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JEE-Mains – 2016 (Hints & Solutions) Narayana IIT/PMT Academy

Narayana Group of Educational Institutions

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NARAYANA IIT/PMT ACADEMY INDIA

JEE-MAINS-2016 SOLUTIONS

This booklet contains 43 printed pages

PAPER -1 : PHYSIC, MATHEMATICS & CHEMISTRY Test Booklet Code

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Booklet, take out the Answer Sheet and fill in the particulars carefully. 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 questions paper A, B, C consisting of Physics, Mathematics and

Chemistry having 30 questions in each part of equation weightage. Each question is allotted 4 (four) marks for correct response.

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PHYSICS 1. A particle perform simple harmonic motion with amplitude A. It’s speed is trebled at the

instant that it is at a distance 2A3

from equilibrium position. The new amplitude of the

motion is :

(1) A 3 (2) 7A3

(3) A 413

(4) 3A

Ans. 2 Sol: Applying conservation of energy

2 2

2 2 2 2 21

1 2A 1 A 1m 9 m A m A2 3 2 3 2

17AA3

2. For a common emitter configuration, if and have their usual meanings, the incorrect relationship between and is :

(1) 1

(2) 2

21

(3) 1 1 1

(4) 1

Ans. 4

Sol: 1

,so1 1

is incorrect

Also, as >> 1. Therefore 2

21

1

3. A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90s, 91s, 95 s and 92 s. If The minimum division in the measuring clock is 1s, then the reported mean time should be :

(1) 92 1.8 s (2) 92 3 s (3) 92 2s (4) 92 5.0 s Ans. 3

Sol: Mean value of time 90 91 95 92 92s4

Also absolute errors are 92 90 2 92 91 92 95 92 92 0

mean absolute error = 2 1 3 0 1.54

s but since the least count of the clock is 1s

so it can’t count 1.5s. hence it will count 2 s as error. (92 2) s

4. If a, b, c, d are inputs to a gate and x is its output, then, as per the following time grapy, the gate is :



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(1) OR (2) NAND (3) NOT (4) AND Ans. 1 Sol: In OR gate, if all the inputs are low (zero), only. Then the output is low (zero). Hence the correct option is (1). 5. A particle of mass m is moving along the side of a square of side ‘a’, with a uniform

speed v in the x-y plane as shown in the figure.

Which of the following statements is false for the angular momentum L

about the origin?

(1) 2 ˆL mv a k2

when the particle is moving from B to C

(2) mv ˆL Rk2

when the particle is moving form D to A.

(3) mv ˆL R k2

when the particle is moving from A to B.

(4) R ˆL mv a k2

when the particle is moving from C to D.

Ans. 2, 4 Sol: L r p

Hence option 2&4 are incorrect. 6. Choose the correct statement: (1) In frequency modulation the amplitude of the high frequency carrier wave is made to

vary in proportion to the amplitude of the audio single. (2) In frequency modulation the amplitude of the high frequency carrier wave is made to

vary in proportion to the frequency of the audio signal. (3) In amplitude modulation the amplitude of the high ‘frequency carrier wave is made to

very in proportion to the amplitude of the audio signal. (4) In amplitude modulation the frequency of the high frequency carrier wave is made to

vary in proportion to the amplitude of the audio signal.



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Ans. 3

Sol: Carrier wave

Audio signal

Modulated wave In amplitude modulation the amplitude of the high frequency carrier wave is made to

vary in proportion to the amplitude of audio signal. 7. Radiation of wavelength , is incident on a photocell. The fastest emitted electron has

speed v. If the wavelength is changed to 34 , the sped of the fastest emitted electron will

be:

(1) 124v

3

(2) 123v

4

(3) 124v

3

(4) 124v

3

Ans. 3

Sol: 2o

hc 1w mv2

…(1)

2o 1

4 hc 1w mv3 2

…(ii)

From (i) & (ii)

2 2o o 1

4 1 1w mv w mv3 2 2

2 2o1

w1 4 1mv mv2 3 3 2

14v v3

.

8. Two identical wires A and B, each of length ‘I’, carry the same current I. Wire A is bent into a circle of radius of radius R and wire B is bent to form a square of side ‘a’. If BA and BB are the values of magnetic field at the centres of the circle and square

respectively, then the ratio A

B

BB

is :

(1) 2

16 (2)

2

8 2 (3)

2

8 (4)

2

16 2

Ans. 2

Sol: o oA

I IB

2.2



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oB

I cos45 cos 45B 4

48

= oI 2.8l

2

A

B

BB 8 2

.

9. A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now:

(1) 2f (2) f (3) f2

(4) 3f4

Ans. 2

Sol: 1v vf f

242

2 10. The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure ),

have volume charge density A ,r

where A is a constant and r is the distance from the

centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is

(1)

2 2

2Qa b

(2) 2

2Q

(3) 2

Q

(4) 2 2

Qb a

Ans. 3 Sol: Using gauss theorem for dotted Gaussian surface

r2 2

ao

1 AE.4 r Q .4 r drr

rQ

ab

2 2

2 2o o

Q 1E .2 A r a4 r 4 r

= 2

2 2o o o

Q A Aa4 r 2r

For E to be independent of r



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2

2 2 2o o

Q Aa Q0 A4 r 2 a

11. An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to

(1) 0.044 H (2) 0.065 H (3) 80 H (4) 0.08 H Key. (2)

Sol. Resistance of the lamp = 80 810

L

~(220 V, 50 Hz)

Lamp

Now VR = 80 V Here 2 2

L RV V V 2 2L220 V 80

VL = 10 420 V Now VR = IR = 80 ... (1) VL = IXL = 10 420 ... (2) On dividing (1) and (2),

L

R 80X 10 420

8 82 fL 420

On solving, L = 0.065 H.

