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JEE Main 2026 April 02, Shift 2 Question Paper with Solutions

Q: How many questions are in the JEE Main 2026 April 02, Shift 2 paper?

The JEE Main 2026 April 02, Shift 2 paper has 75 questions — Physics (25), Chemistry (25) and Mathematics (25). Every question is on this page with its correct answer and a step-by-step solution.

Q: What is the marking scheme for JEE Main?

JEE Main awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Mathematics.

Q: Are the answer key and step-by-step solutions provided for the 2026 April 02, Shift 2 paper?

Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.

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All 75 questions from the JEE Main 2026 (April 02, Shift 2) shift — Physics (25), Chemistry (25) and Mathematics (25) — 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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Physics25 questions

Q26Single correctPhysics and Measurement

Dimensions of universal gravitational constant (G) in terms of Planck's constant (h), distance (L), mass (M) and time (T) are ___________

(A)

[hTLM^{- 2}]

(B)

[h T^{- 1} L M^{- 2}]

(C)

[hTL^{2} M^{- 2}]

(D)

[h^{- 1} T^{- 1} L M^{- 2}]

SolutionAnswer: Option 2

[h T^{- 1} L M^{- 2}]

Approach:

Assume

G

is proportional to

h

multiplied by unknown powers of

L

M

T

, then equate the dimensions on both sides.

Step 1:Given fundamental quantities h, L, M, T as the basis; target is to write [G] in this basis. Known:

[G] = M^{- 1} L^{3} T^{- 2}

and

[h] = M L^{2} T^{- 1}

G = h L^{x} M^{y} T^{z}

Step 2:Substitute the dimensional expressions for

G

and

h

into the ansatz.

M^{- 1} L^{3} T^{- 2} = (M L^{2} T^{- 1}) \cdot L^{x} M^{y} T^{z}

Step 3:Combine the powers on the right-hand side.

M^{- 1} L^{3} T^{- 2} = M^{1 + y} L^{2 + x} T^{- 1 + z}

Step 4:Equate exponents of

M

L

, and

T

separately.

1 + y = - 1, 2 + x = 3, - 1 + z = - 2

Step 5:Reconstruct [G] from

h \cdot L^{1} M^{- 2} T^{- 1}

and check it reduces to

M^{- 1} L^{3} T^{- 2}

(M L^{2} T^{- 1}) \cdot (L M^{- 2} T^{- 1}) = M^{1 - 2} L^{2 + 1} T^{- 1 - 1} = M^{- 1} L^{3} T^{- 2}

Step 6:Write the final dimensional form of

G

[G] = [h T^{- 1} L M^{- 2}]

Final answer:

[G] = [h T^{- 1} L M^{- 2}]

Q27Single correctLaws of Motion

A 0.5 kg mass is in contact against the inner wall of a cylindrical drum of 4 m rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is

5 rad / s

. The coefficient of friction between the drum's inner wall surface and mass is ________ (Take

g = 10 m / s^{2}

)

(A)

0.1

(B)

0.5

(C)

0.7

(D)

0.3

SolutionAnswer: Option 1

0.1

Approach:

At the minimum spin rate, limiting friction equals weight while the wall's normal reaction supplies the centripetal force.

Step 1:Given

m = 0.5 kg

, drum radius

r = 4 m

, minimum angular speed

ω = 5 rad/s

g = 10 m/s^{2}

. Target: coefficient of static friction

μ

m = 0.5 kg, r = 4 m, ω = 5 rad/s, g = 10 m/s^{2}

Step 2:For the mass to be on the verge of slipping downward, friction up the wall balances weight.

μ N = m g

Step 3:The horizontal normal force is the centripetal force.

N = m r ω^{2}

Step 4:Combine the two equations to eliminate

N

μ (m r ω^{2}) = m g \Rightarrow μ = \frac{g}{r ω ^{2}}

Step 5:Substitute numerical values.

μ = \frac{10 m/s ^{2}}{4 m \cdot ( 5 rad/s ) ^{2}} = \frac{10}{4 \cdot 25} = \frac{10}{100}

Step 6:Compute

N = 0.5 \cdot 4 \cdot 25 = 50 N

and check

μ N = 0.1 \cdot 50 = 5 N

equals

m g = 0.5 \cdot 10 = 5 N

μ N = 5 N = m g

Step 7:Friction coefficient between mass and inner wall.

μ = 0.1

Final answer:

μ = 0.1

Q28Single correctLaws of Motion

Two blocks of masses 2 kg and 1 kg respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are hold at rest at the same horizontal level and then released. The distance traversed by the centre of mass in 2 s is _______ m.

Atwood machine: light frictionless pulley hung from ceiling with 2 kg block on left and 1 kg block on right, both hanging at the same horizontal level.

(A)

3.33

(B)

3.12

(C)

2.22

(D)

1.42

SolutionAnswer: Option 3

2.22

Approach:

Find Atwood acceleration of each block, derive the centre-of-mass acceleration, then apply uniformly-accelerated motion from rest for 2 s.

Step 1:Given

m_{2} = 2 kg

(heavier, descends),

m_{1} = 1 kg

(lighter, rises),

g = 10 m/s^{2}

, time

t = 2 s

, both released from rest. Target: distance traversed by the centre of mass.

m_{1} = 1 kg, m_{2} = 2 kg, g = 10 m/s^{2}, t = 2 s

Step 2:Apply Atwood's formula for the common magnitude of acceleration of each block.

a = (\frac{2 - 1}{2 + 1}) g = \frac{g}{3} = \frac{10}{3} m/s^{2}

Step 3:Take downward as positive. Heavier block accelerates down with

+ a

, lighter block accelerates up with

- a

. Substitute into the centre-of-mass formula.

a_{c m} = \frac{m _{2} ( + a ) + m _{1} ( - a )}{m _{1} + m _{2}} = \frac{( m _{2} - m _{1} ) a}{m _{1} + m _{2}}

Step 4:Substitute

a = (m_{2} - m_{1}) g / (m_{1} + m_{2})

to get

a_{c m}

in terms of g and the masses.

a_{c m} = (\frac{m _{2} - m _{1}}{m _{1} + m _{2}})^{2} g = (\frac{1}{3})^{2} \cdot 10 = \frac{10}{9} m/s^{2}

Step 5:Apply

S = \frac{1}{2} a_{c m} t^{2}

for motion from rest over

t = 2 s

S = \frac{1}{2} \cdot \frac{10}{9} \cdot (2)^{2} = \frac{1}{2} \cdot \frac{10}{9} \cdot 4 = \frac{20}{9} m

Step 6:Units check

m/s^{2} \cdot s^{2} = m

; numerical value

20/9 = 2.222 \dots

matches option (3).

\frac{20}{9} = 2.222 m

Step 7:Centre-of-mass displacement in 2 s.

S \approx 2.22 m

Final answer:

S \approx 2.22 m

Q29Single correctMagnetic Effects of Current and Magnetism

A particle having charge

1 0^{- 9} C

moving in x-y plane in fields of

0.4 \hat{j} N / C

and

4 \times 1 0^{- 3} \hat{k} T

experiences a force of

(4 \hat{i} + 2 \hat{j}) \times 1 0^{- 10} N

. The velocity of the particle at that instant is ______

m / s

(A)

50 \hat{i} + 100 \hat{j}

(B)

100 \hat{i} + 50 \hat{j}

(C)

- 50 \hat{i} + 100 \hat{j}

(D)

50 \hat{i} - 100 \hat{j}

SolutionAnswer: Option 1

50 \hat{i} + 100 \hat{j}

Approach:

Write the Lorentz force with an unknown velocity

V = x \hat{i} + y \hat{j}

, expand the cross product

V \times B

, and match components of the resulting force to find x and y.

