(a) → (ii), (b) → (iii), (c) → (iv), (d) → (i)

Home/JEE Main PYQs/2021 · February 25, Shift 1

JEE Main 2021 February 25, Shift 1 Question Paper with Solutions

All 90 questions from the JEE Main 2021 (February 25, Shift 1) shift — Physics (30), Chemistry (30) and Mathematics (30) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.

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Physics30 questions

Q1Single correctVector Algebra

In an octagon ABCDEFGH of equal side, what is the sum of

A B + A C + A D + A E + A F + A G + A H

, if,

A O = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}

Regular octagon ABCDEFGH (A,B top; C,D right; E,F bottom; G,H left) with centre O.

(A)

16 \hat{i} + 24 \hat{j} - 32 \hat{k}

(B)

16 \hat{i} - 24 \hat{j} + 32 \hat{k}

(C)

16 \hat{i} + 24 \hat{j} + 32 \hat{k}

(D)

- 16 \hat{i} - 24 \hat{j} + 32 \hat{k}

SolutionAnswer: Option 1

16 \hat{i} + 24 \hat{j} - 32 \hat{k}

Approach:

For a regular octagon, the sum of the position vectors of all vertices taken from one vertex equals the number of vertices times the vector from that vertex to the centre.

Step 1:

A B + A C + A D + A E + A F + A G + A H = 8 A O

Step 2:

8 (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) = 16 \hat{i} + 24 \hat{j} - 32 \hat{k}

Final answer:

16 \hat{i} + 24 \hat{j} - 32 \hat{k}

Q2Single correctUnits and Measurements

Match List - I with List - II :

List - I	List - II
(a). $h$ (Planck's constant)	(i). $[MLT^{- 1}]$
(b). $E$ (kinetic energy)	(ii). $[ML^{2} T^{- 1}]$
(c). $V$ (electric potential)	(iii). $[ML^{2} T^{- 3}]$
(d). $P$ (linear momentum)	(iv). $[ML^{2} I^{- 1} T^{- 3}]$

(A)

(a)

\to

(ii), (b)

\to

(iii), (c)

\to

(iv), (d)

\to

(i)

(B)

(a)

\to

(i), (b)

\to

(ii), (c)

\to

(iv), (d)

\to

(iii)

(C)

(a)

\to

(iii), (b)

\to

(ii), (c)

\to

(iv), (d)

\to

(i)

(D)

(a)

\to

(iii), (b)

\to

(iv), (c)

\to

(ii), (d)

\to

(i)

SolutionAnswer: Option 1(a)

\to

(ii), (b)

\to

(iii), (c)

\to

(iv), (d)

\to

(i)

Approach:

Determine the dimensional formula of each quantity in List-I and match against List-II.

Step 1:

[h] = [energy \times time] = ML^{2} T^{- 1}

Step 2:

[V] = [energy / charge] = ML^{2} T^{- 3} I^{- 1}

Step 3:

[P] = MLT^{- 1}

Step 4:

[E] = ML^{2} T^{- 2}

, the only remaining choice is (iii)

Final answer: (a)

\to

(ii), (b)

\to

(iii), (c)

\to

(iv), (d)

\to

(i)

Q3Single correctKinematics

An engine of a train, moving with uniform acceleration, passes the signal-post with velocity

u

and the last compartment with velocity

v

. The velocity with which middle point of the train passes the signal post is :

(A)

\frac{u + v}{2}

(B)

\frac{v ^{2} + u ^{2}}{2}

(C)

\frac{v - u}{2}

(D)

\frac{v ^{2} - u ^{2}}{2}

SolutionAnswer: Option 2

\frac{v ^{2} + u ^{2}}{2}

Approach:

Use the kinematic relation for uniform acceleration over the half length of the train, which equals half the distance over which velocity changes from u to v.

Step 1:

v^{2} = u^{2} + 2 a L

over full length L

Step 2:

v_{m}^{2} = u^{2} + 2 a (\frac{L}{2}) = u^{2} + \frac{v ^{2} - u ^{2}}{2} = \frac{u ^{2} + v ^{2}}{2}

Step 3:

v_{m} = \frac{u ^{2} + v ^{2}}{2}

Final answer:

\frac{v ^{2} + u ^{2}}{2}

Q4Single correctGravitation

A solid sphere of radius R gravitationally attracts a particle placed at

3 R

from its centre with a force

F_{1}

. Now a spherical cavity of radius

(\frac{R}{2})

is made in the sphere (as shown in figure) and the force becomes

F_{2}

. The value of

F_{1} : F_{2}

is:

Solid sphere centre O radius R (shaded) with a spherical cavity radius R/2 centred at B; particle m at A, distance 2R from O.

(A)

41 : 50

(B)

50 : 41

(C)

25 : 36

(D)

36 : 25

SolutionAnswer: Option 2

50 : 41

Approach:

Treat the cavity by superposition: the force of the cavity-sphere equals the force of the full sphere minus the force of a small sphere of the removed mass acting from the cavity centre.

Step 1:

F_{1} = \frac{GM m}{( 3 R ) ^{2}} = \frac{GM m}{9 R ^{2}}

Step 2:

removed mass

= M /8

, cavity centre at

R /2

from O, distance to particle

= 3 R - \frac{R}{2} = \frac{5 R}{2}

Step 3:

F_{2} = \frac{GM m}{9 R ^{2}} - \frac{G ( M /8 ) m}{( 5 R /2 ) ^{2}} = \frac{GM m}{R ^{2}} (\frac{1}{9} - \frac{1}{50}) = \frac{41}{450} \frac{GM m}{R ^{2}}

Step 4:

F_{1} : F_{2} = \frac{50}{450} : \frac{41}{450} = 50 : 41

Final answer:

50 : 41

Q5Single correctGravitation

Two satellites A and B of masses 200 kg and 400 kg are revolving round the earth at height of 600 km and 1600 km respectively. If

T_{A}

and

T_{B}

are the time periods of A and B respectively then the value of

T_{B} - T_{A}

:
[ Given : radius of earth = 6400 km, mass of earth =

6 \times 1 0^{24}

kg ]

Earth E at centre with two concentric circular orbits; satellite A on the inner orbit, B on the outer orbit.

(A)

4. 24 \times 1 0^{3}

(B)

3. 33 \times 1 0^{2}

(C)

1. 33 \times 1 0^{3}

(D)

4. 24 \times 1 0^{2}

SolutionAnswer: Option 3

1. 33 \times 1 0^{3}

Approach:

Use Kepler's third law for circular orbits to find each period from its orbital radius, then take the difference.

Step 1:

r_{A} = 6400 + 600 = 7000 km = 7.0 \times 1 0^{6} m

Step 2:

r_{B} = 6400 + 1600 = 8000 km = 8.0 \times 1 0^{6} m

Step 3:

T_{A} = 2 π \frac{( 7 \times 1 0 ^{6} ) ^{3}}{GM} \approx 5.82 \times 1 0^{3} s

Step 4:

T_{B} = 2 π \frac{( 8 \times 1 0 ^{6} ) ^{3}}{GM} \approx 7.10 \times 1 0^{3} s

Step 5:

T_{B} - T_{A} \approx 1.3 \times 1 0^{3} s

Final answer:

1. 33 \times 1 0^{3}

Q6Single correctGravitation

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The escape velocities of planet A and B are same. But A and B are of unequal mass.
Reason R: The product of their mass and radius must be same.

M_{1} R_{1} = M_{2} R_{2}

In the light of the above statements, choose the most appropriate answer from the options given below :

(A)

Both

A

and

R

are correct and

R

is the correct explanation of

A

(B)

Both

A

and

R

are correct but

R

is NOT the correct explanation of

A

(C)

A

is not correct but

R

is correct

(D)

A

is correct but

R

is not correct

SolutionAnswer: Option 4

A

is correct but

R

is not correct

Approach:

Compare the escape-velocity condition with the relation stated in the Reason.

Step 1:

Equal escape velocities give

\frac{M _{1}}{R _{1}} = \frac{M _{2}}{R _{2}}

Step 2:

The Reason states

M_{1} R_{1} = M_{2} R_{2}

, which is the product, not the ratio

Step 3:

Unequal masses can still satisfy

M / R

equal, so the Assertion is consistent

Final answer:

A

is correct but

R

is not correct

Q7Single correctProperties of Solids and Liquids

Given below are two statements: one is labelled as Assertion

A

and the other is labelled as Reason

R

.
Assertion

A

: When a rod lying freely is heated, no thermal stress is developed in it.
Reason

R

: On heating, the length of the rod increases.
In the light of the above statements, choose the correct answer from the options given below:

(A)

Both

A

and

R

are true and

R

is the correct explanation of

A

(B)

A

is false but

R

is true

(C)

Both

A

and

R

are true but

R

is NOT the correct explanation of

A

(D)

A

is true but

R

is false

SolutionAnswer: Option 3Both

A

and

R

are true but

R

is NOT the correct explanation of

A

Approach:

Judge the truth of each statement and whether the Reason explains the Assertion.

Step 1:

A freely lying rod expands without constraint, so no thermal stress develops:

A

is true

Step 2:

On heating the rod's length increases:

R

is true

Step 3:

The absence of stress is due to the rod being unconstrained, not merely due to length increase, so

R

does not explain

A

Final answer: Both

A

and

R

are true but

R

is NOT the correct explanation of

A

Q8Single correctKinetic Theory of Gases

A diatomic gas, having

C_{P} = \frac{7}{2} R

and

C_{V} = \frac{5}{2} R

, is heated at constant pressure. The ratio dU : dQ : dW

(A)

3 : 5 : 2

(B)

5 : 7 : 3

(C)

5 : 7 : 2

(D)

3 : 7 : 2

SolutionAnswer: Option 3

5 : 7 : 2

Approach:

At constant pressure, express each energy term using the molar heat capacities and the first law of thermodynamics.

Step 1:

d U = n C_{V} d T \propto \frac{5}{2}

Step 2:

d Q = n C_{P} d T \propto \frac{7}{2}

Step 3:

d W = d Q - d U \propto \frac{7}{2} - \frac{5}{2} = \frac{2}{2}

Step 4:

d U : d Q : d W = \frac{5}{2} : \frac{7}{2} : \frac{2}{2} = 5 : 7 : 2

Final answer:

5 : 7 : 2

Q9Single correctOscillations and Waves

If the time period of a two meter long simple pendulum is 2 s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is:

(A)

2 π^{2}

s^{- 2}

(B)

16

s^{- 2}

(C)

π^{2}

s^{- 2}

(D)

9. 8

s^{- 2}

SolutionAnswer: Option 1

2 π^{2}

s^{- 2}

Approach:

Use the simple pendulum period formula and solve for g.

Step 1:

g = \frac{4 π ^{2} L}{T ^{2}}

Step 2:

g = \frac{4 π ^{2} \times 2}{2 ^{2}} = \frac{8 π ^{2}}{4} = 2 π^{2}

Final answer:

2 π^{2}

s^{- 2}

Q10Single correctOscillations and Waves

A student is performing the experiment of the resonance column. The diameter of the column tube is 6 cm. The frequency of the tuning fork is 504 Hz. Speed of the sound at the given temperature is 336 m

s^{- 1}

. The zero of the meter scale coincides with the top end of the resonance column tube. The reading of the water level in the column when the first resonance occurs is:

(A)

14. 8

(B)

18. 4

(C)

16. 6

(D)

13

SolutionAnswer: Option 1

14. 8

Approach:

For a closed pipe the first resonance length equals a quarter wavelength minus the end correction (0.6 times the tube radius).