12. ‘n’ moles of an ideal gas undergoes a process A B as shown in the figure. The maximum temperature of the gas during the process will be

(1) 0 09 P V2nR

(2) 0 09 P VnR

(3) 0 09 P V4nR

(4) 0 03P V2nR

Key. (3) Sol. The equation of the process is y = mx + c

oo

o

PP V 3PV



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Now, PV = nRT

P = nRTV

oo

o

PnRT V 3PV V

2oo

o

PnRT V 3P VV

2o o

o

P 3PT V VnRV nR

… (1)

Now for maximum T,

dT 0dV

o o

o

2P 3PdT V 0dV nRV nR

o3V V2

Putting in (1), we get

2o oo o

o

P 3P9 3T V VnRV 4 nR 2

= o o o oP V P V9 92 nR 4 nR

o omax

P V9T4 nR

13. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8 × 107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. [Take g = 9.8 ms–2]

(1) 9.89 × 10–3 kg (2) 12.89 × 10–3 kg (3) 2.45 × 10–3 kg (4) 6.45 × 10–3 kg Key. (2) Sol. Total potential energy U = mgh × 1000 = 10 × 9.8 × 1 × 1000 Since 20% efficiency rate, hence

3.8 × 107 × 20 m100

= 9.8 × 104

m = 4

6

9.8 103.8 2 10

= 12.89 × 10–3 kg

14. A point particle of mass m, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals . The



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particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR.

The values of the coefficient of friction and the distance x(= QR), are respectively close to

(1) 0.29 and 3.5 m (2) 0.29 and 6.5 m (3) 0.2 and 6.5 m (4) 0.2 and 3.5 m Key. (1)

Sol. o

hPQ 2h 4msin 30

Given that work done against friction along track PQ and QR are equal. mg cos 30o 4 = mg x

4 3 x2

x = 2 3 2 1.732 = 3.46 3.5 m Loss in PE = work done against friction along PQ and QR

mg 2 = mg 32

4 + mg (3.5)

On solving = 0.29

15. The temperature dependence of resistances of Cu and undoped Si in the temperature range 300–400 K, is best described by

(1) Linear increase for Cu, exponential decrease for Si (2) Linear decrease for Cu, linear decrease for Si (3) Linear increase for Cu, linear increase for Si (4) Linear increase for Cu, exponential increase for Si Key. (1) Sol. For metals like Cu, resistance increases linearly with temperature.



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T

R

For intrinsic semiconductors like Si, resistance decreases exponentially with temperature

increases.

T

R

16. Arrange the following electromagnetic radiations per quantum in the order of increasing energy

A : Blue light B : Yellow light C : X-ray D : Radiowave (1) C, A, B, D (2) B, A, D, C (3) D, B, A, C (4) A, B, D, C Key. (2) Sol. yellow blue Radiowave X ray As E = hc/ yellow blue radiowave X rayE E E E

17. A galvanometer having a coil resistance of 100 gives a full scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full scale deflection for a current of 10 A, is

(1) 0.1 (2) 3 (3) 0.01 (4) 2 Key. (3) Sol. G = 100 Ig = 10-3 A The shunt resistance (S) required to convert the galvanometer to an ammeter of full scale

deflection current of 10 A is

S = 3

g3

g

I G 0 100I I 10 10

110 110 100

0.01



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18. Half-lives of two radioactive elements A and B are 20 minutes and 40 minutes, respectively. Initially, the samples have equal number of nuclei. After 80 minutes, the ratio of decayed numbers of A and B nuclei will be

(1) 1 : 4 (2) 5 : 4 (3) 1 : 16 (4) 4 : 1 Key. (2)

Sol. No. of atoms left after n half lives is 0n

N2

(where n is the number of half lives)

Number of atoms of A left = o4

N2

= oN16

(4 half lives t = 80 min, t1/2 = 20 min, n = 80/20 = 4)

Number of atoms of B left = o 02

N2 4

(Number of half lives = 2)

Hence ratio of decayed number of A and B will be

o o

o

o oo

N 15NN 516 16N 3N 4N4 4

19. Identify the semiconductor devices whose characteristics are given below, in the order (a), (b), (c), (d)

(1) Solar cell, Light dependent resistance, Zener diode, simple diode (2) Zener diode, Solar cell, Simple diode, Light dependent resistance (3) Simple diode, Zener diode, Solar cell, Light dependent resistance (4) Zener diode, Simple diode, Light dependent resistance, Solar cell Key. (3) Sol. Theoretical

20. A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 F and 9 F capacity), at a point distant 30 m from it, would equal



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4 F3 F

9 F

8V

+ –

2 F

(1) 420 N/C (2) 480 N/C (3) 240 N/C (4) 360 N/C Key. (1) Sol.

4 F

3 F

9 F4 F 12 F

V1 V2

8 V Now V1 + V2 = 8V and 4V1 = 12V2 On solving, V1 = 6 V, V2 = 2 V Hence Q1 = 24 C, Q2 = 18 c Total Q = 42 C

Hence, E = 2o

1 Q4 r

= 9 6

2

9 10 42 1030

= 420 N/C

21. A satellite revolving in a circular orbit at a height ‘h’ from the earth’s surface (radius of earth R; h << R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth’s gravitational field, is close to: (Neglect the effect of atmosphere.)

(1) / 2gR (2) 2 1gR (3) 2gR (4) gR

Key (2)

Sol: oGMV

R

2e

GMVR

Increase e oV V

2 1 GMR

2

2 1 gRR h

2 1 gR



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22. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a time sheet of Aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

(1) 0.70 mm (2) 0.50 mm (3) 0.75 mm (4) 0.80 mm Key (4)

Sol: L.C. 0.5 0.0150

mm

error is negative error = 5 × 0.01 = 0.05 Reading = 0.5 + 25 × 0.01 + 0.05 = 0.5 + 0.25 + 0.05 = 0.80 mm 23. A roller is made by joining together two cones at their vertices O. It is kept on two rails

AB and CD which are placed asymmetrically (see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD (see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to:

A

B

C

D

O

(1) go straight (2) turn left and right alternately (3) turn left (4) turn right Key (4) Sol:

N 1 N y 1N y 2 N 2

N x 1 N x2

A

B

C

D

Initially C.M. will move straight, hence its distances from left rail will decrease.



