set b · 2018. 4. 9. · 25 8. unpolarized light of intensity i passes through an ideal polarizer...

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SET B

PART A – PHYSICS

1. It is found that if a neutron suffers an elastic collinear with deuterium at rest, fractional loss of its energy is

𝑃𝑑 ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is 𝑃𝑐. The values of 𝑃𝑑

and 𝑃𝑐 are respectively.

(A) (0, 0)

(B) (0, 1)

(C) (.89, .28)

(D) (.28, .89)

Solution: (C)

By momentum conservation

𝑢 = 𝑉1 + 2𝑉2

𝑒 = 1 =𝑉2 − 𝑉1

𝑢

𝑢 = 𝑉2 − 𝑉1

𝑉2 =2𝑢

3 ; 𝑉1 = −

𝑢

3

Initial energy =1

2𝑢2

Final energy =1

2× 1 ×

𝑢2

9

=𝑢2

18

Fractional change =𝑢2

2−

𝑢2

18𝑢2

2

= 0.88

𝑢 = 𝑉1 + 12𝑉2

1 =𝑉2 − 𝑉1

𝑢

𝑢 = 𝑉2 − 𝑉1

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𝑉2 =2𝑢

13

𝑉1 = −11𝑢

13

Change in energy = 𝑢2 − (11𝑢

13)2× 100

= 0.294

2. The mass of a hydrogen molecule is 3.32 × 10−27𝑘𝑔. If 1023 hydrogen molecules strike, per second, a fixed

wall of area 2 𝑐𝑚2 at an angle of 45𝑜 to the normal, and rebound elastically with a speed of 103 𝑚 𝑠⁄ , then the

pressure on the wall is nearly:

(A) 2.35 × 102 𝑁 𝑚2⁄ (B) 4.70 × 102 𝑁 𝑚2⁄

(C) 2.35 × 103 𝑁 𝑚2⁄ (D) 4.70 × 103 𝑁 𝑚2⁄

Solution: (C)

𝐹 = 2𝑚𝑛𝑉𝑐𝑜𝑠𝜃

𝑃 =2𝑚𝑛𝑉𝑐𝑜𝑠𝜃

𝐴

𝑃 =2 × 3.32 × 10−27 × 1023 × 103

√2 × 2 × 10−4

𝑃 = 2.35 × 103 𝑁 𝑚2⁄

3. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical

container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of





cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the

fractional decrement in the radius of the sphere, (𝑑𝑟

𝑟), is:

(A) 𝑚𝑔

3𝑘𝐴

(B) 𝑚𝑔

𝐾𝑎

(C) 𝐾𝑎

𝑚𝑔

(D) 𝐾𝑎

3𝑚𝑔

Solution: (A)

𝑃 = 𝐾Δ𝑉

𝑉

𝑃 = 𝐾. 3 (Δ𝑅

𝑅)

𝑚𝑔

3𝐾𝑎= (

Δ𝑅

𝑅)

4. Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10Ω. The internal

resistances of two batteries are 1Ω and 2Ω respectively. The voltage across the load lies between:

(A) 11.4 and 1Ω V

(B) 11.7 V and 11.8 V

(C) 11.6 V and 11.7 V

(D) 11.5 V and 11.6 V

Solution: (D)

𝑉 − 12

1+

𝑉 − 13

2+

𝑉

10= 0

𝑉 +𝑉

2+

𝑉

10⇒ 12 +

13

2

𝑉 ⇒12 +

132

1 +12 +

110

= 11.562





5. A particle is moving in a circular path of radius a under the action of an attractive potential 𝑈 = −𝑘

2𝑟2. Its

total energy is:

(A) Zero

(B) −3

2

𝑘

𝑎2

(C) –𝑘

4𝑎2

(D) 𝑘

2𝑎2

Solution: (A)

𝑉 = −𝐾

2𝑟2

𝐹 = −𝑟𝑢

𝑟𝑟=

2𝐾

2𝑟3=

𝐾

𝑟3

𝑚𝑣2

𝑟=

𝐾

𝑟3

𝑉2 =𝐾

𝑚𝑟2

1

2𝑚𝑣2 =

𝐾

2𝑟2

𝐸 =𝐾

2𝑟2−

𝐾

2𝑟2= 0

6. Two masses 𝑚1 = 5 𝑘𝑔 and 𝑚2 = 10 𝑘𝑔 , connected by an inextensible string over a frictionless pulley, are

moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m

that should be put on top of 𝑚2 to stop the motion is:

(A) 43.3 𝑘𝑔

(B) 10.3 𝑘𝑔

(C) 18.3 𝑘𝑔

(D) 27.3 𝑘

Solution: (Bouns)

Does not match.

In this question coefficient of friction b/w masses 𝑚1 and 𝑚2 is unknown. Without knowing that the problem

cannot be solved.

Even is as assume that friction b/w 𝑚1 and 𝑚2 is sufficient to prevent slipping.





Answer does not match.

7. If the series limits frequency of the Lyman series is 𝑣𝐿′ then the series limits frequency of the P fund series

is:

(A) 𝑣𝐿

16

(B) 𝑣𝐿

25

(C) 25𝑣𝐿

(D) 16𝑣𝐿

Solution: (B)

1

𝜆= 𝑅𝐻𝑍2 (

1

𝑛12 −

1

𝑛22)

𝑣 = 𝑓𝜆

𝑣 = 𝐶𝑅𝐻𝑍2 (1

𝑛12 −

1

𝑛22)

𝑣1 =𝐿𝑦𝑚𝑎𝑛

𝐶𝑅𝐻𝑍2 (1

12 −1

∞2)

P-fund

𝑣2 = 𝐶𝑅𝐻𝑍2 (1

𝑆2−

1

∞2)

𝑉2 =𝑉1

25

8. Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is placed

behind A. The intensity of light beyond B is found to be 1

2. Now another identical polarizer C is placed between

A and B. The intensity beyond B is now found to be 1

8. The angle between polarizer A and C is

(A) 45𝑜

(B) 60𝑜

(C) 0𝑜

(D) 30𝑜

Solution: (A)

When unpolarized light passes through a polarizer.

Intensity becomes 𝐼

2

𝐼

2cos2 𝑎 =

𝐼

8

cos2 𝑎 =1

4

𝑎 =𝜋

3

9. An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let

𝜆𝑛, 𝜆𝑔 be the de Broglie wavelength of the electron in the 𝑛𝑡ℎ state and the ground state respectively. Let ∧𝑛 be





the wavelength of the emitted photon in the transition from the 𝑛𝑡ℎ state to the ground state. For large n, (A, B

are constants)

(A) ∧𝑛2≈ 𝐴 + 𝐵𝜆𝑛

2 (B) ∧𝑛2≈ 𝜆 (C) ∧𝑛≈ 𝐴 +

𝐵

𝜆𝑛2 (D) ∧𝑛≈ 𝐴 + 𝐵𝜆𝑛

Solution: (C)

∧𝑛≈ 𝐴 +𝐵

𝜆𝑛2

10. The reading of the ammeter for a silicon diode in the given circuit is:

(A) 11.5 𝑚𝐴

(B) 13.5 𝑚𝐴

(C) 0

(D) 15 mA





Solution: (D)

The diode is in forward bias.

So its resistance = 0

Therefore, 𝑖 =𝑉

𝑅

𝑖 =3

200= 15 𝑚𝐴

11. An electron, a proton and an alpha particle having the same kinetic energy are moving in circular orbits of

radii 𝑟𝑒 , 𝑟𝑝, 𝑟𝛼 respectively in a uniform magnetic field B. The relation between 𝑟𝑒 , 𝑟𝑝, 𝑟𝛼 is:

(A) 𝑟𝑒 < 𝑟𝑝 < 𝑟𝛼 (B) 𝑟𝑒 < 𝑟𝛼 < 𝑟𝑝

(C) 𝑟𝑒 > 𝑟𝑝 = 𝑟𝛼 (D) 𝑟𝑒 < 𝑟𝑝 = 𝑟𝛼

Solution: (D)

𝑟 =𝑚𝑉

𝑞𝐵=

√2𝑚𝐾

𝑞𝐵

𝑟 ∝√𝑚

𝑞

𝑟𝑒 =√𝑚𝑒

𝑞

𝑟𝑎 =√𝑢𝑚

2𝑞

𝑟𝑝 =√𝑚

𝑞

=2√𝑚

2𝑞

𝑟𝑒 = 𝑟𝑎 < 𝑟𝑝

12. A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20V. If a dielectric material

of dielectric constant 𝐾 =5

3 is inserted between the plates, the magnitude of the induced charge will be:

A) 2.4 nC

(B) 0.9 nC

(C) 1.2 nC

(D) 0.3 nC

Solution: (D)

𝑄1 = 𝐶𝑉1Q

𝑄 = 𝐾𝐶𝑉

𝑄 =5

3× 90 × 10−12 × 20

5 × 30 × 20 = 0.3 𝑛𝐶

13. For an RLC circuit driven with voltage of amplitude 𝑣𝑚 and frequency 𝜔0 =1

√𝐿𝐶 the current exhibits

resonance. The quality factor, Q is given by





(A) 𝑅

(𝜔0𝐶)

(B) 𝐶𝑅

𝜔0

(C) 𝜔0𝐿

𝑅

(D) 𝜔0𝑅

𝐿

Solution: (C)

Quality factor =1

𝑅√

𝐿

𝐶

But, 𝜔0 =1

√𝐿𝐶

So, 𝜔0𝐿

𝑅=

1

𝑅√

𝐿

𝐶

14. A telephonic communication service is working at carrier frequency of 10 GHz. Only 10% of it is utilized

for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a

bandwidth of 5 kHz?

