PHYSICS Solution 2.

JEE Main 2015

PHYSICS

1. As an electron makes a transition from an excited state to the ground state of a hydrogen – like atom/ion: (1) kinetic energy, potential energy and total energy decrease (2) kinetic energy decreases, potential energy increases but total energy remains same (3) kinetic energy and total energy decrease but potential energy increases (4) its kinetic energy increases but potential energy and total energy decreases Solution As electron transits from excited state to ground state, Total energy decreases

Since, total energy = (−) Kinetic energy Potential energy

2

Therefore, total energy and potential energy decreases and kinetic energy increases.

Hence, the correct option is (4).

2. The period of oscillation of a simple pendulum is L

T 2 .g

Measured value of L is 20.0 cm known

to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1s resolution. The accuracy in the determination of g is: (1) 3% (2) 1% (3) 5% (4) 2% Solution

As 2L

Tg

2

2

4

Lg

T

2

g L T

g L T

0.1 1

220 90

g

g

0.027

g

g

3%

g

g


3. A long cylindrical shell carries positive surface charge σ in the upper half and negative surface charge –σ in the lower half. The electric field lines around the cylinder will look like figure given in: (figures and schematic and not drawn to scale)

(a) (b)

(c) (d) Solution Electric field lines originate from (+) charge and terminate at (−) charge.


4. A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz. The frequencies of the resultant signal is/are: (1) 2005 kHz, and 1995 kHz (2) 2005 kHz, 2000 kHz and 1995 kHz (3) 2000 kHz and 1995 kHz (4) 2 MHz only Solution

fr = fc ± fm

Therefore, frequency content of Resultant wave will have frequencies 1995 kHz, 2000 kHz and 2005 kHz


5. Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be

considered as an ideal gas of photons with integral energy per unit volume 4Uu t

V and pressure

1.

3

Up

V

If the shell now undergoes an adiabatic expansion the relation between T and R is:

(1) 3RT e

(2) 1

TR

(3) 3

1T

R

(4) RT e Solution For Adiabatic process

dU = −PdV

3

dU UP

dV V

1

3

dU dV

U V

1log log log

3

U V C

e e e

1/3log ( ) log Ce eU V

1/3UV C

As 4U

TV

Therefore, V = k VT4

V4/3T4 = C

V1/3T = C

1/3

1 1

( ) T

V R


6. An inductor (L = 0.03H) and a resistor (R = 0.15 kΩ) are connected in series to a battery of 15V EMF in a circuit shown below. The key K1 has been kept closed for a long time. Then at t = 0, K1 is opened and

key K2 is closed simultaneously. At t = 1ms, the current in the circuit will be: 5(e 150)

(1) 67 mA (2) 6.7 mA (3) 0.67 mA (4) 100 mA Solution

it = 0 when K1 is open and K2 is closed 0t

Ei

R

/1 ms

t

t

Ei e

R where

L

R

3 3

2

10 0.15 10

3 10( 1 ms)

t

Ei e

R

55

3 3

15 1000.67 mA

0.15 10 10

ee


7. A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM. If the Young’s modulus of the material of the

wire is Y when 1

Y is equal to:

(g = gravitational acceleration)

(1) 2

1MT Mg

T A

(2) 2

1 MT A

T Mg

(3)

2

1M

T A

T Mg

(4) 2

1MT A

T Mg

Solution

2l

Tg

When mass M is added to bob. Then 2

M

l lT

g

Since,

MglY

A l

Mg ll

A Y

2

m

Mg ll

A YT

g

1MT Mg

T AY

2

11MTA

Y Mg T


8. A red LED emits light at 0.1 watt uniformly around it. The amplitude of the electric field of the light at a distance of 1 m from the diode is : (1) 2.45 V/m (2) 5.48 V/m (3) 7.75 V/m (4) 1.73 V/m Solution

Pressure = Energy Density = Power

Area c

20 2

1

2 4

PE

r c

2

20(4 )

PE

r c

3

2 0.1E

99 102 81 3 10

6 V/mE

2.45 V/mE


9. Two coaxial solenoids of different radii carry current I in the same direction. Let 1F

be the magnetic

force on the inner solenoid due to the outer one and 2F

be the magnetic force on the outer solenoid due to

the inner one. Then:

(1) 1F

is radially inwards and 2F

is radially outwards

(2) 1F

is radially inwards and 2F 0

(3) 1F

is radially outwards and 2F 0

(4) 1 2F F 0

Solution Since, both solenoids are in equilibrium, so

Net Force = 0

Therefore, F1 = F2 = 0


10. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as Vq, where V is the volume of the gas. The value of q is:

P

V

C

C

(1) 3 5

6

(2) 1

2

(3) 1

2

(4) 3 5

6

Solution

Average time between collision = rms

Mean free path

V

Now, Mean free path = 2

V

r N

rms

3RTV

M

AV

CVt

T

2where

3

MC

r N R

Now, 2 2

2

AV

C VT

t

For adiabatic process, TVr−1 = constant(K)

2 2 1

2

r

AV

C V VK

t

2 1

2

r

AV

C VK

t

1

2

r

AVt V

1

2

rq


11. An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below:

If a student plots graphs of the square of maximum charge 2

MaxQ on the capacitor with time (t) for two

different values L1 and L2 (L1 > L2) of L then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)

(a) (b)

(c) (d) Solution Since (L1 > L2)

Therefore, rate of energy dissipation through R from L1 is less as compared to L2.