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Normal reaction at left contact N1 will increase. Horizontal component of left N1x will also increase and continue to increase. Hence the roller will tend to move towards left direction.

24. Hysteresis loops for two magnetic material A and B are given below:

(A) (B)

B

H

B

H

These materials are used to make magnets for electric generators, transformer core and

electromagnet core. The it is proper to use: (1) A for transformers and B for electric generators (2) B for electromagnets and transformers (3) A for electric generators and transformers (4) A for electromagnets and B for electric generators Key (2) Sol: Hysteresis loss for electromagnets and transformers are less because it can magnetize or

demagnetize easily 25. The box of a pin hole camera, of length L, has a hole of radius a. It is assumed that when

the hole is illuminated by a parallel beam of light of wavelength the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin) when:

(1) a L and min 4b L (2) 2

aL

and min 4b L

(3) 2

aL

and 2

min2b

L

(4) a L and 2

min2b

L

Key (2) Sol:

bqa

sinb n

tan sinb aL



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a b a

L

2ab a L 2 0a ab L 2 4b L min 4b L

24a L a L

2

aL

26. A uniform straight of length 20 m is suspended from a rigid support. A short wave pulse

is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is:

(take 210g ms ) (1) 2 2s (2) 2s (3) 2 2s (4) 2s Key (1)

Sol: xgv xg

dx xgdt

0

l dx g tx

1/20

lx dx g t

1/22x g t

2 20 tg

2 2 t 27. An ideal gas undergoes a quasi static reversible process in which its molar heat capacity

C remains constant. If during this process the relation of pressure P and volume V is given by nPV constant, then n is given by (Here Cp and CV are molar specific heat at constant pressure and constant volume, respectively):

(1) p

V

C Cn

C C

(2) V

P

C CnC C

(3) P

V

CnC

(4) P

V

C CnC C

Key (4) Sol: nPV constant Then processC for the above polytropic process is known



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1process V

RC C Cn

1V

RC Cn

1

V

RnC C

Finally after solving

V

V

C C Rn

C C

[as P VC C R ]

P

V

C CnC C

28. An observer looks at a distance tree of height 10m with a telescope of magnifying power

of 20. To the observer the tree appears: (1) 20 times taller (2) 20 times nearer (3) 10 times taller (4) 10 times nearer Key (2) Sol: In case of telescopes, magnifying power = 20 means that the object appears 20 times

nearer. 29. In an experiment for determination of refractive index of glass of a prism by i , plot, it

was found that a ray incident at angle 35, suffers a deviation of 40 and that it emerges at angle 79. In that case which of the following is closest to the maximum possible value of the refractive index?

(1) 1.7 (2) 1.8 (3) 1.5 (4) 1.6 Key (4) Sol:

35° r 1 r2

eA

35i 40 79e i = 35o

e = 79o

o40

o4 i e A 74

with above set of data, as refractive index increases emerging ray will have tendency to suffer TIR.

For i = 35o



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o o

o1 1

sin 35 sin 90sin r sin 74 r

Using reversibility principle For i = 79o

o o

o2 2

sin 79 sin 90sin r sin 74 r

From 2nd case o2r 37

1.63 Hence Answer is (4)

30. A pendulum clock loses 12 s a day if the temperature is 40C and gains 4s a day if the temperature is 20C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion () of the metal of the pendulum shaft are respectively:

(1) 30C; = 1.85 × 103/C (2) 55C; = 1.85 × 102/C (3) 25C; = 1.85 × 105/C (4) 60C; = 1.85 × 104/C Key (3)

Sol: 1 12 2

T l TT l

1 11 1 24 36002 2

T T TT

112 12 3600T .........(i)

24 12 3600T ........(ii)

1

2

0320

T TT T

3 60 40T T 25T C 51.85 10 / C

Topic & Level

Q.No. Topic Level 01. S.H.M. Medium 02. Electronics Easy 03. Error analysis Easy 04. Electronics Easy 05. Rotational Medium 06. Electronics Easy 07. Modern Physics Medium 08. Magnetism Easy 09. Sound Easy 10. Electrostatics Difficult 11. AC Medium 12. Thermodynamics Difficult 13. Work Power Energy Medium



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14. Work Power Energy Medium 15. Current Electricity Easy 16. EMW Easy 17. Current electricity Medium 18. Modern Physics (Radio activity) Medium 19. Electronics Medium 20. Electrostatics Medium 21. Gravitation Easy 22. Error Analysis Medium 23. Rotation Difficult 24. Magnetism Easy 25. Optics Difficult 26. Optics Difficult 27. Heat & Thermodynamics Difficult 28. Optics Easy 29. Optics Difficult 30. Heat & Thermodynamics Difficult

PART : B : MATHEMATICS 31. The area (in sq. units) of the region 2 2 2x, y ; y 2x and x y 4x, x 0, y 0 is:

(1) 4 23

(2) 2 22 3 (3) 4

3 (4) 8

3

Ans: (4) Sol. 2 2 2 2y 2x, x y 4x x 2x x 0, 2

Required area = 2

2

0

4x x 2x dx

= 22

0

4 x 2 2 x dx

= 2

2 1 3/2

0

x 2 x 2 2 24x x 2sin x2 2 3

= 80 0 0 23 2

= 83

32. If 1f x 2f 3x, x 0,x

and S x R :f x f x ; then S :

(1) contains exactly two elements (2) contains more than two elements (3) is an empty set (4) contains exactly one element

x=2

X

Y

O(0,0)

y =2x2



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Ans: (1) Sol.