(A) 2 × 105

(B) 2 × 106

(C) 2 × 103

(D) 2 × 104

Solution: (A)

No. of channel ⇒106

5×103 = 2 × 105

15. A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The

density of granite is 2.7 × 103 𝑘𝑔 𝑚3⁄ and its Young’s modulus is 9.27 × 1010 𝑃𝑎. What will be the

fundamental frequency of the longitudinal vibrations?

(A) 10 kHz

(B) 7.5 kHz

(C) 5 kHz

(D) 2.5 kHz

Solution: (C)

𝐿 =𝜆

2

𝑣 = √𝑇

𝑒

𝑣 = 2𝐿√𝑇

𝑒





𝑣 =1

2×0.6√

9.27×1010

2.7×103 ≈ 4.88 × 103𝐻𝑍

16. Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown. The

moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is:

(A) 73

2𝑀𝑅2

(B) 181

2 𝑀𝑅2

(C) 19

2𝑀𝑅2

(D) 55

2𝑀𝑅2

Solution: (B)

MI of system about O

1

2𝑀𝑅2 + 6 ×

1

2𝑀𝑅2 + 6 × 𝑀 × (2𝑅)2

⇒ 𝐼1 =7

2𝑀𝑅2 + 24𝑀𝑅2

MI of system about P

⇒ 𝐼1 + 7𝑀𝐷2

⇒ 7

2𝑀𝑅2 + 24𝑀𝑅2 + 7𝑀 × (3𝑅)2

⇒ 181

2𝑀𝑅2

17. Three concentric metal shells A, B and C of respective radii a, b and c (a < b <c) have surface charge

densities +𝜎,−𝜎 and +𝜎 respectively. The potential of shell B is:

(A) 𝜎

𝜀0[𝑏2−𝑐2

𝑏+ 𝑎]

(B) 𝜎

𝜀0[𝑏2−𝑐2

𝑐+ 𝑎]

(C) 𝜎

𝜀0[𝑎2−𝑏2

𝑎+ 𝑐]

(D) 𝜎

𝜀0[𝑎2−𝑏2

𝑏+ 𝑐]

Solution: (D)





Potential 1

4𝜋𝜀0 𝜎4𝜋𝑅2

𝑑

=𝜎𝑅2

𝜀0𝑑

Required potential of b

=𝜎𝑎2

𝜀0𝑏−

𝜎𝑏2

𝜀0𝑏+

𝜎𝑐2

𝜀0𝑐

=𝜎

𝜀0(𝑎2−−𝑏2

𝑏+ 𝑐)

18. In a potentiometer experiment, it is found that no current passes through the galvanometer when the

terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance

of 5 Ω, balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the

cell.

(A) 2Ω

(B) 2.5 Ω

(C) 1 Ω

(D) 1.5 Ω

Solution: (D)

1.5Ω

19. An EM wave from air enters a medium. The electric fields are �⃗� 1= 𝐸01𝑥 cos [2𝜋𝑣 (𝑧

𝑐− 𝑡)] in air and �⃗� 2 =

𝐸02�̂� cos[𝑘(2𝑧 − 𝑐𝑡)] in medium, where the wave number k and frequency v refer to their values in air. The





medium is non-magnetic. If 𝜖𝑟1 and 𝜖𝑟2 refer to relative permittivities of air and medium respectively, which of

the following, option is correct?

(A) 𝜖𝑟1

𝜖𝑟2 =

1

4

(B) 𝜖𝑟1

𝜖𝑟2 =

1

2

(C) 𝜖𝑟1

𝜖𝑟2 = 4

(D) 𝜖𝑟1

𝜖𝑟2 = 2

Solution: (A)

𝜖𝑟1

𝜖𝑟2 =

1

4

20. The angular width of the central maximum in a single slit diffraction pattern is 60𝑜. The width of the slit is

1𝜇𝑚. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it,





Young’s fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe

width is 1 cm, what id slit separation distance?

(i.e. distance between the centres of each slit.)

(A) 75 𝜇𝑚

(B) 100 𝜇𝑚

(C) 25 𝜇𝑚

(D) 50 𝜇𝑚

Solution: (C)

25 𝜇𝑚

21. A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of

1012 𝑠𝑒𝑐⁄ . What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver =

108 and Avagadro number = 6.02 × 1023𝑔𝑚 𝑚𝑜𝑙𝑒−1)

(A) 2.2 𝑁 𝑚⁄

(B) 5.5 𝑁 𝑚⁄

(C) 6.4 𝑁 𝑚⁄

(D) 7.1 𝑁 𝑚⁄

Solution: (C)

𝑣 = 1012𝑠−1

𝑤 = √𝑘

𝑚

𝑇 = 2𝜋√𝑘

𝑚

𝑣 =1

2𝜋√

𝑘

𝑚

𝑣 =1

𝜋√

𝑘

𝑚

𝑣 =1

2𝜋√

𝐾 × 6.02 × 1023

108 × 10−3

𝐾 =4𝜋2 × 1024 × 108 × 10−3

6.02 × 1023

≈ 7.1 𝑁 𝑚⁄

22. From a uniform circular disc of radius R and mass 9 M, a small disc of radius 𝑅

3 is removed as shown in the

figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and

passing through centre of disc is:





(A) 10 𝑀𝑅2

(B) 37

9𝑀𝑅2

(C) 4 𝑀𝑅2

(D) 40

9 𝑀𝑅2

Solution: (C)

By parallel axis theorem,

MI of smaller disc =𝑀(

𝑅

3)2

2+ 𝑀(

3𝑅

3)2

𝑀𝐼 =𝑀𝑅2

18+

4𝑀𝑅2

9

𝑀𝐼 =𝑀𝑅2

2

𝑀𝐼 = 𝑀𝐼(𝐵𝑖𝑔𝑔𝑒𝑟) − 𝑀𝐼(𝑆𝑚𝑎𝑙𝑙𝑒𝑟)

=9𝑀𝑅2

2−

𝑀𝑅2

2

= 4𝑀𝑅2

23. In a collinear collision, a particle with an initial speed 𝑣0 strikes a stationary particle of the same mass. If

the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative

velocity between the two particles, after collision, is:

(A) 𝑣0

2

(B) 𝑣0

√2

(C) 𝑣0

4

(D) √2 𝑣0

Solution: (D)

𝐾𝐸2 =3

2𝑚𝑣0

2

Conservation of momentum,





𝑣1 + 𝑣2 = 2𝑣0

Conservation of energy

1

2𝑚𝑉1

2 +1

2𝑚𝑉2

2 =3

2𝑚𝑉0

2

𝑉12 + 𝑉2

2 = 3𝑉02

𝑉12 + 𝑉2

2 + 2𝑉1𝑉2 = 4𝑉0

2𝑉1𝑉2 = 𝑉02

(𝑉1 − 𝑉2)2 ⇒ 𝑉1

2 + 𝑉22 − 2𝑉1𝑉2

⇒ 3𝑉02 − 2𝑉0

2

⇒ √2𝑉0

24. The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre of the

loop is 𝐵1. When the dipole moment is doubled by keeping the current constant, the magnetic field at the centre

of the loop is 𝐵2 . The ration 𝐵1

𝐵2 is:

(A) √2

(B) 1

√2

(C) 2

(D) √3

Solution: (A)

Dipole moment = IA

For 𝐿𝑜𝑜𝑝1

𝑚 = 𝐼𝜋𝑅2

𝐵1 =𝜇0𝑖

2𝑟

For 𝐿𝑜𝑜𝑝2

If current is constant m is doubled.