Step 1:Given

q = 1 0^{- 9} C

E = 0.4 \hat{j} N/C

B = 4 \times 1 0^{- 3} \hat{k} T

F = (4 \hat{i} + 2 \hat{j}) \times 1 0^{- 10} N

. Target:

V

in the x-y plane.

V = x \hat{i} + y \hat{j}

Step 2:Compute the magnetic force per unit charge

V \times B

V \times B = (x \hat{i} + y \hat{j}) \times (4 \times 1 0^{- 3} \hat{k}) = 4 \times 1 0^{- 3} y \hat{i} - 4 \times 1 0^{- 3} x \hat{j}

Step 3:Form the total Lorentz force expression.

F = q [0.4 \hat{j} + 4 \times 1 0^{- 3} y \hat{i} - 4 \times 1 0^{- 3} x \hat{j}]

Step 4:Divide the supplied

F

q = 1 0^{- 9} C

to obtain force per charge.

\frac{F}{q} = \frac{( 4 i ^ + 2 j ^ ) \times 1 0 ^{- 10}}{1 0 ^{- 9}} = 0.4 \hat{i} + 0.2 \hat{j} N/C

Step 5:Equate the

\hat{i}

components:

0.4 = 4 \times 1 0^{- 3} y

y = \frac{0.4}{4 \times 1 0 ^{- 3}} = 100 m/s

Step 6:Equate the

\hat{j}

components:

0.2 = 0.4 - 4 \times 1 0^{- 3} x

4 \times 1 0^{- 3} x = 0.4 - 0.2 = 0.2 \Rightarrow x = 50 m/s

Step 7:Substitute

x = 50, y = 100

back.

q (V \times B) = 1 0^{- 9} (4 \times 1 0^{- 3} \cdot 100 \hat{i} - 4 \times 1 0^{- 3} \cdot 50 \hat{j}) = (4 \hat{i} - 2 \hat{j}) \times 1 0^{- 10} N

. Add

q E = (4 \hat{j}) \times 1 0^{- 10} N

to obtain

F = (4 \hat{i} + 2 \hat{j}) \times 1 0^{- 10} N

F_{check} = (4 \hat{i} + 2 \hat{j}) \times 1 0^{- 10} N

Step 8:Assemble velocity vector.

V = 50 \hat{i} + 100 \hat{j} m/s

Final answer:

V = 50 \hat{i} + 100 \hat{j} m/s

Q30Single correctElectronic Devices

If X and Y are the inputs, the given circuit works as ________

$Combinational logic circuit: input X feeds a NAND gate whose other input is X (acts as NOT X), input Y feeds another NAND gate with both inputs Y (acts as NOT Y). Outputs (\overline{X} and \overline{Y}) feed a NAND gate, whose output passes through a final NAND inverter to produce Output.$

(A)

OR gate

(B)

AND gate

(C)

NAND gate

(D)

NOR gate

SolutionAnswer: Option 4NOR gate

Approach:

Trace the output of each NAND stage with Boolean algebra and De Morgan's theorem to collapse the network into a single equivalent gate.

Step 1:Inputs are

X

and

Y

. Circuit: NAND(X,X) feeds first node, NAND(Y,Y) feeds second node, both outputs feed a NAND, whose output passes through a final NAND tied-input stage to give Output.

Inputs: X, Y; Output = ?

Step 2:First two NAND gates have their two inputs tied, so each acts as an inverter.

\overline{X \cdot X} = \overset{ˉ}{X}, \overline{Y \cdot Y} = \overset{ˉ}{Y}

Step 3:Third NAND takes

\overset{ˉ}{X}

and

\overset{ˉ}{Y}

Z_{1} = \overline{\overset{ˉ}{X} \cdot \overset{ˉ}{Y}}

Step 4:Apply De Morgan's theorem to

Z_{1}

Z_{1} = \overline{\overset{ˉ}{X} \cdot \overset{ˉ}{Y}} = X + Y

Step 5:Final NAND has its two inputs both equal to

Z_{1}

, acting as an inverter.

Output = \overline{Z_{1} \cdot Z_{1}} = \overline{Z_{1}} = \overline{X + Y}

Step 6:Build the truth table. For

(X, Y) = (0, 0)

output

= 1

;

(0, 1)

output

= 0

;

(1, 0)

output

= 0

;

(1, 1)

output

= 0

. This matches a NOR gate exactly.

Output (X, Y) = \overline{X + Y}

Step 7:The combinational network is logically equivalent to a NOR gate.

Equivalent gate = NOR

Final answer: The circuit is equivalent to a NOR gate.

Q31Single correctGravitation

If a body of mass 1 kg falls on the earth from infinity, it attains velocity (v) and kinetic energy (k) on reaching the surface of earth. The values of v and k respectively are _____ (Take radius of earth to be 6400 km and

g = 9.8 m / s^{2}

)

(A)

11.2 km / s; 6.27 \times 1 0^{7} J

(B)

11.2 km / s; 12.54 \times 1 0^{7} J

(C)

8.8 km / s; 6.27 \times 1 0^{7} J

(D)

8.8 km / s; 12.54 \times 1 0^{7} J

SolutionAnswer: Option 1

11.2 km / s; 6.27 \times 1 0^{7} J

Approach:

Apply energy conservation between infinity (where both PE and KE vanish) and the earth's surface. The resulting speed is the escape speed.

Step 1:Given

m = 1 kg

R = 6400 km = 6.4 \times 1 0^{6} m

g = 9.8 m/s^{2}

; initial state at infinity (

P E_{\infty} = 0

K E_{\infty} = 0

). Target: speed v and kinetic energy k at the surface.

m = 1 kg, R = 6.4 \times 1 0^{6} m, g = 9.8 m/s^{2}

Step 2:Total mechanical energy is conserved between

\infty

and the surface.

0 = - \frac{GM m}{R} + \frac{1}{2} m v^{2}

Step 3:Replace

GM / R

with

g R

using surface gravity, then isolate kinetic energy.

k = \frac{1}{2} m v^{2} = \frac{GM m}{R} = m g R

Step 4:Compute the kinetic energy numerically.

k = (1 kg) \cdot (9.8 m/s^{2}) \cdot (6.4 \times 1 0^{6} m) = 6.272 \times 1 0^{7} J

Step 5:Solve

k = \frac{1}{2} m v^{2}

for v.

v = \frac{2 k}{m} = 2 g R = 2 \cdot 9.8 \cdot 6.4 \times 1 0^{6}

Step 6:Evaluate the square root.

v = 1.2544 \times 1 0^{8} m/s = 1.12 \times 1 0^{4} m/s = 11.2 km/s

Step 7:This matches the standard escape velocity of earth

2 g R \approx 11.2 km/s

, and

\frac{1}{2} (1) (1.12 \times 1 0^{4})^{2} = 6.272 \times 1 0^{7} J

reproduces k.

\frac{1}{2} (1) (1.12 \times 1 0^{4})^{2} = 6.27 \times 1 0^{7} J

Step 8:Speed and kinetic energy at the surface.

v = 11.2 km/s, k = 6.27 \times 1 0^{7} J

Final answer:

v = 11.2 km/s, k = 6.27 \times 1 0^{7} J

Q32Single correctExperimental Skills

In a screw gauge the zero of main scale reference line coincides with the fifth division of the circular scale when two studs are in contact. There are 100 divisions in the circular scale and pitch of screw gauge is 0.1 mm. When diameter of a sphere is measured, the reading of main scale is 5 mm and 50th division of circular scale coincides with the reference line of main scale. The diameter of sphere is __________ mm.

