure (C) concentration of NH3 does not change with

pressure (D) concentration of hydrogen is less than that of

nitrogen

6. At constant temperature, the equilibrium constant (kp) for the decomposition reaction N2O4 2NO2 is expressed by kp = (4x2P)/(1–x2), where P = pressure, x = extent of decomposition. Which one of the following statement is true ?

(A) kp increases with increase of P (B) kp increases with increase of x (C) kp increases with decrease of x (D) kp remains constant with change in p and x



The IUPAC has set guidelines for logical and methodical naming of organic compounds. The complex substituents are written in small brackets and their numbering is done separately. The bivalent radicals are named by adding 'idene' to the name of alkyl group. In polyfunctional compounds all lower priority groups are written in prefix. Now name the following compounds.

7. CHCHCH2OH

(CH3)2CHOOC Br

is -

(A) 3-(3'-isopropoxycarbonyl cyclopentylidene propane-1-ol

(B)3-(2'-bromo-3'-hydroxypropylidene) cyclopentane carboxylate

(C) Iso-propyl-3-(2'-bromo-3'-hydroxy propylidenyl) cyclopentane carboxylate

(D) Iso-propyl-3-(2'-bromo-3' hydroxypropylidene) cyclopentane carboxylate

8.

CH3CH2O CH3CH2

C2H5

is -

(A) 2-(3'-Ethylphenyl)-1-(4'-ethoxyphenyl) ethane (B) 1-Ethyl-3-(2'-(4''-ethoxyphenyl) ethyl) benzene (C) 1-(3'-Ethylphenyl)-2-(4'-ethoxylphenyl) ethane (D) None of these

9. Cl

O OHN

O

is -

(A) 3-chlorocarbonyl-6-(N, N-diethylamino) hex-4-ene-1-oic acid

(B)4-chlorocarbonyl-3-(N, N-diethylamino) butanoic acid

(C) 3-chlorocarbonyl-3-(3-N, N-diethylamino prop-1'-enyl) butane-1-oic acid

(D) 3-chlorocarbonylmethyl-6- (N, N-diethylamino) hex-4-en-1-oic acid


The property of hydrides of p-block elements mostly depend on

(i) Electronegatively difference between central atom and hydrogen

(ii) Size of central atom (iii) Number of valence electrons in central atom. some

undergo hydride in which central atom is less electronegative, react with OH– to given hydrogen while acidic property of hydride in a period depends on electronegativity of central atom i.e. more electronegative is the atom, more acidic is hydride. In a group, acidic property is proportional to size of central atom.

Some electron deficient hydride behaves as lewis acid while only one hydride of an element in p-plock behaves as lewis base with lone pair of electrons. Hydrides in which central atom's electronegativity to close to hydrogen has no reaction with water.

10. The hydride which do not react with water is - (A) NH3 (B) PH3 (C) B2H6

(D) AsH3

11. Which one undergoes spontaneous combustion with exposure to air ?

(A) PH3 (B) P2H4 (C) N2H4 (D) NH3

12. Which one is strongest base ? (A) OH– (B) HS– (C) HSe–

(D) HTe–


ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T



13. Match the extraction processes listed in Column-I with metals listed in Column-II :

Column-I Column-II (A) Self reduction (P) Lead (B) Carbon reduction (Q) Silver (C) Complex formation (R) Copper and displacement by metal (D) Decomposition of (S) Boron iodide (T) Au

14. Match statements in Column-I with appropriate solution in Column-II

Column-I Column-II (A) A solution having (P) 5M HCl pH less then 7 (B) A solution having (Q) 1M NaCl pH more than 7 (C) A solution having (R) 0.1 M Na2CO3 pH almost equal to 7

(D) solution having (S) 0.1 M CaCl2 negative value of pH (T) 1M H2SO4 This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 OMR has 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

15. A solution is prepared by mixing 50 mL 0.1 M HCl with 50 mL 2.9 M CH3CH2COOH and 100 mL 0.2 M CH3CH2COONa. Find pH of the resulting solution. Ka for CH3CH2COOH is 1 × 10–5.