12. From a solid sphere of mass M and radius R, a spherical portion of radius 2

R is removed, as shown in

the figure. Taking gravitational potential V = 0 at ,r the potential at the centre of the cavity thus

formed is: (G = gravitational constant)

(1) GM

R

(2) 2

3

GM

R

(3) 2GM

R

(4) 2

GM

R

Solution VNet = VM − VM/8

2 2 2

3 3

3 /8 30

2 4 2 /8 4

GM R R GM R

R R

2 2

3 3

11 3

8 8

GMR GMR

R R

8

8

GM GM

R R

Hence, the correct option is (1). 13. A train is moving on a straight track with speed 20 m s−1. It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound = 320 m s−1) close to: (1) 12% (2) 18% (3) 24% (4) 6% Solution When Train approaches

1

320 1000 3201000

320 20 300f

When train goes away

2

320 1000 3201000

320 20 340f

Therefore, 1 2

1 1

340 300 40

340 340

f ff

f f

1

0.117 12%f

f

Hence, the correct option is (1). 14.

Given in the figure are two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks of 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is : (1) 80 N (2) 120 N (3) 150 N (4) 100 N Solution

For A For B

F = N N N F

fAB = 20 N fBW = fAB + 100

fBW = 120 N

Hence, the correct option is (2). 15. Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and its height is h then z0 is equal to:

(1) 3

4

h

(2) 5

8

h

(3) 23

8

h

R

(4) 2

4

h

R

Solution From base its h/4

Therefore, 04

h

Z h3

4

h

Hence, the correct option is (1). 16. A rectangular loop of sides 10 cm and 5 cm carrying a current I of 12 A is placed in different orientations as shown in the figures below:

(a) (b)

(c) (d) If there is a uniform magnetic field of 0.3 T in the positive z direction, in which orientations the loop would be in (i) stable equilibrium and (ii) unstable equilibrium? (1) (a) and (c), respectively (2) (b) and (d), respectively (3) (b) and (c), respectively (4) (a) and (b), respectively Solution

By Right hand thumb Rule angle between and

M B is 0° in option (b) and 180° in option (d)

Hence, the correct option is (2). 17.

In the circuit shown, the current in the 1Ω resistor is : (1) 0A (2) 0.13 A, from Q to P (3) 0.13 A, from P to Q (4) 1.3 A, from P to Q Solution Apply Kirchoff’s Law

6

05 1 3

V q V V

3V – 27 + 15V + 5V + 30 = 0

V = −0.13V

Therefore, (1 )

0.130.13 A

1 i (from Q to P)

Hence, the correct option is (2). 18. A uniformly charged solid sphere of radius R has potential V0 (measured with respect to) on its

surface. For this sphere the equipotential surface with potentials 0 0 0 03 5 3, , and

2 4 4 4

V V V V have radius R1, R2,

R3 and R4 respectively. Then (1) R1 ≠ 0 and (R2 – R1) > (R4 – R3) (2) R1 = 0 and R2< (R4 – R3) (3) 2R < R4 (4) R1 = 0 and R2 > (R4 – R3) Solution For N conducting sphere

2 2in 3

32

KV R x

R

0

KV

R

(at surface)

out

KV

r

where (r > R) and in 0 outV V V

Now

0in

3

2

VV R1 = 0

0in

5

4

VV 2

2

RR

0out

3

4

VV R4 = 4R

Hence, the correct options are (2) and (3). 19. In the given circuit, charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF. Q2 as a function of ‘C’ is given properly by: (figures are drawn schematically and are not to scale)

(a) (b)

(c) (d) Solution

1 1 1

3

effC C

3

3

eff

CC

C

Therefore, 3

3

CEQ

C

23

across F

CEV

C

Thus, 2

2

3

CEQ

C


20. A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in the y direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to: (1) 50% (2) 56% (3) 62% (4) 44% Solution

Apply Momentum Conservation

2ˆ ˆ2 3 m ci mvj mv

2 ˆ ˆ( )3

v

v i j

Therefore, energy loss 2

2 21 1 1 84 2 3

2 2 2 9

vm v mv m

2

2 4 53

3 3

mv mvmv

Therefore, fraction loss 2

2

5 556%

3(3 ) 9

mv

mv

Hence, the correct option is (2). 21. Monochromatic light is incident on a glass prims of angle A. If the refractive index of the material of the prism is μ, a ray, incident at an angle θ, on the face AB would get transmitted through the face AC of the prism provided:

(1) 1 1 1sin sin sinA

(2) 1 1 1cos sin sinA

(3) 1 1 1cos sin sinA

(4) 1 1 1sin sin sinA

Solution

For face AB

1 sin θ = μ sin r

For face AC

μ sin (A – r) < 1 sin 90

(A – r) < sin−1 1

1 1sin .r A

1 1sin sin sinr A

1sin 1sin sinA

1 1 1sin sin sinA

Hence, the correct option is (4). 22. From a solid sphere of mass M and radius R and cube of maximum possible volume is cut. Moment of inertia of cube about an axis passing through its centre and perpendicular to one of its faces is:

(1) 2

16 2

MR

(2) 24

9 3

MR

(3) 24

3 3

MR

(4) 2

32 2

MR

Solution

For Maximum Volume

2 3R a 3

34/3cube

MaM

R

3 8 2

4 3 3 3cube

M MM

2 2 22 4 4

6 3 6 3 9 3

cubecube

M a M R MRI

Hence, the correct option is (2). 23. Match List – I (Fundamental Experiment) with List – II (its conclusion) and select the correct option from the choices given below the list :

List – I List – II (A) Franck-Hertz Experiment (i) Particle nature of light (B) Photo-electric, experiment (ii) Discrete energy levels of

atom (C) Davison – Germer Experiment (iii) Wave nature of electron (iv) Structure of atom

(1) (A) – (ii) (B) – (iv) (C) – (iii) (2) (A) – (ii) (B) – (i) (C) – (iii) (3) (A) – (iv) (B) – (iii) (C) – (ii) (4) (A) – (i) (B) – (iv) (C) – (iii) Solution Frank-Hertz experiment → Discrete energy level of Atom

Photoelectric effect → for particle Nature of tyth

Davison – Germer experiment → for wave nature of

Hence, the correct option is (2). 24. When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 × 10−4 m s−1. If the electron density in the wire is 8 × 1028 m−3, the resistivity of the material is close to: (1) 1.6 × 10−7 Ω m (2) 1.6 × 10−6 Ω m (3) 1.6 × 10−5 Ω m (4) 1.6 × 10−8 Ω m Solution

Pl

V IR IA

d

PlV nAV e

A

V = PlnVde

4 19

28

5 10 10

0.1 2.5 8 10 1.6d

VP

lnV e

P = 1.6 × 10−5 Ω m

Hence, the correct option is (3). 25. For a simple pendulum, a graph is plotted between its kinetic energy (KE) and potential energy (PE) against its displacement d. Which one of the following represents these correctly? (graphs are schematic and not drawn to scale)

(a) (b)

(c) (d) Solution KE + PE = TE = Constant and at mean position KE is max and PE = 0

At extreme position, KE = 0 and PE = Max.