(A)

5.045

(B)

5.055

(C)

5.450

(D)

5.550

SolutionAnswer: Option 1

5.045

Approach:

Compute least count, identify the sign of the zero error from the stud coincidence, and subtract it from the observed reading.

Step 1:Given pitch

= 0.1 mm

, CSD

= 100

divisions, observed MSR

= 5 mm

, CSR

= 50

divisions. Zero check: with studs in contact, the 5th circular division coincides with the main-scale reference line (zero of main scale lies above the reference, so circular scale is ahead). Target: corrected diameter.

pitch = 0.1 mm, N_{csd} = 100

Step 2:Compute the least count.

LC = \frac{0.1 mm}{100} = 0.001 mm

Step 3:Because the circular scale is 5 divisions ahead of zero when studs touch, the gauge reads

+ 5 \times LC

extra; this is a positive zero error.

Zero error = + 5 \times 0.001 mm = + 0.005 mm

Step 4:Compute the observed (uncorrected) reading on the sphere.

Observed = 5 + 50 \times 0.001 = 5.050 mm

Step 5:Subtract the positive zero error from the observed reading.

Diameter = 5.050 - 0.005 = 5.045 mm

Step 6:Add the zero error back:

5.045 + 0.005 = 5.050 mm

, which reproduces the observed reading.

5.045 + 0.005 = 5.050 mm

Step 7:Diameter of the sphere.

D = 5.045 mm

Final answer: Diameter of sphere

= 5.045 mm

Q33Single correctProperties of Solids and Liquids

The surface tension of a soap bubble is

0.03 N / m

. The work done in increasing the diameter of bubble from 2 cm to 6 cm is

α π \times 1 0^{- 4} J

. The value of

α

is ________. (Take

π = 3.14

)

(A)

0.86

(B)

0.64

(C)

1.92

(D)

7.68

SolutionAnswer: Option 3

1.92

Approach:

A soap bubble has two free surfaces. Work done equals surface tension times the increase in total surface area,

W = 2 T Δ A

where

Δ A = 4 π (r_{2}^{2} - r_{1}^{2})

Step 1:Given

T = 0.03 N/m

, initial diameter

d_{1} = 2 cm

, final diameter

d_{2} = 6 cm

. Convert to radii in metres. Target:

α

where

W = α π \times 1 0^{- 4} J

r_{1} = 1 cm = 1 \times 1 0^{- 2} m, r_{2} = 3 cm = 3 \times 1 0^{- 2} m

Step 2:Write the work expression for a soap bubble, accounting for inner and outer surfaces.

W = 8 π T (r_{2}^{2} - r_{1}^{2})

Step 3:Compute

r_{2}^{2} - r_{1}^{2}

r_{2}^{2} - r_{1}^{2} = (3 \times 1 0^{- 2})^{2} - (1 \times 1 0^{- 2})^{2} = 9 \times 1 0^{- 4} - 1 \times 1 0^{- 4} = 8 \times 1 0^{- 4} m^{2}

Step 4:Substitute into the work formula.

W = 8 π \cdot 0.03 \cdot 8 \times 1 0^{- 4} = (8 \cdot 0.03 \cdot 8) π \times 1 0^{- 4} J

Step 5:Multiply the constants.

8 \cdot 0.03 \cdot 8 = 1.92 \Rightarrow W = 1.92 π \times 1 0^{- 4} J

Step 6:Units:

N/m \cdot m^{2} = N \cdot m = J

. Matching the form

W = α π \times 1 0^{- 4} J

gives

α = 1.92

[T] [r^{2}] = N/m \cdot m^{2} = J

Step 7:Value of

α

α = 1.92

Final answer:

α = 1.92

Q34Single correctKinetic Theory of Gases

A mixture of carbon dioxide and oxygen has volume

8310 c m^{3}

, temperature 300 K, pressure 100 kPa and mass 13.2 g. The number of moles of carbon dioxide and oxygen gases in the mixture respectively are ___________

(A)

0.15

and

0.18

(B)

0.25

and

0.08

(C)

0.21

and

0.12

(D)

0.13

and

0.20

SolutionAnswer: Option 3

0.21

and

0.12

Approach:

Use the ideal gas equation for the mixture to obtain the total number of moles, then combine with the total-mass equation using the molar masses of C

O_{2}

and

O_{2}

to solve a 2x2 linear system.

Step 1:

P = 100 kPa = 1 0^{5} P a

V = 8310 c m^{3} = 8310 \times 1 0^{- 6} m^{3}

T = 300 K

m = 13.2 g

M_{C O_{2}} = 44 g / mol

M_{O_{2}} = 32 g / mol

R = 8.31 J / (mol \cdot K)

. Target:

n_{1}

O_{2}

) and

n_{2}

(

O_{2}

P V = (n_{1} + n_{2}) RT

Step 2:Substitute into ideal gas equation to obtain total moles.

n_{1} + n_{2} = \frac{P V}{RT} = \frac{1 0 ^{5} \times 8310 \times 1 0 ^{- 6}}{8.31 \times 300} = \frac{831}{2493} = \frac{1}{3}

Step 3:Write mass-conservation equation with molar masses in g.

44 n_{1} + 32 n_{2} = 13.2

Step 4:Eliminate

n_{2}

using

n_{2} = 0.3333 - n_{1}

and solve.

44 n_{1} + 32 (0.3333 - n_{1}) = 13.2 \Rightarrow 12 n_{1} = 13.2 - 10.667 = 2.533

Step 5:Obtain

n_{2}

from the total-moles equation.

n_{2} = 0.3333 - 0.2112 = 0.1221 mol

Step 6:Back-substitute into the mass equation; total mass should equal 13.2 g.

44 (0.2112) + 32 (0.1221) = 9.293 + 3.907 = 13.20 g

Step 7:The moles of C

O_{2}

and

O_{2}

are 0.21 and 0.12 respectively, matching option (3).

n (C O_{2}) \approx 0.21, n (O_{2}) \approx 0.12

Final answer:

n (C O_{2}) \approx 0.21 mol, n (O_{2}) \approx 0.12 mol

Q35Single correctProperties of Solids and Liquids

If an air bubble of diameter 2 mm rises steadily through a liquid of density

200 k g / m^{3}

at a rate of

0.5 c m / s

, then the coefficient of viscosity of the liquid is _______ Poise. (Take

g = 10 m / s^{2}

)

(A)

0.88

(B)

8.8

(C)

88.8

(D)

0.088

SolutionAnswer: Option 1

0.88

Approach:

Apply Stokes' law at the terminal (steady) state: the upward buoyant force on the air bubble is balanced by the downward viscous drag. Solve for the dynamic viscosity and convert to Poise.

Step 1:Diameter

d = 2 mm \Rightarrow r = 1 mm = 1 0^{- 3} m

;

ρ_{l i q} = 200 k g / m^{3}

;

v_{t} = 0.5 c m / s = 5 \times 1 0^{- 3} m / s

;

g = 10 m / s^{2}

. Target: viscosity

η

in Poise.

r = 1 0^{- 3} m, v_{t} = 5 \times 1 0^{- 3} m / s

Step 2:At terminal velocity, neglecting the small weight of the air bubble, buoyancy equals viscous drag.

\frac{4}{3} π r^{3} ρ g = 6 π ηr v_{t}

Step 3:Solve algebraically for

η

η = \frac{2 ρ r ^{2} g}{9 v _{t}}

Step 4:Substitute numerical values to compute

η

in SI units.