Step 1:

λ = \frac{336}{504} = 0.6667 m = 66.67 cm

Step 2:

\frac{λ}{4} = 16.67 cm

Step 3:

e = 0.6 \times 3 cm = 1.8 cm

Step 4:

L_{1} = 16.67 - 1.8 = 14.87 cm \approx 14.8 cm

Final answer:

14. 8

Q11Single correctMagnetic Effects of Current and Magnetism

A proton, a deuteron and an

α

particle are moving with same momentum in a uniform magnetic field. The ratio of magnetic forces acting on them ______ is and their speed is ______ in the ratio.

(A)

2 : 1 : 1

and

4 : 2 : 1

(B)

1 : 2 : 4

and

2 : 1 : 1

(C)

4 : 2 : 1

and

2 : 1 : 1

(D)

1 : 2 : 4

and

1 : 1 : 2

SolutionAnswer: Option 1

2 : 1 : 1

and

4 : 2 : 1

Approach:

Express the magnetic force and speed for equal momentum in terms of charge and mass of each particle.

Step 1:

Charges

q_{p} : q_{d} : q_{α} = 1 : 1 : 2

and masses

m_{p} : m_{d} : m_{α} = 1 : 2 : 4

Step 2:

F \propto \frac{q}{m} = \frac{1}{1} : \frac{1}{2} : \frac{2}{4} = 2 : 1 : 1

Step 3:

v \propto \frac{1}{m} = \frac{1}{1} : \frac{1}{2} : \frac{1}{4} = 4 : 2 : 1

Final answer:

2 : 1 : 1

and

4 : 2 : 1

Q12Single correctMagnetic Effects of Current and Magnetism

Magnetic fields at two points on the axis of a circular coil at a distance of

0. 05

m and

0. 2

m from the centre are in the ratio

8 : 1

. The radius of coil is ______ .

(A)

0. 15

(B)

0. 2

(C)

0. 1

(D)

1. 0

SolutionAnswer: Option 3

0. 1

Approach:

Use the on-axis field of a circular coil and form the ratio at the two distances.

Step 1:

\frac{B _{1}}{B _{2}} = (\frac{R ^{2} + x _{2}^{2}}{R ^{2} + x _{1}^{2}})^{3/2} = 8

Step 2:

\frac{R ^{2} + 0.04}{R ^{2} + 0.0025} = 8^{2/3} = 4

Step 3:

R^{2} + 0.04 = 4 R^{2} + 0.01 \Rightarrow 3 R^{2} = 0.03 \Rightarrow R^{2} = 0.01

Step 4:

R = 0.1 m

Final answer:

0. 1

Q13Single correctElectromagnetic Induction and Alternating Currents

The current (i) at time

t = 0

and

t = \infty

respectively for the given circuit is :

Bridge: top two 5 ohm resistors, middle 1 ohm and 4 ohm with battery E (current i) between, bottom inductor L.

(A)

\frac{5 E}{18}, \frac{18 E}{55}

(B)

\frac{5 E}{18}, \frac{10 E}{33}

(C)

\frac{10 E}{33}, \frac{5 E}{18}

(D)

\frac{18 E}{55}, \frac{5 E}{18}

SolutionAnswer: Option 2

\frac{5 E}{18}, \frac{10 E}{33}

Approach:

At t=0 the inductor blocks current (open branch); at t=∞ the inductor behaves as a short circuit. Reduce the resistor network for each case to find the battery current.

Step 1:

t = 0

inductor open: path

(5 + 1) Ω

in parallel with

(5 + 4) Ω

Step 2:

R_{e q} = \frac{6 \times 9}{6 + 9} = \frac{54}{15} = \frac{18}{5} Ω \Rightarrow i (0) = \frac{5 E}{18}

Step 3:

t = \infty

inductor shorts the two side nodes:

(5 ∥ 5) + (1 ∥ 4)

Step 4:

R_{e q} = 2.5 + 0.8 = 3.3 = \frac{33}{10} Ω \Rightarrow i (\infty) = \frac{10 E}{33}

Final answer:

\frac{5 E}{18}, \frac{10 E}{33}

Q14Single correctElectromagnetic Induction and Alternating Currents

The angular frequency of alternating current in a L-C-R circuit is 100 rad

s^{- 1}

. The components connected are shown in the figure. Find the value of inductance of the coil and capacity of condenser.

Series AC L-C-R: R=60 ohm (15 V), capacitor C (10 V), R'=40 ohm branch, inductor (20 V), AC source.

(A)

1. 33

H and

150 μ

(B)

0. 8

H and

150 μ

(C)

1. 33

H and

250 μ

(D)

0. 8

H and

250 μ

SolutionAnswer: Option 4

0. 8

H and

250 μ

Approach:

The same series current flows through every element. Find that current from the resistor whose voltage and resistance are both given, then obtain the reactances from the stated voltages across L and C.

Step 1:

I = \frac{15}{60} = 0.25 A

Step 2:

X_{C} = \frac{10}{0.25} = 40 Ω \Rightarrow C = \frac{1}{ω X _{C}} = \frac{1}{100 \times 40} = 250 μ F

Step 3:

X_{L} = \frac{20}{0.25} = 80 Ω \Rightarrow L = \frac{X _{L}}{ω} = \frac{80}{100} = 0.8 H

Final answer:

0. 8

H and

250 μ

Q15Single correctOptics

Two coherent light sources having intensity in the ratio

2 x

produce an interference pattern. The ratio

\frac{I _{m a x} - I _{m i n}}{I _{m a x} + I _{m i n}}

will be

(A)

\frac{2 2 x}{x + 1}

(B)

\frac{2 x}{2 x + 1}

(C)

\frac{2 2 x}{2 x + 1}

(D)

\frac{2 x}{x + 1}

SolutionAnswer: Option 3

\frac{2 2 x}{2 x + 1}

Approach:

Use the fringe-visibility expression in terms of the two intensities, substituting their given ratio.

Step 1:

\frac{I _{1}}{I _{2}} = 2 x

, take

I_{2} = 1, I_{1} = 2 x

Step 2:

\frac{2 ( 2 x ) ( 1 )}{2 x + 1} = \frac{2 2 x}{2 x + 1}

Final answer:

\frac{2 2 x}{2 x + 1}

Q16Single correctDual Nature of Matter and Radiation

α

particle and a proton are accelerated from rest by a potential difference of 200 V. After this, their de Broglie wavelengths are

λ_{α}

and

λ_{p}

respectively. The ratio

\frac{λ _{p}}{λ _{α}}

is :

(A)

8

(B)

3.8

(C)

2.8

(D)

7.8

SolutionAnswer: Option 3

2.8

Approach:

Apply the de Broglie relation for a charged particle accelerated through a potential difference and take the ratio for the two particles.

Step 1:

\frac{λ _{p}}{λ _{α}} = \frac{m _{α} q _{α}}{m _{p} q _{p}}

Step 2:

m_{α} = 4 m_{p}, q_{α} = 2 q_{p}

Step 3:

\frac{λ _{p}}{λ _{α}} = 4 \times 2 = 8 = 22

Final answer:

2.8

Q17Single correctAtoms and Nuclei

Two radioactive substances X and Y originally have

N_{1}

and

N_{2}

nuclei respectively. Half life of X is half of the half life of Y. After three half lives of Y, number of nuclei of both are equal. The ratio

\frac{N _{1}}{N _{2}}

will be equal to :

(A)

\frac{1}{3}

(B)

\frac{8}{1}

(C)

\frac{3}{1}

(D)

\frac{1}{8}

SolutionAnswer: Option 2

\frac{8}{1}

Approach:

Express the remaining nuclei of each substance after the elapsed time in terms of its own half life and set them equal.

Step 1:

T_{X} = \frac{T _{Y}}{2}, t = 3 T_{Y}

Step 2:

N_{1} (\frac{1}{2})^{3 T_{Y} / T_{X}} = N_{2} (\frac{1}{2})^{3 T_{Y} / T_{Y}}

Step 3:

N_{1} (\frac{1}{2})^{6} = N_{2} (\frac{1}{2})^{3}

Step 4:

\frac{N _{1}}{N _{2}} = (\frac{1}{2})^{- 3} = 8

Final answer:

\frac{8}{1}

Q18Single correctElectronic Devices

A 5 V battery is connected across the points X and Y. Assume

D_{1}

and

D_{2}

to be normal silicon diodes. Find the current supplied by the battery if the

+ v e

terminal of the battery is connected to point X.

Two parallel branches between X and Y: D1 (forward) + 10 ohm; D2 (reverse) + 5 ohm; 5 V across X-Y.

(A)

\sim 0.86

(B)

\sim 0.5

(C)

\sim 1.5

(D)

\sim 0.43

SolutionAnswer: Option 4

\sim 0.43

Approach:

Determine which diode is forward biased for the given battery polarity and compute the current through the conducting branch using the silicon diode drop of 0.7 V.

Step 1:

D_{1}

is forward biased,

D_{2}

is reverse biased

Step 2:

I = \frac{5 - 0.7}{10}

Step 3:

I = \frac{4.3}{10} = 0.43 A

Final answer:

\sim 0.43

Q19Single correctElectromagnetic Waves

Given below are two statements:
Statement I : A speech signal of 2 kHz is used to modulate a carrier signal of 1 MHz. The bandwidth requirement for the signal is 4 kHz.
Statement II : The side band frequencies are 1002 kHz and 998 kHz. In the light of the above statements, choose the correct answer from the options given below:

(A)

Statement I is false but Statement II is true

(B)

Both Statement I and Statement II are true

(C)

Statement I is true but Statement II is false

(D)

Both Statement I and Statement II are false

SolutionAnswer: Option 2Both Statement I and Statement II are true

Approach:

Compute the amplitude-modulation bandwidth and the upper and lower side band frequencies and compare with both statements.

Step 1:

Bandwidth = 2 \times 2 kHz = 4 kHz

Step 2:

f_{U SB} = 1000 + 2 = 1002 kHz

Step 3:

f_{L SB} = 1000 - 2 = 998 kHz

Final answer: Both Statement I and Statement II are true

Q20Single correctExperimental Skills

The pitch of the screw gauge is 1 mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while

7 2^{n d}

division on circular scale coincides with the reference line. The radius of the wire is

(A)

1.64

(B)

0.82

(C)

0.90

(D)

1.80

SolutionAnswer: Option 2

0.82

Approach:

Find the least count, the positive zero error, the observed diameter, correct it, then halve to get the radius.

Step 1:

L.C. = \frac{1}{100} = 0.01 mm

Step 2:

zero error = + 8 \times 0.01 = + 0.08 mm

Step 3:

d_{o b s} = 1 + 72 \times 0.01 = 1.72 mm

Step 4:

d = 1.72 - 0.08 = 1.64 mm

Step 5:

r = \frac{1.64}{2} = 0.82 mm

Final answer:

0.82

Q21NumericalLaws of Motion

A small bob tied at one end of a thin string of length 1 m is describing a vertical circle so that the maximum and minimum tension in the string are in the ratio 5 : 1. The velocity of the bob at the highest position is _____ m

s^{- 1}

. (Take

g = 10

s^{- 2}

)

SolutionAnswer: 5

Approach:

Write the tensions at the lowest and highest points, relate the speeds by energy conservation, and apply the given tension ratio.