16. Calculate the change in pressure (in atm) when 2 mole of NO and 16 gram O2 in a 6.25 litre originally at 27ºC react to produce the maximum quantity of NO2 possible according to the equation –

2NO(g) + O2(g) → 2NO2(g)

(Take R = 121 ltr. Atm/mol-K)

17. the number of isomers for the compound with molecular formula C2BrCIFI is.

18. In P4O10 each P atom is linked with ...........O atoms.

19. 0.15 mole of Pyridinium chloride has been added

into 500 cm3 of 0.2 M pyridine solution. Calculate pH (Kb for pyridine = 1.5 × 10–9 M)

MATHEMATICS


1. If A =

γγγβββ

αα

––2

0 is an orthogonal matrix, then

the number of possible triplets (α,β,γ) is - (A) 8 (B) 6 (C) 4

(D) 2

2. If ∑=

αn

1nn = an2 + + bn, where a, b are constants and

α1, α2, α3 ∈ 1, 2, 3........9 and 25α1, 37α2, 49α3 be

three digit number then

ααα

ααα

321

321

493725975 is

equal to - (A) α1 + α2 + α3 (B) α1 – α2 + α3 (C) 7

(D) 0

3. Consider the matrix A, B, C, D with order 2 × 3, 3 × 4, 4 × 4, 4 × 2 respectively. Let x = (αABγC2D)3 where α & γ are scalars. Let |x| = k|ABC2D|3, then k is -

(A) αγ (B)α2γ2 (C) α4γ4

(D) α6γ6

4. Three numbers are selected at random from the set 1, 2, 3...........N, one by one without replacement. If the first number is known to be smaller than second, then the probability that third selected number lies between the first two numbers is -


(A) 21 (B)

31 (C)

61

(D) 81

5. A bag 'A' contains 2 white and 3 red balls, another

bag 'B' contains 4 white and 5 red balls. If one ball is drawn at random from one of the bag and it is found to be red, the probability that it was drawn from the bag B is -

(A) 24589 (B)

5225 (C)

25693

(D) 66324

6. Vectors →a ,

→b and

→c with magnitude 2, 3 & 4

respectively are coplanar. A unit vector →d is

perpendicular to all of them. If (→a ×

→b ) × (

→c ×

→d ) =

6i^ –

3j^ +

3k^ and the angle between

→a and

→b is 30º,

then î.c→

+ ^j.c→

+ ^k.c→

is equal to -

(A) 35 (B)

95 (C)

125

(D) 185



There are four boxes A1, A2, A3 and A4. Box Ai has i cards and on each card a number is printed, the numbers are from 1 to i, A box is selected randomly,

the probability of selection of box Ai is 101 and then

a card is drawn. Let Ei represents the event that a card with number 'i' is drawn.

7. P(E1) is equal to -

(A) 51 (B)

101 (C)

52

(D) 41

8. P(A3/E2) is equal to -

(A) 41 (B)

31 (C)

21

(D) 32

9. Expectation of the number on the card is - (A) 2 (B) 2.5 (C) 3

(D) 3.5


consider the equation of two straight lines

→r =

î3 +

^j5 +

^k7 + λ(

î – ^j2 + ^k ) and

71x + =

6–1y + =

11z + .

