LEADER & ENTHUSIAST COURSE

Test Type : ALL INDIA OPEN TEST Test Pattern : JEE-Main

TEST DATE : 26 - 03 - 2017

TARGET : JEE (MAIN) 2017

LEADER & ENTHUSIAST COURSE

Paper Code : 0000CT103116005Form Number :

CLASSROOM CONTACT PROGRAMME(Academic Session : 2016 - 2017)

HIN

DI

Corporate Office : CAREER INSTITUTE, “SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-5156100 [email protected] www.allen.ac.in

Your Target is to secure Good Rank in JEE (Main) 2017

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Test Booklet with Blue/Black Ball Point Pen. Use of pencil

is strictly prohibited.

2. The candidates should not write their Form Number

anywhere else (except in the specified space) on the Test

Booklet/Answer Sheet.

3. The test is of 3 hours duration.

4. The Test Booklet consists of 90 questions. The maximum

marks are 360.

5. There are three parts in the question paper A,B,C

consisting of Physics, Chemistry and Mathematics

having 30 questions in each part of equal weightage.

Each question is allotted 4 (four) marks for correct

response.

6. One Fourth mark will be deducted for indicated incorrect

response of each question. No deduction from the total

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1.

2.

3. 3 4. 90360

5. A, B, C 30 4

6.

7.

8.

9.

10.

11.

ALL INDIA OPEN TEST/LEADER & ENTHUSIAST COURSE/JEE (Main)/26-03-2017

H-1/390000CT103116005

PART A - PHYSICS

BEWARE OF NEGATIVE MARKING

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

1. A solid sphere of radius r is floating at theinterface of two immiscible liquids ofdensities

1 and

2 (

2 >

1), half of its volume

lying in each. The height of the upper liquidcolumn from the interface of the two liquidsis h. The force exerted on the sphere by theupper liquid is (atmospheric pressure = p

0

& acceleration due to gravity is g) :-

Solidsphere

h1

2

2 >

(1) p0r2 + (h – 2/3r)r2

1g

(2) (h – 2/3r)r21g

(3) 2/3 r3 1g

(4) p0 × r2

2. A transformer may be used to providemaximum power transfer between two ACcircuits that have different impedances Z

1

and Z2. The ratio of turns N

1/N

2 needed to

meet this condition is given by :-

(1) 1 1

2 2

N Z

N Z (2) 1 2

2 1

N Z

N Z

(3) 1 2

2 1

N Z

N Z (4) 1 1

2 2

N Z

N Z

1. r 1

2 (

2 >

1)

h (= p

0 g )

Solidsphere

h1

2

2 >

(1) p0r2 + (h – 2/3r)r2

1g

(2) (h – 2/3r)r21g

(3) 2/3 r3 1g

(4) p0 × r2

2. AC Z

1 Z

2

N1/N

2

(1) 1 1

2 2

N Z

N Z (2) 1 2

2 1

N Z

N Z

(3) 1 2

2 1

N Z

N Z (4) 1 1

2 2

N Z

N Z

0000CT103116005H-2/39


3. A satellite of mass m is at a distance of 'a'

from a star of mass M. The speed of the

satellite is u. Suppose the law of universal

gravity is 2.1

MmF G

r instead of

2

MmF G

r , find the speed of the satellite

when it is at a distance b from the star.

(1) 2

1.1 1.1

1 1u 2GM

b a

(2) 2

1.1 1.1

2 1 1u GM

1.1 b a

(3) 2

1.1 1.1

2 1 1u GM

2.1 b a

(4) 2 2 1 1

u GM1.1 b a

3. m M

'a' u

2

MmF G

r

2.1

MmF G

r b

(1) 2

1.1 1.1

1 1u 2GM

b a

(2) 2

1.1 1.1

2 1 1u GM

1.1 b a

(3) 2

1.1 1.1

2 1 1u GM

2.1 b a

(4) 2 2 1 1

u GM1.1 b a


H-3/390000CT103116005

4. As shown, a narrow beam of light is incidentonto a semi-circular glass cylinder of radiusR. Light can exit the cylinder when the beamis at the centre. When the beam is movedparallarly to a distance d from the centralline, no light can exit the cylinder from itslower surface. Find the refractive index ofthe glass.

R

d

(1) R

d(2)

d

R

(3) 2 2

R

R d(4)

2 2R d

R

5. As shown, the circuit is made of 8 differentresistors. It is found that when R

1 = 4,

the resistance between A and B is 2. Nowreplace R

1 by a 6 resistor, what is the

resistance between A and B?

R1

A

B

(1) 1 (2) 2 (3) 6 (4) 2.4

4. R d

R

d

(1) R

d(2)

d

R

(3) 2 2

R

R d(4)

2 2R d

R

5. 8 R

1 = 4A B

2R1 6

A B

R1

A

B

(1) 1 (2) 2 (3) 6 (4) 2.4

0000CT103116005H-4/39


6. Which of the following is the most precise

device for measuring length :-

(1) a vernier callipers with 20 divisions on

the sliding scale

(2) a screw gauge of pitch 1 mm and 100

divisions on the circular scale

(3) an optical instrument that can measure

length to within a wavelength of light

(4) Precison can not be changed by changing

the instrument.

7. The following diagram is an arrangementto view the interference pattern producedby a wedge of liquid between a microscopeslide and a glass block. The interferencepattern obtained is made of equally spacedparallel fringes. The fringe separation(or fringe width) may be decreased by someof the following actions (Consider one changeat a time).

glass block

glass plate45° to the incident light

wedge of liquidmicroscope

slide

uniformmonochromatic

light

travelingmicroscope

(1) by increasing the angle of the liquidwedge.

(2) by using a liquid of smaller refractiveindex.

(3) by using a thicker glass block.(4) by using a longer liquid wedge of the

same angle.

6. (1)20

(2) 1 mm 100

(3)

(4)

7.

glass block

glass plate45° to the incident light

wedge of liquidmicroscope

slide

uniformmonochromatic

light

travelingmicroscope

(1) (2) (3) (4)


H-5/390000CT103116005

8. A parallel plate capacitor with a dielectric slabcompletely occupying the space between theplates is charged by a battery and thendisconnected. The slab is pulled out with aconstant speed. Which of the following curvesrepresent qualitatively the variation of thecapacitance C of the system with time?

(1) C

t

(2) C

t

(3) C

t

(4) C

t

9. A conducting rod with resistance r per unitlength is moving inside a vertical magnetic

field B at constant speed v on two

horizontal parallel ideal conductor rails.The ends of the rails are connected to aresistor R. The separation between therails is d. The rod maintains a tilted angle to the rails. Find the external forcerequired to keep the rod moving.