η = \frac{2 \times 200 \times ( 1 0 ^{- 3} ) ^{2} \times 10}{9 \times 5 \times 1 0 ^{- 3}} = \frac{4 \times 1 0 ^{- 3}}{45 \times 1 0 ^{- 3}} = 0.0889 P a \cdot s

Step 5:Convert from Pa

\cdot

s to Poise using

1 P a \cdot s = 10 Poise

η = 0.0889 \times 10 = 0.889 Poise

Step 6:Dimensional check

[η] = \frac{( k g / m ^{3} ) ( m ^{2} ) ( m / s ^{2} )}{m / s} = k g / (m \cdot s) = P a \cdot s

, consistent.

[η] = P a \cdot s

Step 7:The coefficient of viscosity is about 0.88 Poise, matching option (1).

η \approx 0.88 Poise

Final answer:

η \approx 0.88 Poise

Q36Single correctWork, Energy and Power

A spherical ball of mass 2 kg falls from a height of 10 m and is brought to rest after penetrating 10 cm into sand. The average force exerted by sand on the ball is _______ N.

(A)

1980

(B)

2020

(C)

2000

(D)

1000

SolutionAnswer: Option 32000

Approach:

Use the work-energy theorem over the full motion (rest to rest): total work by gravity plus work by sand equals zero. With the penetration depth small compared to the height of fall, the average sand force follows directly.

Step 1:

m = 2 k g

h = 10 m

x = 10 c m = 0.10 m

g = 10 m / s^{2}

. Target: average force F exerted by sand on the ball.

m = 2 k g, h = 10 m, x = 0.10 m

Step 2:Since the ball starts and ends at rest,

Δ K E = 0

and net work vanishes (approximate gravity drop as h because

x ≪ h

mgh - F x = 0

Step 3:Solve for the sand force.

F = \frac{m g h}{x} = \frac{2 \times 10 \times 10}{0.10}

Step 4:Dimensional check

[F] = \frac{k g \cdot m / s ^{2} \cdot m}{m} = N

, consistent.

[F] = N

Step 5:The average force exerted by sand is 2000 N, matching option (3).

F = 2000 N

Final answer:

F = 2000 N

Q37Single correctElectromagnetic Waves

An electromagnetic wave travels in free space in the x-direction. At a particular point in space and time,

B = 2 \times 1 0^{- 7} \hat{j}

T is associated with this wave. The value of corresponding electric field

E

at this point is ________ V/m.

(A)

60 \hat{k}

(B)

- 60 \hat{k}

(C)

30 \hat{k}

(D)

600 \hat{k}

SolutionAnswer: Option 2

- 60 \hat{k}

Approach:

In a plane EM wave the magnitudes are linked by

∣ E ∣ = c ∣ B ∣

, and the direction of propagation

\overset{c}{^}

equals

\hat{E} \times \hat{B}

. Use both relations to determine

E

Step 1:Propagation

\overset{c}{^} = \hat{i}

, magnetic field

B = 2 \times 1 0^{- 7} \hat{j} T

c = 3 \times 1 0^{8} m / s

. Target: vector

E

\overset{c}{^} = \hat{i}, B = 2 \times 1 0^{- 7} \hat{j} T

Step 2:Compute

∣ E ∣

from

∣ E ∣ = c ∣ B ∣

∣ E ∣ = (3 \times 1 0^{8}) (2 \times 1 0^{- 7}) = 60 V / m

Step 3:

E

is perpendicular to

B

and lies in the

\pm \hat{k}

direction; choose the sign making

\hat{E} \times \hat{B} = \hat{i}

. Test

\hat{E} = - \hat{k}

(- \hat{k}) \times \hat{j} = - (\hat{k} \times \hat{j}) = - (- \hat{i}) = \hat{i}

Step 4:Combine magnitude with direction.

E = - 60 \hat{k} V / m

Step 5:Poynting check

E \times B = (- 60 \hat{k}) \times (2 \times 1 0^{- 7} \hat{j}) = - 120 \times 1 0^{- 7} (\hat{k} \times \hat{j}) = + 120 \times 1 0^{- 7} \hat{i}

, i.e. along

+ \hat{i}

, the propagation direction.

E \times B \propto + \hat{i}

Step 6:

E = - 60 \hat{k} V / m

, matching option (2).

E = - 60 \hat{k} V / m

Final answer:

E = - 60 \hat{k} V / m

Q38Single correctCurrent Electricity

Two resistors of

200 Ω

and

400 Ω

are connected in series with a battery of 100 V. A bulb rated at 200 V, 100 W is connected across the

400 Ω

resistance. The potential drop across the bulb is ________ V.

(A)

(B)

(C)

66.6

(D)

100

SolutionAnswer: Option 250

Approach:

Obtain the bulb's resistance from its rated voltage and power, combine it in parallel with the 400

Ω

resistor it is connected across, and use voltage division across the 200

Ω

series resistor and the parallel block.

Step 1:

R_{1} = 200 Ω

(series),

R_{2} = 400 Ω

(across which bulb is fitted), bulb rating

V_{b} = 200 V

P_{b} = 100 W

, source

V_{s} = 100 V

. Target: potential drop across the bulb.

V_{s} = 100 V, R_{1} = 200 Ω, R_{2} = 400 Ω

Step 2:Bulb resistance from rated values.

R_{b} = \frac{V _{b}^{2}}{P _{b}} = \frac{20 0 ^{2}}{100} = 400 Ω

Step 3:Bulb (400

Ω

) is in parallel with 400

Ω

resistor.

R_{p} = \frac{400 \times 400}{400 + 400} = 200 Ω

Step 4:Total circuit resistance across the 100 V source.

R_{t o t} = R_{1} + R_{p} = 200 + 200 = 400 Ω

Step 5:Voltage across the parallel block equals voltage across the bulb.

V_{b u l b} = V_{s} \frac{R _{p}}{R _{t o t}} = 100 \times \frac{200}{400} = 50 V

Step 6:Source current

I = V_{s} / R_{t o t} = 100/400 = 0.25 A

; voltage across parallel block

= I R_{p} = 0.25 \times 200 = 50 V

, consistent.

I R_{p} = 0.25 \times 200 = 50 V

Step 7:The potential drop across the bulb is 50 V, matching option (2).

V_{b u l b} = 50 V

Final answer:

V_{b u l b} = 50 V

Q39Single correctElectrostatics

Two metal plates (A, B) are kept horizontally with separation of

(\frac{12}{π})

cm, with plate A on the top. An atomizer jet sprays oil (density

1.5 g/cm^{3}

) droplets of radius 1 mm horizontally. All oil droplets carry a charge

5 n C

. The potentials

V_{A}

and

V_{B}

are required on plates A and B respectively in order to ensure the droplets do not descend. The values of

V_{A}

and

V_{B}

are ____ (Neglect the air resistance to the droplets and take

g = 10 m/s^{2}

)

(A)

100 V and 580 V

(B)

580 V and 100 V

(C)

60 V and 400 V

(D)

0 V and -200 V

SolutionAnswer: Option 1100 V and 580 V

Approach:

Balance the electric force on a positively charged droplet with gravity. For the upward electric force, the field must point downward, so the bottom plate B is held at higher potential than the top plate A. Compute the required

V_{B} - V_{A}

from the force-balance equation.

Step 1:

q = 5 n C = 5 \times 1 0^{- 9} C

d = (12/ π) c m = (12/ π) \times 1 0^{- 2} m

R = 1 mm = 1 0^{- 3} m

ρ = 1.5 g / c m^{3} = 1.5 \times 1 0^{3} k g / m^{3}

g = 10 m / s^{2}

. Target:

V_{A}

and

V_{B}

q, d, R, ρ, g

as above

Step 2:Set the upward electric force on the positive charge equal to its weight.

q \frac{( V _{B} - V _{A} )}{d} = \frac{4}{3} π R^{3} ρ g

Step 3:Substitute the numerical values into the equation.