10. The equation of the line of shortest distance between the given lines is -

(A) →r = (

î3 +

^j5 +

^k7 ) – δ(

î + ^j + ^k )

(B) →r = (

î + ^j + ^k ) + δ(3

î + 5 ^j + 7 ^k )

(C) →r = (

î3 +

^j5 +

^k7 ) – δ(4

î + 6 ^j + 8 ^k )

(D) →r = (

î4 +

^j6 +

^k8 ) – δ(

î + ^j + ^k )

11. The length of shortest distance between the lines is - (A) 29 (B) 2 29

(C) 3 29

(D) 4 29

12. The point of intersection of the lines is -

(A)

34–,

353,

310– (B)

34,

353–,

310

(C) (–10, 53, –4) (D) None of these


ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T


13. Match the following : Column-I Column-II (A) Through (γ, γ + 1) there (P) 2 < γ < 8 can not be more than one normal to parabola y2 = 4x if (B) If two cicles (x–1)2 + (y–1)2 = γ2 (Q) – 2 < γ < 2 and x2 + y2 – 8x + 2y + 8 = 0 intersect at two points, then


(C) Point (γ, γ + 1) lies inside the (R) γ < – 2 circle x2 + y2 = 1 for

(D) Both equation x2 + y2 + 2γx + 4 = 0 (S) – 1 < γ < 0 and x2 + y2 – 4γy + 8 = 0 represent real circles if (T) γ > 8

14. Consider the matrix A =

1–14–3

; B =

1032/1

.

Let P be an orthogonal matrix and Q = PAPT, Rk = PTQk.P, S = PBPT & Tk = PTSkP where k ∈ N

Column-I Column-II

(A) ∑=

5

1kka , where ak represents the element (P) –9

of first row & first column in matrix Rk

(B) ∑=

3

1kkb , where bk represents the element (Q) 10

of second row & second column in matrix Rk

(C) ∑∞

=1kkX , where Xk represents the element (R) 35

of first row & first column in matrix Tk

(D) ∑=

10

1kky , where yk represents the element (S) 1

of second row and second column in Matrix TK (T) 15 This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 OMR has 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

15. There are N + 1 identical boxes each containing N wall clocks. The first box contains zero defective clocks. The second box contains one defective and (N – 1) effective clocks, in general rth box contains (r – 1) defective and (N – r + 1) effective clocks (1 ≤ r ≤ N + 1). Thus, the (N + 1)th box contains all defective clocks. A wall clock is selected and found an effective one. The probability that it is from kth

box is NN

K–N2 +

γ+γγ find γ.

16. If a determinant of order 3 × 3 is formed by using the numbers 1 or – 1then it minimum value of determinant is – γ find the value of γ.

17. If equation of the plane through the straight line

21–x =

3–2y + =

5z and perpendicular to the plane

x – y + z + 2 = 0 ia ax – by + cz + 4 = 0, then find the

value of 342

c10b10a10 23 ++ .

18. If A =

23–41–12

32–3. Solve the system of equations

204012303

zyx

=

418

+

y3zy2

, then find the value

of x + 2y +

3z

19. Find the coefficient of x in the determinant

332313

322212

312111

bababa

bababa

bababa

)x1()x1()x1()x1()x1()x1()x1()x1()x1(

+++++++++

where ai, bj ∈N

How to Handle Difficult People

A bully at your work is difficult for you to face. He is demanding you do part of his job without pay or credit. How do you handle it?

Your neighbors are constantly fighting. They wake you up in the middle of the night with their screams and curses. What do you say to them?

Your father is unhappy about your career choice. He constantly criticizes your work and points out what he thinks you should do. How do you deal with him?


PHYSICS


1. The molar heat capacity for a process is :

C = γ–1

R + Tα , then process equation is -

(A) Ve–(α/RT)T = constant (B) Ve (α/R)T = constant (C) VT = constant (D) Veα/RT = constant 2. Three black bodies 1,2,3 have radius r1 < r2 < r3.