1f x 2f 3xx

… (1)

Replace 1xx

1 3f 2f xx x

1 62f 4f xx x

… (2)

63f x 3xx

From (1) and (2)

2f x xx

2f x xx

2 2f x f x x xx x

24 2x x 2 x 2x

33. The integral

12 9

35 3

2x 5x dxx x 1

is equal to:

(1)

5

25 3

x C2 x x 1

(2)

10

25 3

x C2 x x 1

(3)

5

25 3

x Cx x 1

(4)

10

25 3

x C2 x x 1

Where C is an arbitrary constant. Ans: (4) Sol.

12 9

35 3

2x 5xdx

x x 1

= 3 6

3

2 5

2 5 dxx x

1 11x x



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Let 2 5

1 11 tx x

3 6

2 5 dx dtx x

3 2

dt 1I Ct 2t

= 2

2 5

1 C1 12 1x x

=

10

25 3

x C2 x x 1

34. For x R, f x log 2 sin x and g x f f x , then (1) g ' 0 cos log 2 (2) g is differentiable at x = 0 and

g ' 0 sin log 2 (3) g is not differentiable at x 0 (4) g ' 0 cos log 2 Ans: (4) Sol. f x log 2 sin x g x f f x

g x log 2 sin log 2 sin x

g x log 2 sin log 2 sin x in neighbourhood of x 0

g ' 0 cos log 2 sin 0 . cos 0 = cos log 2

35. The centres of those circles which touch the circle, 2 2x y 8x 8y 4 0 , externally and also touch the x-axis, lie on:

(1) a hyperbola (2) a parabola (3) a circle (4) an ellipse which is not a circle Ans: (2) Sol. Circles touch externally and required circle touches x-axis. Centre (h, k) and radius = | k | 1 2 1 2c c r r

2 2 2h 4 k 4 6 | k |

2 2 2h 2.4h 16 k 8k 16 36 k 12| k | 2h 8h 8k 12 | k | 4 2x 8x 8y 12 y 4 It represents two-semi parabola

36. The sum of all real values of x satisfying the equation 2x 4x 602x 5x 5 1

is: (1) 6 (2) 5 (3) 3 (4) -4



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Ans: (3) Sol.

2x 4x 602x 5x 5 1

ba 1 , when a is 1 b R a is not zero b = 0 a is -1 b even integer

2x 5x 5 1 x 1, 4

2x 4x 60 0 ; x 10,6 ; a 0 2x 5x 5 1 , 2x 4x 60 even integer x 2, 3 at x = 2, 2x 4x 60 48 ( even integer ) at x = 3, 9 12 60 odd solutions are x 1, 4, 10, 6, 2

sum = 3 37. If the nd th th2 , 5 and 9 terms of a non-constant A.P., are in G.P., then the common ratio of

this G.P. is:

(1) 1 (2) 74

(3) 5 (4) 4

3

Ans: (4) Sol. Given terms of A.P. are 2T a d ; 5T a 4d ; 9T a 8d If 2 5 9T , T & T are in G.P.

2a 4d a d a 8d 2 2 2a 16d 8ad a ad 8ad 8d 28d ad d 8d a 0 8d a d 0

Now common ratio is a 4d 12d 4a d 9d 3

38. The eccentricity of the hyperbola whose length of the latus rectum is equal to 8 and the

length of its conjugate axis is equal to half of the distance between its foci, is :

(1) 23

(2) 3 (3) 43

(4) 43

Ans: (1)



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Sol.

Given 22b 8 & 2b ae

a

2 2 24b a e ; 2 2 2 2 24a e 1 a e e3

39. If the number of terms in the expansion of n

2

2 41 , x 0,x x

is 28, then the sum of

the coefficients of all the terms in this expansion, is : (1) 243 (2) 729 (3) 64 (4) 2187 Ans: (??) Sol. Remark : Here total number of terms will be 2n + 1 which cannot be equal to 28. 40. The Boolean Expression p q q p q is equivalent to : (1) p q (2) p q (3) p q (4) p q Ans: (1) Sol.

p q p q p q p q p q q p q q p q p q

T T F F F F T T T

T F F T T F T T T

F T T F F T T T T

F F T T F F F F F So p q q p q is equivalent to p q

41. Consider

1 1 sin xf (x) tan1 sin x

, x 0,2

A normal to y = f(x) at x = 6 also passes through the point:

(1) , 06

(2) , 04

(3) (0, 0) (4) 20,3

Key. (4)

Sol. y = 1

x xcos sin2 2tan x xcos sin2 2



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y = 1 xtan tan4 2

y = x4 2

dydx

= 12

Hence, slope of normal –2 Equation of normal becomes

y4 12

= –2 x

6

The point 20,3

satisfying the equation.

42. 1/n

2nn

(n 1)(n 2)....3nlimn

is equal to:

(1) 2

9e

(2) 3log 3 2 (3) 4

18e

(4) 2

27e

Key. (4)

Sol. Let, y = 1/n

2n

(n 1)(n 2)....3nn

log y = 1 1 2 2nlog 1 1 .... 1n n n n

log y = 2n

r 1

1 rlog 1n n

nlim log y

= 2n

n r 1

1 rlim log 1n n

nlim log y

= 2

0log(1 x)dx = 3log 3 2

y = 3log3 2e = 2

27e

43. If one of the diameters of the circle, given by the equation, x2 + y2 – 4x + 6y – 12 = 0, is a chord of a circle S, whose centre is at (–3, 2), then the radius of S is:

(1) 5 (2) 10 (3) 5 2 (4) 5 3 Key. (4)



-23-

AC = 50 25 = 75

44. Let two fair six-faced dice A and B be thrown simultaneously. If E1 is the event that die A shows up four, E2 is the event that die B shows up two and E3 is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true?