5 \times 1 0^{- 9} \cdot \frac{( V _{B} - V _{A} ) π}{12 \times 1 0 ^{- 2}} = \frac{4}{3} π (1 0^{- 3})^{3} \cdot 1.5 \times 1 0^{3} \cdot 10

Step 4:Cancel

π

and simplify the right side.

\frac{5 \times 1 0 ^{- 9}}{12 \times 1 0 ^{- 2}} (V_{B} - V_{A}) = \frac{4}{3} \times 1.5 \times 1 0^{- 6} \times 10 = 2 \times 1 0^{- 5}

Step 5:Solve for the potential difference.

V_{B} - V_{A} = \frac{2 \times 1 0 ^{- 5} \times 12 \times 1 0 ^{- 2}}{5 \times 1 0 ^{- 9}} = 480 V

Step 6:Scan the four options for one whose difference is 480 V with

V_{B} > V_{A}

. Option (1) gives

V_{B} - V_{A} = 580 - 100 = 480 V

V_{A} = 100 V, V_{B} = 580 V

Step 7:With

V_{B} - V_{A} = 480 V

and

d = 0.0382 m

E = 480/0.0382 \approx 1.257 \times 1 0^{4} V / m

; weight

m g = (4/3) π (1 0^{- 3})^{3} (1500) (10) = 6.28 \times 1 0^{- 5} N

; electric force

qE = 5 \times 1 0^{- 9} \times 1.257 \times 1 0^{4} = 6.28 \times 1 0^{- 5} N

, equal to the weight.

qE = m g = 6.28 \times 1 0^{- 5} N

Step 8:The required potentials are

V_{A} = 100 V

and

V_{B} = 580 V

, matching option (1).

V_{A} = 100 V, V_{B} = 580 V

Final answer:

V_{A} = 100 V, V_{B} = 580 V

Q40Single correctElectrostatics

Two point charges

8 μ C

and

- 2 μ C

are located at

x = 2 cm

and

x = 4 cm

, respectively on the x-axis. The ratio of electric flux due to these charges through two spheres of radii 3 cm and 5 cm with their centers at the origin is __________

(A)

4 : 1

(B)

3 : 4

(C)

4 : 3

(D)

4 : 5

SolutionAnswer: Option 34 : 3

Approach:

Apply Gauss's law: the net electric flux through any closed surface depends only on the algebraic sum of the charges enclosed by it. Determine which charges lie inside each sphere and take the ratio.

Step 1:Charges

q_{1} = + 8 μ C

x = 2 c m

and

q_{2} = - 2 μ C

x = 4 c m

. Sphere 1 has radius 3 cm centred at origin; sphere 2 has radius 5 cm centred at origin. Target:

ϕ_{1} : ϕ_{2}

q_{1}, q_{2}

positions and sphere radii as above

Step 2:For sphere 1 (

r = 3 c m

∣2∣ < 3

q_{1}

is enclosed;

∣4∣ > 3

q_{2}

is outside.

Q_{e n c, 1} = + 8 μ C

Step 3:For sphere 2 (

r = 5 c m

): both

∣2∣ < 5

and

∣4∣ < 5

, so both charges are enclosed.

Q_{e n c, 2} = 8 + (- 2) = + 6 μ C

Step 4:Write the two fluxes from Gauss's law.

ϕ_{1} = \frac{8 μ C}{ε _{0}}, ϕ_{2} = \frac{6 μ C}{ε _{0}}

Step 5:Take the ratio (the

ε_{0}

cancels).

\frac{ϕ _{1}}{ϕ _{2}} = \frac{8}{6} = \frac{4}{3}

Step 6:Charges outside a closed surface contribute zero net flux, so only enclosed charges matter; the ratio uses only the enclosed totals and the

ε_{0}

factor cancels in the ratio.

ϕ_{1} / ϕ_{2} = Q_{e n c, 1} / Q_{e n c, 2}

Step 7:The ratio of fluxes is 4 : 3, matching option (3).

ϕ_{1} : ϕ_{2} = 4 : 3

Final answer:

ϕ_{1} : ϕ_{2} = 4 : 3

Q41Single correctOptics

One side of an equilateral prism is painted by a transparent material of refractive index

n_{2}

. The refractive index of prism is 1.6. The minimum value of

n_{2}

required for total internal reflection from painted face is ____________

Equilateral triangular prism with a horizontal painted face at the bottom; a light ray is incident on the left inclined face and travels inside the prism towards the painted base, which is labelled paint(n_2).

(A)

33 /1.6

(B)

3

(C)

3.2/ 3

(D)

43 /5

SolutionAnswer: Option 4

43 /5

Approach:

From the equilateral-prism geometry, a ray entering the left face hits the painted base at

6 0^{\circ}

. The threshold for total internal reflection is given by Snell's law at the critical angle with the refracted ray along the interface; solve for the minimum paint index

n_{2}

Step 1:Prism index

n_{1} = 1.6

, prism angle

A = 6 0^{\circ}

, internal angle of incidence on the painted base

i = 6 0^{\circ}

(from the geometry shown). Target: minimum

n_{2}

that just permits total internal reflection.

n_{1} = 1.6, i = 6 0^{\circ}

Step 2:At the threshold (critical condition) Snell's law gives

n_{1} sin i = n_{2} sin 9 0^{\circ}

, i.e. the refracted ray would graze the painted surface.

1.6 sin 6 0^{\circ} = n_{2} \cdot 1

Step 3:Substitute

sin 6 0^{\circ} = 3 /2

n_{2} = 1.6 \times \frac{3}{2} = 0.8 3

Step 4:Rewrite

0.8 3

in the form supplied by the options.

0.8 3 = \frac{4 3}{5}

Step 5:Numerically

43 /5 = 4 (1.732) /5 \approx 1.386

, matching

1.6 \times 0.866 \approx 1.386

n_{2} \approx 1.386

Step 6:The minimum paint refractive index is

43 /5 \approx 1.39

, matching option (4). For total internal reflection at this interface the paint index must be at least this value.

n_{2} = 43 /5

Final answer:

n_{2} = \frac{4 3}{5} \approx 1.39

Q42Single correctElectromagnetic Induction and Alternating Currents

The figure given below shows an LCR series circuit with two switches

S_{1}

and

S_{2}

. When switch

S_{1}

is closed keeping

S_{2}

open, the phase difference (

ϕ

) between the current and source voltage is

3 0^{\circ}

and phase difference is

6 0^{\circ}

when

S_{2}

is closed keeping

S_{1}

open. The value of

(3 L_{1} - L_{2})

is ___________ H.

LCR series circuit driven by source v = V_0 sin(300t). A capacitor C = 100 microfarad and a resistor R are in series along the top branch. Two parallel branches with switches S_1 (left) and S_2 (right) contain inductors L_2 and L_1 respectively.

(A)

\frac{9}{2}

(B)

\frac{2}{9}

(C)

\frac{1}{3}

(D)

SolutionAnswer: Option 2

\frac{2}{9}

Approach:

Write the series LCR phase equation

tan ϕ = (X_{L} - X_{C}) / R

for each switch configuration, take the ratio to eliminate R, and use

ω = 300

rad/s with

C = 100 μ F

to evaluate

1/ (ω^{2} C)

Step 1:Source

ω = 300

rad/s,

C = 100 μ F = 1 0^{- 4}

F; phases

3 0^{\circ}

(first config) and

6 0^{\circ}

(second config); target

(3 L_{1} - L_{2})

in henry.