RB

v

v

d

(1)

2 2B d vF

R dr(2)

2 2B d vF

R dr / sin

(3)

2 2B d v / sinF

R dr / sin

(4)

2 2 2B d v / cosF

R dr / cos

8. C

(1) C

t

(2) C

t

(3) C

t

(4) C

t

9. r B v

R d

RB

v

v

d

(1)

2 2B d vF

R dr(2)

2 2B d vF

R dr / sin

(3)

2 2B d v / sinF

R dr / sin

(4)

2 2 2B d v / cosF

R dr / cos

0000CT103116005H-6/39


10. Heart-lung machines and artifical kidneymachines employ blood pumps. Amechanical pump can mangle blood cells.Figure represents an electromagneticpump. The blood is confined to anelectrically insulating tube, represented asa rectangle of width and height h. Twoelectrodes fit into the top and the bottomof the tube. The potential differencebetween them establishes an electriccurrent through the blood, with currentdensity J over a section of length L. Aperpendicular magnetic field exists in thesame region. The section of liquid in themagnetic field experiences a pressureincrease given by :-

J

B

L

h

(1) JLB(2) JhB

(3) J hB

L

(4) No pressure increase occurs.

10. h L J

J

B

L

h

(1) JLB

(2) JhB

(3) J hB

L

(4)


H-7/390000CT103116005

11. To produce a uniform magnetic field directed

parallel to a diameter of a cylindrical region,

one can use the saddle coils illustrated in

figure. The loops are wrapped over a

somewhat flattened tube. Assume the

straight sections of wire are very long. The

end view of the tube shows how the windings

are applied. The overall current distribution

is the superposition of two overlapping,

circular cylinders of uniformly distributed

current, one toward you and one away from

you. The current density J is the same for

each cylinder. The position of the axis of one

cylinder is described by a position vector a

relative to the other cylinder. The magnetic

field inside the hollow tube is.

I

I

a

(a) (b)

x

y

z

(1) 0Ja

2 along x-axis

(2) 0Ja

2 along y-axis

(3) µ0Ja along z-axis

(4) µ0Ja along y-axis

11.

saddle coils

J

I

I

a

(a) (b)

x

y

z

(1) 0Ja

2, x-

(2) 0Ja

2, y-

(3) µ0Ja, z-

(4) µ0Ja, y-

0000CT103116005H-8/39


12. In figure suppose the transmission axes ofthe left and right polarizing disks areperpendicular to each other. Also, let thecenter disk be rotated on the common axiswith an angular speed . That if unpolarizedlight is incident on the left disk with anintensity I

max the intensity of the beam

emerging from the right disk is given by :-

Imax

Transmission axis

= t

Transmissionaxis

I

(1) I = 1

16I

max(1 – cos 4t)

(2) I = 1

16I

max(1 – cos 2t)

(3) I = 1

8I

max(1 – cos 4t)

(4) I = 1

8I

max(1 – cos 2t)

13. A mathematical representation ofelectromagnetic wave is given by the twoequations E = E

max cos (kx – t) and

B = Bmax

cos (kx – t), where Emax

is theamplitude of the electric field and B

max is

the amplitude of the magnetic field. Whatis the intensity in terms of E

max and

universal constants µ0,

0.

(1) 20max

0

1I E

2

(2) 20

max

0

1I E

2

(3) 20max

0

I 2 E

(4) 20max

0

I 2 E

12. I

max

Imax

Transmission axis

= t

Transmissionaxis

I

(1) I = 1

16I

max(1 – cos 4t)

(2) I = 1

16I

max(1 – cos 2t)

(3) I = 1

8I

max(1 – cos 4t)

(4) I = 1

8I

max(1 – cos 2t)

13.

E = Emax

cos (kx – t) B = Bmax

cos (kx – t)

Emax

Bmax E

max

µ0,

0

(1) 20max

0

1I E

2

(2) 20

max

0

1I E

2

(3) 20max

0

I 2 E

(4) 20max

0

I 2 E


H-9/390000CT103116005

14. A light balloon filled with helium of density

He

is tied to a light string of length L. The

string is tied to the ground forming an

"inverted" simple pendulum (figure). If the

balloon is displaced slightly from

equilibrium as in figure and released, the

period of the motion is. (Take the density

of air to be air

. Assume the air applies a

buoyant force on the balloon but does not

otherwise affect its motion.)

He

L

g

Air

(1) air

He

LT 2

g

(2) air He

He

LT 2

g

(3) He

air

LT 2

g

(4) He

air He

LT 2

g

14. He

L

air

He

L

g

Air

(1) air

He

LT 2

g

(2) air He

He

LT 2

g

(3) He

air

LT 2

g

(4) He

air He

LT 2

g

0000CT103116005H-10/39


15. A space station consists of two living

modules attached to a central hub on

opposite sides of the hub by long corridors

of equal length. Each living module

contains N astronauts of equal mass. The

mass of the space station is negligible

compared to the mass of the astronauts,

and the size of the central hub and living

modules is negligible compared to the

length of the corridors. At the beginning

of the day, the space station is rotating so

that the astronauts feel as if they are in

gravitational field of strength g. Two

astronauts, one from each module, climb

into the central hub, and the remaining

astronauts now feel a gravitational of

strength g'. What is the ratio g'/g in terms

of N?

(1) N

N 1 (2)

N

N 1

(3) N 1

N

(4)

2N

N 1

15.

N

g

g'

N

g'/g

(1) N

N 1 (2)

N

N 1

(3) N 1

N

(4)

2N

N 1


H-11/390000CT103116005

16. One insulated conductor from a household

extension cord has a mass per unit length of

µ. A section of this conductor is held under

tension between two clamps. A subsection

is located in a magnetic field of magnitude

B directed perpendicular to the length of the

cord. When the cord carries an AC current

of "i" at a frequency of f, it vibrates in

resonance in its simplest standing-wave

vibration state. Determine the relationship

that must be satisfied between the

separation d of the clamps and the tension

T in the cord.

(1) T = 4µf2d2 (2) T = 2µf2d2

(3) 2 2f d

T2

(4)

2 2f dT

4

17. The muon has the same charge as an

electron but a mass that is 207 times

greater. The negatively charged muon can

bind to a proton to form a new type of

hydrogen atom. How does the binding

energy EBµ

of the muon in the ground state

of a muonic hydrogen atom compare with

the binding energy EBe

of an electron in the

ground state of a conventional hydrogen

atom ?