𝑟 =𝑟

√2

𝐵2 =𝜇0𝑖𝑟

√2

𝐵1

𝐵2= √2

25. The density of a material in the shape of a cube is determined by measuring three sides of the cube and its

mass. If the relative errors in measuring the mass and length are respectively 1.5% and 1%, the maximum error

in determining the density is:

(A) 4.5%

(B) 6%

(C) 2.5%

(D) 3.5%

Solution: (A)





𝑒 ⇒𝑀

𝐿3

Δ𝑒

𝑒⇒

Δ𝑀

𝑀+

3Δ𝐿

𝐿

Δ𝑒

𝑒× 100 ⇒ 1.5 + 3 × 1

% relative error = 4.5%

26. On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10cm. The

resistance of their series combination is 1𝑘Ω. How much was the resistance on the left slot before

interchanging the resistances?

(A) 550 Ω (B) 910 Ω (C) 990 Ω (D) 505 Ω

Solution: (A)

𝑎 + 𝑏 = 1000

𝑎

𝑏=

𝑙

100 − 𝑙

𝑏

𝑎=

𝑙 − 10

110 − 𝑙

1 =;(𝑙−10)

(100−𝑙)(110−𝑙)

27. . In an a.c circuit, the instantaneous e.m.f. and current are given by

𝑒 = 100 sin 30𝑡

𝑟 = 20 sin (30𝑡 −𝜋

4)

In one cycle of a.c., the average power consumed by the circuit and the wattles current are, respectively:

(A) 50

√2, 10

(B) 50,0

(C) 50,0

(D) 1000

√2, 10

Solution: (D)

𝑒 = 100 sin 30𝑡

𝑟 = 20 sin (30𝑡 −𝜋

4)

𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑝𝑜𝑤𝑒𝑟 = 𝑉𝑟𝑚𝑠 𝐼𝑟𝑚𝑠𝑐𝑜𝑠𝜙

=100

√2×

20

√2×

1

√2

=1000

√2





28. All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

(A)

(B)

(C)

(D)

Solution: (C)

29. Two moles of an ideal monoatomic gas occupies a volume V at 27𝑜𝐶. The gas expands adiabatically to a

volume 2V. Calculate (𝑎) the final temperature of the gas and (𝑏) change in its internal energy.

(A) (𝑎) 189 𝐾 (𝑏) − 2.7 𝑘𝐽

(B) (𝑎) 195 𝐾 (𝑏) 2.7 𝑘𝐽

(C) (𝑎)189 𝐾 (𝑏) 2.7 𝑘𝐽

(D) (𝑎)195 𝐾 (𝑏) − 27 𝑘𝐽

Solution: (A)





𝑇𝑉𝛾−1 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝛾 =5

3

300(𝑉)23 = 𝑇(2𝑉)

23

𝑇 = 189𝐾

Δ𝑢 = 𝑛𝑐𝑉𝑎𝑇

= 2 ×3

2× 8.314 × 11

= 2.7 𝑘𝐽(−𝑣𝑒)

30. A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely

proportional to the 𝑛𝑡ℎ power of R. If the period of rotation of the particle is T, then:

(A) 𝑇 ∝ 𝑅𝑛+1

2

(B) 𝑇 ∝ 𝑅𝑛 2⁄

(C) 𝑇 ∝ 𝑅3 2⁄ for any n

(D) ) 𝑇 ∝ 𝑅𝑛+1

2

Solution: (A)

𝐹 =𝐾

𝑟𝑛

𝑚𝑉2

𝑟=

𝐾

𝑟𝑛

𝑉2 ∝𝐾

𝑟𝑛−1

𝑉 ∝1

𝑟𝑛−12

𝑇 =2𝜋𝑟

𝑉∝

𝑟

𝑉

𝑇 ∝𝑟

1

𝑟𝑛−12

𝑇 ∝ 𝑅𝑛+12

PART B - MATHEMATICS

31. If the tangent at (1, 7) to the curve 𝑥2 = 𝑦 − 6 touches the circle 𝑥2 + 𝑦2 + 16𝑥 + 12𝑦 + 𝑐 = 0 then the

value of c is:

(A) 85

(B) 95

(C) 195

(D) 185

Solution: (B)





2𝑥 =𝑑𝑦

𝑑𝑥

𝑑𝑦

𝑑𝑥|(1,7)

= 2 × 1 = 2

𝑦 − 7

𝑥 − 1= 2

𝑦 = 2𝑥 + 5

Distance from centre =

[−6 − 2(−8) − 5

√12 + 22]

|5

√5| = √5

Radius of circle = √82 + 62 − 𝑐

√100 − 𝑐 = √5

𝑐 = 95

32. If 𝐿1 is the line of intersection of the planes 2𝑥 − 2𝑦 + 3𝑧 − 2 = 0, 𝑥 − 𝑦 + 𝑧 + 1 = 0 and 𝐿2 is the line of

intersection of the planes 𝑥 + 2𝑦 − 𝑧 − 3 = 0, 3𝑥 − 𝑦 + 2𝑧 − 1 = 0, then the distance of the origin from the

plane, containing the lines 𝐿1 and 𝐿2 is:

(A) 1

2√2

(B) 1

√2

(C) 1

4√2

(D) 1

3√2

Solution: (D)

Dr’S of 𝐿1

|𝑖 𝑗 𝑘2 −2 31 −1 1

|

𝑖(1) − 𝑗 (−1) + 𝑘(0)

(1, 1, 0)

But 𝑦 = 0

2𝑥 + 3𝑧 = 2𝑥 + 𝑧 = −1

] → (-5, 0, 4) point

Equation of 𝐿1

𝑥 + 5

1=

𝑦

1=

𝑧 − 4

0= 𝛼

Similarly for 𝐿2





𝑥 −75

3=

𝑦

−5=

𝑧 +85

−7= 𝛽

Point of intersection

(-1, 4, 4)

�̂� (Plane) |𝑖 𝑗 𝑘1 1 03 −5 −7

| = (−7, 7,−2)

Equation of plane

−7𝑥 + 7𝑦 − 8𝑧 = 𝐶.

(-1, 4, 4)

𝑐 = 3

Equation of plane

7𝑥 − 7𝑦 + 2 𝑧 + 3 = 0

Distance from origin

=3

√162=

3

9√2=

1

3√2

33. If 𝛼, 𝛽 ∈ 𝐶 are the distinct roots, of the equation 𝑥2 − 𝑥 + 1 = 0, then 𝛼101 + 𝛽107 is equal to:

(A) 1 (B) 2 (C) −1 (D) 0

Solution: (A)

𝑥2 − 𝑥 + 1 = 0

Roots are −𝑤,−𝑤2

𝛼101 + 𝛽107 = (−𝑤)101 + (−𝑤2)107

= −𝑤101 − 𝑤214

= −𝑤2 − 𝑤

= 1

As (1 + 𝑤 + 𝑤2 = 0)

34. Tangents are drawn to the hyperbola 4𝑥2 − 𝑦2 = 36 at the points P and Q. If these tangents intersect at the

point 𝑇(0, 3) then the area (in sq. units) of Δ𝑃𝑇𝑄 is:

(A) 60√3

(B) 36√5

(C) 45√5

(D) 54√3

Solution: (C)

4𝑥2 − 𝑦2 = 36

𝑥2

9−

𝑦2

36= 1

Equation of tangent: 𝑥𝑥1

9−

𝑦𝑦1

36= 1

𝑥 = 0, 𝑦 = 3

⇒ 𝑥1 × 0

9−

𝑦1 × 3

36= 1

⇒ 𝑦1 = −12





𝑥1 = √36 + (144)2

4= ± 3√5

ℎ = 3 + 12 = 15

𝑏 = 6√5

Area =1

2× 6√5 + 15 = 45√5

35. If the curves 𝑦2 − 6𝑥, 9𝑥2 + 𝑏𝑦2 = 16 intersect each other at right angles, then the value of b is:

(A) 4

(B) 9

2

(C) 6

(D) 7

2

Solution:

𝑦2 = 6𝑥

9𝑥2 + 𝑏𝑦2 = 16

2𝑦1

𝑑𝑦

𝑑𝑥= 6

⇒ 𝑑𝑦

𝑑𝑥=

3

𝑦1

18𝑥1 + 2𝑏𝑦1

𝑑𝑦

𝑑𝑥= 0

⇒ 𝑑𝑦

𝑑𝑥=

−9𝑥1

𝑏𝑦1

(3

𝑦1) × (

−9𝑥1

𝑏𝑦1) = −1

27𝑥1 = 𝑏𝑦12

𝑦12 = 6𝑥1

27𝑥1 = 𝑏(6𝑥1)