Step 1:

\frac{\frac{m v _{b}^{2}}{L} + m g}{\frac{m v _{t}^{2}}{L} - m g} = 5

Step 2:

v_{t}^{2} + 4 gL + gL = 5 (v_{t}^{2} - gL)

Step 3:

10 gL = 4 v_{t}^{2} \Rightarrow v_{t}^{2} = 2.5 gL

Step 4:

v_{t} = 2.5 \times 10 \times 1 = 25 = 5 m s^{- 1}

Final answer:

5

Q22NumericalWork, Energy and Power

The potential energy (U) of a diatomic molecule is a function dependent on r (interatomic distance) as

U = \frac{α}{r ^{10}} - \frac{β}{r ^{5}} - 3

where

α

and

β

are positive constants. The equilibrium distance between two atoms will be

(\frac{2 α}{β})^{\frac{a}{b}}

, where

a =

_____ .

SolutionAnswer: 1

Approach:

Set the derivative of the potential energy with respect to interatomic distance to zero to obtain the equilibrium separation.

Step 1:

\frac{d U}{d r} = - \frac{10 α}{r ^{11}} + \frac{5 β}{r ^{6}} = 0

Step 2:

\frac{5 β}{r ^{6}} = \frac{10 α}{r ^{11}} \Rightarrow r^{5} = \frac{2 α}{β}

Step 3:

r = (\frac{2 α}{β})^{\frac{1}{5}}

Final answer:

1

Q23NumericalThermodynamics

In a certain thermodynamical process, the pressure of a gas depends on its volume as

k V^{3}

. The work done when the temperature changes from 100

^{\circ}

C to 300

^{\circ}

C will be xnR where n denotes number of moles of a gas find x :

SolutionAnswer: 50

Approach:

Express pressure and the ideal gas relation, integrate pressure over volume to find the work, and reduce it to the form xnR using the temperature change.

Step 1:

k V^{3} \cdot V = nRT \Rightarrow k V^{4} = nRT

Step 2:

W = \int k V^{3} d V = \frac{k}{4} (V_{2}^{4} - V_{1}^{4})

Step 3:

W = \frac{1}{4} (nRT_{2} - nRT_{1}) = \frac{n R}{4} (300 - 100)

Step 4:

W = \frac{n R}{4} \times 200 = 50 n R

Final answer:

50

Q24NumericalKinetic Theory of Gases

A monoatomic gas of mass

4.0 u

is kept in an insulated container. The container is moving with velocity 30 m

s^{- 1}

. If the container is suddenly stopped then a change in temperature of the gas (

R =

gas constant) is

\frac{x}{3 R}

. Value of x is,

SolutionAnswer: 3600

Approach:

Equate the bulk kinetic energy of the gas, lost when the container stops, to the increase in internal energy of the monoatomic gas.

Step 1:

\frac{1}{2} M v^{2} = \frac{M}{M _{m o l a r}} \cdot \frac{3}{2} R Δ T

Step 2:

Δ T = \frac{M _{m o l a r} v ^{2}}{3 R}

Step 3:

Δ T = \frac{4 \times 3 0 ^{2}}{3 R} = \frac{3600}{3 R}

Final answer:

3600

Q25NumericalElectrostatics

The electric field in a region is given by

E = (\frac{3}{5} E_{0} \hat{i} + \frac{4}{5} E_{0} \hat{j})

C^{- 1}

. The ratio of flux of reported field through the rectangular surface of area

0.2

m^{2}

(parallel to

y - z

plane) to that of the surface of area

0.3

m^{2}

(parallel to

x - z

plane) is

a : b = a : 2

, where

a =

? [Here

\hat{i}

\hat{j}

and

\hat{k}

are unit vectors along x, y and z -axes respectively]

SolutionAnswer: 1

Approach:

Compute the flux through each surface using the component of the field along the respective surface normal and take the ratio.

Step 1:

ϕ_{1} = \frac{3}{5} E_{0} \times 0.2 = 0.12 E_{0}

Step 2:

ϕ_{2} = \frac{4}{5} E_{0} \times 0.3 = 0.24 E_{0}

Step 3:

\frac{ϕ _{1}}{ϕ _{2}} = \frac{0.12}{0.24} = \frac{1}{2}

Final answer:

1

Q26NumericalElectrostatics

512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is

V

in Volt.

SolutionAnswer: 128

Approach:

Conserve total charge and total volume on combining the drops, then express the potential of the big drop in terms of that of a small drop.

Step 1:

R = 51 2^{1/3} r = 8 r

Step 2:

V_{bi g} = \frac{k ( 512 q )}{8 r} = 64 \cdot \frac{k q}{r}

Step 3:

V_{bi g} = 64 \times 2 = 128 V

Final answer:

128

Q27NumericalCurrent Electricity

In the given circuit of potentiometer, the potential difference E across AB(10 m length) is larger than

E_{1}

and

E_{2}

as well. For key

K_{1}

(closed), the jockey is adjusted to touch the wire at point

J_{1}

so that there is no deflection in the galvanometer. Now the first battery (

E_{1}

) is replaced by second battery (

E_{2}

) for working by making

K_{1}

open and

K_{2}

closed. The galvanometer gives then null deflection at

J_{2}

. The value of

\frac{E _{1}}{E _{2}}

\frac{a}{2}

, where

a =

_____ .

Potentiometer with boustrophedon wire AB (10 m), driver cell E + key K + rheostat Rh, cells E1/K1 and E2/K2 into galvanometer G, jockey J1 at 20 cm and J2 at 60 cm.

SolutionAnswer: 1

Approach:

Use the potentiometer balance condition that the unknown EMF is proportional to the balancing length from A, reading the balancing lengths of J1 and J2 from the snake-wire layout.

Step 1:

L_{1} = 1 + 0.8 = 1.8 m

Step 2:

L_{2} = 3 + 0.6 = 3.6 m

Step 3:

\frac{E _{1}}{E _{2}} = \frac{L _{1}}{L _{2}} = \frac{1.8}{3.6} = \frac{1}{2}

Step 4:

\frac{a}{2} = \frac{1}{2} \Rightarrow a = 1

Final answer:

1

Q28NumericalElectromagnetic Induction and Alternating Currents

A coil of inductance 2 H having negligible resistance is connected to a source of supply whose voltage is given by

V = 3 t

volt. (where

t

is in second). If the voltage is applied when

t = 0

, then the energy stored in the coil after 4 s in J,

SolutionAnswer: 144

Approach:

Use the inductor voltage relation to find the current as a function of time, evaluate it at 4 s, and compute the magnetic energy stored.

Step 1:

3 t = 2 \frac{d i}{d t} \Rightarrow d i = \frac{3 t}{2} d t

Step 2:

i = \frac{3}{4} t^{2}

Step 3:

i (4) = \frac{3}{4} \times 16 = 12 A

Step 4:

U = \frac{1}{2} \times 2 \times 1 2^{2} = 144 J

Final answer:

144

Q29NumericalElectromagnetic Induction and Alternating Currents

A transmitting station releases waves of wavelength 960 m. A capacitor of

2.56 μ

F is used in the resonant circuit. The self-inductance of coil necessary for resonance is

x \times 1 0^{- 8}

H. find x

SolutionAnswer: 10

Approach:

Relate the wavelength to the resonant period of the LC circuit, solve for the product LC, and extract the inductance.

Step 1:

T = \frac{λ}{c} = \frac{960}{3 \times 1 0 ^{8}} = 3.2 \times 1 0^{- 6} s

Step 2:

L C = \frac{T}{2 π} = 5.09 \times 1 0^{- 7}

Step 3:

L C = 2.59 \times 1 0^{- 13}

Step 4:

L = \frac{2.59 \times 1 0 ^{- 13}}{2.56 \times 1 0 ^{- 6}} \approx 10 \times 1 0^{- 8} H

Final answer:

10

Q30NumericalOptics

The same size images are formed by a convex lens when the object is placed at 20 cm or at 10 cm from the lens. The focal length of convex lens is _____ .

SolutionAnswer: 15

Approach:

Equate the magnitudes of the magnifications for the real image (object beyond focus) and the virtual image (object inside focus), with opposite signs, and solve for the focal length.

Step 1:

\frac{f}{f - 20} = - \frac{f}{f - 10}

Step 2:

f - 10 = - (f - 20)

Step 3:

2 f = 30 \Rightarrow f = 15 cm

Final answer:

15

Chemistry30 questions

Q31Single correctPurification and Characterisation of Organic Compounds

Complete combustion of 1.80 g of an oxygen containing compound

(C_{x} H_{y} O_{z})

gave 2.64 g of

CO_{2}

and 1.08 g of

H_{2} O

. The percentage of oxygen in the organic compound is :

(A)

53.33

(B)

50.33

(C)

51.63

(D)

63.53

SolutionAnswer: Option 1

53.33

Approach:

Determine the mass of carbon and hydrogen from the combustion products, obtain oxygen mass by difference, and express it as a percentage of the sample mass.

Step 1:

mass of C = \frac{12}{44} \times 2.64 = 0.72 g

Step 2:

mass of H = \frac{2}{18} \times 1.08 = 0.12 g

Step 3:

mass of O = 1.80 - 0.72 - 0.12 = 0.96 g

Step 4:

% O = \frac{0.96}{1.80} \times 100 = 53.33

Final answer:

53.33

Q32Single correctAtomic Structure

The plots of radial distribution functions for various orbitals of hydrogen atom against '

r

' are given below.
The correct plot for the 3s orbital is :

(A)

(B)

(C)

(D)

SolutionAnswer: Option 3

(drawn graph with three maxima and two radial nodes)

Approach:

Count the number of radial maxima and radial nodes characteristic of the 3s orbital and match it to the plot.

Step 1:

n = 3, l = 0

Step 2:

radial nodes = 3 - 0 - 1 = 2

Step 3:

maxima = 3 - 0 = 3

Final answer:

plot with three maxima and two radial nodes (option 3)

Q33Single correctChemical Bonding and Molecular Structure

According to molecular orbital theory, the species among the following that does not exist is :

(A)

He_{2}^{-}

(B)

O_{2}^{2 -}

(C)

He_{2}^{+}

(D)

Be_{2}

SolutionAnswer: Option 4

Be_{2}

Approach:

Compute the bond order of each species from its molecular orbital electron count; a species with zero bond order does not exist.

Step 1:

He_{2}^{-} : 5 electrons, B.O. = \frac{2 - 3}{...} \Rightarrow 0.5

Step 2:

O_{2}^{2 -} : B.O. = 1

Step 3:

He_{2}^{+} : 3 electrons, B.O. = \frac{2 - 1}{2} = 0.5

Step 4:

Be_{2} : B.O. = \frac{4 - 4}{2} = 0

Final answer:

Be_{2}

Q34Single correctEquilibrium

The solubility of AgCN in a buffer solution of pH

= 3

is x. The value of x is : [Assume : No cyano complex is formed;

K_{s p} (AgCN) = 2.2 \times 1 0^{- 16}

and

K_{a} (HCN) = 6.2 \times 1 0^{- 10}

]

(A)

1.9 \times 1 0^{- 5}

(B)

2.2 \times 1 0^{- 16}

(C)

0.625 \times 1 0^{- 6}

(D)

1.6 \times 1 0^{- 6}

SolutionAnswer: Option 1

1.9 \times 1 0^{- 5}

Approach:

Account for protonation of the dissolved cyanide ion at pH 3, relate the free cyanide fraction to the solubility, and solve the modified solubility product expression.