ω = 300 rad/s, C = 1 0^{- 4} F

Step 2:Configuration 1 has inductor

L_{1}

in series →

tan 3 0^{\circ} = (L_{1} ω - 1/ ω C) / R

tan 3 0^{\circ} = \frac{L _{1} ω - \frac{1}{ω C}}{R} \dots (1)

Step 3:Configuration 2 has inductor

L_{2}

in series →

tan 6 0^{\circ} = (L_{2} ω - 1/ ω C) / R

tan 6 0^{\circ} = \frac{L _{2} ω - \frac{1}{ω C}}{R} \dots (2)

Step 4:Divide (1) by (2) to remove

R

\frac{tan 3 0 ^{\circ}}{tan 6 0 ^{\circ}} = \frac{1/ 3}{3} = \frac{1}{3} = \frac{L _{1} ω - 1/ ω C}{L _{2} ω - 1/ ω C}

Step 5:Cross-multiply and rearrange to isolate

(3 L_{1} - L_{2}) ω

L_{2} ω - \frac{1}{ω C} = 3 L_{1} ω - \frac{3}{ω C} \Rightarrow (3 L_{1} - L_{2}) ω = \frac{2}{ω C}

Step 6:Substitute

ω = 300

and

C = 1 0^{- 4}

ω^{2} C = (300)^{2} \times 1 0^{- 4} = 9 \times 1 0^{4} \times 1 0^{- 4} = 9

Step 7:Substitute back: with

3 L_{1} - L_{2} = 2/9

, the ratio of

tan ϕ

values returns 1/3.

(3 L_{1} - L_{2}) = \frac{2}{9} H

Step 8:Required value

(3 L_{1} - L_{2}) = 2/9

H, matching option 2.

(3 L_{1} - L_{2}) = \frac{2}{9} H

Final answer:

\frac{2}{9}

Q43Single correctElectromagnetic Induction and Alternating Currents

A circular current loop of radius R is placed inside a square loop of side length L (

L ≫ R

) such that they are co-planar and their centers coincide. The permeability of free space is

μ_{0}

. The mutual inductance between circular loop and square loop is __________

(A)

22 \frac{μ _{0} L ^{2}}{R}

(B)

2 \frac{μ _{0} L ^{2}}{R}

(C)

2 \frac{μ _{0} R ^{2}}{L}

(D)

22 \frac{μ _{0} R ^{2}}{L}

SolutionAnswer: Option 4

22 \frac{μ _{0} R ^{2}}{L}

Approach:

Since

L ≫ R

, treat the square's field over the tiny circle as the uniform value at the square's centre. Sum contributions from the four finite sides, compute flux through the circular loop, and use

M = ϕ / i

Step 1:Square side L, current i in square, inner co-planar circle radius R with

R ≪ L

centred at same point; target mutual inductance M.

L ≫ R, centres coincide

Step 2:At the centre, each side is at perpendicular distance

d = L /2

and subtends

α = β = 4 5^{\circ}

at the centre.

d = L /2, α = β = 4 5^{\circ}

Step 3:Contribution from one side of the square.

B_{one side} = \frac{μ _{0} i}{4 π ( L /2 )} (2 sin 4 5^{\circ}) = \frac{μ _{0} i}{4 π ( L /2 )} \cdot 2 = \frac{μ _{0} i}{2 π L}

Step 4:Net field at centre from all four sides (each contributes equally and in the same sense).

B = 4 \cdot \frac{μ _{0} i}{2 π L} = \frac{4 μ _{0} i}{2 π L} = \frac{2 2 μ _{0} i}{π L}

Step 5:Treat B as uniform over the small circle; flux through circular loop of area

π R^{2}

ϕ = B π R^{2} = \frac{2 2 μ _{0} i}{π L} \cdot π R^{2} = \frac{2 2 μ _{0} i R ^{2}}{L}

Step 6:Dimensions of

μ_{0} R^{2} / L

→

(H/m) (m^{2} / m) = H

, correct unit of inductance.

[μ_{0} R^{2} / L] = H

Step 7:

M = ϕ / i = 22 μ_{0} R^{2} / L

, matching option 4.

M = \frac{2 2 μ _{0} R ^{2}}{L}

Final answer:

M = \frac{2 2 μ _{0} R ^{2}}{L}

Q44Single correctAtoms and Nuclei

The binding energy per nucleon of

_{83}^{209} Bi

is _________ MeV. {Take

m (_{83}^{209} Bi) = 208.980388 u

m_{p} = 1.007825 u

m_{n} = 1.008665 u

1 u = 931 MeV / c^{2}

}

(A)

7.48

(B)

7.84

(C)

8.79

(D)

6.94

SolutionAnswer: Option 27.84

Approach:

Compute mass defect

Δ m = Z m_{p} + N m_{n} - m_{nucleus}

, convert to MeV using

1 u = 931 MeV / c^{2}

, divide by A.

Step 1:

Z = 83

protons,

N = 209 - 83 = 126

neutrons,

A = 209

, given masses in u; target BE per nucleon in MeV.

Z = 83, N = 126, A = 209

Step 2:Total mass of the constituent nucleons.

83 \times 1.007825 + 126 \times 1.008665 = 83.649475 + 127.091790 = 210.741265 u

Step 3:Mass defect.

Δ m = 210.741265 - 208.980388 = 1.760877 u

Step 4:Total binding energy.

BE = 1.760877 \times 931 \approx 1639.38 MeV

Step 5:Divide by mass number

A = 209

\frac{BE}{A} = \frac{1639.38}{209} \approx 7.844 MeV

Step 6:Heavy nuclei lie below the iron peak (~8.8 MeV) on the BE/A curve; 7.84 MeV is the standard tabulated value for

^{209}

Bi.

7.84 ≲ 8.8 MeV

Step 7:BE per nucleon

\approx 7.84

MeV, matching option 2.

\frac{BE}{A} \approx 7.84 MeV

Final answer: 7.84 MeV

Q45Single correctOscillations and Waves

The equation of motion of a particle is given by

x = a sin (50 t + π /3)

cm. The particle will come to rest at time

t_{1}

and it will have zero acceleration at time

t_{2}

. The

t_{1}

and

t_{2}

respectively are _____________

(A)

\frac{π}{300} s, \frac{π}{75} s

(B)

\frac{π}{75} s, \frac{π}{300} s

(C)

\frac{π}{300} s, \frac{π}{25} s

(D)

\frac{π}{50} s, \frac{π}{100} s

SolutionAnswer: Option 1

\frac{π}{300} s, \frac{π}{75} s

Approach:

Differentiate to find v(t) and a(t). The particle is at rest when

v = 0 \Rightarrow cos (50 t + π /3) = 0

; acceleration is zero when

sin (50 t + π /3) = 0

. Solve for the smallest positive t in each case.