(1) E1 and E3 are independent (2) E1, E2 and E3 are independent (3) E1 and E2 are independent (4) E2 and E3 are independent Key. (2)

P(E1) = 636

= 16

P(E2) = 636

= 16

P(E3) = 1836

= 12

1 2P(E E ) = 136

2 3P(E E ) = 112

1 3P(E E ) = 112

and 1 2 3P(E E E ) = 0

Hence, E1, E2 and E3 are not independent.

45. A value of q for which 2 3i sin1 2isin q q

is purely imaginary, is

(1) 1 3sin4

(2) 1 1sin3

(3) 3 (4)

6

Key. (2)

Sol. Z = 2 3i sin1 2isin q q

For Z to be purely imaginary, Z = –Z

2 3isin1 2isin q q

= 2 3i sin1 2isin q q



-24-

sin q = 13

q = 1 1sin3

46. If the sum of the first ten terms of the series

2 2 2 2

23 2 1 41 2 3 4 4 ......5 5 5 5

is 165

m, then m is equal to

(1) 100 (2) 99 (3) 102 (4) 101 Key. (4)

Sol. Here, Tn = 28 (n 1)4

5

Tn = 216 (n 1)25

Sn = 16 (n 1)(n 2)(2n 3) 125 6

S10 = 16 11 12 23 125 6

= 16 [506 1]25

= 16 50525

= 165

× 101 m = 101

47. The system of linear equations

x y z = 0

x y z = 0

x + y – z = 0

has a non-trivial solution for (1) exactly two values of (2) exactly three values of (3) infinitely many values of (4) exactly one value of Key. (2) Sol. D = 0

1 1

1 11 1

= 0

21( 1) ( 1) 1( 1) = 0



-25-

31 1 = 0 3 = 0 2( 1) = 0 = 0, 1, –1 (2) exactly three value of .

48. If the line, x 3 y 2 z 42 1 3

lies in the plane, x my z 9 , then 2 2m is equal

to (1) 5 (2) 2 (3) 26 (4) 18 Key. (2) As per given condition, 2 m = 3 … (i) and 3 2m = 5 … (ii) = 1 and m = –1 Hence, 2 + m2 = 2

49. If all the words (with or without meaning having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is

(1) 52nd (2) 58th (3) 46th (4) 59th Key. (2) Sol. A, L, L, M, S

No. of words starting with A = 4!2!

= 12

No. of words starting with L = 4! = 24

No. of words starting with M = 4!2!

= 12

No. of words starting with S A = 3!2!

= 3

No. of words starting with S L = 3! = 6 and next word formed will be SMALL.

Hence, rank of the word SMALL = 12 + 24 + 12 + 3 + 6 + 1 = 58

50. If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true?

(1) 3a2 – 34a + 91 = 0 (2) 3a2 – 23a + 44 = 0 (3) 3a2 – 26a + 55 = 0 (4) 3a2 – 32a + 84 = 0 Key. (4)

Sol. x = 16 a4

Variance = (3.5)2 = 12.25



-26-

Variance = 2i 2x

(x)n

2ix = 134 + a2

12.25 = 22134 a 16 a

4 4

494

= 2 2536 4a 256 a 32a

16

196 = 280 + 3a2 – 32a 3a2 – 32a + 84 = 0 51. A wire of length 2 units is cut into two parts which are bent respectively to form a square

of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then :

(1) x = 2r (2) 2x = 4 (3) 2x ( 4)r (4) (4 )x r Key. (1) Sol. 4x 2 r 2 2x r 1 ...(i) 2 2A x r

2

2 1 2xA x

22 1x 1 2x

For maximum / minimum

dA 0dx

12x 4 1 2x 0

2x4

2

2d A 82 0dx

For 2x4

Area is minimum

21 214r

( 4)

x 2r

x = 2r



-27-

52. Let 1

2 2xx 0

p lim 1 tan x

then log p is equal to :

(1) 12

(2) 14

(3) 2 (4) 1

Key. (1)

Sol. 1

2 2xx 0

P lim 1 tan x

2

x 0

tan xlim2xe

2

x 0

1 tan xlim2 xe

12e

1log p2

53. Let p be the point on the parabola, y2 = 8x which is at a minimum distance from the centre

C of the circle, x2 + (y + 6)2 = 1. Then the equation of the circle, passing through C and having its centre at P is :

(1) 2 2 xx y 2y 24 04

(2) x2 + y2 – 4x + 9y + 18 = 0

(3) x2 + y2 – 4x + 8y + 12 = 0 (4) x2 + y2 – x + 4y – 12 = 0 Key. (3) Sol.

Equation of normal at P y + tx = 4t + 2t3 As it passes through (0, -6) -6 = 4t + 2t3 or 2t3 + 4t + 6 = 0



-28-

or t3 + 2t + 3 = 0 or (t + 1) (t2 – t + 3) = 0 t = - 1 P (2, 4) Equation of required circle is (x – 2)2 + (y + 4)2 = 4 + 4 or x2 + y2 – 4x + 8y + 12 = 0 54. If a curve y = f(x) passes through the point (1, -1) and satisfies the differential equation,

y(1 + xy) dx = x dy, then 1f2

is equal to :

(1) 25

(2) 45

(3) 25

(4) 45

Key. (2) Sol. y (1 + xy ) dx = x dy xy2 dx = - (ydx – x dy)

2ydx xdy xxdx d

yy

Integrating both sides

2x x C

2 y

As curve passes through the point (1, -1)

1 1 C2

1C2

2x x 1

y 2 2

2

xy1 x2 2

2

xf (x)1 x2 2

11 2f

1 122 8



-29-

142

5 58

55. Let a, b and c be three unit vectors such that 3a b c b c

2 . If b

is not parallel to

c , then the angle between a and b

is :

(1) 23 (2) 5

6 (3) 3

4 (4)

2

Key. (2)