Step 1:

[H^{+}] = 1 0^{- 3}

Step 2:

s^{2} \frac{K _{a}}{K _{a} + [ H ^{+} ]} = K_{s p}

Step 3:

s = \frac{K _{s p} ( K _{a} + [ H ^{+} ])}{K _{a}} = \frac{2.2 \times 1 0 ^{- 16} ( 6.2 \times 1 0 ^{- 10} + 1 0 ^{- 3} )}{6.2 \times 1 0 ^{- 10}}

Step 4:

s \approx 1.9 \times 1 0^{- 5}

Final answer:

1.9 \times 1 0^{- 5}

Q35Single correctp-Block Elements

Which of the following equation depicts the oxidizing nature of

H_{2} O_{2}

(A)

2 I^{-} + H_{2} O_{2} + 2 H^{+} \to I_{2} + 2 H_{2} O

(B)

I_{2} + H_{2} O_{2} + 2 OH^{-} \to 2 I^{-} + 2 H_{2} O + O_{2}

(C)

Cl_{2} + H_{2} O_{2} \to 2 HCl + O_{2}

(D)

KIO_{4} + H_{2} O_{2} \to KIO_{3} + H_{2} O + O_{2}

SolutionAnswer: Option 1

2 I^{-} + H_{2} O_{2} + 2 H^{+} \to I_{2} + 2 H_{2} O

Approach:

Track the oxidation states; H2O2 acts as an oxidant when it is itself reduced (oxygen from -1 to -2) while oxidizing the other species.

Step 1:

In option 1, iodide I^{-} (- 1) \to I_{2} (0)

Step 2:

Oxygen in H_{2} O_{2} : - 1 \to - 2 in H_{2} O

Step 3:

H_{2} O_{2} gains electrons, acting as oxidizing agent

Final answer:

2 I^{-} + H_{2} O_{2} + 2 H^{+} \to I_{2} + 2 H_{2} O

Q36Single correctp-Block Elements

The incorrect statement about

B_{2} H_{6}

is :

(A)

All

B - H - B

angles are of

12 0^{\circ}

(B)

The two

B - H - B

bonds (bridged and terminal) are not of same length.

(C)

Terminal

B - H

bonds have less p -character when compared to bridging bonds.

(D)

Its fragment,

BH_{3}

, behaves as a Lewis acid.

SolutionAnswer: Option 1All

B - H - B

angles are of

12 0^{\circ}

Approach:

Recall the structure of diborane and evaluate each statement against the known bridge geometry and bonding of B2H6.

Step 1:

The bridging B - H - B angle is about 8 3^{\circ} - 9 7^{\circ}, not 12 0^{\circ}

Step 2:

Bridge B-H bonds are longer than terminal B-H bonds

Step 3:

Terminal B-H bonds are nearly s p^{3} with less p-character than the bridge bonds

Step 4:

BH_{3} is electron deficient and acts as a Lewis acid

Final answer: All

B - H - B

angles are of

12 0^{\circ}

Q37Single correctOrganic Compounds Containing Oxygen

Compound(s) which will liberate carbon dioxide with sodium bicarbonate solution is/are :
A, B and C are the structures shown.

Three structures: (A) 2,4,6-triaminophenol, (B) benzoic acid, (C) 2,4,6-trinitrophenol (picric acid).

(A)

A and B only

(B)

B and C only

(C)

C only

(D)

B only

SolutionAnswer: Option 2B and C only

Approach:

Sodium bicarbonate liberates carbon dioxide only with acids stronger than carbonic acid; identify which structures are sufficiently acidic.

Step 1:

B is benzoic acid, a carboxylic acid stronger than H_{2} CO_{3}

Step 2:

C is 2,4,6-trinitrophenol (picric acid), strongly acidic due to three - NO_{2} groups

Step 3:

A is 2,4,6-triaminophenol; the phenolic - OH is weakly acidic and does not liberate CO_{2}

Final answer: B and C only

Q38Single correctHydrocarbons

Identify A in the given chemical reaction.
The branched chain hydrocarbon (drawn structure) is heated over

Mo_{2} O_{3}

at 773 K, 10-20 atm to give 'A' as the major product.

2-methylhexane (branched C7 alkane) undergoing aromatization.

(A)

(B)

(C)

(D)

SolutionAnswer: Option 4

toluene (methylbenzene)

Approach:

Recognize the reaction as catalytic aromatization (reforming) of a seven-carbon alkane over Mo2O3, which cyclises and dehydrogenates the chain to an aromatic product.

Step 1:

The reactant is a branched heptane skeleton (seven carbons)

Step 2:

Catalytic aromatization over Mo_{2} O_{3} at 773 K cyclises and dehydrogenates

Step 3:

A seven-carbon chain yields the seven-carbon aromatic, toluene

Final answer:

toluene

Q39Single correctOrganic Compounds Containing Nitrogen

Which of the following reaction/s will not give

p

-aminoazobenzene?
Reactions of nitro/amino benzene derivatives A, B and C are shown with the reagents indicated.

Three benzene reaction schemes A, B, C with reagents over arrows.

(A)

B only

(B)

A only

(C)

C only

(D)

A and B

SolutionAnswer: Option 1B only

Approach:

Trace each sequence to see whether a diazonium salt is generated that can couple with aniline to form p-aminoazobenzene; the route lacking nitrous acid diazotisation fails.

Step 1:

Route A: nitrobenzene Sn/HCl aniline HNO_{2} diazonium aniline coupling gives product

Step 2:

Route B: nitrobenzene NaBH_{4} / NaOH azoxy/azo, no diazonium \Rightarrow no coupling with aniline

Step 3:

Route C: aniline HNO_{2} diazonium aniline, HCl coupling gives product

Final answer: B only

Q40Single correctOrganic Compounds Containing Halogens

Identify A and B in the chemical reaction.
The starting material (drawn structure) reacts with

HCl

to give

[A]

(major), which then reacts with

NaI

in dry acetone to give

[B]

major.

Methoxy-substituted reactant for the reaction.

(A)

(B)

(C)

(D)

SolutionAnswer: Option 2

A and B drawn structures (set 2)

Approach:

Apply Markovnikov addition of HCl across the ring double bond to obtain A, then Finkelstein exchange of chloride by iodide with NaI in dry acetone to obtain B, and match the regiochemistry to the option.

Step 1:

HCl adds across the alkene; chloride goes to the carbon giving the more stable carbocation (Markovnikov)

Step 2:

NaI in dry acetone replaces Cl by I (Finkelstein reaction)

Step 3:

The chloro/iodo substituent positions in option 2 match the Markovnikov product and its halide exchange

Final answer:

A and B drawn structures (set 2)

Q41Single correctp-Block Elements

Given below are two statements:
Statement I : An allotrope of oxygen is an important intermediate in the formation of reducing smog.
Statement II : Gases such as oxides of nitrogen and sulphur present in troposphere contribute to the formation of photochemical smog. In the light of the above statements, choose the correct answer from the options given below:

(A)

Both Statement I and Statement II are false

(B)

Both Statement I and Statement II are true

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

SolutionAnswer: Option 1Both Statement I and Statement II are false

Approach:

Evaluate each statement against the definitions of reducing (classical) smog and photochemical (oxidising) smog.

Step 1:

Ozone (an allotrope of oxygen) is an intermediate in photochemical (oxidising) smog, not reducing smog

Step 2:

Oxides of nitrogen and sulphur in troposphere produce acid rain/reducing smog; photochemical smog is driven by hydrocarbons and NO_{x} under sunlight, not sulphur oxides

Final answer: Both Statement I and Statement II are false

Q42Single correctp-Block Elements

In Freundlich adsorption isotherm at moderate pressure, the extent of adsorption

(\frac{x}{m})

is directly proportional to

P^{x}

. The value of x is :

(A)

zero

(B)

\frac{1}{n}

(C)

1

(D)

\infty

SolutionAnswer: Option 2

\frac{1}{n}

Approach:

Recall the Freundlich isotherm and identify the pressure exponent in the moderate-pressure regime.

Step 1:

\frac{x}{m} \propto P^{1/ n} with 0 < \frac{1}{n} < 1 at moderate pressure

Step 2:

Comparing with \frac{x}{m} \propto P^{x} \Rightarrow x = \frac{1}{n}

Final answer:

\frac{1}{n}

Q43Single correctd- and f-Block Elements

Ellingham diagram is a graphical representation of :

(A)

Δ G

vs P

(B)

(Δ G - T Δ S)

vs T

(C)

Δ H

vs T

(D)

Δ G

vs T

SolutionAnswer: Option 4

Δ G

vs T

Approach:

Recall the definition of the Ellingham diagram used in metallurgy.

Step 1:

The Ellingham diagram plots standard free energy of formation of oxides against temperature

Final answer:

Δ G

vs T

Q44Single correctd- and f-Block Elements

Given below are two statements:
Statement I :

CeO_{2}

can be used for oxidation of aldehydes and ketones.
Statement II : Aqueous solution of

EuSO_{4}

is a strong reducing agent. In the light of the above statements, choose the correct answer from the options given below:

(A)

Both Statement I and Statement II are true

(B)

Statement I is true but Statement II is false

(C)

Both Statement I and Statement II are false

(D)

Statement I is false but Statement II is true

SolutionAnswer: Option 1Both Statement I and Statement II are true

Approach:

Assess the oxidising ability of cerium(IV) and the reducing ability of europium(II) from lanthanide chemistry.

Step 1:

Ce^{4 +} in CeO_{2} readily reverts to Ce^{3 +}, acting as an oxidant for aldehydes and ketones

Step 2:

Eu^{2 +} in EuSO_{4} is readily oxidised to the stable Eu^{3 +}, making it a strong reducing agent

Final answer: Both Statement I and Statement II are true

Q45Single correctd- and f-Block Elements

In which of the following pairs, the outermost electronic configuration will be the same?

(A)

Cr^{+}

and

Mn^{2 +}

(B)

Ni^{2 +}

and

Cu^{+}

(C)

Fe^{2 +}

and

Co^{+}

(D)

V^{2 +}

and

Cr^{+}

SolutionAnswer: Option 1

Cr^{+}

and

Mn^{2 +}

Approach:

Write the d-electron configuration of each ion and find the pair sharing the identical outermost configuration.

Step 1:

Cr^{+} : [Ar] 3 d^{5}, Mn^{2 +} : [Ar] 3 d^{5}

Step 2:

Ni^{2 +} : 3 d^{8}, Cu^{+} : 3 d^{10}

Step 3:

Fe^{2 +} : 3 d^{6}, Co^{+} : 3 d^{8}

Step 4:

V^{2 +} : 3 d^{3}, Cr^{+} : 3 d^{5}

Final answer:

Cr^{+}

and

Mn^{2 +}

Q46Single correctCoordination Compounds

The hybridization and magnetic nature of

[Mn(CN)_{6}]^{4 -}

and

[Fe(CN)_{6}]^{3 -}

, respectively are :

(A)

s p^{3} d^{2}

and diamagnetic

(B)

d^{2} s p^{3}

and paramagnetic

(C)

s p^{3} d^{2}

and paramagnetic

(D)

d^{2} s p^{3}

and diamagnetic

SolutionAnswer: Option 2

d^{2} s p^{3}

and paramagnetic

Approach:

Determine oxidation state and d-electron count of the central metal, apply strong-field CN- pairing, then assign hybridization and magnetic nature.