(1) B BeE E (2) B BeE 200 E

(3) B BeE 100 E (4) B BeE E / 200

16.

dk æO;eku µ

B

f

AC "i"

d T

(1) T = 4µf2d2 (2) T = 2µf2d2

(3) 2 2f d

T2

(4)

2 2f dT

4

17.

207

EBµ

EBe

(1) B BeE E (2) B BeE 200 E

(3) B BeE 100 E (4) B BeE E / 200

0000CT103116005H-12/39


18. As shown in the figure, a particle A of mass

2m and carrying charge q is connected by a

light rigid rod of length L to another particle

B of mass m and carrying charge

–q. The system is placed in an electric field

E

. The electric force on a charge q in an

electric field E

is F qE

. After the system

settles into equilibrium, one particle is

given a small push in the transverse

direction so that the rod makes a small

angle 0 with the electric field. Find

maximum tension in the rod.

L

A (2m, +q)

B(m, –q)

(1) qE + qE02 (2)

20qE

qE4

(3)

20qE

qE3

(4)

20qE

qE6

18. 2m q A

L m

–q B

E

E

q

F qE

0

L

A (2m, +q)

B(m, –q)

(1) qE + qE02 (2)

20qE

qE4

(3)

20qE

qE3

(4)

20qE

qE6


H-13/390000CT103116005

19. You are driving in your car listening to music

on the radio. Your car is equipped with radio

that can recieved radio singals of frequency

(f1 3 MHz), other of frequency (f

2 30 MHz)

and third of frequency (f3 = 3 GHz). You

enter a tunnel with a circular opening of

diameter 10m. Which kind of radio signal

will you be able to receive the longest as you

continue to travel in the tunnel ?

(1) Radio signal of frequency f1



(4) Distance upto which signal can penetratein tunnel is independent of frequency.

20. The diagram below shows an object placed at adistance 2f in front of a convex lens of focallength f. Which of the following may representsthe positions of the image when (i) a parallelsides glass block, or (ii) a converging lens of longfocal length is inserted between the object andthe convex lens?

P Q R S

f

2f 2f

Glass block Converginginserted lens inserted

(1) Between R and S between R and S(2) Between R and S further away from S(3) Further away from S further away from S

(4) Further away from S between R and S

19. (f

1 3 MHz),

(f2 30 MHz) (f

3 = 3 GHz)

10m (1) f

1

(2) f2

(3) f3

(4)

20. f 2f (i)(ii)

P Q R S

f

2f 2f

(1) R S R S (2) R S S (3) S S (4) S R S

0000CT103116005H-14/39


21. Draw the output signals C1 and C

2 in the

given combination of gates. (Upto 5 sec)

A

B

A

B

C1

C2

0 1 2 3 4 5

0 1 2 3 4 5

A

B

t(s)

t(s)

(1)

0 1 2 3 4 5

0

0

1

1

2

2

3

3

4

4

5

5

A

B

t(s)

t(s)

t(s)

0 1 2 3 4 5 t(s)

C1

C2

(2)

0 1 2 3 4 5

0

0

1

1

2

2

3

3

4

4

5

5

A

B

t(s)

t(s)

t(s)

0 1 2 3 4 5 t(s)

C1

C2

(3)

0 1 2 3 4 5

0

0

1

1

2

2

3

3

4

4

5

5

A

B

t(s)

t(s)

t(s)

0 1 2 3 4 5 t(s)

C1

C2

(4)

0 1 2 3 4 5

0

0

1

1

2

2

3

3

4

4

5

5

A

B

t(s)

t(s)

t(s)

0 1 2 3 4 5 t(s)

C1

C2

21. C1 C

2

(5 sec )

A

B

A

B

C1

C2

0 1 2 3 4 5

0 1 2 3 4 5

A

B

t(s)

t(s)

(1)

0 1 2 3 4 5

0

0

1

1

2

2

3

3

4

4

5

5

A

B

t(s)

t(s)

t(s)

0 1 2 3 4 5 t(s)

C1

C2

(2)

0 1 2 3 4 5

0

0

1

1

2

2

3

3

4

4

5

5

A

B

t(s)

t(s)

t(s)

0 1 2 3 4 5 t(s)

C1

C2

(3)

0 1 2 3 4 5

0

0

1

1

2

2

3

3

4

4

5

5

A

B

t(s)

t(s)

t(s)

0 1 2 3 4 5 t(s)

C1

C2

(4)

0 1 2 3 4 5

0

0

1

1

2

2

3

3

4

4

5

5

A

B

t(s)

t(s)

t(s)

0 1 2 3 4 5 t(s)

C1

C2


H-15/390000CT103116005

22. A long solenoid has 1000 turns per metre

and carries a current of 1 A. It has a soft iron

core of µT = 1000. The core is heated beyond

the Curie temperature, TC.

(1) The H field in the solenoid is (nearly)

unchanged but the B field decreases

drastically.

(2) The H and B fields in the solenoid are

nearly unchanged.

(3) The magnetisation in the core reverses

direction.

(4) The magnetisation in the core increases

by a factor of about 108.

23. A train, standing in a station yard, blows a

whistle of frequency 400 Hz in still air. The

wind starts blowing in the direction from

the yard to the station with a speed of

10m/s. Given that the speed of sound in

still air is 340m/s. Mark the INCORRECT

statement :

(1) The frequency of sound as heard by an

observer standing on the platform is

400Hz.

(2) The speed of sound for the observer

standing on the platform is 350m/s.

(3) The frequency of sound as heard by the

observer standing on the platform will

increase.

(4) The wavelength of sound as received

by the observer standing on the

platform will increase.

22. 1000

1 A

µT = 1000

TC

(1) H

B

(2) H B

(3)

(4) 108

23.

400 Hz

10m/s

340m/s

(1)

400Hz

(2)

350m/s

(3)

(4)

0000CT103116005H-16/39


24. Suppose that two heat

Engine 1 W1

Qh

Th

Qm

Qm

Tm

Engine 2 W2

Qc

Tc

engines are connected

in series, such that the

heat released by the

first engine is used as

the heat absorbed by

the second engine, as

shown in figure. The

efficiencies of the

engines are 1 and

2,

respectively. The net

efficiency of the

combination is given

by :

(1) net

= 1 +

2(2)

1 +

2 – 1 2

(3) net

= 1 +

2–

1

2(4)

1 +

2 – 2

1

2

25. A crew of scientists has built a new spacestation. The space station is shaped like awheel of radius R, with essentially all itsmass M at the rim. When the crew arrives,the station will be set rotating at a rate thatcauses an object at the rim to have radialacceleration g, thereby simulating Earth'ssurface gravity. This is accomplished by twosmall rockets, each with thrust T newtons,mounted on the station's rim. How long atime t does one need to fire the rockets toachieve the desired condition?