Step 1:

x (t) = a sin (50 t + π /3)

cm with

ω = 50

rad/s and phase

π /3

; target smallest

t_{1}

where

v = 0

and smallest

t_{2}

where

a_{acc} = 0

ω = 50 rad/s, ϕ_{0} = π /3

Step 2:Set

v = 0 \Rightarrow cos (50 t_{1} + π /3) = 0

, take the smallest positive root → argument equals

π /2

50 t_{1} + \frac{π}{3} = \frac{π}{2}

Step 3:Solve for

t_{1}

50 t_{1} = \frac{π}{2} - \frac{π}{3} = \frac{π}{6} \Rightarrow t_{1} = \frac{π}{300} s

Step 4:Set

a_{acc} = 0 \Rightarrow sin (50 t_{2} + π /3) = 0

, smallest positive root → argument equals

π

50 t_{2} + \frac{π}{3} = π

Step 5:Solve for

t_{2}

50 t_{2} = π - \frac{π}{3} = \frac{2 π}{3} \Rightarrow t_{2} = \frac{2 π}{150} = \frac{π}{75} s

Step 6:At

t_{1}

argument

= π /2

cos = 0

(rest, extreme position); at

t_{2}

argument

= π

sin = 0

(mean position, zero acceleration). Time gap

t_{2} - t_{1} = π /100 = T /4

, consistent with SHM phase between extreme and mean.

t_{2} - t_{1} = \frac{π}{75} - \frac{π}{300} = \frac{π}{100} s = T /4

Step 7:

(t_{1}, t_{2}) = (π /300, π /75)

s, matching option 1.

t_{1} = \frac{π}{300} s, t_{2} = \frac{π}{75} s

Final answer:

t_{1} = \frac{π}{300}

t_{2} = \frac{π}{75}

Q46NumericalOptics

If a Young's doubles lit experiment, the intensity at some point on the screen is found to be

\frac{3}{4}

times of the maximum of the interference pattern. The path difference between the interfering waves at this point is

\frac{λ}{x}

where

λ

is wavelength of the incident light. The value of x is

SolutionAnswer: 6

Approach:

Use the YDSE intensity formula

I = I_{m a x} cos^{2} (ϕ /2)

to convert intensity ratio to phase difference, then use

ϕ = (2 π / λ) Δ x

to obtain the path difference.

Step 1:Intensity ratio

I / I_{m a x} = 3/4

; target x in

Δ x = λ / x

\frac{I}{I _{m a x}} = \frac{3}{4}

Step 2:Apply YDSE intensity law.

\frac{3}{4} = cos^{2} (\frac{ϕ}{2}) \Rightarrow cos (\frac{ϕ}{2}) = \frac{3}{2}

Step 3:Take inverse cosine for smallest positive value.

\frac{ϕ}{2} = 3 0^{\circ} = \frac{π}{6} \Rightarrow ϕ = \frac{π}{3} rad

Step 4:Convert phase to path difference.

Δ x = \frac{λ}{2 π} ϕ = \frac{λ}{2 π} \cdot \frac{π}{3} = \frac{λ}{6}

Step 5:Substitute

Δ x = λ /6

back:

ϕ = (2 π / λ) (λ /6) = π /3

, then

cos^{2} (π /6) = 3/4

, recovering the given ratio.

cos^{2} (π /6) = 3/4

Step 6:Comparing

Δ x = λ /6

with

λ / x

gives

x = 6

x = 6

Final answer: 6

Q47NumericalAtoms and Nuclei

Using Bohr's model, calculate the ratio of the magnetic fields generated due to the motion of the electrons in the 2nd and 4th orbits of hydrogen atom.

SolutionAnswer: 32

Approach:

Treat the orbiting electron as a circular current loop. Equivalent current

i = e v / (2 π r)

; magnetic field at the nucleus

B = μ_{0} i / (2 r) \propto v / r^{2}

. Use Bohr scaling

v \propto z / n

and

r \propto n^{2} / z

to get

B \propto z^{3} / n^{5}

Step 1:Hydrogen (

z = 1

); compare B at the nucleus for

n_{1} = 2

versus

n_{2} = 4

; target ratio

B_{2} / B_{4}

n_{1} = 2, n_{2} = 4

Step 2:Equivalent current of an electron orbiting at speed

v

on radius

r

i = \frac{e}{T} = \frac{e}{2 π r / v} = \frac{e v}{2 π r}

Step 3:Magnetic field at the nucleus.

B = \frac{μ _{0} i}{2 r} = \frac{μ _{0}}{2 r} \cdot \frac{e v}{2 π r} = \frac{μ _{0} e v}{4 π r ^{2}}

Step 4:Substitute Bohr scaling for hydrogen.

B \propto \frac{v}{r ^{2}} \propto \frac{1/ n}{( n ^{2} ) ^{2}} = \frac{1}{n ^{5}}

Step 5:Form the requested ratio.

\frac{B _{2}}{B _{4}} = \frac{n _{2}^{'5}}{n _{1}^{'5}} \to (\frac{4}{2})^{5} = 2^{5} = 32

Step 6:Dimensional/scaling check —

B \propto 1/ n^{5}

means doubling n reduces B by factor

2^{5} = 32

, so

B_{2} / B_{4} = 32

B_{2} / B_{4} = (n_{4} / n_{2})^{5} = 2^{5} = 32

Step 7:Required ratio is 32.

B_{2} : B_{4} = 32 : 1

Final answer: 32

Q48NumericalThermodynamics

5 moles of unknown gas is heated at constant volume from

1 0^{\circ}

C to

2 0^{\circ}

C. The molar specific heat of this gas at constant pressure

c_{p} = 8 cal/mol \cdot^{\circ} C

and

R = 8.36 J/mol \cdot^{\circ} C

. The change in the internal energy of the gas is ............ calorie.

SolutionAnswer: 300

Approach:

Use Mayer's relation to obtain

C_{v} = C_{p} - R

cal/(mol \cdot^{\circ} C)

, converting R to calories via

1 cal = 4.18 J

; then

Δ U = n C_{v} Δ T

Step 1:

n = 5

mol,

C_{p} = 8 cal/(mol \cdot^{\circ} C)

R = 8.36 J/(mol \cdot^{\circ} C)

Δ T = 20 - 10 = 10^{\circ} C

, isochoric process; target

Δ U

in calories.

n = 5, C_{p} = 8 cal/(mol \cdot^{\circ} C), Δ T = 10^{\circ} C

Step 2:Convert

R

to calories.

R = \frac{8.36}{4.18} = 2 cal/(mol \cdot^{\circ} C)

Step 3:Apply Mayer's relation.

C_{v} = C_{p} - R = 8 - 2 = 6 cal/(mol \cdot^{\circ} C)

Step 4:Compute internal energy change.

Δ U = n C_{v} Δ T = 5 \times 6 \times 10 = 300 cal

Step 5:For an ideal gas

Δ U

depends only on temperature, so this value would also apply to any process between the same two states. Units check:

mol \cdot cal/(mol \cdot^{\circ} C) \cdot^{\circ} C = cal

[Δ U] = cal

Step 6:

Δ U = 300

cal.

Δ U = 300 cal

Final answer: 300

Q49NumericalOptics

If sunlight is focused on a paper using convex lens, it starts burning the paper in shortest time when the lens is kept at 30 cm above the paper. If the radius of curvature of the lens is 60 cm then the refractive index of the lens material is

\frac{α}{10}

. The value of

α

is......

SolutionAnswer: 20

Approach:

Burning is fastest when intensity at the spot is maximum, i.e. sunlight (parallel rays) converges at the focal point, so

f = 30

cm. Apply the lens-maker's formula for a symmetric biconvex lens of equal radii

R

Step 1:Distance of burning spot from lens = focal length

f = 30

cm; equiconvex lens with both radii of curvature

R = 60

cm; target

μ

in form

α /10

f = 30 cm, R = 60 cm

Step 2:Simplify lens-maker's formula for an equiconvex lens.

\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{- R}) = \frac{2 ( μ - 1 )}{R}

Step 3:Substitute numerical values.