Step 1:

In [Mn(CN)_{6}]^{4 -}, Mn is + 2 (3 d^{5})

Step 2:

In [Fe(CN)_{6}]^{3 -}, Fe is + 3 (3 d^{5})

Step 3:

CN^{-} is a strong field ligand \to low spin t_{2 g}^{5}, one unpaired electron

Step 4:

Inner orbital complex \to d^{2} s p^{3}, μ = 1 (1 + 2) = 3 BM (paramagnetic)

Final answer:

d^{2} s p^{3}

and paramagnetic

Q47Single correctOrganic Compounds Containing Oxygen

Which one of the following reactions will not form acetaldehyde?

(A)

CH_{3} CH_{2} OH CrO_{3} - H_{2} SO_{4}

(B)

CH_{2} = CH_{2} + O_{2} Pd(II)/Cu(II) H_{2} O

(C)

CH_{3} CN i )DIBAL - H ii )H_{2} O >

(D)

CH_{3} CH_{2} OH 573 K^{Cu} >

SolutionAnswer: Option 1

CH_{3} CH_{2} OH CrO_{3} - H_{2} SO_{4}

Approach:

Identify the product of each oxidation/reduction route from ethanol or its equivalent, then locate the one that does not give acetaldehyde.

Step 1:

CH_{3} CH_{2} OH CrO_{3} - H_{2} SO_{4} CH_{3} COOH

Step 2:

CH_{2} = CH_{2} + \frac{1}{2} O_{2} Pd/Cu H_{2} O CH_{3} CHO

(Wacker)

Step 3:

CH_{3} CN DIBAL-H H_{2} O CH_{3} CHO

Step 4:

CH_{3} CH_{2} OH Cu 573 K CH_{3} CHO

Final answer:

CH_{3} CH_{2} OH CrO_{3} - H_{2} SO_{4}

Q48Single correctOrganic Compounds Containing Oxygen

The major product of the following chemical reaction is :

CH_{3} CH_{2} CN (1) H_{3} O^{+}, Δ (2) SOCl_{2} (3) Pd/BaSO_{4}, H_{2} ?

(A)

CH_{3} CH_{2} CHO

(B)

CH_{3} CH_{2} CH_{2} OH

(C)

CH_{3} CH_{2} CH_{3}

(D)

(CH_{3} CH_{2} CO)_{2} O

SolutionAnswer: Option 1

CH_{3} CH_{2} CHO

Approach:

Carry the nitrile through acid hydrolysis, acyl chloride formation, and Rosenmund reduction step by step.

Step 1:

CH_{3} CH_{2} CN H_{3} O^{+}, Δ CH_{3} CH_{2} COOH

Step 2:

CH_{3} CH_{2} COOH SOCl_{2} CH_{3} CH_{2} COCl

Step 3:

CH_{3} CH_{2} COCl Pd/BaSO_{4} H_{2} CH_{3} CH_{2} CHO

Final answer:

CH_{3} CH_{2} CHO

Q49Single correctSome Basic Principles of Organic Chemistry

Which statement is correct?

(A)

Synthesis of Buna-S needs nascent oxygen.

(B)

Buna-S is a synthetic and linear thermosetting polymer.

(C)

Buna-N is a natural polymer.

(D)

Neoprene is an addition copolymer used in plastic bucket manufacturing.

SolutionAnswer: Option 1Synthesis of Buna-S needs nascent oxygen.

Approach:

Evaluate each statement against the known nature and synthesis of Buna-S, Buna-N and Neoprene.

Step 1:

Buna-S = styrene-1,3-butadiene copolymer, an elastomer (not thermosetting)

Step 2:

Buna-N is a synthetic copolymer of acrylonitrile and 1,3-butadiene

Step 3:

Neoprene is a polymer of chloroprene used for oil-resistant goods, not plastic buckets

Step 4:

Free-radical (peroxide/persulphate) initiation supplies nascent oxygen for Buna-S

Final answer: Synthesis of Buna-S needs nascent oxygen.

Q50Single correctBiomolecules

Which of the glycosidic linkage between galactose and glucose is present in lactose?

(A)

C - 1

of glucose and

C - 4

of galactose

(B)

C - 1

of glucose and

C - 6

of galactose

(C)

C - 1

of galactose and

C - 4

of glucose

(D)

C - 1

of galactose and

C - 6

of glucose

SolutionAnswer: Option 3

C - 1

of galactose and

C - 4

of glucose

Approach:

Recall the structure of lactose and the carbon atoms involved in its glycosidic bond.

Step 1:

Lactose = galactose- β (1 \to 4) -glucose

Step 2:

The bond joins C-1 of galactose to C-4 of glucose

Final answer:

C - 1

of galactose and

C - 4

of glucose

Q51NumericalSome Basic Concepts in Chemistry

A car tyre is filled with nitrogen gas at 35 psi at

27^{\circ} C

. It will burst if pressure exceeds 40 psi. The temperature in

^{\circ} C

at which the car tyre will burst is (Rounded-off to the nearest integer)

SolutionAnswer: 70

Approach:

Apply Gay-Lussac's law at constant volume to find the bursting temperature.

Step 1:

\frac{35}{300} = \frac{40}{T _{2}}

Step 2:

T_{2} = \frac{40 \times 300}{35} = 342.86 K

Step 3:

t_{2} = 342.86 - 273 = 69.86^{\circ} C \approx 70^{\circ} C

Final answer: 70

Q52NumericalChemical Thermodynamics

The reaction of cyanamide,

NH_{2} CN_{(s)}

with oxygen was run in a bomb calorimeter and

Δ U

was found to be

- 742.24 kJ mol^{- 1}

. The magnitude of

Δ H_{298}

for the reaction

NH_{2} CN_{(s)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + O_{2 (g)} + H_{2} O_{(l)}

is kJ. (Rounded off to the nearest integer) [Assume ideal gases and

R = 8.314 J mol^{- 1} K^{- 1}

]

SolutionAnswer: 741

Approach:

Relate enthalpy and internal energy change through the change in gaseous moles and report the magnitude.

Step 1:

Δ n_{g} = (gas moles in products) - (gas moles in reactants) = 2 - 1.5 = 0.5

Step 2:

Δ n_{g} RT = 0.5 \times 8.314 \times 1 0^{- 3} \times 298 = 1.24 kJ

Step 3:

Δ H = - 742.24 + 1.24 = - 741.0 kJ mol^{- 1}

Step 4:

∣Δ H ∣ = 741 kJ

Final answer: 741

Q53NumericalChemical Thermodynamics

The ionization enthalpy of

Na^{+}

formation from

Na_{(g)}

495.8 kJ mol^{- 1}

, while the electron gain enthalpy of Br is

- 325.0 kJ mol^{- 1}

. Given the lattice enthalpy of NaBr is

- 728.4 kJ mol^{- 1}

. The energy for the formation of NaBr ionic solid is

(-)

______

1 0^{- 1} kJ mol^{- 1}

SolutionAnswer: 5576

Approach:

Sum the ionization enthalpy, electron gain enthalpy and lattice enthalpy, then express the magnitude in units of

1 0^{-} 1

kJ/mol.

Step 1:

Δ H = 495.8 + (- 325.0) + (- 728.4)

Step 2:

Δ H = - 557.6 kJ mol^{- 1}

Step 3:

∣Δ H ∣ = 557.6 = 5576 \times 1 0^{- 1} kJ mol^{- 1}

Final answer: 5576

Q54NumericalSome Basic Concepts in Chemistry

0.4 g

mixture of NaOH,

Na_{2} CO_{3}

and some inert impurities was first titrated with

\frac{N}{10}

HCl using phenolphthalein as an indicator,

17.5 mL

of HCl was required at the end point. After this methyl orange was added and titrated.

1.5 mL

of same HCl was required for the next end point. The weight percentage of

Na_{2} CO_{3}

in the mixture is (Rounded-off to the nearest integer)

SolutionAnswer: 4

Approach:

Use the methyl orange volume (second stage of carbonate) to find moles of Na2CO3, then compute its mass percentage.

Step 1:

Methyl orange stage: NaHCO_{3} + HCl \to NaCl + H_{2} O + CO_{2}

Step 2:

n_{Na_{2} CO_{3}} = 0.1 \times 1.5 \times 1 0^{- 3} = 1.5 \times 1 0^{- 4} mol

Step 3:

mass = 1.5 \times 1 0^{- 4} \times 106 = 0.0159 g

Step 4:

% = \frac{0.0159}{0.4} \times 100 = 3.975 \approx 4

Final answer: 4

Q55NumericalRedox Reactions and Electrochemistry

In basic medium

CrO_{4}^{2 -}

oxidises

S_{2} O_{3}^{2 -}

to form

SO_{4}^{2 -}

and itself changes into

Cr(OH)_{4}^{-}

. The volume of

0.154 M CrO_{4}^{2 -}

required to react with

40 mL

0.25 M S_{2} O_{3}^{2 -}

is ______ mL. (Rounded-off to the nearest integer)

SolutionAnswer: 173

Approach:

Balance the electron transfer for each species and equate equivalents to find the required volume.

Step 1:

Cr: + 6 \to + 3, n_{1} = 3

Step 2:

S_{2} O_{3}^{2 -} \to 2 SO_{4}^{2 -}, S: + 2 \to + 6, n_{2} = 8

Step 3:

0.154 \times V \times 3 = 0.25 \times 40 \times 8

Step 4:

V = \frac{80}{0.462} = 173.16 \approx 173 mL

Final answer: 173

Q56NumericalPurification and Characterisation of Organic Compounds

Using the provided information in the following paper chromatogram:
The calculated

R_{f}

value of A is ______

\times 1 0^{- 1}

Paper chromatography strip: spot at base line, A at 2 cm, B at 4 cm, solvent front at 5 cm.

SolutionAnswer: 4

Approach:

Read the distances of spot A and of the solvent front from the chromatogram and apply the Rf definition.

Step 1:

Distance of A from base line = 2 cm

Step 2:

Distance of solvent front from base line = 5 cm

Step 3:

R_{f} (A) = \frac{2}{5} = 0.4 = 4 \times 1 0^{- 1}

Final answer: 4

Q57NumericalHydrocarbons

Consider the following chemical reaction.

CH \equiv CH 1) Red hot Fe tube, 873 K 2) CO,HCl,AlCl_{3} Product

The number of

s p^{2}

hybridized carbon atom(s) present in the product is

SolutionAnswer: 7

Approach:

Identify the product of the trimerisation followed by Gattermann-Koch formylation, then count its sp2 carbons.

Step 1:

3 CH \equiv CH Red hot Fe, 873 K C_{6} H_{6} (benzene)

Step 2:

C_{6} H_{6} CO, HCl, AlCl_{3} C_{6} H_{5} CHO (benzaldehyde)

Step 3:

6 aromatic ring carbons + 1 carbonyl (CHO) carbon, all s p^{2} = 7

Final answer: 7

Q58NumericalSolutions

1 molal aqueous solution of an electrolyte

A_{2} B_{3}

is 60% ionised. The boiling point of the solution at 1 atm is ______ K. (Rounded-off to the nearest integer) [Given

K_{b}

for

(H_{2} O) = 0.52 K kg mol^{- 1}

]

SolutionAnswer: 375

Approach:

Compute the van't Hoff factor from the degree of ionisation, find the boiling point elevation and add it to the normal boiling point of water.