(1)

3gR Mt

2T (2)

gRM

t2T

(3) gR M

tT

(4) gRM

tT

24.

Engine 1 W1

Qh

Th

Qm

Qm

Tm

Engine 2 W2

Qc

Tc

1

2

(1)

net =

1 +

2

(2) 1 +

2 – 1 2

(3) net

= 1 +

2–

1

2

(4) 1 +

2 – 2

1

2

25. R M g T t

(1)

3gR Mt

2T (2)

gRM

t2T

(3) gR M

tT

(4) gRM

tT


H-17/390000CT103116005

26. Suppose all the people in the world line up

at the equator and all start running at speed

vrel

relative to surface of earth along

equatorial circle. Initial angular velocity

of earth = 0. Moment of inertia of earth I

E,

moment of inertia of all people = IP, radius

of earth = R :

(1) There will be no change in angular

velocity of rotation of earth.

(2) If people run due east change in

angular velocity of earth will be

P rel

0

P E

I v

I I R

(3) If people run due west change in


P rel

0

P E

I v

I I R

(4) If people run due west change in


P rel

0

P E

I v

I I R

26.

vrel

= 0

IE, = I

P

= R

(1)

(2)

P rel

0

P E

I v

I I R

(3)

P rel

0

P E

I v

I I R

(4)

P rel

0

P E

I v

I I R

0000CT103116005H-18/39


27. Figure (a) shows plot of voltage across the

capacitor as a function of the driving

frequency for a sinusoidally driven

electromagnetic oscillator LCR circuit.

Figure (b) shows phase angle (phase

difference between voltage and current) vs

/0 graph for same circuit, for three

different quality factor graph 1, 2, 3 of

figure (a) and each one can be matched by

one of graphs a, b, c also of figure (b) :-

(1)

(2)

(3)

/ 0

V

/ 0

( )

+–2

– –2

abc

figure (a) figure (b)

(1) Graph (3) corresponds to graph (a)

(2) Graph (1) corresponds to graph (c)

(3) The circuit of graph 1 has high quality

factor.

(4) The circuit of graph 3 has high quality

factor.

27. (a)

LCR

(b)

(

/0

(a)

1, 2, 3 (b) a, b, c

(1)

(2)

(3)

/ 0

V

/ 0

( )

+–2

– –2

abc

figure (a) figure (b)

(1) (3), (a)

(2) (1), (c)

(3) 1

(4) 3


H-19/390000CT103116005

28. A circular conducting loop of radius R carries

a current I. Another straight infinite

conductor carrying current I passes through

the diameter of this loop as shown in the

figure. The magnitude of force exerted by

the straight conductor on the loop is :-

RO

I

(1) µ0I2 (2) µ

0I2

(3)

20I

2(4)

20I

29. A plastic circular disc of radius R is placed

on a thin oil film, spread over a flat

horizontal surface. The torque required to

spin the disc about its central vertical axis

with a constant angular velocity is

proportional to :-

(1) R2 (2) R3

(3) R4 (4) R6

28. R I

I

RO

I

(1) µ0I2 (2) µ

0I2

(3)

20I

2(4)

20I

29. R

(1) R2 (2) R3

(3) R4 (4) R6

0000CT103116005H-20/39


30. The graph in figure shows how the

count-rate A of a radioactive source as

measured by a Geiger counter varies with

time t. The relationship between A and t

is : (Assume ln 12 = 2.6)

0 10 20 30Time t (in sec)

1

2

3

In A

(1) A = 2.6 e–10t

(2) A = 12 e–10t

(3) A = 2.6 e–0.1t

(4) A = 12 e–0.1t

30.

(Geiger counter)

A t

A t

(ln 12 = 2.6)

0 10 20 30Time t (in sec)

1

2

3

In A

(1) A = 2.6 e–10t

(2) A = 12 e–10t

(3) A = 2.6 e–0.1t

(4) A = 12 e–0.1t


H-21/390000CT103116005

PART B - CHEMISTRY

31. Solubility (in molarity) of sparingly soluble

salts MX, MX2 and MX

3 in water are same.

The order of Ksp

of MX, MX2 and MX

3 is

(1) Ksp

(MX) = Ksp

(MX2) = K

sp(MX

3)

(2) Ksp

(MX) < Ksp

(MX2) < K

sp(MX

3)

(3) Ksp

(MX) = 1

2K

sp(MX

2) =

1

3 K

sp(MX

3)

(4) Ksp

(MX) > Ksp

(MX2) > K

sp(MX

3)

32. In a mixture of H and He+ gases, all

H atoms and He+ ions are excited to their

first excited states, Subsequently on

de-excitation in H atoms, energy is

transfered to excited He+ ion. The quantum

number "n" of He+ ion after the energy

transferred is.

(1) 4 (2) 6 (3) 3 (4) 5

33. 100 mL of a given H2O

2 sample in water

(density 1.08 g/mL) required 5 ml of M/25

KMnO4 in acidic medium for its oxidation.

Select the correct statement regarding

H2O

2 in the sample :

(1) Molarity of H2O

2 in the solution = 0.5

(2) Moles of H2O


(3) Molality of H2O


(4) Moles of H2O

2 in the solution = 5 × 10–4

31. MX, MX2 MX

3

() MX, MX2 MX

3

Ksp -

(1) Ksp

(MX) = Ksp

(MX2) = K

sp(MX

3)

(2) Ksp

(MX) < Ksp

(MX2) < K

sp(MX

3)

(3) Ksp

(MX) = 1

2K

sp(MX

2) =

1

3 K

sp(MX

3)

(4) Ksp

(MX) > Ksp

(MX2) > K

sp(MX

3)

32. H He+ H He+ H He+ He+ "n"

(1) 4 (2) 6 (3) 3 (4) 5

33. H2O

2 100 ml 1.08 g/mL

5 ml M/25 KMnO

4

H2O

2 :

(1) H2O

2 = 0.5

(2) H2O

2 = 0.5

(3) H2O

2 = 0.5

(4) H2O

2 = 5 × 10–4

0000CT103116005H-22/39


34. The correct statement(s) regarding defects

in solids is/are

(1) Frenkel defect is usually favoured by

small difference in sizes of cations and

anions

(2) Frenkel defect is a dislocation effect

(3) Paramagnetic substance are repelled

by magnetic field

(4) Schottky defect has no effect on the

physical properties of solids

35. For the first order decompsition reaction

of N2O

5, it is found that -

(a) 2N2O

5 4NO

2(g) + O

2(g)

– 2 52 5

d[N O ]k[N O ]

dt

(b) N2O

5 2NO

2(g) + 1/2O

2(g)

– 2 52 5

d[N O ]k'[N O ]

dt

which of the following is true ?