𝑥1 ≠ 0

𝑏 =27

6=

9

2

36. If the system of linear equations

𝑥 + 𝑘𝑦 + 3𝑧 = 0

3𝑥 + 𝑘𝑦 − 2𝑧 = 0





2𝑥 + 4𝑦 − 3𝑧 = 0

has a non-zero solution (𝑥, 𝑦, 𝑧), then 𝑥𝑧/𝑦2 is equal to:

(A) −30

(B) 30

(C) −10

(D) 10

Solution: (D)

|1 𝑘 33 𝑘 −22 4 −3

| = 0

(−3𝑘 + 8) − 3(−3𝑘 − 12) + 2(−5𝑘) = 0

−4𝑘 + 44 = 0 𝑘 = 11

𝑥 + 11𝑦 + 3𝑧 = 0 …. (i)

3𝑥 + 11𝑦 − 2𝑧 = 0 ….. (ii)

2𝑥 + 4𝑦 − 3𝑧 = 0 … (iii)

Equation (ii) and (i)

2𝑥 − 5𝑧 = 0

𝑥 =5𝑧

2= −5𝑦

Put it in

2𝑧 + 4𝑦 = 0

𝑧 = −2𝑦

𝑥𝑧

𝑦2=

(−2𝑦) × (−5𝑦)

𝑦2= 10

37. Let 𝑆 = {𝑥 ∈ 𝑅 ∶ 𝑥 ≥ 0 and 2|√𝑥 − 3| + √𝑥 (√𝑥 − 6) + 6 = 0}. Then S:

(A) Contains exactly two elements

(B) Contains exactly four elements

(C) Is an empty set

(D) Contains exactly one element

Solution: (A)

2|√𝑥 − 3| + √𝑥(√𝑥 − 6) + 6 = 0

Let √𝑥 = 𝑡 > 0

2|𝑡 − 3| + 𝑡(𝑡 − 6) + 6 = 0

= 2|𝑡 − 3| = −(𝑡2 − 6𝑡 + 6) = −(𝑡2 − 6𝑡 + 9) + 3





Has 2 solutions for t = 2 solutions for x

38. If sum of all the solutions of the equation 8 cos𝑥 . (cos (𝜋

6+ 𝑥) . cos (

𝜋

6− 𝑥) −

1

2) = 1 in [0, 𝜋] is 𝑘𝜋, then

k is equal to:

(A) 8

9 (B)

20

9 (C)

2

3 (D)

13

9

Solution: (A)

8 cos 𝑥 . (cos (𝜋

6+ 𝑥) . cos (

𝜋

6− 𝑥) −

1

2) = 1

We would use the formula

cos(𝐴 + 𝐵) . cos(𝐴 − 𝐵) = cos2 𝐴 − sin2 𝐵

Now, cos (𝜋

6+ 𝑥) . cos (

𝜋

6− 𝑥) = cos2 𝜋

6− sin2 𝑥

=3

4− sin2 𝑥

8 cos 𝑥 (3

4− sin2 𝑥 −

1

2) = 1

8 cos 𝑥 (1

4− sin2 𝑥) = 1

8 cos 𝑥 (1

4− (1 − cos2 𝑥)) = 1

8 cos 𝑥 (cos2 𝑥 −3

4) = 1

8 cos3 𝑥 − 6 cos 𝑥 = 1

2(4 cos3 𝑥 − 3 cos𝑥) = 1

2 cos 3𝑥 = 1

cos 3𝑥 =1

2

3𝑥 = 2𝑛𝜋 ±𝜋

3

𝑥 =2𝑛𝜋

3±

𝜋

9

𝑥 ∈ [0, 𝜋]

∴ 𝑥 =𝜋

9,7𝜋

9

Sum =8𝜋

9 𝑘 =

8

9





39. A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed and

this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at

random from the bag, then the probability that this drawn ball is red, is:

(A) 1

5 (B)

3

4 (C)

3

10 (D)

2

5

Solution: (D)

P(red, red) + p(black, red)

4

(4 + 6)×

6

(4 + 6 + 2)+

6

6 + 4×

4

6 + 4 + 2

=24 + 24

10 × 12=

48

10 × 12=

2

5

40. Let 𝑓(𝑥) = 𝑥2 +1

𝑥2 and 𝑔(𝑥) = 𝑥 −1

𝑥, 𝑥 ∈ 𝑅 − {−1, 0, 1}. If ℎ(𝑥) =

𝑓(𝑥)

𝑔(𝑥), then the local minimum value of

ℎ(𝑥)is:

(where C is a constant of integration)

(A) −2√2 (B) 2√2 (C) 3 (D) −3





Solution: (B)

𝑓(𝑥) = 𝑥2 +1

𝑥2 = (𝑥 −1

𝑥)2+ 2

𝑔(𝑥) = 𝑥 −1

𝑥

𝑥 ∈ 𝑅 − {−1. 0, 1}

Let 𝑥 −1

𝑥= 𝑡

ℎ(𝑥) =𝑓(𝑥)

𝑔(𝑥)=

𝑡2 + 2

𝑡= 𝑡 +

2

𝑡

𝑝(𝑥) = 𝑥 −1

𝑥

𝑝′(𝑥) = 1 +1

𝑥2

𝑡 = 𝑥 −1

𝑥

Since 𝑥 ≠ (−1, 0, 1)

𝑡 ∈ 𝑅 − {0}

ℎ(𝑥) = 𝑡 +2

𝑡

𝐴𝑀 ÷≥ 𝐺𝑀

Min of ℎ(𝑥) ⇒ 𝑡+

2

𝑡

2 ≥ √𝑡 ×

2

𝑡

𝑡 +2

𝑡≥ 2√2

Local minimum = 2√2

41. Two sets A and B are as under:

𝐴 = {(𝑎, 𝑏) ∈ 𝑅 × 𝑅 ∶ |𝑎 − 5| < 1 and |𝑏 − 5| < 1};

𝐵 = {(𝑎, 𝑏) ∈ 𝑅 × 𝑅: 4(𝑎 − 6)2 + 9(𝑏 − 5)2 ≤ 36}. Then:

(A) 𝐴 ∩ 𝐵 = 𝜙 (an empty set)

(B) Neither 𝐴 ⊂ 𝐵 nor 𝐵 ⊂ 𝐴

(C) 𝐵 ⊂ 𝐴

(D) 𝐴 ⊂ 𝐵

Solution: (D)

𝐴 = {(𝑎, 𝑏) ∈ 𝑅 × 𝑅 ∶ |𝑎 − 5| < 1 and |𝑏 − 5| < 1};

(𝑎 − 6)2

9+

(𝑏 − 5)2

4≤ 1





4

9+

1

4≤ 1 (4, 4) lies in B

(4,6) lies in B

(6, 4) lies in B (6,3) lies in A

42. The Boolean expression ~(𝑝 ∨ 𝑞) ∨ (~𝑝 ∧ 𝑞) is equivalent to:

(A) 𝑞 (B) ~𝑞 (C) ~𝑝 (D) 𝑝

Solution: (C)

~(𝑝 ∨ 𝑞) = (~𝑃 ∧ ~𝑞)⋁(~𝑝⋀𝑞)

= ~𝑝⋀(~𝑞⋁𝑞)

= ~𝑝⋀𝑞 = ~𝑝





43. Tangent and normal are drawn at 𝑃(16, 16) on the parabola 𝑦2 = 16𝑥, which intersect the axis of the

parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and ∠𝐶𝑃𝐵 = 𝜃,

then a value of tan 𝜃 is:

(A) 3 (B) 3

4 (C)

1

2 (D) 2

Solution: (D)

Equation of tangent = 𝑡𝑦 = 𝑥 + 𝑎𝑡2

Equation of normal = 𝑦 + 𝑥𝑡 = 2𝑎𝑡 + 𝑎𝑡3

Tangent = 2𝑦 = 𝑥 + 16 (−16,0)

Normal = 𝑦 + 2𝑥 = 48 (24,0)

Centre – (24−16

2, 0)

Centre = (4,0)

tan 𝜃 = |

43 + 2

1 − 2 ×43

| = |10

−5| = 2

44. If |𝑥 − 4 2𝑥 2𝑥2𝑥 𝑥 − 4 2𝑥2𝑥 2𝑥 𝑥 − 4

| = (𝐴 + 𝐵𝑥) (𝑥 − 𝐴)2, then the ordered pair (A, B) is equal to:

(A) (−4,5) (B) (4, 5) (C) (−4,−5) (D) (−4, 3)

Solution: (A)

|𝑥 − 4 2𝑥 2𝑥2𝑥 𝑥 − 4 2𝑥2𝑥 2𝑥 𝑥 − 4

| = (𝐴 + 𝐵𝑥) (𝑥 − 𝐴)2





Put 𝑥 = 0 ⇒ |−4 0 00 −4 00 0 −4

| = 𝐴3 ⇒ 4

|𝑥 − 4 2𝑥 2𝑥2𝑥 𝑥 − 4 2𝑥2𝑥 2𝑥 𝑥 − 4

| = (𝐵𝑥 − 4)(𝑥 + 4)2

|

|1 −

4

𝑥2 2

2 1 −4

𝑥2

2 2 1 −4

𝑥

|

|= (𝐵 −

4

𝑥) (1 +

4

𝑥)2

Put 𝑥 → ∞ ⇒ |1 2 22 1 22 2 1

| = 𝐵 ⇒ 𝐵 = 5

Ordered pair (A, B) is (-4, 5)

45. The sum of the co-efficient of all odd degree terms in the expansion of (𝑥 + √𝑥3 − 1)5+

(𝑥 − √𝑥3 − 1)5, (𝑥 > 1) is:

(A) 1 (B) 2 (C) −1 (D) 0

Solution: (B)

(𝑥 + √𝑥3 − 1)5+ (𝑥 − √𝑥3 − 1)

5𝑥 > 1

𝑓(𝑥) = 2[𝑥5 + 5𝐶3𝑥3(𝑥3 − 1) + 5𝐶3𝑥(𝑥

3 − 1)2]

For sum of add power coefficient it

Put 𝑓(1)−𝑓(−1)

2

𝑓(1) = 2[10]

𝑓(−1) = 2[−1 + 5𝐶3 × 20 − 5𝐶3 × 4]

𝑓(−1) = 2[−1 + 20 − 20] = −2

𝑓(1) − 𝑓(−1)

2=

2 − (−2)

2= 2

46. . Let 𝑎1, 𝑎2, 𝑎3, …… , 𝑎49 be in A.P. such that ∑ 𝑎4𝑘+1 = 41612𝑘=0 and 𝑎9 + 𝑎43 = 66. If 𝑎1

2 + 𝑎22 + ⋯+

𝑎172 = 140𝑚, then m is equal to:

(A) 34 (B) 33 (C) 66 (D) 68

Solution: (A)

Let 𝑎1 = 1 and d = common difference.

𝑎𝑛 = 1 + (𝑛 − 1)𝑑

Given, ∑ 𝑎4𝑘+1 = 41612𝑘=0

𝑎1 + 𝑎5 + 𝑎9 + ⋯𝑎49 = 416





𝑎 + 𝑎 + 4𝑑 + 𝑑 + 8𝑑 + ⋯+ 𝑑 + 48𝑑 = 416

13𝑎 +4 × 23 × 13𝑑

2= 416

𝑎 + 24𝑑 = 32 … (i)

𝑎9 + 𝑎43 = 66

𝑎 + 8𝑑 + 𝑎 + 42𝑑 = 66

2𝑎 + 50𝑑 = 66

𝑎 + 25𝑑 = 33 … (ii)

On solving we get,

𝑑 = 1, 𝑎 = 8

Now, 𝑎12 + 𝑎2

2 + ⋯𝑎172 = 140𝑚

𝑎2(𝑎 + 𝑑)2 + ⋯(𝑎 + 16𝑑)2

⇒ 𝑎2 + 𝑎2 + 𝑑2 + 2𝑑 + 𝑎2 + 4𝑑2 + 4𝑑 + ⋯𝑎2 + 256𝑑2 + 32𝑑

⇒ 17𝑎2 + 𝑑2(1 + 4 + 9 + ⋯256) + 2𝑑(1 + 2 + ⋯16)

⇒ 17𝑎2 + 𝑑2 ×16 × 17 × 33

6+ 2𝑑 (

16 × 17

2)

⇒ 17 × 64 + 1496 + 272 × 8

⇒ 1088 + 1496 + 272 × 8

= 4760

47. A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is

the origin and the rectangle OPRQ is completed, then the locus of R is:

(A) 3𝑥 + 2𝑦 = 𝑥𝑦 (B) 3𝑥 + 2𝑦 = 6𝑥𝑦

(C) 2𝑥 + 3𝑦 = 𝑥𝑦 (D) 2𝑥 + 3𝑦 = 𝑥𝑦

Solution: (A)

Let the equation of line be 𝑥

𝑝+

𝑦

𝑞= 1

(p, 0), (O,q) are the points & Q representing passes through (2, 3)

⇒2

𝑝+

3

𝑞= 1

Q locus of R is (p, q)

𝑝 = 𝑥, 𝑞 = 𝑦

4

2𝑥+

6

2𝑦= 1

𝑦 + 6𝑥 = 2𝑥𝑦

= 2𝑦 + 𝑥 = 𝑥𝑦

48. The values of ∫sin2 𝑥

1+2𝑥

𝜋

2

−𝜋

2

𝑑𝑥 is:

(A) 4𝜋 (B) 𝜋

4 (C)

𝜋

8 (D)

𝜋

2

Solution: (B)

𝐼 = ∫sin2 𝑥

1+2𝑥

𝜋

2

−𝜋

2

𝑑𝑥 ……(i)





∫𝑓(𝑥)

𝑏

𝑎

𝑑𝑥

∫𝑓(𝑥) = ∫(𝑎 + 𝑏 − 𝑥)

𝐼 = ∫sin2(−𝑥)

1+2−𝑥

𝜋

2

−𝜋

2

𝑑𝑥

𝐼 = ∫2𝑥 sin2 𝑥

1+2𝑥

𝜋

2

−𝜋

2

𝑑𝑥 …..(ii)

Equations (i) and (ii)

2𝐼 = ∫sin2 𝑥 (1 + 2𝑥)

(1 + 2𝑥)

𝜋2

−𝜋2

𝑑𝑥

2𝐼 = 2 ∫ sin2 𝑥

𝜋2

−𝜋2

𝑑𝑥

𝐼 = ∫ cos2 𝑥 𝑑𝑥

𝜋2

0

2𝐼 = ∫(sin2 𝑥 + cos2 𝑥)

𝜋2

0

𝑑𝑥

2𝐼 = ∫ 1 𝑑𝑥

𝜋2

0

𝐼 =𝜋

2×

1

2=

𝜋

4

49. Let 𝑔(𝑥) = cos 𝑥2, 𝑓(𝑥) = √𝑥, and 𝛼, 𝛽(𝛼 < 𝛽) be the roots of the quadratic equation 18𝑥2 − 9𝜋𝑥 +

𝜋2 = 0. Then the area (in sq. units) bounded by the curve 𝑦 = (𝑔𝑜𝑓) (𝑥) and the lines 𝑥 = 𝛼, 𝑥 = 𝛽 and 𝑦 =

0, is:

(A) ) 1

2(√3 − √2)

(B) 1

2(√2 − 1)

(C) 1

2(√3 − 1)

(D) 1

2(√+1)

Solution: (C)

Given 𝑔(𝑥) = cos𝑥2 , 𝑓(𝑥) = √𝑥

18𝑥2 − 9𝜋𝑥 + 𝜋2 = 0

The roots : 𝑥 =9𝜋± √81𝜋2−72𝜋2

2×18





𝑥 =9𝜋 ± √9𝜋2

36

𝑥 =(9 ± 3)𝜋

36

𝑥 =𝜋

3,𝜋

6

𝑦 = 𝑔𝑜𝑓𝑙𝑢𝑠

As cos(√𝑥)2

= cos 𝑥

Area = ∫ cos𝑥 𝑑𝑥𝜋

3𝜋

6

= [sin 𝑥]𝜋6

𝜋

3

⇒ √3

2−

1

2

50. For each 𝑡 ∈ 𝑅, let [𝑡] be the greatest integer less than or equal to t. Then lim𝑥→0+

𝑥 ([1

𝑥] + [

2

𝑥] + ⋯+ [

15

𝑥])

(A) is equal to 120

(B) does not exist (in R)

(C) Is equal to 0

(D) IS equal to 15

Solution: (B)

lim𝑥→0+

𝑥 ([1

𝑥] + [

2

𝑥] + ⋯[

15

𝑥])

lim𝑥→0+

𝑥 (1

𝑥+

2

𝑥+ ⋯

15

𝑥)