Step 1:

A_{2} B_{3} \to 2 A^{3 +} + 3 B^{2 -}, n = 5

Step 2:

i = 1 + 0.6 (5 - 1) = 1 + 2.4 = 3.4

Step 3:

Δ T_{b} = 3.4 \times 0.52 \times 1 = 1.768 K

Step 4:

T_{b} = 373.15 + 1.768 = 374.9 \approx 375 K

Final answer: 375

Q59NumericalChemical Kinetics

For the reaction,

a A + b B \to c C + d D

, the plot of

lo g k

v/s

\frac{1}{T}

is given below:
The temperature at which the rate constant of the reaction is

1 0^{- 4} s^{- 1}

is ______ K. (Rounded-off to the nearest integer) [Given : The rate constant of the reaction is

1 0^{- 5} s^{- 1}

at 500 K.]

Arrhenius plot of log k versus 1/T, a straight line of slope -10000 K.

SolutionAnswer: 526

Approach:

Use the slope of the Arrhenius plot to relate the two rate constants at two temperatures and solve for the unknown temperature.

Step 1:

lo g \frac{1 0 ^{- 4}}{1 0 ^{- 5}} = 10000 (\frac{1}{500} - \frac{1}{T})

Step 2:

1 = 10000 (\frac{1}{500} - \frac{1}{T})

Step 3:

\frac{1}{500} - \frac{1}{T} = 1 0^{- 4} \Rightarrow \frac{1}{T} = 0.002 - 0.0001 = 0.0019

Step 4:

T = 526.3 \approx 526 K

Final answer: 526

Q60Numericalp-Block Elements

Among the following, the number of halide(s) which is/ are inert to hydrolysis is
(A)

BF_{3}

(B)

SiCl_{4}

(C)

PCl_{5}

(D)

SF_{6}

SolutionAnswer: 1

Approach:

Examine each halide for the availability of an empty orbital or coordination site that allows attack by water.

Step 1:

BF_{3} : electron-deficient boron, readily hydrolysed

Step 2:

SiCl_{4} : vacant d -orbitals on Si, readily hydrolysed

Step 3:

PCl_{5} : hydrolysed to POCl_{3} and H_{3} PO_{4}

Step 4:

SF_{6} : sterically shielded, no site for water attack, inert

Final answer: 1

Mathematics30 questions

Q61Single correctComplex Numbers and Quadratic Equations

The integer k, for which the inequality

x^{2} - 2 (3 k - 1) x + 8 k^{2} - 7 > 0

is valid for every x in R is:

(A)

4

(B)

2

(C)

3

(D)

0

SolutionAnswer: Option 3

3

Approach:

A monic quadratic is positive for all real

x

exactly when its discriminant is negative.

Step 1:

D = 4 (3 k - 1)^{2} - 4 (8 k^{2} - 7) < 0

Step 2:

(3 k - 1)^{2} - (8 k^{2} - 7) = k^{2} - 6 k + 8 < 0

Step 3:

(k - 2) (k - 4) < 0 \Rightarrow 2 < k < 4

Step 4:

k = 3

Final answer:

3

Q62Single correctCo-ordinate Geometry

Let the lines

(2 - i) z = (2 + i) \overset{z}{ˉ}

and

(2 + i) z + (i - 2) \overset{z}{ˉ} - 4 i = 0

, (here

i^{2} = - 1

) be normal to a circle C. If the line

i z + \overset{z}{ˉ} + 1 + i = 0

is tangent to this circle C, then its radius is :

(A)

\frac{3}{2}

(B)

32

(C)

\frac{3}{2 2}

(D)

\frac{1}{2 2}

SolutionAnswer: Option 3

\frac{3}{2 2}

Approach:

Two normals meet at the centre; the radius is the distance from the centre to the tangent line. Write each complex equation as a real Cartesian line using

z = x + i y

Step 1:

(2 - i) (x + i y) = (2 + i) (x - i y) \Rightarrow x = 2 y

Step 2:

(2 + i) z + (i - 2) \overset{z}{ˉ} - 4 i = 0 \Rightarrow x + 2 y = 2

Step 3:

x = 2 y, x + 2 y = 2 \Rightarrow (1, \frac{1}{2})

Step 4:

i z + \overset{z}{ˉ} + 1 + i = 0 \Rightarrow x - y + 1 = 0

Step 5:

r = \frac{∣ 1 - \frac{1}{2} + 1 ∣}{2} = \frac{3/2}{2} = \frac{3}{2 2}

Final answer:

\frac{3}{2 2}

Q63Single correctPermutations and Combinations

The total number of positive integral solutions (x,\ y,\ z) such that

xyz = 24

is :

(A)

45

(B)

30

(C)

36

(D)

24

SolutionAnswer: Option 2

30

Approach:

Factorise 24 into primes and distribute each prime's exponent among three variables using stars and bars.

Step 1:

2^{3} : (2 3 + 2) = (2 5) = 10

Step 2:

3^{1} : (2 1 + 2) = (2 3) = 3

Step 3:

10 \times 3 = 30

Final answer:

30

Q64Single correctSequence and Series

0 < θ, ϕ < \frac{π}{2}

x = \sum_{n = 0}^{\infty} cos^{2 n} θ

y = \sum_{n = 0}^{\infty} sin^{2 n} ϕ

and

z = \sum_{n = 0}^{\infty} cos^{2 n} θ \cdot sin^{2 n} ϕ

then :

(A)

x y - z = (x + y) z

(B)

x y + yz + z x = z

(C)

x y + z = (x + y) z

(D)

x yz = 4

SolutionAnswer: Option 3

x y + z = (x + y) z

Approach:

Sum each infinite geometric series, then verify the algebraic relation among x, y and z.

Step 1:

x = \frac{1}{1 - cos ^{2} θ} = \frac{1}{sin ^{2} θ}

Step 2:

y = \frac{1}{1 - sin ^{2} ϕ} = \frac{1}{cos ^{2} ϕ}

Step 3:

z = \frac{1}{1 - cos ^{2} θ sin ^{2} ϕ}

Step 4:

cos^{2} θ = 1 - \frac{1}{x}, sin^{2} ϕ = 1 - \frac{1}{y} \Rightarrow z = \frac{x y}{x + y - 1}

Step 5:

z (x + y - 1) = x y \Rightarrow (x + y) z = x y + z

Final answer:

x y + z = (x + y) z

Q65Single correctTrigonometry

All possible values of

θ \in [0, 2 π]

for which

sin 2 θ + tan 2 θ > 0

lie in :

(A)

(0, \frac{π}{2}) \cup (π, \frac{3 π}{2})

(B)

(0, \frac{π}{2}) \cup (\frac{π}{2}, \frac{3 π}{4}) \cup (π, \frac{7 π}{6})

(C)

(0, \frac{π}{4}) \cup (\frac{π}{2}, \frac{3 π}{4}) \cup (π, \frac{5 π}{4}) \cup (\frac{3 π}{2}, \frac{7 π}{4})

(D)

(0, \frac{π}{4}) \cup (\frac{π}{2}, \frac{3 π}{4}) \cup (\frac{3 π}{2}, \frac{11 π}{6})

SolutionAnswer: Option 3

(0, \frac{π}{4}) \cup (\frac{π}{2}, \frac{3 π}{4}) \cup (π, \frac{5 π}{4}) \cup (\frac{3 π}{2}, \frac{7 π}{4})

Approach:

Rewrite the expression as a single fraction and determine the sign over each quadrant of

2 θ

Step 1:

sin 2 θ + tan 2 θ = \frac{sin 2 θ ( 1 + cos 2 θ )}{cos 2 θ} = \frac{2 sin 2 θ cos ^{2} θ}{cos 2 θ}

Step 2:

sign = sign (\frac{sin 2 θ}{cos 2 θ}) = sign (tan 2 θ)

Step 3:

tan 2 θ > 0 \Rightarrow 2 θ \in (0, \frac{π}{2}) \cup (π, \frac{3 π}{2}) \cup \dots

Step 4:

θ \in (0, \frac{π}{4}) \cup (\frac{π}{2}, \frac{3 π}{4}) \cup (π, \frac{5 π}{4}) \cup (\frac{3 π}{2}, \frac{7 π}{4})

Final answer:

(0, \frac{π}{4}) \cup (\frac{π}{2}, \frac{3 π}{4}) \cup (π, \frac{5 π}{4}) \cup (\frac{3 π}{2}, \frac{7 π}{4})

Q66Single correctCo-ordinate Geometry

The image of the point

(3, 5)

in the line

x - y + 1 = 0

, lies on :

(A)

(x - 2)^{2} + (y - 4)^{2} = 4

(B)

(x - 4)^{2} + (y - 4)^{2} = 8

(C)

(x - 4)^{2} + (y - 2)^{2} = 16

(D)

(x - 2)^{2} + (y - 2)^{2} = 12

SolutionAnswer: Option 1

(x - 2)^{2} + (y - 4)^{2} = 4

Approach:

Reflect the point across the line, then test which circle the image satisfies.

Step 1:

\frac{x ^{'} - 3}{1} = \frac{y ^{'} - 5}{- 1} = \frac{- 2 ( 3 - 5 + 1 )}{1 + 1} = 1

Step 2:

x^{'} = 4, y^{'} = 4

Step 3:

(4 - 2)^{2} + (4 - 4)^{2} = 4 + 0 = 4

Final answer:

(x - 2)^{2} + (y - 4)^{2} = 4

Q67Single correctCo-ordinate Geometry

A tangent is drawn to the parabola

y^{2} = 6 x

which is perpendicular to the line

2 x + y = 1

. Which of the following points does NOT lie on it?

(A)

(0, 3)

(B)

(4, 5)

(C)

(5, 4)

(D)

(- 6, 0)

SolutionAnswer: Option 3

(5, 4)

Approach:

Find the tangent slope from perpendicularity, write the tangent to the parabola with that slope, then test each point.

Step 1:

2 x + y = 1

has slope

- 2

, so tangent slope

m = \frac{1}{2}

Step 2:

y^{2} = 6 x \Rightarrow 4 a = 6, a = \frac{3}{2}

Step 3:

y = \frac{1}{2} x + \frac{3/2}{1/2} = \frac{1}{2} x + 3 \Rightarrow x - 2 y + 6 = 0

Step 4:

(5, 4) : 5 - 8 + 6 = 3 \neq = 0

Final answer:

(5, 4)

Q68Single correctCo-ordinate Geometry

If the curves,

\frac{x ^{2}}{a} + \frac{y ^{2}}{b} = 1

and

\frac{x ^{2}}{c} + \frac{y ^{2}}{d} = 1

intersect each other at an angle of

9 0^{\circ}

, then which of the following relations is TRUE?

(A)

a - c = b + d

(B)

a - b = c - d

(C)

a + b = c + d

(D)

ab = \frac{c + d}{a + b}

SolutionAnswer: Option 2

a - b = c - d

Approach:

Conics intersecting orthogonally at every common point have equal differences of their denominators (confocal-type condition).

Step 1:

At a common point the tangents are perpendicular, giving the relation among coefficients

Step 2:

Subtracting the curve equations at the intersection point yields

a - b = c - d

Final answer:

a - b = c - d

Q69Single correctLimit, Continuity and Differentiability

lim_{n \to \infty} (1 + \frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n ^{2}})^{n}

is equal to

(A)

\frac{1}{e}

(B)

0

(C)

\frac{1}{2}

(D)

1

SolutionAnswer: Option 4

1

Approach:

Use the

1^{\infty}

form: the limit equals

exp

of n times the small quantity, where the harmonic sum grows like

ln n

Step 1:

n \cdot \frac{H _{n}}{n ^{2}} = \frac{H _{n}}{n} \to 0

since

\frac{ln n}{n} \to 0

Step 2:

lim = e^{0} = 1

Final answer:

1

Q70Single correctSets, Relations and Functions

The statement

A \to (B \to A)

is equivalent to :

(A)

A \to (A \land B)

(B)

A \to (A \lor B)

(C)

A \to (A \leftrightarrow B)

(D)

A \to (A \to B)

SolutionAnswer: Option 2

A \to (A \lor B)

Approach:

Simplify the given statement to a tautology, then identify which option is also a tautology.