Emissive powers of black bodies at Temperature T are E1, E2, E3, Then correct relation between them is -

(A) E1 < E2 < E3 (B) E1 > E2 > E3 (C) E1 = E2 = E3 (D) E1 > E2 < E3

3. A taut string for which µ = 5 × 10–2 kg/m is under tension of 80 N. How much is the average rate of transport of potential energy if the frequency is 60 Hz and amplitude 6 cm - (Given 4π2 = 39.5)

(A) 52 W (B) 256 W (C) 512 W

(D) 215 W

4. In case of standing waves - (A) At nodes particles displacement is time

dependent (B) At antinodes displacement of particle may or

may not be zero (C) Wave does not travel but energy is transmitted (D) Components waves traveling in same direction

having same amplitude and same frequency are superimposed

5. For an ideal gas graph is shown for three processes. Processes 1, 2 and 3 are respectively –

1Temperature change

2

3

Work done (magnitude)

(A) Isochoric, isobaric, adiabatic (B) Isochoric, adiabatic, isobaric (C) isobaric, adiabatic, isochoric (D) Adiabatic, isobaric, isochoric

IIT-JEE 2012

XtraEdge Test Series # 3

Time : 3 Hours Syllabus : Physics : Calorimetry, K.T.G.,Thermodynamics, Heat Transfer, Thermal expansion, Transverse wave, Sound wave, Doppler's effect. Chemistry : Chemical Equilibrium, Acid Base, Ionic Equilibrium, Classification & Nomenclature, Isomerism , Hydrogen Family, Boron Family & Carbon Family, S-block elements. Mathematics: Point, Straight line, Circle, Parabola, Ellipse, Hyperbola, Vector, 3-D

Instructions : Section - I • Question 1 to 6 are multiple choice questions with only one correct answer. +5 marks will be awarded for correct answer

and -2 mark for wrong answer. Section - II • Question 7 to 12 are passage based single correct type questions. +3 marks will be awarded for correct answer and

-1 mark for wrong answer. Section - III • Question 13 to 14 are Column Match type questions 8 marks will be awarded for correct answer and 0 mark for wrong

answer. Section - IV • Question 15 to 19 are numerical response questions (with single digit Answer). 3 marks will be awarded for correct answer

and 0 mark for wrong answer.

Based on New Pattern


6. Figure shows cyclic process. From c to b, 40 J is transferred as heat from b to a, 130 J is transferred as heat, and work done is 80 J from a to c, 400 J is transferred as heat then –

P

V a b

c

(A) Work done in process a to c is 310 J (B) Net work done is cycle is 230 (C) Net change in internal energy in cycle is 130 J (D) None of these



Many waveforms are described in terms of combinations of travelling waves. Superposition principle is used to analyse such wave combinations. Two pulses travelling on same string are described by-

y1 = 2)t4–x3(

52 +

, y2 = 2)6–t4x3(

5–2 ++

7. The direction in which each pulse is travelling is - (A) y1 is in positive x-axis, y2 is in positive x-axis (B) y1 is in negative x-axis, y2 is in negative x-axis (C) y1 is in positive x-axis, y2 is in negative x-axis (D) y1 is in negative x-axis, y2 is in positive x-axis

8. The time when the two waves cancel everywhere - (A) 1 sec (B) 0.5 sec (C) 0.25 sec (D) 0.75 sec

9. The point where two waves always cancel- (A) 0.25 m (B) 0.5 m (C) 0.75 m

(D) 1 m


One mole of monoatomic gas is taken through above cyclic process. TA = 300 K

Process AB is defined as PT = constant P

C B 3P0

P0

T 10. Work done in process AB is - (A) 400 R (B) – 400 R (C) 200 R

(D) – 300 R

11. Change in internal energy in process CA (A) 900 R (B) 300 R (C) 1200 R (D) zero 12. Heat transferred in the process BC is - (A) 1000 R (B) 500 R (C) 2000 R (D) 1500 R This section contains 2 questions (Questions 13, 14). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer. 13. Match the standing waves formed in column-II due to

plane progressive waves in Column-I and also with conditions in Column-I

Column-I Column-II (A) Incident wave is (P) y = 2A cos kx sin ωt y = A sin (kx – ωt) (B) Incident wave is (Q) y = 2A sin kx cos ωt y = A cos (kx – ωt) (C) x = 0 is rigid support (R) y = 2A sin kx cos ωt (D) x = 0 is flexible support (S) y = 2A cos kx cos ωt (T) None of these 14. Column-I Column-II (A) Specific heat capacity S (P) l1 – l2 = constant for l1α1 = l2α2 (B) Two metals (l1, α1) and (Q) Y is same (l2, α2) are heated uniformly (C) Thermal stress (R) S = ∞ for ∆T = 0 (D) Four wires of same (S) Y α ∆t material (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