[1

𝑥] − [

2

𝑥]… . [

15

𝑥] → these terms will be neglected as lim

𝑥→0+𝑥 {[

1

𝑥] − [

2

𝑥]… . [

15

𝑥]} lying between 0 and 1

lim𝑥→0+

𝑥 (1 + 2 + ⋯15

𝑥)

1 + 2 + ⋯15 =15 × 16

2

= 120

51. If ∑ (𝑥𝑖 − 5)9𝑖=1 = 9 and ∑ (𝑥𝑖 − 5)29

𝑖=1 = 45, then the standard deviation of the 9 items 𝑥1, 𝑥2, … . , 𝑥9 is:

(A) 2 (B) 3 (C) 9 (D) 4

Solution: (A)

∑(𝑥𝑖 − 5)

9

𝑖=1

= 9

Mean =9

9+ 5 = 6

𝑥 = 6

∑ (𝑥𝑖 − 5)29𝑖=1 = 459

∑𝑥2

9

𝑖=1

− 10∑𝑥

9

𝑖=1

; +25 × 9 = 45





∑𝑥12

9

𝑖=1

= 45 − 25 × 9 + 10 × (9 + 5 × 9)

= 360

𝜎 = √∑ (𝑥;−6)29

𝑖=1

9

𝜎 = √∑ 𝑥29

𝑖=1 −12∑ 𝑥9𝑖=1 ;+36+9

9

𝜎 = √360 − 12 × 54 + 36 + 9

9

= 2

52. The integral ∫sin2 𝑥 cos2 𝑥

(sin5 𝑥+cos3 𝑥 sin2 𝑥+sin3 𝑥 cos2 𝑥+cos5 𝑥)2𝑑𝑥 is equal to:

(A) 1

1+cot3 𝑥+ 𝐶

(B) −1

1+cot3 𝑥+ 𝐶

(C) 1

3(1+tan3 𝑥)+ 𝐶

(D) −1

3(1+tan3 𝑥)+ 𝐶

Solution: (D)

∫sin2 𝑥 cos2 𝑥

(sin5 𝑥 + cos3 𝑥 sin2 𝑥 + sin3 𝑥 cos2 𝑥 + cos5 𝑥)2𝑑𝑥


(sin2 𝑥 (sin2 𝑥 + cos2 𝑥) + cos3 𝑥 (cos2 𝑥 + sin2 𝑥))2𝑑𝑥


(sin3 𝑥 + cos3 𝑥)2𝑑𝑥


(cos3 𝑥)2(tan3 𝑥 + 1)2𝑑𝑥

∫tan2 𝑥 sec2 𝑥

(tan3 𝑥 + 1)2𝑑𝑥

Put tan3 𝑥 + 1 = 𝑡

𝑑𝑡 = 3 tan2 𝑥 sec2 𝑥 𝑑𝑥

⇒ ∫𝑑𝑡

3(𝑡)2

⇒ −1

3𝑡+ 𝐶

⇒ −1

3(tan3 𝑥 + 1)+ 𝐶

53. Let 𝑆 = {𝑡 ∈ 𝑅: 𝑓(𝑥) = |𝑥 − 𝜋| ∙ (𝑒|𝑥| − 1) sin|𝑥| is not differentiable at 𝑡}. Then the set S is equal to:

(A) {𝜋}

(B) {0, 𝜋}

(C) 𝜙 (an empty set)





(D) {0}

Solution: (C)

𝑓(𝑥) = (𝑥 − 𝜋). (𝑒|𝑥| − 1). 𝑠𝑖𝑛|𝑥|

𝑓′(𝜋 + ℎ) = lim𝑥→ℎ+𝜋

𝑓(𝜋 + ℎ) − 𝑓(𝜋)

ℎ=

|ℎ| sin(𝜋 + ℎ) (𝑒𝜋 − 1)

ℎ= 0

𝑓′(𝜋 + ℎ) = lim𝑥→ℎ+𝜋

𝑓(𝜋) − 𝑓(𝜋 − ℎ)

ℎ=

|ℎ| sin(𝜋 + ℎ) (𝑒𝜋 − 1)

ℎ= 0

Differentiable at 𝜋

𝑓′(0 + ℎ) = lim𝐿→0

𝑓(ℎ)−𝑓(0)

ℎ= |𝜋| ×

|ℎ|𝑠𝑖𝑛|ℎ|

ℎ= 0

𝑓′(0 − ℎ) = lim𝐿→0

𝑓(0) − 𝑓(ℎ)

ℎ= |𝜋| ×

|ℎ|𝑠𝑖𝑛|ℎ|

ℎ= 0

Differentiable at 0

F(x) is d9fferentiable everywhere

𝑆 = 𝜙

54. Let 𝑦 = 𝑦(𝑥) be the solution of the differential equation

sin 𝑥𝑑𝑦

𝑑𝑥+ 𝑦 cos = 4𝑥, 𝑥 ∈ (0, 𝜋). If 𝑦 (

𝜋

2) = 0, then 𝑦 (

𝜋

6) is equal to:

(A) −8

9𝜋2 (B) −

4

9𝜋2 (C)

4

9√3𝜋2 (D)

−8

9√3𝜋2

Solution: (A)

𝑑𝑦

𝑑𝑥+ 𝑦𝑐𝑜𝑡 𝑥 =

𝑥

𝑠𝑖𝑛𝑥 𝑥 ∈ (0, 𝜋)

𝐼. 𝐹 = 𝑒∫𝑐𝑜𝑡𝑥 = 𝑒ln 𝑠𝑖𝑛𝑥 = 𝑠𝑖𝑛𝑥

∴ 𝑦 𝑠𝑖𝑛𝑥 = ∫𝑥

𝑠𝑖𝑛𝑥. 𝑠𝑖𝑛𝑥 𝑑𝑥

𝑦𝑠𝑖𝑛𝑥 = 2𝑥2 + 𝑥

𝑦 (𝜋

2) = 0

0 × 1 = 3 × (𝜋

2)2

+ 𝑐

= 𝑐 = −𝜋2

2

(𝜋

6) =

2 × (𝜋6)

2− (

𝜋2)

2

12

= (𝜋2

18) − (

𝜋2

2) × 2

=𝜋2

9− 𝜋2

= −8𝜋2

9

55. Let �⃗� be a vector coplanar with the vectors 𝑎 = 2𝑖̂ + 3𝑗̂ − �̂� and �⃗� = 𝑗̂ + �̂�. If �⃗� is perpendicular to 𝑎 and

�⃗� ∙ �⃗� = 24, then |�⃗� |2 is equal to:

(A) 256 (B) 84 (C) 336 (D) 315

Solution: (C)





�⃗� = 𝑥𝑖̂ + 𝑦𝑗̂ + 𝑧�̂�

|𝑥 𝑦 𝑧2 3 −10 1 1

| = 0

4𝑥 − 2𝑦 + 2𝑧 = 0

4𝑦 − 2𝑧 = 0

2𝑥 − 𝑦 + 𝑧 = 0

𝑦 = 27

�⃗� . 𝑎 = 0 2𝑥 + 3𝑦 − 𝑧 = 0

𝑦 + 𝑧 = 24

𝑧 = 8

𝑦 = 16

𝑥 = 4

∴ 42 + 82 + 162

= 336

56. The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane, 𝑥 +

𝑦 + 𝑧 = 7 is:

(A) 1

3

(B) √2

3

(C) 2

√3

(D) 2

3

Solution: (C)

𝐴(5,−1,4) ; 𝑏(4, −1, 3)

Vector joining A, B

𝐴𝐵 = 1𝑖̂ + 0𝑗̂ + 1�̂�

Normal vector 𝑖̂ + 𝑗̂ + �̂�

|𝐴𝐵| = √2

|𝐴𝐵1| =1 + 1

√3=

2

√3

𝐴𝐵11 = √2 −4

3= √

2

3





57. PQR is a triangular park with PQ = PR = 200m. A T.V. tower stands at the mid-point of QR. If the angles of

elevation of the top of the tower at P, Q and R are respectively 45𝑜, 30𝑜 and 30𝑜, then the height of the tower

(in m) is:

(A) 100√3

(B) 50√2

(C) 100

(D) 50

Solution (C)

tan 30 =2𝑅

𝑄𝑅

𝑄𝑅 =2ℎ

(1

√3)

= 2ℎ√3

tan 45𝑜 =ℎ

𝑃𝑀 𝑃𝑀 = ℎ

So, 4ℎ2 = 2002

ℎ2 + 3ℎ2 = 2002

2ℎ = 200

ℎ = 100

58. From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and

arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is:

(A) At least 500 but less than 750

(B) At least 750 but less than 1000

(C) At least 1000

(D) Less than 500

Solution: (C)





𝑁1, 𝑁2, 𝐷, 𝑁3, 𝑁4

N → Novels

D→ Dictionary

D can be chosen in 3 ways

𝑁1 → 6

𝑁2 → 5

𝑁3 → 4

𝑁4 → 3

Total ways = 3 × 6 × 5 × 4 × 3 = 1080

59. Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series

12 + 2 ⋅ 22 + 32 + 2 ⋅ 42 + 52 + 2 ⋅ 62 + ⋯.