Hence, the correct option is (1). 26. Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed of 10 m/s and 40 m/s respectively. Which of the following graph best represents the time variation of relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g = 10 m/s2) (The figures are schematic and not drawn to scale)

(a) (b)

(c) (d) Solution For Ball (1)

2

1 102

gt

y t

For Ball (2) 2

2 402

gt

y t

(y2 – y1) = 30t

Therefore, straight line till Ball (2) Reaches ground then only ball (1) will travel with equation 2

102

gt

y t so parabola graph afterwards.

Hence, the correct option is (2). 27. A solid body of constant heat capacity 1 J/°C is being heated by keeping it in contact with reservoirs in two ways: (i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same amount of heat. (ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat. In both the cases body is brought from initial temperature 100°C to final temperature 200°C. Entropy change of the body in the two cases respectively is : (1) ln2, ln2 (2) ln2, 2ln2 (3) 2ln2, 8ln2 (4) ln2, 4ln2 Solution Case (I):

Since, dT

S CT

Therefore, 423 473

373 423

4731 ln

373

dT dTS

T T

Case (II): 385.5 398 410.5 423 435.5 448 460.5 473

373 385.5 398 410.5 423 435.5 448 460.5

1dT dT dT dT dT dT dT dT

ST T T T T T T T

=

473ln

373

Hence, NO option is correct.

28. Assuming human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that human eye can resolve at 500 nm wavelength is: (1) 30 μm (2) 100 μm (3) 300 μm (4) 1 μm Solution

Since 1.22

D

9

4

1.22 500 10

25 10 2

5524.4 10

12.2 102

Minimum separation = (25 × 10−2)θ

2 524.425 10 10

2

= 30 μm

Hence, the correct option is (1). 29. Two long current carrying thin wires, both with current I, are held by insulating threads of length L and are in equilibrium as shown in the figure, with threads making an angle ‘θ’ with the vertical. If wires have mass λ per unit length then the value of I is: (g = gravitational acceleration)

(1) 2sincoso

gL

(2) 0

2 tangL

(3) 0

tangL

(4) 0

sincos

gL

Solution At equilibrium,

T cos θ = Mg

T sin θ = FM

tan MF

Mg

2( / ) .

tan( )/ 2 2 sin

MF l I

Mg l L g

2.

tan4 sin

I

Lg

0

4 sin tanLgI

2sincos

LgI

Hence, the correct option is (1). 30. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygen’s principle leads us to conclude that as it travels, the light beam : (1) goes horizontally without any deflection (2) bends downwards (3) bends upwards (4) becomes narrower Solution Since the light beam is parallel to the ground there will not be any change in the refractive index of its

path. So neither refraction nor total internal reflection will take place. As a result the light beam will go

horizontally without any deflection.


CHEMISTRY

31. Which of the following is the energy of a possible excited state of hydrogen? (1) –6.8 eV (2) –3.4 eV (3) +6.8 eV (4) +13.6 eV Solution

Energy of hydrogen atom 2

213.6 eVn

ZE

n

where Z = 1 and n = 2. Substituting, we get 1

13.6 3.4 eV4

E

Hence, the correct option is (2) 32. In the following sequence of reactions:

4 2 2

4

KMnO SOCl H /Pd

BaSOToluene A B C

the product C is

(1) C6H5CH3 (2) C6H5CH2OH (3) C6H5CHO (4) C6H5COOH Solution

Hence, the correct option is (3) 33. Which compound would give 5-keto-2-methyl hexanal upon ozonolysis?

Solution

Hence, the correct option is (1) 34. The ionic radii (in Å) of N3, O2 and F are, respectively, (1) 1.36, 1.71 and 1.40 (2) 1.71, 1.40 and 1.36 (3) 1.71, 1.36 and 1.40 (4) 1.36, 1.40 and 1.71 Solution All of these are isoelectronic species, that is, they contain the same number of electrons. As the negative charge increases, ionic radius increases in the order N3 > O2 > F. Hence, the correct option is (2) 35. The color of KMnO4 is due to (1) d–d transition (2) L M charge transfer transition (3) –* transition (4) M L charge transfer transition Solution Color of KMnO4 is due to charge transfer from O2 to Mn. Hence, the correct option is (2)