Sol. 3(a.c)b (a.b)c b c2

3a.b2

3a . b cos2

q

31.1.cos2

q

56

q

56. If 5a b

A3 2

and A adj A = A AT, then 5a + b is equal to

(1) 4 (2) 13 (3) -1 (4) 5 Key. (4)

Sol. 5a b

A3 2

A (Adj A) = A AT T

2| A | .I A A

or 1 0 5a b 5a 3

(10a 3b)0 1 3 2 b 2

or 2 210a 3b 0 25a b 15a 2b

0 10a 3b 15a 2b 13

15a – 2b = 0 ... (i) 10 a + 3b = 13 ...(ii) and 10a + 3b = 25a2 + b2 ...(iii) from (i) and (ii)



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2a b 35

which satisfies (iii) equation also

25a b 5 3 55

57. A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a

certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30°. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60°. Then the time taken (in minutes) by him, from B to reach the pillar, is :

(1) 20 (2) 5 (3) 6 (4) 10 Key. (2)

Sol.

Let velocity is V, ABV10

o

BOBO BO PO cot 60t 10 10V AB AB

oPOAO

tan 30

AB + BO = PO Cot 30° AB = PO Cot 30° - PO Cot 60°

o

BO o oPOcot 60t 10

POcot 30 POcot 60

13 1013

3

= 5 minutes

58. The distance of the point (1, -5, 9) from the plane x – y + z = 5 measured along the line x = y = z is :

(1) 103

(2) 203

(3) 3 10 (4) 10 3

Key. (4) Sol.



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A(1,-5, 9)

B

x 1 y 5 z 91 1 1

x-y+z=5

Co-ordinates of point B can be taken as (r + 1, r – 5, r + 9) As this lie on plane x – y + z – 5 = 0 (r + 1) – (r – 5) + (r + 9) – 5 = 0 or r + 10 = 0 r = - 10 B (-9, -15, -1)

2 2 2AB (1 9) ( 5 15) (9 1)

3 100 10 3 59. Two sides of a rhombus are along the lines, x – y + 1 = 0 and 7x – y – 5 = 0. If its

diagonals intersect at (-1, -2), then which one of the following is a vertex of this rhombus ?

(1) 1 8,3 3

(2) 10 7,3 3

(3) (-3, -9) (4) (-3, -8)

Key. (1)

Sol. Equation of AC

1y 2 (x 1)2

x + 2y + 5 = 0

Solving, this equation with 7x – y – 5 = 0 we get 1 8C ,3 3

60. If 0 x 2 , then the number of real values of x, which satisfy the equation cos x + cos 2x + cos 3x + cos 4x = 0, is (1) 7 (2) 9 (3) 3 (4) 5 Key. (1) Sol. cos x + cos 2x + cos 3 x + cos 4x = 0 or 2 cos 2x cos x + 2 cos 3x cos x = 0 or 2 cos x (cos 2x + cos 3x) = 0



-32-

or 5x x2cos x 2cos cos 02 2

or 5x x4cos x cos cos 02 2

3 3 7 9x , , , , , ,5 2 5 2 5 5

Topic & Difficulty Level

Q.No. Topic Difficulty Level 31. Area under the curve Average 32. Function Difficult 33. Indefinite Integral Difficult 34. L.C.D. Difficult 35. Circle Easy 36. Basic Mathematics Difficult 37. Sequence series Easy 38. Hyperbola Easy 39. Binomial Theorem Wrong question 40. Reasoning Average 41. Application of derivative Average 42. Definite Integrals Difficult 43. Circle Easy 44. Probability Average 45. Complex Numbers Easy 46. Sequence series Average 47. Matrices & determinants Easy 48. 3D-geometry Easy 49. Permutation & Combination Easy 50. Statistics Easy 51. Application of derivative Average 52. L.C.D. Average 53. Parabola Average 54. Differential Equation Average 55. Vectors Easy 56. Matrices & determinants Average 57. Heights & Distances Average 58. 3D-geometry Average 59. Straight Lines Average 60. Trigonometric equation Average

CHEMISTRY 61. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure p1 and

temperature T1 are connected through a narrow tube of negligible volume as shown in the



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figure below. The temperature of one of the bulbs is then raised to T2. The final pressure pf is :

(1) 2i

1 2

T2pT T

(2) 1 2i

1 2

T T2pT T

(3) 1 2i

1 2

T TpT T

(4) 1i

1 2

T2pT T

Ans. (1) Sol. Initially there were equal moles of ideal gas in both bulbs. Now when temperature of second bulb is raised to T2 and both are still connected it

means momentarily (until both the bulbs comes to thermal equilibrium) some moles of gas must have been got shifted to first bulb to equalize pressure of both bulbs. Suppose n1 and n2 are the moles of gases present in first and second bulbs then

1 2f 1 1 f 2 2 i 1

n np V n RT p V n RT p V RT

2

i f f

1 1 2

2p V p V p VRT RT RT

Or 2f i

1 2

Tp 2pT T

62. Which one of the following statements about water is FALSE? (1) There is extensive intramolecular hydrogen bonding in the condensed phase. (2) Ice formed by heavy water sinks in normal water. (3) Water is oxidized to oxygen during photosynthesis. (4) Water can act both as an acid and as a base. Ans. (1) Sol. Water molecules have extensive intermolecular hydrogen bonding and no intramolecular

hydrogen bonding. 63. In the Hofmann bromamide degradation reaction, the number of moles of NaOH and Br2

used per mole of amine produced are : (1) Two moles of NaOH and two moles of Br2 (2) Four moles of NaOH and one mole of Br2 (3) One mole of NaOH and one mole of Br2 (4) Four moles of NaOH and two moles of Br2 Ans. (2) Sol. 2 2 2 3 2R CONH Br 4NaOH RNH 2NaBr NaCO 2H O 64. Which of the following atoms has the highest first ionization energy? (1) K (2) Sc (3) Rb (4) Na Ans. (2) Sol. The first ionization energy of d-block elements are greater than the first ionization energy

of alkali metals. Therefore, The order is 1IE : Na : = 495.8 kJ/mol IE1 : Sc : = 631.0 kJ/mol