(1) k = k' (2) k > 2k'

(3) k2 = k' (4) 2k = k'

34.

(1)

(2) (dislocation)

(3)

(4)

35. N2O

5

(a) 2N

2O

5 4NO

2(g) + O

2(g)

– 2 52 5

d[N O ]k[N O ]

dt

(b) N2O

5 2NO

2(g) + 1/2O

2(g)

– 2 52 5

d[N O ]k'[N O ]

dt

?

(1) k = k' (2) k > 2k'

(3) k2 = k' (4) 2k = k'


H-23/390000CT103116005

36. Which of the following is true

(1) The ideal behaviour of a liquid solution

is due to the fact that the different

molecules present in it do not interact

with one another

(2) Henry's laws deals with the variation

of solubility of gas with temperature

(3) In case of negative deviation from

Raoult's law, maximum boiling point

azeotrope is formed

(4) The addition of a nonvolatile solute to

a volatile solvent decreases the boiling

point of the latter

37. On diluting the concentration of Cl– ions

by ten times in a calomel electrode, its

reduction potential at 298K is -

(1) increased by 0.059 V

(2) decreased by 0.059 V

(3) increased by 0.0295 V

(4) decreased by 0.0295 V

38. The enthalpies of combustion of S(s) and

H2(g) are –300 and –290 kcal mol–1.

Given SO3(g) + H

2O() H

2SO

4(l)

H = –130 kcal mol–1

SO2(g) + 1/2O

2(g) SO

3(g)


the enthalpy of formation of H2SO

4(l)

would be-

(1) –300 kcal mol–1 (2) –130 kcal mol–1

(3) –820 kcal mol–1 (4) –560 kcal mol–1

36.

(1)

(2)

(3)

(4)

37. Cl– 298K

(1) 0.059 V

(2) 0.059 V

(3) 0.0295 V

(4) 0.0295 V

38. S(s) H2(g) –300

–290 kcal mol–1 SO

3(g) + H

2O() H

2SO

4(l)


SO2(g) + 1/2O

2(g) SO

3(g)


H2SO

4(l) -

(1) –300 kcal mol–1 (2) –130 kcal mol–1

(3) –820 kcal mol–1 (4) –560 kcal mol–1

0000CT103116005H-24/39


39. If fGº [X ()] = –65 kcal mol–1 and

fGº

[X (g)] = – 60.4 kcal mol–1, the vapour

pressure of water at 500K would be about

- Given : lna = 2.3 loga

(1) 0.01 atm (2) 100 atm

(3) 0.1 atm (4) 10 atm

40. Positive deviation from ideal behaviour is

due to the fact that

(1) Molecular attractions play dominant

role

(2) Molecular volume plays dominant role

(3) Vander waal constant 'a' dominates over 'b'

(4) Molecular attractions cancel the effect

of molecular volume

41. Find the number of electrons in ground

state of Pd with l = 2 :-

(1) 18 (2) 20 (3) 8 (4) 10

42. Find the number of CORRECT order

against mentioned property :-

(i) D2O > H

2O (number of electrons)

(ii) D2O > H

2O (|H

f|)

(iii) D2O > H

2O (H

vaporization)

(iv) D2O > H

2O (density)

(v) D2O > H

2O (Molecular mass)

(1) 2 (2) 5 (3) 4 (4) 3

39. fGº [X ()] = –65 kcal mol–1

fGº

[X (g)] = – 60.4 kcal mol–1, 500K - :lna = 2.3 loga

(1) 0.01 atm (2) 100 atm

(3) 0.1 atm (4) 10 atm

40.

(1)

(2)

(3) 'a' , 'b'

(4)

41. Pd l = 2 :-(1) 18 (2) 20 (3) 8 (4) 10

42. :-

(i) D2O > H

2O ()

(ii) D2O > H

2O (|H

f|)

(iii) D2O > H

2O (H

)

(iv) D2O > H

2O ()

(v) D2O > H

2O ()

(1) 2 (2) 5 (3) 4 (4) 3


H-25/390000CT103116005

43. A mixture of chlorides of Cu, Cd, Cr and

Fe was dissolved in water acidified with

dil. HCl and H2S gas was passed. It was

filtered, and boiled later adding two drops

of HNO3. To this solution NH

4Cl and

NaOH were added in excess and filtered.

The filtrate shall give test for :-

(1) Sodium and iron ion

(2) Copper and chromium ion

(3) Sodium and chromium ion

(4) Iron and cadmium ion

44. Consider the following orders :-

(I) Li > Na (Reducing nature)

(II) BF3 > BCl

3 > BBr

3 > BI

3

(Rate of hydrolysis)

(III) B > C > N > O > F (1st ionization energy)

(IV) CCl4 > SiCl

4 (Boiling point)

The CORRECT order(s) are :-

(1) I and IV (2) II and III

(3) I, II and IV (4) I and III

45. The magnetic moments of following,

arranged in decreasing order will be :-

(i) Co3+ (Octahedral complex with a strong

field ligand)

(ii) Co3+ (Octahedral complex with a weak

field ligand like F

– )

(iii) Co2+ (Tetrahedral complex)

(iv) Co2+ (Square planar complex)

(1) i > ii > iii > iv (2) iii > ii > iv > i

(3) ii > iii > iv > i (4) ii > iv > iii > i

43. Cu, Cd, Cr Fe HCl H

2S

HNO3

NH4Cl NaOH

:-(1) (2) (3) (4)

44. :-(I) Li > Na ()(II) BF

3 > BCl

3 > BBr

3 > BI

3

( )(III) B > C > N > O > F (1st )(IV) CCl

4 > SiCl

4 ()

:-(1) I IV (2) II III

(3) I, II IV (4) I III

45. :-(i) Co3+ ( )

(ii) Co3+ (F

– )

(iii) Co2+ ( )

(iv) Co2+ ()

(1) i > ii > iii > iv (2) iii > ii > iv > i(3) ii > iii > iv > i (4) ii > iv > iii > i

0000CT103116005H-26/39


46. Hydrolysis of which of the following

carbides does not take place :-

(1) Al4C

3(2) Mg

2C

3

(3) SiC (4) CaC2

47. Which of the following pairs both the

complexes show optical activity ?