36. Assertion: Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen. Reason: The reaction between nitrogen and oxygen requires high temperature. (1) Both assertion and reason are correct, but the reason is not the correct explanation for the assertion (2) The assertion is incorrect, but the reason is correct (3) Both the assertion and reason are incorrect (4) Both assertion and reason are correct, and the reason is the correct explanation for the assertion Solution Here, both assertion and reason are true because nitrogen does not react with oxygen in the atmosphere as it is highly stable and requires very high temperature. Also, the reaction to form oxides takes place at higher temperature. Hence, the correct option is (4) 37. Which of the following compounds is not an antacid? (1) Cimetidine (2) Phenelzine (3) Ranitidine (4) Aluminium hydroxide Solution This is because phenelzine is a tranquilizer, whereas rests all are antacids. Hence, the correct option is (2) 38. In the context of the Hall-Heroult process for the extraction of Al, which of the following statement is false? (1) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity (2) Al3+ is reduced at the cathode to form Al (3) Na3AlF6 serves as the electrolyte (4) CO and CO2 are produced in this process Solution Al2O3 acts as an electrolyte and Na3AlF6 acts as a solvent, rest all the statements are true. Hence, the correct option is (3) 39. Match the catalysts to the correct processes: Catalyst Process a. TiCl3 (i) Wacker process b. PdCl2 (ii) Ziegler-Natta polymerization c. CuCl2 (iii) Contact process d. V2O5 (iv) Deacon’s process (1) a(ii), b(i), c(iv), d(iii) (2) a(ii), b(iii), c(iv), d(i) (3) a(iii), b(i), c(ii), d(iv) (4) a(iii), b(ii), c(iv), d(i) Solution TiCl3 used as a catalyst in Zeigler Natta polymerization. PdCl2 used in Wacker’s process. CuCl2 is used in Deacon’s process. V2O5 is a catalyst used in contact process. Hence, the correct option is (1)

40. In the reaction

the product E is

Solution

Hence, the correct option is (2) 41. Which polymer is used in the manufacture of paints and lacquers? (1) Glyptal (2) Polypropene (3) Polyvinyl chloride (4) Bakelite Solution Glyptal is used in the manufacture of paints and lacquers. Hence, the correct option is (1). 42. The number of geometric isomers that can exist for square planar [Pt (Cl) (py) (NH3)(NH2OH)]+ is (py = pyridine) (1) 3 (2) 4 (3) 6 (4) 2 Solution Type Mabcd has 3 geometrical isomers.

Hence, the correct option is (1) 43. Higher order (>3) reactions are rare due to (1) Increase in entropy and activation energy as more molecules are involved (2) Shifting of equilibrium towards reactants due to elastic collisions (3) Loss of active species on collision (4) Low probability of simultaneous collision of all the reacting species Solution Higher order reactions are rare because the probability of collision of more than three species is very rare. Hence, the correct option is (4) 44. Which among the following is the most reactive? (1) Br2 (2) I2 (3) ICl (4) Cl2 Solution Interhalogen compounds are more reactive than halogens due to low bond dissociation enthalpy. Hence, the correct option is (3) 45. Two faraday of electricity is passed through a solution of CuSO4. The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 amu) (1) 63.5 g (2) 2 g (3) 127 g (4) 0 g Solution According to Faraday’s law of electrolysis and according to the reaction, 2Cu 2 Cue

2F will deposit 63.5 g of Cu. Hence, the correct option is (1) 46. 3 g of activated charcoal was added to 50 mL of acetic acid solution (0.06 N) in a flask. After an hour it was filtered and the strength of the filtrate was found to be 0.042 N. The amount of acetic acid adsorbed (per gram of charcoal) is (1) 36 mg (2) 42 mg (3) 54 mg (4) 18 mg Solution Initial mmoles of CH3COOH = 0.06 50 Final mmoles of CH3COOH = 0.042 50

So, the mass of CH3COOH adsorbed per gram of charcoal = 3 30.06 0.042 50 10 60 1( )

18 mg0

3

Hence, the correct option is (4) 47. The synthesis of alkyl fluorides is best accomplished by (1) Sandmeyer’s reaction (2) Finkelstein reaction (3) Swarts reaction (4) Free radical fluorination

Solution Alkyl fluorides are best prepared by Swarts reaction

R X AgF R F AgX

Hence, the correct option is (3) 48. The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (mol. wt. 206). What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin?

(1) 1

206 (2)

2

309 (3)

1

412 (2)

1

103

Solution According to the chemical reaction,

2

8 7 3 8 7 3 21/412 mo1/ m l l206 o

(2C H SO Na Ca C H SO C 2 a) a N

Maximum uptake of Ca2+ ions = 1 1

206 2 412

Hence, the correct option is (3) 49. Which of the vitamins given below is water soluble? (1) Vitamin D (2) Vitamin E (3) Vitamin K (4) Vitamin C Solution Vitamins B and C are water soluble, while vitamins A, D, E and K are fat soluble (or water insoluble). Hence, the correct option is (4) 50. The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is (1) Ion–dipole interaction (2) London force (3) Hydrogen bond (4) Ion–ion interaction Solution Hydrogen bond is a special case of dipole–dipole interaction, so for stationary polar molecules,

3

1F

r

Hence, the correct option is (3) 51. The following reaction is performed at 298 K.

2 22NO(g) O (g) 2NO (g)

The standard free energy of formation of NO(g) is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of NO2(g) at 298 K? (Kp = 1.6 1012)

(1) 86600 + R(298) ln(1.6 1012) (2) 12ln(1.6 10 )

86600(298)R

(3) 120.5[2 86,600 (298)ln (1.6 10 )]R (4) 12(298) ln(1.6 10 ) 86600R

Solution

2 2

o o o o(NO ) (NO) (O ) 2 [2 ] 2 [2 86,600 0]G G G G x

Also, ∆Go = RT ln Kp

12

12

(298)ln(1.6 10 ) 2 2 86,600

0.5[2 86600 (298)ln(1.6 10 ]

R x

x R

Hence, the correct option is (3) 52. Which of the following compounds is not colored yellow? (1) K3[Co(NO2)6] (2) (NH4)3[As(Mo3O10)4] (3) BaCrO4 (4) Zn2[Fe(CN)6] Solution Zn2[Fe(CN)6] is white in color as it does not have unpaired electrons. Hence, the correct option is (4) 53. In Carius method of estimation of halogens, 250 mg of an organic compound gave 141 mg of AgBr. The percentage of bromine in the compound is (at mass Ag = 108; Br = 80) (1) 36 (2) 48 (3) 60 (4) 24 Solution

Atomic mass of Br Weight of AgBr 100Percentage of Br

Molecular mass of AgBr Weight of organic substance

80 141 100 24%

188 250

Hence, the correct option is (4) 54. Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å. The radius of sodium atom is approximately (1) 3.22 Å (2) 5.72 Å (3) 0.93 Å (4) 1.86 Å Solution