-34-

So, Scandium has highest ionization energy. 65. The concentration of fluoride, lead, nitrate and iron in a water sample from an

underground lake was found to be 1000 ppb, 40 ppb, 100 ppm and 0.2 ppm, respectively. This water is unsuitable for drinking due to high concentration of :

(1) Nitrate (2) Iron (3) Fluoride (4) Lead Ans. (1) Sol. The permissible limit for Nitrate to be present in drinking water is 50 parts per million. 66. The heats of combustion of carbon and carbon monoxide are –393.5 and –283.5 kJ mol-1,

respectively. The heat of formation (in kJ) of carbon monoxide per mole is (1) –676.5 (2) –110.5 (3) 110.5 (4) 676.5 Ans. (2) Sol. Equation I s 2 g 2 gC O CO H 393.5

Equation II g 2 g 2 g1CO O CO H 283.52

Subtracting equation II from equation I we get

s 2 g g1C O CO2

H 393.5 283.5 110.0 kJ / mol

67. The equilibrium constant at 298 K for a reaction A B C D . If the initial concentration of all the four species were 1 M each, then equilibrium concentration of D (in mol L-1) will be :

(1) 1.818 (2) 1.182 (3) 0.182 (4) 0.818 Ans. (1) Sol. A B C D Initial 1 1 1 1 Equilibrium 1-x 1-x 1+x 1+x

c

C DK 100

A B

1 x 1 x

1001 x 1 x

1 x 101 x

1 x 10 10x 11x 9

9x 0.818

11

(D) at equilibrium = 1 + x = 1 + 0.818 = 1.818 68. The absolute configuration of



-35-

H

CO2H

OH

H Cl

CH3 (1) (2S, 3S) (2) (2R, 3R) (3) (2R, 3S) (4) (2S, 3R) Ans. (4)

Sol.

H

H Cl

OH

COOH

CH3

1

2

3

4

‘S’

‘R’

69. For a linear plot of log (x/m) versus log p in a Freundlich adsorption isotherm, which of

the following statements is correct ? (k and n are constants) (1) Only 1/n appears as the slope (2) log (1/n) appears as the intercept (3) Both k and 1/n appear in the slope term (4) 1/n appears as the intercept Ans. (1) Sol. Freundlich adsorption 1/n isotherm equation is

1/nx kpm where k & n are constants.

So, x 1log log k log pm n (y = mx+c form)

So slope = 1/n and intercept = log k 70. The distillation technique most suited for separating, glycerol from spent-lye in the soap

industry is : (1) Steam distillation (2) Distillation under reduced pressure (3) Simple distillation (4) Fractional distillation Ans. (2) Sol. Glycerol boils at pretty high temperature of 290oC. To prevent it from decomposition it is

made to distill at lower temperature by reducing external pressure during its isolation from spent-lye.

71. Which of the following is an anionic detergent? (1) Cetytrimethyl ammonium bromide (2) Glyceryl oleate (3) Sodium stearte (4) Sodium lauryl sulphate Key: (4) Sol: Anionic detergents mainly contains 4SO as polar group.



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72. The species in which the N atom is in a state of sp hybridization is (1) 3NO (2) 2NO (3) 2NO (4) 2NO Key: (3) Sol: In 2NO , Number of valence electrons = 16 No lone pairs on central atom So, it is sp hybridized. 73. Thiol group is present in (1) Cysteine (2) Methionine (3) Cytosine (4) Cystine Key: (1) Sol:

SH OHNH2

O

Cysteine 74. Which one of the following ores is best concentrated by froth floatation method? (1) Galena (2) Malachite (3) magnetite (4) Siderite Key: (1) Sol: Galena (PbS) is a sulphide ore hence it is concentrated by froth flotation process. 75. Which of the following statement about low density polythene is FALSE? (1) Its synthesis requires dioxygen or a peroxide initiator as a catalyst (2) It is used in the manufacture of buckets, dust-bins etc. (3) Its synthesis requires high pressure (4) It is a poor conductor of electricity Key: (2) Sol: High density polythene is used for manufacturing buckets and dustbins. 76. Which of the following compounds is metallic and ferromagnetic? (1) VO2 (2) MnO2 (3) TiO2 (4) CrO2 Key: (4) Sol: CrO2 is attracted very strongly by magnetic field. 77. The product of the reaction given below is

2 2 3

1. NBS/h2. H O/K CO X

(1)

O

(2)

CO2H



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(3) (4)

OH

Key: (4) Sol:

2 2 3

1. NBS/h2. H O/K CO

Br

2 3 2K CO /H ONucleophilic substitution byOH

OH

78. The hottest region of Bunsen flame shown in the figure below is

(1) region 3 (2) region 4 (3) region 1 (4) region 2 Key: (4) Sol: Hottest part of the flame is at the tip part of the inner curve (in blue flame). 79. At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing

20% O2 by volume for complete combustion. After combustion the gases occupy 330 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is

(1) C4H8 (2) C4H10 (3) C3H6 (4) C3H8 Key: (None of them is correct) Remark : The given data does not correspond to any one of the options given. 80. The pair in which phosphorous atoms have a formal oxidation state of +3 is (1) Orthophosphorous and hypophosphoric acids (2) Pyrophosphorous and pyrophosphoric acids (3) Orthophosphorous and pyrophosphorous acids (4) Pyrophosphorous and hypophosphoric acids Key: (3) Sol: Orthophosphorous acid H3PO3 and pyrophosphorous acids H4P2O5 81. The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate : (1) 3 2CH CH OH CH (2) 3 2CH CHCl CH



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(3) 3 2CH CH CH OH (4) 3 2CH CH CH Cl

Key. (4)

Sol. Cl2 Cl is added first, hence possible intermediate among the given option will be

3 2H C CH CH|Cl

82. 2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields:

(a) 3

2 5 2 3

3

CH|

C H CH C OCH|

CH

(b) 2 5 2 2

3

C H CH C CH|

CH

(c) 2 5 3

3

C H CH C CH|

CH

(1) (c) only (2) (a) and (b) only (3) All of these (4) (a) and (c) Key. (3) Sol. Under given condition both SN and elimination is possible.