(1) Cis-[Cr(C2O

4)2Cl

2]3–, Trans-[Co(NH

3)4Cl

2]

(2) [Pt Cl(dien)]Cl , [Ni Cl2 Br

2]2–

(3) [Co(EDTA)]– , [Co(en)3]Cl

3

(4) [Co(NO3)3(NH

3)

3] , Cis-[Pt(en)

2Cl

2]

48. When alloy of silver and lead is rich in

silver :-

(1) Distillation is used

(2) Cupellation process is used

(3) Parke's method is used

(4) Belt's method is used

49. Which of the following is INCORRECT :-

Molecule Type of overlapping

in bridge bond

(1) B2H

6sp3–s–sp3

(2) Al2Cl

6sp3–sp3–sp3

(3) (BeCl2)

2sp2–sp2–sp2

(4) Al2(CH

3)

6sp3–sp3–sp3

50. Total number of lone pair(s) in the XeO2F

2 :-

(1) 1 (2) 2

(3) 3 (4) None of these

46. :-

(1) Al4C

3(2) Mg

2C

3

(3) SiC (4) CaC2

47.

?

(1) Cis-[Cr(C2O

4)2Cl

2]3–, Trans-[Co(NH

3)4Cl

2]

(2) [Pt Cl(dien)]Cl , [Ni Cl2 Br

2]2–

(3) [Co(EDTA)]– , [Co(en)3]Cl

3

(4) [Co(NO3)3(NH

3)3] , Cis-[Pt(en)

2Cl

2]

48. :-(1) (2) (3) (4)

49. :-

(1) B2H

6sp3–s–sp3

(2) Al2Cl

6sp3–sp3–sp3

(3) (BeCl2)

2sp2–sp2–sp2

(4) Al2(CH

3)

6sp3–sp3–sp3

50. XeO2F

2 :-

(1) 1 (2) 2

(3) 3 (4)


H-27/390000CT103116005

51. Barbiturates are hypnotic agents. Veronal

is derivative of barbituric acid. It can be

synthesis by following sequence, then

structure of veronal (P) is :

CH (CO Et)2 2 2

Urea (P)(i) EtO

–

(ii) Et–I

(i) EtO–

(ii) Et–I

(1)

O

NHHN

OO(2)

HN

O NH

O

(3) NH

O

NH

O

(4)

O

NHHN

OO

52. Which of the following option is correctly

matched ?

List-I List-II

(A) Bithional (P) Analgesics

(B) Morphine (Q) Tranquilizer

(C) Nardil (R) Antihistamine

(D) Terfenadine (S) Antispetic

(1) AS, BP, CR, DQ

(2) AS, BP, CQ, DR

(3) AP, BS, CR, DQ

(4) AR, BP, CQ, DS

51. (P)

CH (CO Et)2 2 2

Urea (P)(i) EtO

–

(ii) Et–I

(i) EtO–

(ii) Et–I

(1)

O

NHHN

OO(2)

HN

O NH

O

(3) NH

O

NH

O

(4)

O

NHHN

OO

52.

-I -II

(A) (P)

(B) (Q)

(C) (R)

(D) (S)

(1) AS, BP, CR, DQ

(2) AS, BP, CQ, DR

(3) AP, BS, CR, DQ

(4) AR, BP, CQ, DS

0000CT103116005H-28/39


53. Polymer (Q) produced in following

sequence is :

O

(i) NH –OH2 PH O2

533 - 543 KQ

(ii) H SO , 2 4

(1) Nylon-6,6 (2) Nylon-4,6

(3) Nylon-6 (4) Terylene

54. Which of the following is non-aromatic ?

(1) Pyrrole (2) Thiophene

(3) Furan (4) Pyran

55. Major product (R) of following reaction is

NH2

HNO2

0-5ºCP

HBF4Q R

NaNO2

Cu ,

(1)

F

(2)

F

NO2

(3)

NO2

(4)

53. (Q)

O

(i) NH –OH2 PH O2

533 - 543 KQ

(ii) H SO , 2 4

(1) -6,6 (2) -4,6

(3) -6 (4)

54.

(1) (2)

(3) (4)

55. (R)

NH2

HNO2

0-5ºC PHBF4

Q RNaNO2

Cu ,

(1)

F

(2)

F

NO2

(3)

NO2

(4)


H-29/390000CT103116005

56. Which of the following option will not

produce (P) as major product on reaction

with dil. H2SO

4 ?

OH

(P)

(1) (2) OH

(3) (4)

57. Choose the incorrect option :

(1) Cyclohexanone forms cyanohydrin in

good yield but 2,2,6-trimethylcyclohexanone does not

(2) Tertiary amines (Me3N) have higher

boiling point than primary amine

(MeNH2)

(3) pKb of aniline is more than that of

methyl amine

(4) PCC can oxidise 1º alcohol to aldehyde

in anhydrous condition

56. H2SO

4

(P)

OH

(P)

(1) (2) OH

(3) (4)

57.

(1)

2,2,6-

(2) (MeNH2)

(Me3N)

(3) pKb

(4) PCC , 1º

0000CT103116005H-30/39


58. Which of the following amino acid contains

alcohol fuctional group ?

(1) Valine (2) Serine

(3) Cysteine (4) Alanine

59. For the following compounds, choose the

incorrect option ?

CH —NH3 2 Ph—NH2

(P) (Q)

(1) (P) is more basic than (Q)

(2) Both (P) & (Q) will give foul smelling

compound with CHCl3, KOH

(3) Both (P) & (Q) will form base soluble

sulphonamide with Hinsberg reagent

(4) Both (P) & (Q) can be obtained by

gabriel phthalimide synthesis

60. Major product (Q) of following sequence is

(i) CH COCl / AlCl3 3

(ii) Zn-Hg / HCl(P)

(i) NBS

(ii) H O / K CO2 2 3

(Q)

(1)

CHO

(2)

O

(3) OH

(4)

OH

58.

(1) (2)

(3) (4)

59.