For bcc arrangement, 3 1.732 4.29

1.86 4 4

Åa

r

Hence, the correct option is (4) 55. Which of the following compounds will exhibit geometrical isomerism? (1) 3-Phenyl-1-butene (2) 2-Phenyl-1-butene (3) 1,1-Diphenyl-1-propane (4) 1-Phenyl-2-butene Solution 1-Phenyl-2-butene

Hence, the correct option is (4)

56. The vapour pressure of acetone at 20°C is 185 torr. When 1.2 g of a non-volatile substance was dissolved in 100 g of acetone at 20°C, its vapour pressure was 183 torr. The molar mass (g mol1) of the substance is (1) 64 (2) 128 (3) 488 (4) 32 Solution Using the colligative property (relative lowering of vapour pressure)

2

0 s 2 1

s

2

2

1

185 183 1.2 5864 g mol

183 100

p p w M

p

MM

M w

Hence, the correct option is (1) 57. From the following statements regarding H2O2, choose the incorrect statement. (1) It decomposes on exposure to light (2) It has to be stored in plastic or wax lined glass bottles in dark. (3) It has to be kept away from dust (4) It can act only as an oxidizing agent Solution The oxidation state of oxygen in H2O2 is 1 (peroxide). So, oxygen can increase and decrease its oxidation number which means it can act as a reducing as well as an oxidizing agent. Hence, the correct option is (4) 58. Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy? (1) BeSO4 (2) BaSO4 (3) SrSO4 (4) CaSO4 Solution BeSO4 is soluble in water as its hydration energy is more than lattice enthalpy. Also, the solubility of sulphates of Group 2 metals in water decreases as we move down the group. Hence, the correct option is (1) 59. The standard Gibbs energy change at 300 K for the reaction 2A B C is 2494.2 J. At a given

time, the composition of the reaction mixture is [A] = 1

2, [B] = 2 and [C] =

1

2. The reaction proceeds in

the : [R = 8.314 J/K/mol, e = 2.718] (1) Reverse direction because Q > KC (2) Forward direction because Q < KC (3) Reverse direction because Q < KC (4) Forward direction because Q > KC Solution

2 2

[C][B] (1/ 2)(2)4

[A] (1/ 2)Q

Using the equation,

o ln 2494.2 8.314 300ln 4G G RT Q

we would get a positive quantity. Now, oClnG RT K . So, Cln( / )G RT Q K .

As ∆G is positive, so Q > KC and thus the reaction shifts in the reverse direction. Hence, the correct option is (1)

60. Which one has the highest boiling point? (1) Ne (2) Kr (3) Xe (4) He Solution London dispersion forces increases from He to Xe because the molecular mass increases, so the boiling point also increases from He to Xe. Hence, the correct option is (3)

Mathematics

61. The sum of coefficients of integral powers of x in the binomial expansion of 50

1 2 x is

(1) 501(3 )

2

(2) 501(3 1)

2

(3) 501(2 1)

2

(4) 501(3 1)

2

Solution 50

50 50 50 2

0

(1 2 ) (1) ( 2 )rr

r

x C x

5050 /2

0

( 2) ( )r rr

r

C x

Therefore, for integer powers of x, r < 0, 2, 4, 6, …, 50

So the required sum of coefficients = 50 50 2 50 4 50 500 2 4 50(2) (2) (2)C C C C (1)

Since, 50 50 50 50 2 50 4 50 500 2 4 50(1 2) (1 2) 2 (2) (2) (2)C C C C (2)

In view of (1) and (2),

Required sum 501[(3) 1]

2


62. Let f(x) be a polynomial of degree four having extreme values at x = 1 and x = 2 If 2x 0

( )lim 1 3,

f x

x

the f(2) is equal to:

(1) −4

(2) 0

(3) 4

(4) −8

Solution

( ) ( 1)( 2)( )f x a x x x k

3 2( ) [ (3 ) (2 3 ) 2 ]f x a x k x k x k

4 3 2(3 ) (2 3 )

( ) 24 3 2

x k x k xf x a kx C

2

2 2

( ) (3 ) (2 3 ) 2

4 3 2

f x x k x k k Ca

x x x

2

2 20 0

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

2x x

f x k x kx Ca

x x

(given)

2

20

[ (2 3 ) 2] 4 2lim 3

2x

a k x kax Ca

x

Ca = 0 (1)

and 0

[ (2 3 ) 2](2 ) 4lim 3

4x

a k x ka

x

ka = 0 (2)

and 0

[(2 3 ) 2](2)lim 3

4x

k

a(2 + 3k) + 2 = 6

2a + 3ak = 4 a = 2 ( ka = 0)

C = 0 and K = 0

4 4

3 2 3 2( ) 2 2 24 2

x xf x x x x x

f(2) = 8 – 16 + 8 = 0


63. The mean of the data set comprising of 16 observations is 16. If one of the observation valued 16 is

deleted and three new observations valued 3, 4 and 5 are added to the data, then the mean of the resultant

data, is

(1) 16.0

(2) 15.8

(3) 14.0

(4) 16.8

Solution We have,

n = 16, 16x

256xi nx

Therefore, new mean, 256 16 3 4 5

18x

= 14


64. The sum of first 9 terms of the series

3 3 3 3 3 31 1 2 1 2 3....

1 1 3 1 3 5

is

(1) 96

(2) 142

(3) 192

(4) 71

Solution The given series:

3 3 3 3 3 31 1 2 1 2 3

1 1 3 1 3 5

….

Therefore, 3 3 3 31 2 3

1 3 5 (2 1)n

nt

n

2

2

( 1)

2

n n

n

Therefore, 29 10

29

1 2

( 1) 1S

4 4n k

nk

1 (10)(11)(21)1 96

4 6


65. Let O be the vertex and Q be any point on the parabola, x2 = 8y. If the point P divides the line segment

OQ internally in the ratio 1:3, then the locus of P is

(1) y2 = x

(2) y2 = 2x

(3) x2 = 2y

(4) x2 = y

Solution

Since P divides OQ internally in the ratio 1:3.