3 3

3

CH O / CH OH3 2 2 2 5 2 3

3 3

Cl H CH| | |

H C CH CH C CH C H CH C OCH| | |H CH CH

2 5 2 2 2 5 3

3 3

C H CH C CH C H CH C CH| |

CH CH

83. Which one of the following complexes shows optical isomerism? (1) trans[Co(en)2Cl2]Cl (2) [Co(NH3)4Cl2]Cl (3) [Co(NH3)3Cl3] (4) cis[Co(en)2Cl2]Cl (en = ethylenediamine) Key. (4)

Sol. Co

Cl

Clen

en

Co

Cl

Clen

en

Enantiomer

84. The main oxides formed on combustion of Li, Na and K in excess of air are, respectively :



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(1) Li2O2, Na2O2 and KO2 (2) Li2O, Na2O2 and KO2 (3) Li2O, Na2O and KO2 (4) LiO2, Na2O2 and K2O Key. (2)

Sol. 2 24Li O 2Li O oxide

2 2 22Na O Na O peroxide

2 2K O KO superoxide

85. 18 g glucose (C6H12O6) is added to 178.2 g water. The vapour pressure of water (in torr) for this aqueous solution is :

(1) 752.4 (2) 759.0 (3) 7.6 (4) 76.0 Key. (1) Sol. We assume that the vapour pressure has to be found out at 100oC. Vapour pressure of pure water at 100oC is 760 Torr. w = 18 gm; W = 178.2 gm

O S

S

18P P n w 180 0.0118 178.2P N n

180 18

O

S

P 1 0.01P

O

S

P 1.01P

S760P 752.471.01

86. The reaction of zinc with dilute and concentrated nitric acid, respectively, produces: (1) NO and N2O (2) NO2 and N2O (3) N2O and NO2 (4) NO2 and NO Key. (3)

Sol. 3 3 2 224Zn s 10HNO dil 4Zn NO 5H O N O

3 3 2 22Zn s 4HNO conc. Zn NO 2H O 2NO

87. Decomposition of H2O2 follows a first order reaction In fifty minutes the concentration of H2O2 decreases from 0.5 to 0.125 M in one such decomposition. When the concentration of H2O2 reaches 0.05 M, the rate of formation of O2 will be :

(1) 2.66 L min–1 at STP (2) 1.34×10–2 mol min–1

(3) 6.93×10–2 mol min–1 (4) 6.93×10–4 mol min–1 Key. (4)



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Sol. 2 2 2 21H O H O O2

t1/2 = 25 min

2rate K H O

3 1

1/ 2

0.693 0.6930.05 0.05 1.386 10 mole lit mint 25

Expression for rate will be,

2 2 2d H O d Orateof reaction 2

dt dt

2 2d O d H O 1dt dt 2

3

11.386 10 mol lit min2

4 16.93 10 mol lit min

88. The pair having the same magnetic moment is : [At. No. : Cr = 24, Mn = 25, Fe = 26, Co = 27] (1) [Mn(H2O) 6]2+ and [Cr(H2O)6]2+ (2) [CoCl4]2– and [Fe(H2O)6]2+

(3) [Cr(H2O)6]2+ and [CoCl4]2– (4) [Cr(H2O)6]2+ and [Fe(H2O)6]2+ Key. (4) Sol. [Cr(H2O)6]2+ and [Fe(H2O)6]2+ has same no. of unpaired electron (i.e. four) in Cr2+ and Fe2+

in respectively. Thus they have same magnetic moment i.e. 24 B.M.

89. Galvanization is applying a coating of: (1) Cu (2) Zn (3) Pb (4) Cr Key. (2) Sol. Galvanization is the process of applying a protective ‘Zn’ coating to steel or iron to prevent

rusting. 90. A stream of electrons from a heated filament was passed between two charged plates kept

at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of h / (where is wavelength associated with electrons wave) is given by

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

Key. (2) Sol. de Broglie equation

h

mv ...(1)

21K.E. mv m2

...(2)



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2mv 2K.E.m

Mv 2K.E.m ...(3)

For a charged particle moving under acceleration by applying potential ‘V’ the EK (kinetic energy) of charged particle will be

KE eV ...(4)

mv 2meV ...(5)

h2meV

...(6)

h 2meV



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Topic & Difficulty Level

Q.No. Topic Difficulty Level 61. Gaseous law Difficult 62. Bonding Easy 63. Amines Difficult 64. Periodic classification Average 65. Environment Average 66. Thermodynamics Easy 67. Chemical equilibrium Easy 68. Stereo chemistry Average 69. Surface chemistry Easy 70. Practical organic chemistry Easy 71. Surface chemistry Easy 72. Chemical bonding Easy 73. Amino acid Average 74. Metallurgy Easy 75. Polymer Easy 76. d-block elements Easy 77. Alkyl halide Average 78. Practical organic chemistry Easy 79. Gas eudiometry Wrong Question 80. d-block elements Easy 81. Hydrocarbon Easy 82. Alkyl halide Average 83. Complex compounds Average 84. s-block Average 85. Liquid solution Easy 86. p-block Easy 87. Chemical Kinetics Difficult 88. Complex Compounds Average 89. Electrochemistry Easy 90. Atomic structure Easy

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