CH —NH3 2 Ph—NH2

(P) (Q)

(1) (Q) (P)

(2) (P) (Q) CHCl3, KOH

(3) (P) (Q)

(4) (P) (Q)

60. (Q)

(i) CH COCl / AlCl3 3

(ii) Zn-Hg / HCl(P)

(i) NBS

(ii) H O / K CO2 2 3

(Q)

(1)

CHO

(2)

O

(3) OH

(4)

OH


H-31/390000CT103116005

PART C - MATHEMATICS

61. If a + 2b + 3c = 6, then the greatest value of

abc2 is (where a,b,c are positive real

numbers)

(1) 9

8(2)

9

16(3)

27

8(4)

27

16

62. A tangent to the ellipse 2 2x y

125 16

intersect the co-ordinate axes at A and B,

then locus of circumcentre of triangle AOB

(where O is origin) is

(1) 2 2

16 251

x y(2) 16x2 + 25y2 = 4

(3) 2 2

25 164

x y(4)

2 2

25 161

x y

63. Given z is a complex number such that

|z| < 2, then the maximum value of

|iz + 6 – 8i| is equal to-

(1) 10 (2) 8 (3) 12 (4) 6

64.

2 2

32

1

2

x nxdx

1 x

equals-

(1) 1 (2) –1 (3)

1 1

3 n22 2

(4) 0

61. a + 2b + 3c = 6 abc2 (a,b,c )

(1) 9

8(2)

9

16(3)

27

8(4)

27

16

62. 2 2x y

125 16

A B AOB (O )

(1) 2 2

16 251

x y(2) 16x2 + 25y2 = 4

(3) 2 2

25 164

x y(4)

2 2

25 161

x y

63. z |z| < 2 |iz + 6 – 8i| (1) 10 (2) 8

(3) 12 (4) 6

64.

2 2

32

1

2

x nxdx

1 x

-

(1) 1 (2) –1 (3)

1 1

3 n22 2

(4) 0

0000CT103116005H-32/39


65. Given equation 4x2 + 4(a – 1)x + (1 – 2a) = 0

has roots sin and

cos 02

, then

maximum value of (a + sin) is-

(1) 2 3

2(2)

1

2

(3) 1 3

2(4)

3

2

66. If A and B are two invertible square

matrices of the same order such that

(A + B)(A – B) = A2– B2, then (A2BA–1B–1)3

is equal to-

(1) B2 (2) A2 (3) A3 (4) A3B3

67. The equation of base BC of an equilateral

triangle is 3x + 4y = 1 and vertex is (–3,2),

then the area of triangle is-

(1) 4 3

75(2)

4

5 3

(3) 8 3

75(4)

16 3

25

68. Number of integral values of for which

ƒ x n 2 cos x 5 is defined for all

x R is-

(1) 5 (2) 4 (3) 6 (4) 11

65. 4x2 + 4(a – 1)x + (1 – 2a) = 0

sin

cos 02(a + sin)

-

(1) 2 3

2(2)

1

2

(3) 1 3

2(4)

3

2

66. AB (A + B)(A – B) = A2– B2 (A2BA–1B–1)3 (1) B2 (2) A2

(3) A3 (4) A3B3

67. BC 3x + 4y = 1 (–3,2) -

(1) 4 3

75(2)

4

5 3

(3) 8 3

75(4)

16 3

25

68.

ƒ x n 2 cos x 5 x R

(1) 5 (2) 4 (3) 6 (4) 11


H-33/390000CT103116005

69. The order and degree of the differential

equation

7/53 23

3 2

d y dy d yy x

dxdx dx are m and

n respectively, then (m + n) is equal to-

(1) 8 (2) 7

(3) 10 (4) none of these

70. A box is to be made with square base and

open top. If the area of material used is

48 sq. meter, then maximum volume of the

box is-

(1) 48m3 (2) 16m3 (3) 32m3 (4) 36m3

71. The function ƒ(x) = 2n|x| – x|x| is

increasing on the interval-

(1) (0,1) (2) (0,) (3) (–1,1) (4) (–1,0)

72. If

5

ii 1

x 10 5 and

5

2

ii 1

x 10 25 ,

then standard deviation of observations

2x1 + 7, 2x

2 + 7, 2x

3 + 7, 2x

4 + 7 and 2x

5 + 7

is equal to-

(1) 8 (2) 16 (3) 4 (4) 2

73. The angle between the line

ˆ ˆˆ ˆ ˆ ˆr i j 2k 2i j k and normal to

the plane ˆˆ ˆr. i j 3k 2 is-

(1)

1 4cos

66(2)

1 4sin

66

(3)

1 4tan

66(4)

1 4cot

66

69.

7/53 23

3 2

d y dy d yy x

dxdx dx

m n (m + n) (1) 8 (2) 7

(3) 10 (4) 70.

48 -(1) 48m3 (2) 16m3 (3) 32m3 (4) 36m3

71. ƒ(x) = 2n|x| – x|x| (1) (0,1) (2) (0,) (3) (–1,1) (4) (–1,0)

72.

5

ii 1

x 10 5

5

2

ii 1

x 10 25

2x1 + 7, 2x2 + 7, 2x3 + 7, 2x4 + 7 2x5 + 7 -

(1) 8 (2) 16 (3) 4 (4) 2

73. ˆ ˆˆ ˆ ˆ ˆr i j 2k 2i j k

ˆˆ ˆr. i j 3k 2

-

(1)

1 4cos

66(2)

1 4sin

66

(3)

1 4tan

66(4)

1 4cot

66

0000CT103116005H-34/39


74. Given the system of equation a(x + y + z)=x

b(x + y + z) = y, c(x + y + z) = z where a,b,c

are non-zero real numbers. If the real

numbers x,y,z are such that xyz 0, then

(a + b + c) is equal to-

(1) 0 (2) –1 (3) 1 (4) 2

75. If

x 0,4

, then the expression

2

cosx

sin x cosx sin x can not take the value

(1) 8 (2) 10 (3) 11 (4) 12

76. If Tr = 2016C

rx2016–r, for r = 0,1,2.....2016, then

(T0 – T

2 + T

4.......+T

2016)2

+ (T1 – T

3 + T

5.......T

2015)2 is equal to-

(1) (x2 + 1)1008 (2) (x + 1)2016

(3) (x2 – 1)2016 (4) (x2 + 1)2016

77.