,4

ax

2

2

2

84 2

aa

y

21

2y x


66. Let α and β be the roots of equation x2 – 6x – 2 = 0. If an = an – βn, for n ≥ 1, then the value of

10 8

9

2

2

a a

a

is equal to:

(1) −6

(2) 3

(3) −3

(4) 6

Solution α, β are roots of x2 – 6x – 2 = 0 α + β = 6, αβ = −2

Now, αn – βn = an

10 10 8 8

10 8

9 99

2 ( ) 2( )

2 2( )

a a

a

8 2 8 2

9 9

( 2) ( 2)

2( )

8 8

9 9

(6 ) (6 )

2( )

[ α2 − 6α – 2 = 0 and β2 − 6β – 2 = 0]

9 9

9 9

6( )3

2( )

Hence, the correct option is (2) 67. If 12 identical balls are to be placed in 3 identical boxes, then the probability that one of the boxes

contains exactly 3 balls is

(1) 10

255

3

(2) 12

1220

3

(3) 11

122

3

(4) 11

55 2

3 3

Solution

Choose 3 balls out of 12 in 123C ways and distribute the remaining 9 balls in two boxes in 29 ways.

However total number of possible ways = (3)12

Therefore required probability 12 9

3

12

(2)

(3)

C

119

12

12 11 10 2 55 2

6 3 3 3

Hence, the correct option is (4) Note: This option is correct only when the boxes are different and not identical.

68. A complex number z is said to be unimodular if |z| = 1. Suppose z1 and z2 are complex numbers such

that 1 2

1 2

2

2

z z

z z is unimodular and z2 is not unimodular. Then the point z1 lies on a:

(1) straight line parallel to y-axis.

(2) circle of radius 2.


(4) straight line parallel to x-axis.

Solution

1 2

1 2

21;

2

z z

z z

[z2] ≠ 1

2 2

1 2 1 22 2z z z z 1 2 1 2 1 2 1 2( 2 )( ) (2 )(2 )z z z z z z z z

2 21 1 2 2 1 2| | 2 2 4 | |z z z z z z 2 2

1 2 1 2 1 24 2 2 | | | |z z z z z z

|z1|2 + 4 |z2|2 = 4|z1|2 |z2|2 |z1|2 (1 − |z2|2) – 4(1 − |z2|2) = 0

(|z1|2 – 4)(1 |z2|2) = 0 |z1|2 = 4 or |z2|2 = 1

But |z2| ≠ 1 |z1|2 = 4

|z1| = 2

z1 lies on a circle of radius 2.


69. The integral 2 4 3/4

d

( 1)x

x x equals:

(1) 1

4 4( 1) x c

(2) 1

4 4( 1) x c

(3)

14 4

4

1

xc

x

(4)

14 4

4

1

xc

x

Solution

2 4 3/4( 1)

dxI

x x

3/4

5

4

11

dx

xx

Put x−4 = t

5

4dx dt

x

3/44(1 )

dtI

t

1/41 (1 )

4 1/4

tC

= −(1+x−4)1/4 + C


70. The number of points, having both co-ordinates as integers that lie in the interior of the triangle with

vertices (0, 0) (0, 41) and (41, 0), is

(1) 861

(2) 820

(3) 780

(4) 901

Solution

Number of points of desired type = 1 + 2 + 3 + … + 39 40 39

7802


71. The distance of the point (1, 0, 2) from the point of intersection of the line 2 1 2

3 4 12

x y z and

the plane x – y + z = 16, is

(1) 8

(2) 3 21

(3) 13

(4) 2 14

Solution

Let 2 1 2

3 4 12

x y z

x = 3λ + 2, y = 4λ – 1, z = 12λ + 2

If above is the point of intersection, then (3λ + 2) – (4λ – 1) + 12λ + 2 = 16

Therefore, λ = 1 point of intersection is P ≡ (5, 3, 14) and Q ≡ (1, 0, 2)

16 9 144 13PQ

Hence, the correct option is (3) 72. The equation of the plane containing the line 2x – 5y + z = 3; x + y + 4z = 5, and parallel to the plane,

x + 3y + 6z = 1, is

(1) x + 3y + 6z = −7

(2) x + 3y + 6z = 7

(3) 2x + 6y + 12z = −13

(4) 2x + 6y + 12z = 13

Solution Since plane is parallel to x + 3y + 6z = 1 direction ratios of normal to required plane are <1, 3, 6>

Also plane contains the line

{2x – 5y + z = 3; x + y + 4z = 5} (1)

Therefore, the required plane contains the line having direction ratios given by 21 7 7

x y z

or <3, 1, −1>

Also point (4, 1, 0) lies on line (1).

Therefore, Equation of the required plane will be

1(x – 4) + (y – 1)(3) + (z – 0)(6) = 0 or x + 3y + 6z = 7


73. The area (in sq. units) of the region described by {(x, y) : y2 ≤ 2x and y ≥ 4x – 1} is

(1) 5

64

(2) 15

64

(3) 9

32

(4) 7

32

Solution R = {(x, y); y2 ≤ 2x and y ≥ 4x – 1}

R = shaded area line parabola( ) ( )B B

A A

y y

y y

x dy x dy

1 1 2

1/2 1/2

1( 1)

5 2

yy dy dy

1 12 3

1/2 1/2

1 1 9sq.units

4 2 2 3 32

y yy

Hence, the correct option is (3) 74. If m is the A.M. of two distinct real numbers l and n (l, n > 1) and G1, G2 and G3 are three geometric

means between l and n, then 4 4 41 2 32G G G equals.