15. If the volume of a block of metal changes by 0.12 % when heat is changed from 40ºC to 60ºC, find the linear expansion coefficient of the metal ?

[Ans. in …… × 10–5/ºK] 16. Calculate the pressure exerted by a mixture of 8 g of

oxygen, 14 g of nitrogen and 22 g of carbon di-oxide in a container of 30 litres at a temperature of 27ºC.

[Ans. in …… × 105 N/m2]

17. A sphere and a cube of same material and total

surface area placed in an evacuated chamber turn by turn and heated to the same temperature. Calculate the ratio of the rate of cooling of spherical to cubical surface. [Ans. in …… × 10–1]

18. Two oscillating waves have a phase difference of 2π

is 25 oscillations. What is the percentage difference in their frequency ?

19. For a certain organ pipe, three successive resonance

observed are 425, 595 and 765 Hz. Taking the speed of sound to be 340 ms–1 , find the length of the pipe, in metre.

CHEMISTRY


1. The IUPAC name of the given compound

CH3–CH–CH–CH–CH–COOH | |

| |

OHC CONH2

Br COCl

is -

(A) 2-Bromo-4-carbamoyl-5-chloroformyl-3-formyl hexanoic acid (B) 5-Bromo-3-carbamoyl-2-chloroformyl-4-formyl hexanoic (C) 4-Formyl-2-chloroformyl-3-carbamoyl-5-bromo hexanoic acid (D) 2-Chloroformyl-3-carbamoyl-4-formyl-5-bromo

hexanoic acid

2. Geometry in the given compound is CH3

CH3

H

H

(A) cis (B) trans (C) cis as well as trans (D) no geometrical isomerism

3. The structure of spiro [3,3] heptane is -

(A) (B)

(C)

(D)

4. The pH of a 10–8 molar solution of HCl in water is - (A) 8 (B) – 8 (C) between 7 and 8 (D) between 6 and 7

5. Consider the following equilibrium in a closed container N2O4(g) 2NO2(g). At a fixed temperature, the volume of the reaction container is halved. For this change, which of the following statements holds true regarding the equilibrium constant (kp) and degree of dissociation (α) ?

(A) neither kp nor α changes (B) both kp and α change (C) kp changes, but α does not change

(D) kp does not change, but α changes

6. For H3PO3 and H3PO4 the correct choice is - (A) H3PO3 is dibasic and reducing (B) H3PO3 is dibasic and non-reducing (C) H3PO4 is tribasic and reducing

(D) H3PO3 is tribasic and non-reducing




In a reversible chemical reaction, the rate of forward reaction decreases and that of backward reaction increases with the passage of time; at equilibrium the rate of forward and backward reaction become same. Let us consider the formation of SO3(g) in the following reversible reaction :

2SO2(g) + O2(g) 2SO3 (g) Following graphs are plotted for this reactions

Conc.

time t1 t2 t3

A. B.C.

t4

7. In the above graph, A,B & C respectively are - (A) SO3, SO2 and O2 (B) SO3, O2 and SO2 (C) SO2, O2 and SO3 (D) O2, SO2 and SO3 8. In the above graph, the equilibrium state is attained at

time - (A) t1 (B) t2 (C) t3

(D) t4 9. Which of the following represent the rates of forward

reaction (rf) and rates of backward reaction (rb) at equilibrium ?