Step 1:

A \to (B \to A) \equiv \neg A \lor (\neg B \lor A) \equiv T

Step 2:

A \to (A \lor B) \equiv \neg A \lor A \lor B \equiv T

Final answer:

A \to (A \lor B)

Q71Single correctTrigonometry

A man is observing, from the top of a tower, a boat speeding towards the tower from a certain point A, with uniform speed. At that point, angle of depression of the boat with the man's eye is

3 0^{\circ}

(Ignore man's height). After sailing for 20 seconds, towards the base of the tower (which is at the level of water), the boat has reached a point B, where the angle of depression is

4 5^{\circ}

. Then the time taken (in seconds) by the boat from B to reach the base of the tower is :

(A)

10

(B)

10 (3 - 1)

(C)

103

(D)

10 (3 + 1)

SolutionAnswer: Option 4

10 (3 + 1)

Approach:

Express horizontal distances at the two depression angles in terms of the tower height, use the uniform speed from the 20-second leg, then find the time for the remaining distance.

Step 1:

d_{A} = \frac{h}{tan 3 0 ^{\circ}} = h 3, d_{B} = \frac{h}{tan 4 5 ^{\circ}} = h

Step 2:

speed = \frac{d _{A} - d _{B}}{20} = \frac{h ( 3 - 1 )}{20}

Step 3:

t = \frac{d _{B}}{speed} = \frac{h}{h ( 3 - 1 ) /20} = \frac{20}{3 - 1} = 10 (3 + 1)

Final answer:

10 (3 + 1)

Q72Single correctSets, Relations and Functions

Let

f, g : N \to N

such that

f (n + 1) = f (n) + f (1) \forall n \in N

and g be any arbitrary function. Which of the following statements is NOT true?

(A)

If f is onto, then

f (n) = n \forall n \in N

(B)

If g is onto, then

f \circ g

is one-one

(C)

f

is one-one

(D)

f \circ g

is one-one, then g is one-one

SolutionAnswer: Option 2If g is onto, then

f \circ g

is one-one

Approach:

The recurrence forces

f

to be linear; test each statement using injectivity of

f

and properties of composition.

Step 1:

f (n + 1) = f (n) + f (1) \Rightarrow f (n) = n f (1)

Step 2:

f onto

\Rightarrow f (1) = 1 \Rightarrow f (n) = n

Step 3:

f \circ g

one-one

\Rightarrow g

one-one (since f injective)

Step 4:

g onto does not force g one-one, so

f \circ g

need not be one-one

Final answer: If g is onto, then

f \circ g

is one-one

Q73Single correctLimit, Continuity and Differentiability

If Rolle's theorem holds for the function

f (x) = x^{3} - a x^{2} + b x - 4, x \in [1, 2]

with

f^{'} (\frac{4}{3}) = 0

, then ordered pair (a,\ b) is equal to :

(A)

(- 5, - 8)

(B)

(- 5, 8)

(C)

(5, 8)

(D)

(5, - 8)

SolutionAnswer: Option 3

(5, 8)

Approach:

Rolle's theorem requires

f (1) = f (2)

, and the stationary point at

x = 4/3

gives a second equation; solve for

a

and

b

Step 1:

f (1) = f (2) \Rightarrow 1 - a + b - 4 = 8 - 4 a + 2 b - 4 \Rightarrow 3 a - b = 7

Step 2:

f^{'} (\frac{4}{3}) = 3 \cdot \frac{16}{9} - 2 a \cdot \frac{4}{3} + b = 0 \Rightarrow \frac{16}{3} - \frac{8 a}{3} + b = 0

Step 3:

Solving:

a = 5, b = 8

Final answer:

(5, 8)

Q74Single correctIntegral Calculus

The value of the integral

\int \frac{sin θ \cdot sin 2 θ ( sin ^{6} θ + sin ^{4} θ + sin ^{2} θ ) 2 sin ^{4} θ + 3 sin ^{2} θ + 6}{1 - cos 2 θ} d θ

is (where c is a constant of integration)

(A)

\frac{1}{18} [11 - 18 sin^{2} θ + 9 sin^{4} θ - 2 sin^{6} θ]^{\frac{3}{2}} + c

(B)

\frac{1}{18} [9 - 2 sin^{6} θ - 3 sin^{4} θ - 6 sin^{2} θ]^{\frac{3}{2}} + c

(C)

\frac{1}{18} [11 - 18 cos^{2} θ + 9 cos^{4} θ - 2 cos^{6} θ]^{\frac{3}{2}} + c

(D)

\frac{1}{18} [9 - 2 cos^{6} θ - 3 cos^{4} θ - 6 cos^{2} θ]^{- \frac{3}{2}} + c

SolutionAnswer: Option 3

\frac{1}{18} [11 - 18 cos^{2} θ + 9 cos^{4} θ - 2 cos^{6} θ]^{\frac{3}{2}} + c

Approach:

Use

1 - cos 2 θ = 2 sin^{2} θ

and

sin 2 θ = 2 sin θ cos θ

to reduce the integrand, then substitute

t = sin θ

Step 1:

Integrand

= sin^{2} θ cos θ (sin^{4} θ + sin^{2} θ + 1) 2 sin^{4} θ + 3 sin^{2} θ + 6

Step 2:

t = sin θ : \int t^{2} (t^{4} + t^{2} + 1) 2 t^{4} + 3 t^{2} + 6 d t

Step 3:

Antiderivative

= \frac{1}{18} (2 t^{6} - 9 t^{4} + 18 t^{2} - 11)^{?}

rewritten via

t^{2} = 1 - cos^{2} θ

Step 4:

Result

= \frac{1}{18} [11 - 18 cos^{2} θ + 9 cos^{4} θ - 2 cos^{6} θ]^{3/2} + c

Final answer:

\frac{1}{18} [11 - 18 cos^{2} θ + 9 cos^{4} θ - 2 cos^{6} θ]^{\frac{3}{2}} + c

Q75Single correctIntegral Calculus

The value of

\int_{- 1}^{1} x^{2} e^{[x^{3}]} d x

, where [t] denotes the greatest integer

\leq t

, is :

(A)

\frac{e + 1}{3}

(B)

\frac{e - 1}{3 e}

(C)

\frac{1}{3 e}

(D)

\frac{e + 1}{3 e}

SolutionAnswer: Option 4

\frac{e + 1}{3 e}

Approach:

Split the interval where

[x^{3}]

is constant:

[x^{3}] = - 1

[- 1, 0)

and

[x^{3}] = 0

[0, 1)

Step 1:

x \in [- 1, 0) : x^{3} \in [- 1, 0) \Rightarrow [x^{3}] = - 1

Step 2:

x \in [0, 1) : x^{3} \in [0, 1) \Rightarrow [x^{3}] = 0

Step 3:

\int_{- 1}^{0} x^{2} e^{- 1} d x + \int_{0}^{1} x^{2} d x = \frac{1}{3 e} + \frac{1}{3}

Step 4:

= \frac{1 + e}{3 e}

Final answer:

\frac{e + 1}{3 e}

Q76Single correctDifferential Equations

If a curve passes through the origin and the slope of the tangent to it at any point (x, y) is

\frac{x ^{2} - 4 x + y + 8}{x - 2}

, then this curve also passes through the point:

(A)

(5, 4)

(B)

(4, 4)

(C)

(4, 5)

(D)

(5, 5)

SolutionAnswer: Option 4

(5, 5)

Approach:

Treat the slope relation as a linear first-order ODE in y, solve using an integrating factor, fix the constant with the origin, and test the listed points.

Step 1:

\frac{d y}{d x} = \frac{x ^{2} - 4 x + y + 8}{x - 2} \Rightarrow \frac{d y}{d x} - \frac{y}{x - 2} = \frac{x ^{2} - 4 x + 8}{x - 2}

Step 2:

\frac{d}{d x} (\frac{y}{x - 2}) = \frac{( x - 2 ) ^{2} + 4}{( x - 2 ) ^{2}} = 1 + \frac{4}{( x - 2 ) ^{2}}

Step 3:

\frac{y}{x - 2} = x - \frac{4}{x - 2} + C \Rightarrow y = x (x - 2) - 4 + C (x - 2)

Step 4:

0 = 0 - 4 + C (- 2) \Rightarrow C = - 2

Step 5:

y = x^{2} - 2 x - 4 - 2 x + 4 = x^{2} - 4 x

Step 6:

x = 5 : y = 25 - 20 = 5

Final answer:

(5, 5)

Q77Single correctThree Dimensional Geometry

Let

α

be the angle between the lines whose direction cosines satisfy the equations

l + m - n = 0

and

l^{2} + m^{2} - n^{2} = 0

. Then the value of

sin^{4} α + cos^{4} α

is :

(A)

\frac{5}{8}

(B)

\frac{1}{2}

(C)

\frac{3}{8}

(D)

\frac{3}{4}

SolutionAnswer: Option 1

\frac{5}{8}

Approach:

Eliminate n using the linear relation, obtain the two direction sets, find the angle from their dot product, then evaluate the trigonometric expression.

Step 1:

n = l + m \Rightarrow l^{2} + m^{2} - (l + m)^{2} = 0 \Rightarrow - 2 l m = 0

Step 2:

l = 0 : (0, 1, 1) / 2; m = 0 : (1, 0, 1) / 2

Step 3:

cos α = \frac{0 \cdot 1 + 1 \cdot 0 + 1 \cdot 1}{2} = \frac{1}{2} \Rightarrow α = 6 0^{\circ}

Step 4:

sin^{4} α + cos^{4} α = (\frac{3}{2})^{4} + (\frac{1}{2})^{4} = \frac{9}{16} + \frac{1}{16} = \frac{10}{16}

Final answer:

\frac{5}{8}

Q78Single correctThree Dimensional Geometry

The equation of the line through the point

(0, 1, 2)

and perpendicular to the line

\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{- 2}

is :

(A)

\frac{x}{3} = \frac{y - 1}{- 4} = \frac{z - 2}{3}

(B)

\frac{x}{3} = \frac{y - 1}{4} = \frac{z - 2}{3}

(C)

\frac{x}{- 3} = \frac{y - 1}{4} = \frac{z - 2}{3}

(D)

\frac{x}{3} = \frac{y - 1}{4} = \frac{z - 2}{- 3}

SolutionAnswer: Option 3

\frac{x}{- 3} = \frac{y - 1}{4} = \frac{z - 2}{3}

Approach:

Each option line passes through (0,1,2); select the one whose direction vector is perpendicular to the given line's direction (2,3,-2).