(1) 1 lm2n

(2) 4 lmn2

(3) 4 l2m2n2

(4) 4 l2mn

Solution

;2

l nm

(1, n > 1) (1)

and l, G1, G2, G3 n are in G.P.

1/4

1 ,n

G ll

1/2

2 ,n

G ll

3/4

3

nG l

l

Therefore, 2 3

4 4 4 41 2 3 2 3

2( ) 2( ) ( )

n n nG G G l

l l l

= nl(n + 1)2 = 4m2nl (from (1))


75. Locus of the image of the point (2, 3) in the line (2x – 3y + 4) + k (x – 2y + 3) = 0. k R, is a :

(1) straight line parallel to y-axis.

(2) circle of radius 2 .


(4) straight line parallel to x-axis.

Solution Note that P(1, 2) is the point of intersection of given pair of lines. Thus we are to find the locus of image of point R(2, 3) on the given line. Let Q(x, y) be the image of R.

Clearly PR = PQ as ∆PRL ~ ∆PQL

(x – 1)2 + (y – 2)2 = (z – 1)2 + (3 – 2)2

(x – 1)2 + (y – 2)2 = 2( 2)

which is a circle of radius 2 .


76. The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latera recta

to the ellipse 2 2

1,9 5

x y is :

(1) 18

(2) 27

2

(3) 27

(4) 27

4

Solution

Equation of given ellipse is 2 2

19 5

x y

a2 = 9, b2 = 5

Now, b2 = a2 (1 – e2) 5 = 9(1 – e2) e = 2/3

One of the end points of the latus recta is P(ae, b2/a) ≡ (2, 5/3)

Equation of tangent to the ellipse at P is 2 5

19 15

x y or 2x + 3y = 9 or 1

9/2 3

x y

Area of quadrilateral 1 9

4 (3) 272 2

sq. units


77. The number of integers greater than 6,000 that can be formed, using the digits 3, 5, 6, 7 and 8, without

repetition, is

(1) 192

(2) 120

(3) 72

(4) 216

Solution Any no. greater than 6000 but less than 10,000 that can be formed using the digits, 3, 5, 6, 7 and 8, without repetition has its thousand place digit 6, 7 or 8 Therefore, for the first left place, number of choices = 3

For second left place, number of choices = 4

For third left place, number of choices = 3

For place number of choices = 2

Therefore, number of 4 digit numbers greater than 6000 = 72

Now if we use all the 5 integers the number obtained is definitely greater than 6000, number of such numbers

= 5! = 120

Therefore, total numbers formed = 72 + 120 = 192

Now if we used all the 5 integers the number obtained is definitely greater than 6000, number of such numbers

= 5! = 120

Therefore, total numbers formed = 72 + 120 = 192


78. Let A and B be two sets containing four and two elements respectively. Then the number of subjects

for the set A × B, each having at least three elements is

(1) 256

(2) 275

(3) 510

(4) 219

Solution n(A) = 4; n(B) = 2 We are to find number of sets of the form {xi, yi): xi A, yi B and i ≥ 3}

Case (i) when set contains exactly one element: i.e. {(x, y)}

x1 has 4 choices and y1 has 2

8 sets of such type

Case (ii) When set contains exactly two elements: i.e. {(x1, y1), (x2, y2)}

No. of such that 82 28C

Case (iii) Set contains no elements: i.e < y or

Therefore, subsets of (A × B) having at least 3 elements = (2)8 – [8 + 28 + 1] = 219


79. Let 1 1 1

2

2tan tan tan ,

1

xy x

x where

1| | .

3x Then a value of y is

(1) 3

2

3

1 3

x x

x

(2) 3

2

3

1 3

x x

x

(3) 3

2

3

1 3

x x

x

(4) 3

2

3

1 3

x x

x

Solution

Since, 1 1 1tan tan tan1

x yx y

xy

for xy < 1.

Now 2

2 2

2 2.

1 1

x xx

x x

2

2 2

1 1 22 2

1 1

x

x x

Further 1

3x 2 1

03

x 210

3x 22

1 13

x

2

1 31

1 2x

2

22 3

1 x

2

20 2 1

1 x

2

2. (0,1)

1

xx

x

1 1

2

2tan ( ) tan

1

xx

x

21

2

2

21

1tan2

11

x

xx

x

31 1

2

3tan tan ( )

1 3

x xy

x

(given)

3

2

3

1 3

x xy

x


80. The integral 4 2

2 2

2

logd

log log(36 12 )

xx

x x x is equal to:

(1) 4

(2) 1

(3) 6

(4) 2

Solution

4 2

2 2

2

log

log log(36 12 )

x dxI

x x x

4 2

2 2

2

log

log log(6 )

xdx

x x

4

2

log

log log(6 )

xdx

x x

(1)

4

2

log(2 4 )

log(2 4 ) log( )

xI dx

x x

By using property ( ) ( )b b

a a

f x dx f a b x dx

4

2

log(6 )

log(6 ) log

xI

x x

(2)

(1) + (2) gives,

4

2

2 1 2I dx I = 1


81. The negation of ~ (~ )s r s is equivalent to:

(1) ( ~ )s r s

(2) ( ~ )s r s

(3) s r

(4) ~s r

Solution Negation of ~ (~ )s r s is given by,

~ [~ (~ )]s r s ~ (~ )s r s

( ~ )

( ) ( ~ )

s r s

s r s s

( ) ( );s r C C = contradiction

( )s r [ p c p ]

Hence, the correct option is (3) 82. If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading

to the foot of the tower, are 30°, 45° and 60° respectively, then the ratio, AB : BC, is

(1) 3 : 2

(2) 1: 3

(3) 2 : 3

(4) 3 :1

Solution

tan 60h

x x = h cot 60° (1)

tan 45h

BC x

BC = h – h cot 60° (2)

and tan 30h

AB h

AB = h cot 30° − h (3)

cot 30 1 3 1 3

:11 cot 60 1

13

AB BC


83. 0

(1 cos 2 )(3 cos )lim

tan 4

x

x x

x x is equal to:

(1) 3

(2) 2

(3) 1

2

(4) 4

Solution

0

(1 cos 2 )(3 cos )lim

tan 4x

x x

x x

2

02

(2sin )(3 cos )lim

tan 44

4

x

x x

xx

x

0

1lim(3 cos ) 2

2 xx


84. Let , anda b c

be three non-zero vectors such that no two of them are collinear and

1( ) | b || c | .