6

1 1

1

sec x cot x dx is equal to

(where [.] denotes greatest integer function)

(1) 12 – sec1 (2) 6 – cot1

(3) 6 – cot1 – sec1 (4) 6 – sec1

78. Given ABC is an equilateral triangle ofside length 1 unit and P be any arbitrarypoint on the circumcircle of triangle ABC,

then 2 2 2

PA PB PC

is equal to -

(1) 3 (2) 1 (3) 2 (4) 2 3

74. a(x + y + z) = x,

b(x + y + z) = y, c(x + y + z) = z a,b,c x,y,z

xyz 0 (a + b + c) (1) 0 (2) –1 (3) 1 (4) 2

75.

x 0,4

2

cosx

sin x cos x sin x,

(1) 8 (2) 10

(3) 11 (4) 12

76. r = 0,1,2.....2016 Tr = 2016Crx2016–r

(T0 – T2 + T4.......+T2016)2

+ (T1 – T

3 + T

5.......T

2015)2

(1) (x2 + 1)1008 (2) (x + 1)2016

(3) (x2 – 1)2016 (4) (x2 + 1)2016

77.

6

1 1

1

sec x cot x dx

([.] )(1) 12 – sec1 (2) 6 – cot1

(3) 6 – cot1 – sec1 (4) 6 – sec1

78. ABC 1 P ABC

2 2 2

PA PB PC

(1) 3 (2) 1 (3) 2 (4) 2 3


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79. If all the letters of the word 'GANGARAM'

be arranged, then number of words in

which exactly two vowels are together but

no two 'G' occur together is-

(1) 1320 (2) 1560 (3) 1800 (4) 1740

80. Statement-I : ~ p q is equivalent to

p ~ q ~ p ~ q .

Statement-II : p p q is a tautology.

(1) Statement-1 is True, Statement-2 is True;

Statement-2 is a correct explanation for

Statement-1.

(2) Statement-1 is True, Statement-2 is True;

Statement-2 is NOT a correct explanation

for Statement-1.

(3) Statement-1 is True, Statement-2 is False.

(4) Statement-1 and Statement-2 both are

False

81. Area of the triangle formed by any tangent

to the curve xy = 100 with the coordinate

axes, is -

(1) 100 (2) 200 (3) 50 (4) 100 2

82. A bag contains 5 distinct Red, 4 distinct

Green and 3 distinct Black balls. Balls are

drawn one by one without replacement,

then the probability of getting a particular

red ball in fourth draw is-

(1) 1

12(2)

223

1188(3)

335

1320(4)

5

12

79. 'GANGARAM' 'G' -(1) 1320 (2) 1560 (3) 1800 (4) 1740

80. -I : ~ p q

p ~ q ~ p ~ q

-II : p p q

(1) -I -II -II -I

(2) -I -II -II

-I

(3) -I -II

(4) -I -II

81. xy = 100

(1) 100 (2) 200 (3) 50 (4) 100 2

82. 5 43

(1) 1

12(2)

223

1188(3)

335

1320(4)

5

12

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83. Minimum integral value of for which

graph of ƒ(x) = ||x – 2| – |– 5 has exactly

four x-intercepts-

(1) 6 (2) 4 (3) 7 (4) 5

84.

2

x 0

1lim x 1 2 3 ......

|x| is equal to

(where [.] denotes greatest integer function)

(1) 0 (2) 1

2

(3) 2 (4) does not exist

85. Given

2

n 1 sgn x xif x 0ƒ x 1 cos x

k if x 0

,

then

(where [.], {.} and sgnx denotes greatest

integer function, fractional part function

and signum function respectively)

(1) ƒ(x) is continuous at x = 0 if k = 2

(2) for k = 1, ƒ(x) has removable

discontinuity at x = 0

(3) for k = 2, ƒ(x) has non-removable

discontinuity at x = 0.

(4) x 0

lim ƒ x exists.

83. ƒ(x) = ||x – 2| – |– 5 x--(1) 6 (2) 4 (3) 7 (4) 5

84.

2

x 0

1lim x 1 2 3 ......

|x|

([.] )

(1) 0 (2) 1

2

(3) 2 (4)

85.

2

n 1 sgn x xx 0ƒ x 1 cos x

k x 0

( [.], {.} sgn )

(1) ƒ(x), x = 0 k = 2

(2) k = 1 ƒ(x) x = 0

(3) k = 2 ƒ(x) x = 0

(4) x 0

lim ƒ x


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86. Let S1,S2 and S3 be three circles of unit

radius which touch each other externally.

The common tangent to each pair of circles

are drawn and extended so that they can

intersect and form a triangle ABC with

circumradius R, then R is equal to

(1) 4 2 3

(2) 1

2 13

(3) 4 1 3

(4) 3 1 3

2

87. Number of rational roots of equation

x2016 – x2015 + x1008 + x1003 + 1 = 0, is equal

to -

(1) 0

(2) 1008

(3) 2015

(4) 2016

86. S1,S2 S3 ABC R R

(1) 4 2 3

(2) 1

2 13

(3) 4 1 3

(4) 3 1 3

2

87. x2016 – x2015 + x1008 + x1003 + 1 = 0 -

(1) 0

(2) 1008

(3) 2015

(4) 2016

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88. Given a1,a

2,a

3.. ... form an increasing

geometric progression with common ratio r

such that log8a1 + log8a2 +.....+ log8a12 = 2014,

then the number of ordered pairs of

integers (a1, r) is equal to -

(1) 44 (2) 45 (3) 46 (4) 47

89. A vertical pole of height h stands on a

building of height H and the angle of

elevation of top of the building from a point

on the ground which is 5 unit away from

the foot of building is , also the pole

subtends an angle at the same point on

the ground. If 2 tancot = 1, then-

(1) H3 + hH2 + 25H – 50h = 0

(2) H3 + hH2 – 25H + 50h = 0

(3) H3 + hH2 + 50H – 25h = 0

(4) H3 + hH2 + 25H + 50h = 0

90. A beam of light travels along the line y = –4

from right to left and strikes a parabola at

point P. If focus and directrix of parabola

are (2, 0) and x = –2 respectively, then the

coordinates of the point where the

reflected beam contact the parabola again

(1) 1,2 2 (2) (2, 4)

(3)

7

42, 2

(4) 3, 24

88. a1,a

2,a

3.....

r log8a1 + log8a2 +.....+ log8a12 = 2014 (a1, r) -(1) 44 (2) 45

(3) 46 (4) 47

89. h H 5 2 tancot = 1 -(1) H3 + hH2 + 25H – 50h = 0

(2) H3 + hH2 – 25H + 50h = 0

(3) H3 + hH2 + 50H – 25h = 0

(4) H3 + hH2 + 25H + 50h = 0

90. y = –4 P (2, 0) x = –2

(1) 1,2 2 (2) (2, 4)

(3)

7

42, 2

(4) 3, 24


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LEADER & ENTHUSIAST COURSE

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