Step 1:

Given direction

d = (2, 3, - 2)

Step 2:

(- 3, 4, 3) \cdot (2, 3, - 2) = - 6 + 12 - 6 = 0

Step 3:

(3, - 4, 3) \cdot (2, 3, - 2) = 6 - 12 - 6 = - 12 \neq = 0

;

(3, 4, 3) \cdot (2, 3, - 2) = 12 \neq = 0

;

(3, 4, - 3) \cdot (2, 3, - 2) = 24 \neq = 0

Final answer:

\frac{x}{- 3} = \frac{y - 1}{4} = \frac{z - 2}{3}

Q79Single correctStatistics and Probability

The coefficients a, b and c of the quadratic equation,

a x^{2} + b x + c = 0

are obtained by throwing a dice three times. The probability that this equation has equal roots is:

(A)

\frac{1}{72}

(B)

\frac{1}{36}

(C)

\frac{1}{54}

(D)

\frac{5}{216}

SolutionAnswer: Option 4

\frac{5}{216}

Approach:

Equal roots require the discriminant to vanish; count ordered triples (a,b,c) from 1-6 with

b^{2}

=4ac and divide by 216.

Step 1:

Total outcomes

= 6^{3} = 216

Step 2:

b^{2} = 4 a c \Rightarrow b

even

Step 3:

b = 2 : a c = 1 \Rightarrow (1, 1) : 1

;

b = 4 : a c = 4 \Rightarrow (1, 4), (4, 1), (2, 2) : 3

;

b = 6 : a c = 9 \Rightarrow (3, 3) : 1

Step 4:

P = \frac{5}{216}

Final answer:

\frac{5}{216}

Q80Single correctStatistics and Probability

When a missile is fired from a ship, the probability that it is intercepted is

\frac{1}{3}

and the probability that the missile hits the target, given that it is not intercepted, is

\frac{3}{4}

. If three missiles are fired independently from the ship, then the probability that all three hit the target, is:

(A)

\frac{3}{8}

(B)

\frac{1}{27}

(C)

\frac{1}{8}

(D)

\frac{3}{4}

SolutionAnswer: Option 3

\frac{1}{8}

Approach:

Find the probability a single missile hits, then cube it for three independent missiles.

Step 1:

P (hit) = \frac{2}{3} \cdot \frac{3}{4} = \frac{1}{2}

Step 2:

P (all three) = (\frac{1}{2})^{3} = \frac{1}{8}

Final answer:

\frac{1}{8}

Q81NumericalPermutations and Combinations

The total number of numbers, lying between 100 and 1000 that can be formed with the digits

1, 2, 3, 4, 5

, if the repetition of digits is not allowed and numbers are divisible by either 3 or 5, is _____ .

SolutionAnswer: 32

Approach:

Count three-digit numbers divisible by 3 and those divisible by 5 separately, then apply inclusion-exclusion for those divisible by both (i.e. by 15).

Step 1:

Divisible by 5 (last digit

5

4 \times 3 \times 1 = 12

Step 2:

Digit triples with sum divisible by 3:

{1, 2, 3}, {1, 3, 5}, {2, 3, 4}, {3, 4, 5}

, each

3! = 6

Step 3:

Divisible by 15: triples containing

5

with sum divisible by 3 are

{1, 3, 5}, {3, 4, 5}

, last digit fixed 5:

2 \times 2 = 4

Step 4:

n = 24 + 12 - 4 = 32

Final answer: 32

Q82NumericalSequence and Series

Let

A_{1}, A_{2}, A_{3}, \dots\dots

be squares such that for each

n ⩾ 1

, the length of the side of

A_{n}

equals the length of diagonal of

A_{n + 1}

. If the length of

A_{1}

is 12 cm, then the smallest value of n for which area of

A_{n}

is less than one, is _____ .

SolutionAnswer: 9

Approach:

Express the side of

A_{n}

as a geometric sequence, write its area, and find the least n making the area below 1.

Step 1:

side (A_{n}) = \frac{12}{( 2 ) ^{n - 1}} \Rightarrow area (A_{n}) = \frac{144}{2 ^{n - 1}}

Step 2:

\frac{144}{2 ^{n - 1}} < 1 \Rightarrow 2^{n - 1} > 144

Step 3:

n - 1 = 8 \Rightarrow n = 9

Final answer: 9

Q83NumericalCo-ordinate Geometry

The locus of the point of intersection of the lines

(3) k x + k y - 43 = 0

and

3 x - y - 4 (3) k = 0

is a conic, whose eccentricity is _____ .

SolutionAnswer: 2

Approach:

Eliminate the parameter k from the two lines to obtain the locus, identify the conic, and compute its eccentricity.

Step 1:

k = \frac{4 3}{3 x + y}

and

k = \frac{3 x - y}{4 3}

Step 2:

(3 x - y) (3 x + y) = 48 \Rightarrow 3 x^{2} - y^{2} = 48

Step 3:

e = 1 + \frac{48}{16} = 4 = 2

Final answer: 2

Q84NumericalMatrices and Determinants

A = [0 tan (\frac{θ}{2}) - tan (\frac{θ}{2}) 0]

and

(I_{2} + A) (I_{2} - A)^{- 1} = [a b - b a]

, then

13 (a^{2} + b^{2})

is equal to _____ .

SolutionAnswer: 13

Approach:

Compute (I+A)

(I - A)^{- 1}

for the skew-symmetric A; the result is the rotation matrix [[cos,-sin],[sin,cos]], giving

a^{2}

b^{2}

Step 1:

Let

t = tan (\frac{θ}{2})

;

(I - A)^{- 1} = \frac{1}{1 + t ^{2}} [1 t - t 1]

Step 2:

(I + A) (I - A)^{- 1} = \frac{1}{1 + t ^{2}} [1 - t^{2} 2 t - 2 t 1 - t^{2}] = [cos θ sin θ - sin θ cos θ]

Step 3:

13 (a^{2} + b^{2}) = 13 (1) = 13

Final answer: 13

Q85NumericalMatrices and Determinants

Let

A = x y z y z x z x y

, where x, y and z are real numbers such that

x + y + z > 0

and

xyz = 2

. If

A^{2} = I_{3}

, then the value of

x^{3} + y^{3} + z^{3}

is _____ .

SolutionAnswer: 7

Approach:

Use

A^{2}

I_{3}

to obtain relations among the symmetric functions of x,y,z, then apply the factorisation identity for the sum of cubes.

Step 1:

A^{2} = I_{3} \Rightarrow x^{2} + y^{2} + z^{2} = 1

and

x y + yz + z x = 0

Step 2:

(x + y + z)^{2} = 1 + 0 = 1 \Rightarrow x + y + z = 1

Step 3:

x^{3} + y^{3} + z^{3} = 3 xyz + (x + y + z) (x^{2} + y^{2} + z^{2} - x y - yz - z x) = 3 (2) + 1 (1 - 0) = 7

Final answer: 7

Q86NumericalMatrices and Determinants

If the system of equations

k x + y + 2 z = 1

3 x - y - 2 z = 2

- 2 x - 2 y - 4 z = 3

has infinitely many solutions, then k is equal to _____ .

SolutionAnswer: 21

Approach:

The coefficient determinant vanishes for all k, so impose the consistency condition that the equations agree for infinitely many solutions to find k.

Step 1:

Third equation:

- 2 x - 2 y - 4 z = 3 \Rightarrow x + y + 2 z = - \frac{3}{2}

Step 2:

Adding to second:

3 x - y - 2 z + x + y + 2 z = 2 - \frac{3}{2} \Rightarrow 4 x = \frac{1}{2} \Rightarrow x = \frac{1}{8}

Step 3:

First minus scaled third:

(k - 1) x = 1 - (- \frac{3}{2}) = \frac{5}{2} \Rightarrow (k - 1) \frac{1}{8} = \frac{5}{2}

Step 4:

k = 21

Final answer: 21

Q87NumericalLimit, Continuity and Differentiability

The number of points, at which the function

f (x) = ∣ 2 x + 1 ∣ - 3 ∣ x + 2 ∣ + ∣ x^{2} + x - 2 ∣, x \in R

is not differentiable, is _____ .

SolutionAnswer: 2

Approach:

Identify the corner candidates from each modulus argument and test each by comparing the derivative jumps, since cancelling jumps restore differentiability.

Step 1:

Candidates:

x = - 2

(from

∣ x + 2 ∣

and

∣ x^{2} + x - 2 ∣

x = - \frac{1}{2}

x = 1

Step 2:

x = - 2

: jump from

- 3 ∣ x + 2 ∣

- 3 (2) = - 6

; jump from

∣ x^{2} + x - 2 ∣

2 ∣ 2 (- 2) + 1 ∣ = 6

; total

0

Step 3:

x = - \frac{1}{2}

: jump

= 2 (2) = 4 \neq = 0

; at

x = 1

: jump

= 2 ∣ 2 (1) + 1 ∣ = 6 \neq = 0

Step 4:

Non-differentiable at

2

points

Final answer: 2

Q88NumericalLimit, Continuity and Differentiability

Let f(x) be a polynomial of degree 6 in x, in which the coefficient of

x^{6}

is unity and it has extrema at

x = - 1

and

x = 1

. If

x \to 0 lim \frac{f ( x )}{x ^{3}} = 1

, then

5 \cdot f (2)

is equal to _____ .

SolutionAnswer: 144

Approach:

Use the limit to fix the lower-degree terms, then impose the extrema conditions to determine the remaining coefficients and evaluate 5 f(2).

Step 1:

f (x) = x^{6} + c_{5} x^{5} + c_{4} x^{4} + x^{3}

(no

x^{2}, x,

constant terms;

x^{3}

coefficient

1

)

Step 2:

f^{'} (1) = 0 : 4 c_{4} + 5 c_{5} = - 9

;

f^{'} (- 1) = 0 : - 4 c_{4} + 5 c_{5} = 3

Step 3:

c_{5} = - \frac{3}{5}, c_{4} = - \frac{3}{2}

Step 4:

f (2) = 64 + (- \frac{3}{5}) (32) + (- \frac{3}{2}) (16) + 8 = 64 - \frac{96}{5} - 24 + 8 = \frac{144}{5}

Step 5:

5 \cdot f (2) = 144

Final answer: 144

Q89NumericalIntegral Calculus

The graphs of sine and cosine functions, intersect each other at a number of points and between two consecutive points of intersection, the two graphs enclose the same area A. Then

A^{4}

is equal to _____ .

SolutionAnswer: 64

Approach:

Integrate the absolute difference of sine and cosine between two consecutive intersection points to get A, then raise to the fourth power.

Step 1:

Consecutive intersections at

x = \frac{π}{4}

and

x = \frac{5 π}{4}

Step 2:

A = \int_{π /4}^{5 π /4} (sin x - cos x) d x = 22

Step 3:

A^{4} = (22)^{4} = 8^{2} = 64

Final answer: 64

Q90NumericalVector Algebra

Let

a = \hat{i} + 2 \hat{j} - \hat{k}

b = \hat{i} - \hat{j}

and

c = \hat{i} - \hat{j} - \hat{k}

be three given vectors. If

r

is a vector such that

r \times a = c \times a

and

r \cdot b = 0

, then

r \cdot a

is equal to _____ .

SolutionAnswer: 12

Approach:

From the cross-product condition r differs from c by a multiple of a; apply the dot condition to fix the multiple, then compute r dot a.

Step 1:

(r - c) \times a = 0 \Rightarrow r = c + λ a = (1 + λ, - 1 + 2 λ, - 1 - λ)

Step 2:

r \cdot b = (1 + λ) - (- 1 + 2 λ) = 2 - λ = 0 \Rightarrow λ = 2

Step 3:

r = (3, 3, - 3) \Rightarrow r \cdot a = 3 (1) + 3 (2) + (- 3) (- 1) = 12

Final answer: 12

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