3a b c a

If θ is the angle between vectors and ,b c

then a value of sin θ is

(1) 2

3

(2) 2

3

(3) 2 3

3

(4) 2 2

3

Solution

1 1 1( ) | b || c | ( ) | b || c | ( ) ( ) | b || c |

3 3 3a b c a c a b a c b a c a b a

1| b || c | ( ) ( )

3c b a c a b

Since anda b

are not collinear,

1

| b || c | ( ) 03

c b

and 0c a

1 1 8 2 2

cos 0 cos sin3 3 3 3


85. If

1 2 2

2 1 2

2

A

a b

is a matrix satisfying the equation AAT = 9I, where I is 3 × 3 identity matrix, then

the ordered pair (a, b) is equal to:

(1) (−1, 1)

(2) (2, 1)

(3) (−2, −1)

(4) (2, −1)

Solution

1 2 2

2 1 2 ; 9

2

TA A A I

a b

1 2 2 1 2 9 0 0

2 1 2 2 1 2 0 9 0

2 2 2 0 0 9

a

a b b

2 2

9 0 ( 4 2 ) 4 0 0

0 9 (2 2 2 ) 0 9 0

( 4 2 ) (2 2 2 ) ( 4 ) 0 0 9

a b

a b

a b a b a b

a + b = −4; a2 + b2 + 4 = 9; 2a – 2b = −2;

a = −2, b = −1 (a, b) ≡ (−2, −1)


86. If the function. 1, 0 3

( )2, 3 5

k x xg x

mx x is differentiable, then the value of k + m is :

(1) 16

5

(2) 10

3

(3) 4

(4) 2

Solution

1; 0 3( )

2; 3 5

k x xg x

mx x

g(3−) = 2k; g(3+) = 3m + 2; g(3) = 2k

2k = 3m + 2 (1)

Also, ; 0 3

( ) 2 1

; 3 5

kx

g x x

m x

(3 ) ;4

kg (3 )g m

4

km k = 4m

Therefore, from (1), 2

,5

m 8

5k k + m = 2


87. The set of all values of λ for which the system of linear equations:

2x1 – 2x2 + x3 = λx1

2x1 – 3x2 + 2x3 = λx2

−x1 + 2x2 = λx3

has a non-trivial solution,

(1) is a singleton.

(2) contains two elements.

(3) contains more than two elements.

(4) is an empty set.

Solution 2x1 – 2x2 + x3 = λx1 2x1 – 3x2 + 2x3 = λx2

−x1 + 2x2 = λx3

has a non-trivial solution if

2 2 1

2 3 2 0

1 2

(2 – λ) [(3λ + λ2) – 4] + 2(−2λ + 2) + 1(4 – 3 – λ) = 0

1, 3


88. The normal to the curve, x2 + 2xy – 3y2 = 0 at (1, 1):

(1) meets the curve again in the second quadrant.

(2) meets the curve again in the third quadrant.

(3) meets the curve again in the fourth quadrant.

(4) does not meet the curve again.

Solution Given curve is, x2 + 2xy – 3y2 = 0

2 2 2 6 0dy dy

x x y ydx dx

(2 6 ) 2( )dy

x y x ydx

( )

1(3 )

dy x y

dx y x

at (1, 1)

1dx

dy at (1, 1)

This means equation of normal to given curve is (y – 1) = −1(x – 1)

x + y = 2

Substitute y = 2 − x in x2 + 2xy – 3y2 = 0

x2 + 2x(2 – x) – 3(x – x)2 = 0

(x – 1) (x – 3) = 0

x = 1 or x = 3

So it will again intersect the given curve at (3, 3), (3, −1), 1

1,3

i.e. again meet in the 1st or 4th quadrant.


89. The number of common tangents to the circles x2 + y2 – 4x – 6y – 12 = 0 and x2 + y2 + 6x + 18y + 26 =

0, is

(1) 2

(2) 3

(3) 4

(4) 1

Solution C1 : x2 + y2 – 4x – 6y – 12 = 0 C2 : x2 + y2 + 6x + 18y + 26 = 0

C1(2, 3); r1 = 5

C2(−3, −9); r2 = 8

and 2 21 2 1 2(2 3) (3 9) 13C C r r

The two circles touch each other externally.

Hence three common tangents can be drawn.


90. Let y(x) be the solution of the differential equation (x log x) d

2 log , ( 1).d

y

y x x xx

Then y(e) is

equal to:

(1) 0

(2) 2

(3) 2e

(4) e

Solution

( log ) 2 log ;dy

x x y x xdx

(x ≥ 1)

1

2log

dyy

dx x x

It is a linear differential equation of 1st order of the form dy

Py Qdx

1

,log

Px x

Q = 2

Note: At x = 1, P is not defined, hence this question is conceptually not correct.

So, 1

logI.F.dxPdx x xe e

ln logdt

tte e t x

Therefore, solution of given differential equation is given by

.(I.F) .(I.F)y Q dx C

(log ) 2 log ,y x x dx C

1

2 (log ) .x x x dx Cx

= 2x log x – 2x + C

when x = 1,

y(log 1) = 2 log 1 – 2 + C

0 = −2 + C C = 2

Note: Since we need to put x = 1 in order to find the value of constant and at x = 1, P is not defined, so this

whole question is conceptually incorrect. Although if we still solve it we get,

y(log x) = 2x log x – 2x + 2 as the general solution.

y(e) = 2e – 2e + 2

y(e) = 2


PHYSICS Solution 2.

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