(A)

rate

of

reac

tion

time

rb rf

(B)

rate

of

reac

tion

time

rbrf

(C)

rate

of

reac

tion

time

rb

rf

(D)

rate

of

reac

tion

time

rbrf


Different spatial arrangements of the atoms that result from restricted rotation about a single bond are conformers. n-Butane has four conformers eclipsed, fully eclipsed, gauche and anti. The stability order of these conformers are as follows: anti > gauche > partial eclipsed > fully eclipsed

Although anti is more stable than gauche but in some cases gauche is more stable than anti.

10. Which one of the following is most stable conformer?

(A)

Cl

ClCH3

CH3 H

H (B)

Cl

H CH3

Cl CH3

H

(C)

Cl

CH3

CH3

H Cl

H

(D)

Cl

CH3

CH3H

Cl

H

11. Which one of the following is the most stable

conformer ?

(A)

CH3

CH3

OH

HO H

H (B)

CH3

OH OH

H CH3

H

(C)

CH3

CH3

OH

H OH

H (D)

OH

OH CH3

H CH3

H

12. Number of possible conformers of n-butane is - (A) 2 (B) 4 (C) 6

(D) infinite This section contains 2 questions (Questions 13, 14). Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column I have to be matched with statements (P, Q, R, S, T) in Column II. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-P, A-S, A-T; B-Q, B-R; C-P, C-Q and D-S, D-T then the correctly bubbled 4 × 5 matrix should be as follows :

ABCD

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T



13. Match the following : Column-I (A) N2(g) + 3H2(g) 2NH3(g) ; ∆H = –ve (B) N2(g) + O2(g) 2NO(g); ∆H = +ve

(C) A(g) + B(g) 2C(g) + D(g); ∆H = +ve (D) PCl5(g) PCl3(g) + Cl2(g); ∆H = +ve Column-II (P) K increases with increase in temperature (Q) K decreases with increase in temperature (R) Pressure has no effect (S) Product concentration, increases due to addition

of inert gas at constant pressure (T) Product concentration, increases due to addition

of inert gas at constant volume

14. Match the following : Column-I Column-II (A) Bi3+ → (BiO)+ (P) Heat (B) [AlO2]– → Al(OH)3 (Q) Hydrolysis (C) [SiO4]4– → [Si2O7]6– (R) Acidification (D) [B4O7]2– → [B(OH)3] (S) Dilution by water (T) Basification This section contains 5 questions (Q.15 to 19). +3 marks will be given for each correct answer and no negative marking. 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 OMR has 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

15. Calculate the pH at which the following conversion (reaction) will be at equilibrium in basic medium

I2 (s) I–(aq.) + IO3– (aq.)

When the equilibrium concentrations at 300 K are [I–] = 0.10 M and [IO3

–] = 0.10 M. Given → ∆Gfº (I–, aq.) = – 50 kJ./mol, ∆Gfº (IO3

–, aq) = – 123.5 KJ/mol, ∆Gfº (H2O, l) = –233 KJ/mol ∆Gfº (OH–, aq.) = – 150 KJ/mol

R(Gas constant) = 325 J/mol–K

Log e = 2.3

16. Number of configurational isomers of 2,3-dibromocinnamic acid is .

17. Consider the reaction AB2(g) ABg + B(g). It the initial pressure of AB2 is 100 torr and equilibrium pressure is 120 torr. The equilibrium constant Kp in terms of torr is.

18. Dissociation of H3PO3 occurs in ......... stages.

19. The number of hydroxyl groups in pyrophosphoric

acid is.

MATHEMATICS


1. A circle C1 of radius b touches the circle x2 + y2 = a2

externally and has its centre on the positive x-axis; another circle C2 of radius c touches the circle C1 externally and has its centre on the positive x-axis. Given a < b < c, then the three circles have a common tangent if a, b, c are in -

(A) A.P. (B) G.P. (C) H.P.