If 𝐵 − 2𝐴 = 100𝜆, then 𝜆 is equal to:

(A) 464 (B) 496 (C) 232 (D) 248

Solution: (D)

𝑆 = 1 + 2.22 + 32 + 3 + 2.42 + 52 + 2.62 + ⋯

𝑆 = 1 + 22 + 32 + 42 + ⋯

𝑆1 =20 × 21 × 41

6+

4 × 10 × 11 × 21

6

𝑆 − 1 𝐴 = 4410

𝐵 =40×41×81

6+

4×20×21×41

6

= 33620

𝐵 − 2𝐴 = 24800

= 100𝜆 = 7𝜆 = 248

60. Let the orthocentre and centroid of a triangle be 𝐴(−3, 5) and 𝐵(3, 3) respectively. If C is the circumcentre

of this triangle, then the radius of the circle having line segment AC as diameter, is:

(A) 3√5

2 (B)

3√5

2 (C) √10 (D) 2√10

Solution: (A)

Centroid divides the line joining O and C in the ratio 2 : 1

A = O, B = G

𝑂𝐺 = √(3 + 3)2 + (5 − 3)2 = √62 + 22 = √40 = 2√10

𝑂𝐶 =3

2× 2√10 = 3√10

Radius =𝑂𝐶

2= 3√

10

4= 3√

5

2





PART C - CHEMISTRY

61. Total number of lone pair of electrons in I3− ions is:

(A) 9 (B) 12 (C) 3 (D) 6

Solution: (A)

62. Which of the following salts is the most basic in aqueous solution?

(A) 𝐹𝑒𝐶𝑙3 (B) 𝑃𝑏(𝐶𝐻3𝐶𝑂𝑂)2

(C) 𝐴𝑙(𝐶𝑁)3 (D) 𝐶𝐻3𝐶𝑂𝑂𝐾

Solution: (D)

𝐶𝐻3𝐶𝑂𝑂𝐾

63. Phenol reacts with methyl chloroformate in the presence of 𝑁𝑎𝑂𝐻 to form product A. A reacts with 𝐵𝑟2 to

form product B. A and Bare respectively:

(A)

(B)

(C)

(D)





Solution: (A)

64. The increasing order of basicity of the following compounds is:

(i)

(ii)

(iii)

(iv)

(A) (ii) < (i) < (iv) < (i)

(B) (iv) < (ii) < (i) <(iii)

(C) (i) < (ii) < (iii) < (iv)

(D) (ii) < (i) < (iii) < (iv)

Solution: (A)

(iii) Is stabilized by resonance when lone pair is donated

(ii) Is most unstable because position charge in double bond is unstable





(i) Is more stable than (d)

65. An alkali is titrated against an acid with methyl orange as indiactor, which of the following is a correct

combination?

(A) Base Acid End pointWeak Strong Yellow to pinkish red

(B) Base Acid End point

Strong Strong Pink to colourless

(C) Base Acid End pointWeak Srtong Colourless to pink

(D) Base Acid End point

Strong strong Pinkish red to yellow

Solution: (A)

Methyl orange is yellow in basic and reddish id acid

66. The trans-alkenes are formed by the reduction of alkynes with:

(A) 𝑁𝑎 𝑙𝑖𝑞⁄ . 𝑁𝐻3 (B) 𝑆𝑛 − 𝐻𝐶𝑙

(C) 𝐻2 − 𝑃𝑑 𝐶, 𝐵𝑎𝑆𝑂4⁄ (D) 𝑁𝑎𝐵𝐻4

Solution: (A)

𝑁𝑎 𝑙𝑖𝑞⁄ . 𝑁𝐻3

67. The ratio of mass percent of C and H of an organic compound (𝐶𝑋𝐻𝑌𝑂𝑍) is 6 : 1. If one molecule of the

above compound (𝐶𝑋𝐻𝑌𝑂𝑍) contains half as much oxygen as required to burn one molecule of compound

𝐶𝑋𝐻𝑌 completely to 𝐶𝑂2 and 𝐻2𝑂. The empirical formula of compound 𝐶𝑋𝐻𝑌𝑂𝑍 is:

(A) 𝐶3𝐻4𝑂2 (B) 𝐶2𝐻4𝑂3 (C) 𝐶3𝐻6𝑂3 (D) 𝐶2𝐻4𝑂

Solution: (B)

𝑚𝑎𝑠𝑠 𝑜𝑓 𝐶

𝑚𝑎𝑠𝑠 𝑜𝑓 𝐻=

6

1

12𝑥 = 6𝑦

𝑥 ∶ 𝑦 = 1 ∶ 2

𝐶𝑦𝐻𝑦 + (

𝑥 + 𝑦22

)𝑂2 → 𝑥𝐶𝑂2 +𝑦

2 𝐻2𝑂

(𝑥 +

𝑦2

2) ×

1

2= 𝑍

Empirical formula = 𝐶2𝐻4𝑂3

68. Hydrogen peroxide oxidises [𝐹𝑒(𝐶𝑁)6]4−to [Fe(𝐶𝑁)6]

3− in acidic medium but reduces

[𝐹𝑒(𝐶𝑁)6]3−to[𝐹𝑒(𝐶𝑁)6]

4− in alkaline medium. The other products formed are, respectively:

(A) 𝐻2𝑂 and (𝐻2𝑂 + 𝑂2)

(B) 𝐻2𝑂 and (𝐻2𝑂 + 𝑂𝐻−)





(C) (𝐻2𝑂 + 𝑂2) and H2𝑂

(D) (𝐻2𝑂 + 𝑂2) and (𝐻2𝑂 + 𝑂𝐻−)

Solution: (A)

𝐻2𝑂 and (𝐻2𝑂 + 𝑂2)

69. The major product formed in the following reaction is:

(A)

(B)

(C)

(D)

Solution: (B)





70. How long (approximate) should water be electrolyzed by passing through 100 amperes current so that the

oxygen released can completely burn 27.66 g of diborane? (Atomic weight of B = 10.8 u)

(A) 3.2 hours

(B) 1.6 hours

(C) 6.4 hours

(D) 0.8 hours

Solution: (A)

𝐵2𝐻6 + 3𝑂2 → 𝐵2𝑂3 + 3𝐻2𝑂

𝐼𝑡

𝐹=number of moles of electron

100𝑡

69500= 3 × 4

𝑡 = 3.2 hours

71. Which of the following lines correctly show the temperature dependence of equilibrium constant, K, for an

exothermic reaction?





(A) C and D

(B) A and D

(C) A and B

(D) B and C

Solution: (C)

In K = −Δ𝐻

𝑅𝑇+ 𝐶

𝐾 = 𝐴𝑒−Δ𝐻𝑅𝑇

72. At 518𝑜𝐶, the rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363

Torr, was 1.00 𝑇𝑜𝑟𝑟 𝑠−1 when 5% had reacted and 0.50 𝑇𝑜𝑟𝑟 𝑠−1 when 33% had reacted. The order of the

reaction is:

(A) 1

(B) 0

(C) 2

(D) 3

Solution: (C)

Rate 𝛼(𝑐𝑜𝑛𝑒)𝑛

Let initial concentration be x

0.95𝑥 → 1 𝑡𝑜𝑟𝑟 𝑠−1

0.67𝑥 → 0.5 𝑡𝑜𝑟𝑟 𝑠−1

(0.95

0.67)𝑛

=1

0.5

⇒ 𝑛 = 2

73. Glucose on prolonged heating with 𝐻𝐼 gives:

(A) Hexanoic acid

(B) 6 – iodohexanal

(C) n – Hexane

(D) 1 – Hexene

Solution: (C)

𝐶6𝐻12𝑂6 + 𝐻𝐼 + Δ→ 𝐶6𝐻14

74. Consider the following reaction and statement:

[𝐶𝑜(𝑁𝐻3)4𝐵𝑟2]+ + 𝐵𝑟− → [𝐶𝑜(𝑁𝐻3)3𝐵𝑟3] + 𝑁𝐻3

(i) Two isomers are produced if the reactant complex ion is a cis – isomer.