(D) None of these

2. P is a point on the axis of the parabola y2 = 4ax; Q and R are the extremities of its latus rectum, A is its vertex. If PQR is an equilateral triangle lying within the parabola and ∠AQP = θ, then cos θ =

(A) 523–2 (B)

589

(C) 32

2–5 (D) None of these

3. The length of the diameter of the ellipse 25x2

+ 9y2

= 1,

perpendicular to the asymptote of the hyperbola

16x2

– 9y2

= 1 passing through the first and third

quadrants is :

(A) 431

100 (B) 481

150

(C) 3

25

(D) 211


4. If →a ,

→b ,

→c are such that [

→a ,

→b ,

→c ] = 1,

→c = γ(

→a ×

→b ),

→a ^

→b <

32π , and |

→a | = 2 , |

→b |

= 3 , |→c | =

31 , then the angle between

→a and

→b

is -

(A) 6π (B)

4π (C)

3π

(D) 2π

5. Equation of a plane which passes through the point of

intersection of lines 3

1–x = 1

2–y = 2

3–z and

13–x =

21–y =

32–z and at greatest distance

from the point (0, 0, 0) is - (A) 4x + 3y + 5z = 25 (B) 4x + 3y + 5z = 50 (C) 3x + 4y + 5z = 49 (D) x + 7y – 5z = 2 6. p1, p2 are lengths of perpendicular from foci on

tangent to ellipse and p3, p4 are perpendiculars from extremities of major axis and p from centre of ellipse

on same tangent, then 243

221

p–ppp–pp equals -

(A) e (B) e (C) e2

(D) None of these



Let C : x2 + y2 = 9, E : 9

x2 +

4y2

= 1, L : y = 2x

7. P is a point on the circle C, the perpendicular PQ to the major axis of the ellipse E meets the ellipse at M,

then PQMQ is equal to -

(A) 1/3 (B) 2/3 (C) 1/2

(D) none of these

8. If L represents the line joining the point P on C to its centre O, then equation of the tangent at M to the ellipse E is -

(A) x + 3y = 53 (B) 4x + 3y = 5

(C) x + 3y + 5 = 0 (D) 4x +3y + 5 = 0

9. Equation of the diameter of the ellipse E conjugate to the diameter represented by L is -

(A) 9x + 2y = 0 (B) 2x + 9y = 0 (C) 4x + 9y = 0

(D) 4x – 9y = 0 Passage: II (Ques. 10 to 12)

In a parallelogram OABC vector →a ,

→b ,

→c are

respectively the position vectors of vertices A, B, C with reference to O as origin. A point E is taken on the side BC which divides it in the ratio of 2 : 1. Also the line segment AE intersects the line besecting the angle O internally in point P. If CP when extended meets AB in point F, then

10. The position vector of point P, is -

(A)

+

+→

→

→

→

→→

→→

|c|

c

|a|

a

|a|2|c|3

|c||a|3

(B)

+

+→

→

→

→

→→

→→

|c|

c

|a|

a

|a|2|c|3

|c||a|

(C)

+

+→

→

→

→

→→

→→

|c|

c

|a|

a

|a|2|c|3

|c||a|2

(D) none of these

11. The position vector of point F is -

(A) →

→

→→

+ c|c|

|a|31a (B)

→

→

→→

+ c|c|

|a|a

(C) →

→

→→

+ c|c|

|a|2a (D) →

→

→→

c|c|

|a|–a

12. The vector →

AF , is given by -

(A) →

→

→

c|c|

|a|31 (B)

→

→

→

c|c|

|a|

(C) →

→

→

c|c|

|a|2 (D) →

→

→

c|c|

|a|–



A B C D

P Q R S T

T S

P

P P Q R

R R

Q Q

S S T

T

P Q R S T

Mark your response in OMR sheet against the question number of that question in section-II. + 8 marks will be given for complete correct answer (i.e. +2 marks for each correct row) and NO NEGATIVE MARKING for wrong answer. 13. Column-I Column-II (A) If lines x + 2y – 1 = 0, ax + y + 3 = 0 (P) 4 and bx – y + 2 = 0 are concurrent, the least distance from origin to (a, b) is S. The value of 58 . S is (B) A,B are two fixed points on a line (Q) 5 L. Let locus of point P such that PA = 2PB be a curve cutting line