(ii) Two isomers are produced if the reactant complex ion is a trans – isomer.

(iii) Only one isomer is produced if the reactant complex ion is a trans – isomer.

(iv) Only one is produced if the reactant complex ion is a cis – isomer.

The correct statements are:

(A) (iii) and (iv)

(B) (ii) and (iv)

(C) (i) and (ii)





(D) (i) and (iii)

Solution: (D)

Trans → only meridonial

Cis → fac and mer

75. The major product of the following reaction is:

(A)

(B)

(C)

(D)

Solution: (D)

76. Phenol on treatment with 𝐶𝑂2 in the presence of NaOH followed by acidification produces compound X as

the major product. X on treatment with (𝐶𝐻3𝐶𝑂)2 in the presence of catalytic amount of 𝐻2𝑆𝑂4 produces:

(A)





(B)

(C)

(D)

Solution: (C)

77. An aqueous solution contains an unknown concentration of 𝐵𝑎2+. When 50 mL of a 1 M solution of

𝑁𝑎2𝑆𝑂4 is added, 𝐵𝑎𝑆𝑂4 just begins to precipitate. The final volume is 500 mL. The solubility product of

𝐵𝑎𝑆𝑂4 is 1 × 10−10. What is the original concentration of 𝐵𝑎2+?

(A) 1.1 × 10−9𝑀





(B) 1.0 × 10−10𝑀

(C) 5 × 10−9𝑀

(D) 2 × 10−9𝑀

Solution: (A)

𝐵𝑎𝑆𝑂4 ⇌ 𝐵𝑎2+ + 𝑆𝑂42−

(𝑥)

0.5

(𝑥+0.05)

0.5= 10−10

X is small

𝑥 = 5 × 10−10

𝑥 + 0.05 ≈ 0.05

Required answer = 5 × 10−10 ÷ 0.95 = 1.1 × 10−9𝑀

78. Which of the following compounds will be suitable for Kjeldahl’s method for nitrogen estimation?

(A)

(B)

(C)

(D)

Solution: (D)





Kjedlahl’s method can’t be used for testing nitrogen in azo nitro and nitrogen in aromatic ring.

79. When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in

excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as an

adsorbent. The metal ‘M’ is:

(A) Al (B) Fe (C) Zn (D) Ca

Solution: (A)

Alumina (𝐴𝑙2𝑂3) is used in chromatography

80. An aqueous solution contains 0.10 𝑀 𝐻2𝑆 and 0.20 M 𝐻𝐶𝑙. If the equilibrium constants for the formation of

𝐻𝑆− from H2S is 1.0 × 10−7 and that of S2− from 𝐻𝑆− ions is 1.2 × 10−13 then the concentration of 𝑆2− ions

in aqueous solution is:

(A) 6 × 10−21

(B) 5 × 10−19

(C) 5 × 10−8

(D) 3 × 10−20

Solution: (D)

𝐻2𝑆 ⇌ 𝐻𝑆− + 𝐻𝐻7

0.1 − 𝑥 𝑥 (𝑥 + 0.2)

𝑥(𝑥 + 0.2)

0.1 − 𝑥= 10−7

𝑥 =1

2× 10−7

𝐻𝑆− ⇌ 𝑆2 + 𝐻7

𝑥 − 𝑦 𝑦 𝑥 + 𝑦 + 0.2

𝑦(𝑥 + 𝑦 + 0.2)

(𝑥 + 𝑦)= 1.2 × 10−13

𝑦 = 3 × 10−20

81. The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is required

to make teeth enamel harder by converting [3𝐶𝑎3(𝑃𝑂4)2 . 𝐶𝑎(𝑂𝐻)2] to:

(A) [3𝐶𝑎3(𝑃𝑂4)2. 𝐶𝑎𝐹2]

(B) [3{𝐶𝑎(𝑂𝐻)2}. 𝐶𝑎𝐹2]





(C) [𝐶𝑎𝐹2]

(D) [3(𝐶𝑎𝐹2). 𝐶𝑎(𝑂𝐻)2]

Solution: (A)

[3𝐶𝑎3(𝑃𝑂4)2. 𝐶𝑎𝐹2]

82. The compound that does not produce nitrogen gas by thermal decomposition is:

(A)𝑁𝐻4𝑁𝑂2 (B) (𝑁𝐻4)2𝑆𝑂4

(C) 𝐵𝑎(𝑁3)2 (D) (𝑁𝐻4)2𝐶𝑟2𝑂7

Solution: (B)

(𝑁𝐻4)2𝑆𝑂4

83. The predominant form of histamine presenr in human blood is (𝑝𝐾𝑎 , Histidine = 60. )

(A)

(B)

(C)

(D)

Solution: (B)





84. The oxidation states of 𝐶𝑟 in [𝐶𝑟(𝐻2𝑂)6]𝐶𝑙3, [𝐶𝑟(𝐶6𝐻6)2], and 𝐾2[𝐶𝑟(𝐶𝑁)2(𝑂)2(𝑂)2(𝑁𝐻3)] respectively

are:

(A) +3, 0 and +6

(B) +3, 0 and +4

(C) +3,+4 and +6

(D) +3,+2 and +4

Solution: (A)

+3, 0 and +6

85. Which type of ‘defect’ has the presence of cations in the interstitial sites?

(A) Frenkel defect

(B) Metal deficiency defect

(C) Schottky defect

(D) Vacancy defect

Solution: (A)

Frenkel defect

86. The combustion of benzene (𝑙) gives 𝐶𝑂2(𝑔) and 𝐻2𝑂(𝑙). Given that heat of combustion of benzene at

constant volume is −3263.9 𝑘𝐽 𝑚𝑜𝑙−1 at 25𝑜𝐶; heat of combustion (𝑖𝑛 𝑘𝐽 𝑚𝑜𝑙−1) of benzene at constant

pressure will be:

(𝑅 = 8.314 𝐽𝐾−1 𝑚𝑜𝑙−1)

(A) 3260

(B) −3267.6 (C) 4152.6 (D) −452.46

Solution: (B)

𝐶6𝐻6 + 𝑂2 → 𝐶𝑂2 + 𝐻2𝑂

Use, Δ𝐻𝑝 = Δ𝐻𝑐 + Δ𝑛𝑔𝑅𝑇

−3267.6 𝑘𝐽 𝑚𝑜𝑙−1

87. Which of the following are Lewis acids?

(A) 𝑃𝐻3 and 𝑆𝑖𝐶𝑙4 (B) 𝐵𝐶𝑙3 and 𝐴𝑙𝐶𝑙3

(C) 𝑃𝐻3 and 𝐵𝐶𝑙3 (D) 𝐴𝑙𝐶𝑙3 and 𝑆𝑖𝐶𝑙4

Solution: (B, D)

Has sextet configured

88. Which of the following compounds contain(s) no covalent bond(s)?

𝐾𝐶𝑙, 𝑃𝐻3, 𝑂2, 𝐵2𝐻6, 𝐻2𝑆𝑂4

(A) 𝐾𝐶𝑙

(B) 𝐾𝐶𝑙, 𝐵2𝐻6

(C) 𝐾𝐶𝑙, 𝐵2𝐻6, 𝑃𝐻3

(D) 𝐾𝐶𝑙, 𝐻2𝑆𝑂4

Solution: (A)

𝐾𝐶𝑙





89. For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

(A) [𝐶𝑜(𝐻2𝑂)4𝐶𝑙2]𝐶𝑙. 2𝐻2𝑂

(B) [𝐶𝑜(𝐻2𝑂)3𝐶𝑙3]. 3𝐻2𝑂

(C) [𝐶𝑜(𝐻2𝑂)6]𝐶𝑙3

(D) [𝐶𝑜(𝐻2𝑂)5𝐶𝑙]𝐶𝑙2. 𝐻2𝑂

Solution: (B)

Melting point is lower for more solute concentration [𝐶𝑜(𝐻2𝑂)3𝐶𝑙3]. 3𝐻2𝑂 doesn’t dissociate into ions and has

least solute conc. Is the answer.

90. According to molecular orbital theory, which of the following will not be a viable molecule?

(A) 𝐻2− (B) 𝐻2

2− (C) 𝐻𝑒22+ (D) 𝐻𝑒2

+

Solution: (B)

In 𝐻22− bond order is zero



set b · 2018. 4. 9. · 25 8. unpolarized light of intensity i passes through an ideal polarizer...

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