L at R and S. If slope of PR is – 21 ,

then slope of PS is (C) Let tangents at P and Q to curve (R) 2 y2 – 4x – 2y + 5 = 0 intersect at T. If S(2, 1) is a point such that SP.QS = 16 then the length ST is

(D) Let the double ordinate PNP' of the (S) π – 1

hyperbola )1–(

y–)1–(

x 2

2

2

ππ = 1 is

preduced both side to meet asymptotes in Q and Q'. The product PQ. PQ' is equal to (T) (π–1)2 14. Column-I Column-II (A) The distance of the point (1, – 2, 3) (P) 1 from the plane x – y + z = 5 measured parallel to the line

21 x =

31 y =

61– z is

(B) The shortest distance between (Q) 6

1

the lines 2

1–x = 3

2–y = 4

3–z

and 3

2–x = 4

4–y = 5

5–x is

(C) The points (0, –1, –1), (4, 5, 1), (R) 4 (3, 9, 4) and (–4, 4, k) are coplanar then k =

(D) The radius of the circular section (S) 3

of the sphere |r|→

= 5 by the plane

→r . (

^i +

^j +

^k ) = 33 is

(T) 2


012

3

4

56

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

15. Two circle of radii 'a' and 'b' touching externally are

inscribed in area bounded by y = 2x–1 and

x-axis. If b = 21 and a =

k1 , then k is ...............

16. If a circle S (x, y) = 0 touches at the point (2, 3) of the line x + y = 5 and S (1, 2) = 0, then 2 × radius of such circle is.

17. Consider two concentric circles C1 : x2 + y2 = 1, C2: x2 + y2 = 4. Tangents are drawn to C1 from any point P on C2. These tangents again meet circle C2 at A & B. It can be proved that locus of point of intersection of tangents drawn to C2 at A and B is a circle, what is the radius of that circle.

18. In a regular tetrahedron let θ be the angle between any edge and a face not containing the edge.

If cos2θ = ba where a, b ∈ I+ also a and b are

coprime, then find the value of 135 (10a + b)

19. Let A (1, 2), B (3, 4) be two point and C (x,y) be a point such that (x – 1) (x – 3) + (y – 2) (y – 4) = 0. If area of ∆ABC is 1 sq, unit. Then maximum number of positions of C in xy plane is.


XtraEdge Test Series ANSWER KEY

PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 11 12 Ans C A B D C D A C B D A B

13 A → P, Q, R, S B → P, Q, R, S C → R, S D → R, S Column Match 14 A → R B → P C → Q D → P, S

Ques 15 16 17 18 19 Numerical Response Ans 8 5 1 4 2

CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 11 12 Ans C B C A A D D B D D B A

13 A → P, R B → P, R C → Q, T D → S Column Match 14 A → P, S, T B → R, S C → Q D → P


MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 11 12 Ans A D D C B D C B A C B D

13 A → Q, R, S B → P C → S D → P, R, T Column Match 14 A → R B → P C → S D → Q


PHYSICS

Ques 1 2 3 4 5 6 7 8 9 10 11 12 Ans D C B B A A C D D B A C

13 A → P, R B → Q, S C → Q, R D → P, S Column Match 14 A → R B → P C → S D → Q


CHEMISTRY

Ques 1 2 3 4 5 6 7 8 9 10 11 12 Ans B B D D D A A C A A C D

13 A → Q B → P, R C → P, S D → P, S Column Match 14 A → Q B → S C → P D → R


MATHEMATICS

Ques 1 2 3 4 5 6 7 8 9 10 11 12 Ans B A B B B C B A B A A A

13 A → Q B → R C → P D → S Column Match 14 A → P B → Q C → R D → R


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