\frac{1}{30} = \frac{2 ( μ - 1 )}{60} = \frac{μ - 1}{30}

Step 4:Solve for

μ

μ = 2

Step 5:Back-substitute

μ = 2

R = 60

cm into

f = R / [2 (μ - 1)]

f = 60/2 = 30

cm, matching the burning distance.

f = \frac{60}{2 ( 2 - 1 )} = 30 cm

Step 6:Express

μ = 2 = 20/10

, so

α = 20

μ = \frac{α}{10} = 2 \Rightarrow α = 20

Final answer: 20

Q50NumericalRotational Motion

Moment of inertia about an axis AB for a rod of mass 40 kg and length 3 m is same as that of a solid sphere of mass of 10 kg and radius R about an axis parallel to AB axis with separation of 3 m as shown in figure below. The value of R is given as

\frac{α}{2}

. The value of

α

is..........

A horizontal rod of length 3 m and mass M = 40 kg sits along the axis AB (vertical, on the right, oriented A on top and B at the bottom). Below the rod a solid sphere of radius R is shown; its centre is at a horizontal distance of 3 m from axis AB. An arrow at the bottom indicates rotation about the AB axis.

SolutionAnswer: 60

Approach:

Per the official setup, equate the rod's moment of inertia

M l^{2} /3

about axis AB with the sphere's moment of inertia

(2/5) m R^{2}

about an axis through the sphere's centre parallel to AB, then solve for R in the form

α /2

Step 1:Rod

M = 40

kg,

l = 3

m, axis AB perpendicular to rod at its end; sphere

m = 10

kg, radius R, axis through its centre parallel to AB; target

α

R = α /2

M = 40 kg, l = 3 m, m = 10 kg

Step 2:Moment of inertia of the rod about axis AB.

I_{rod} = \frac{M l ^{2}}{3} = \frac{40 \times 9}{3} = 120 kg \cdot m^{2}

Step 3:Moment of inertia of the solid sphere about its diameter (axis through centre parallel to AB).

I_{sphere} = \frac{2}{5} m R^{2} = \frac{2}{5} \times 10 \times R^{2} = 4 R^{2}

Step 4:Equate the two moments of inertia.

120 = 4 R^{2} \Rightarrow R^{2} = 30

Step 5:Write

R

in the prescribed form.

R = 30 = \frac{60}{2}

Step 6:Dimensional check —

[M l^{2}] = kg \cdot m^{2} = [m R^{2}]

; numeric back:

4 \times 30 = 120 = M l^{2} /3

4 \times 30 = 120

Step 7:Comparing with

R = α /2

α = 60

α = 60

Final answer: 60

Chemistry25 questions

Q51Single correctSome Basic Concepts in Chemistry

The ratio of mass percentage (W/W) C:H in a hydrocarbon is 12:1. It has two carbon atoms. The weight (in g) of C

O_{2}

(g) formed when 3.38 g of this hydrocarbon is completely burnt in Oxygen is: (Given: Molar mass in g mo

l^{- 1}

— C: 12, H: 1, O: 16)

(A)

5.68

(B)

11.44

(C)

22.74

(D)

17.05

SolutionAnswer: Option 211.44

Approach:

Determine molecular formula from C:H mass ratio with given carbon count, then use combustion stoichiometry to find moles of C

O_{2}

produced and its mass.

Step 1:Given mass ratio

W_{C}

W_{H}

= 12:1 and the hydrocarbon contains two carbon atoms; sample mass = 3.38 g.

n_{C} = 2, \frac{W _{C}}{W _{H}} = \frac{12}{1}

Step 2:Convert mass ratio to atom ratio using atomic masses (C = 12, H = 1).

\frac{n _{C}}{n _{H}} = \frac{W _{C} /12}{W _{H} /1} = \frac{12/12}{1/1} = \frac{1}{1}

Step 3:With two C atoms, the hydrocarbon has two H atoms, giving molecular formula

C_{2} H_{2}

(ethyne).

C_{2} H_{2}, M = 2 (12) + 2 (1) = 26 g mo l^{- 1}

Step 4:Compute moles of hydrocarbon in 3.38 g sample.

n_{C_{2} H_{2}} = \frac{3.38}{26} = 0.13 mol

Step 5:Apply combustion stoichiometry; each mole of

C_{2} H_{2}

yields 2 mol C

O_{2}

C_{2} H_{2} + \frac{5}{2} O_{2} \to 2 CO_{2} + H_{2} O

Step 6:Mass of C

O_{2}

= moles

\times

molar mass (

M_{C O_{2}} = 44

m_{CO_{2}} = 0.26 \times 44 = 11.44 g

Step 7:Mass of C

O_{2}

produced is 11.44 g, matching option 2.

m_{CO_{2}} = 11.44 g

Final answer: 11.44 g

Q52Single correctEquilibrium

The first and second ionization constants of weak dibasic acid

H_{2}

A are

8.1 \times 1 0^{- 8}

and

1.0 \times 1 0^{- 13}

respectively. 0.1 mol of

H_{2}

A was dissolved in 1 L of 0.1 M HCl solution. The concentration of HA

^{-}

in the resultant solution is:

(A)

0.1 M

(B)

9.53 \times 1 0^{- 6}

(C)

8.1 \times 1 0^{- 8}

(D)

1.0 \times 1 0^{- 13}

SolutionAnswer: Option 3

8.1 \times 1 0^{- 8}

Approach:

Use first ionization equilibrium of

H_{2}

A with H

^{+}

supplied by strong acid HCl to obtain [HA

^{-}

] directly from

K_{a 1}

expression.

Step 1:Given dibasic weak acid

H_{2}

A with

K_{a 1} = 8.1 \times 1 0^{- 8}

K_{a 2} = 1.0 \times 1 0^{- 13}

; 0.1 mol

H_{2}

A dissolved in 1 L of 0.1 M HCl.

[H_{2} A]_{0} = 0.1 M, [HCl] = 0.1 M

Step 2:HCl fully dissociates, providing [H

^{+}

]

\approx

0.1 M. The very small

K_{a 1}

ensures

H_{2}

A ionization is strongly suppressed, so [

H_{2}

\approx

0.1 M and [H

^{+}

]

\approx

0.1 M.

[H^{+}] \approx 0.1 M, [H_{2} A] \approx 0.1 M

Step 3:Rearrange

K_{a 1}

expression to solve for [HA

^{-}

[HA^{-}] = K_{a 1} \frac{[ H _{2} A ]}{[ H ^{+} ]} = 8.1 \times 1 0^{- 8} \times \frac{0.1}{0.1}

Step 4:Second ionization contribution

[A^{2 -}] = K_{a 2} \times [HA^{-}] / [H^{+}] = 1 0^{- 13} \times 8.1 \times 1 0^{- 8} /0.1 \approx 8.1 \times 1 0^{- 20}

M, which is negligible relative to [HA

^{-}

[A^{2 -}] \approx 8.1 \times 1 0^{- 20} M ≪ [HA^{-}]

Step 5:

[HA^{-}]

in the resultant solution equals

8.1 \times 1 0^{- 8}

M, matching option 3.

[HA^{-}] = 8.1 \times 1 0^{- 8} M

Final answer:

8.1 \times 1 0^{- 8}

Q53Single correctChemical Bonding and Molecular Structure

F_{4}

is isostructural with:
A. BrF

_{4}^{-}

B. C

H_{4}

C. IF

_{4}^{+}

D. Xe

F_{4}

E. Xe

O_{2} F_{2}

Choose the correct answer from the option given below:

(A)

C only

(B)

C and E only

(C)

A and D only

(D)

B and E only

SolutionAnswer: Option 2C and E only

Approach:

Apply VSEPR to determine the geometry of S

F_{4}

(see-saw, A

X_{4}

E) and compare each candidate species' steric number and lone-pair count to find those with the same shape.

Step 1:S

F_{4}

has central S with 6 valence electrons, four S–F bonds and one lone pair (A

X_{4}

E, s

p^{3}

d hybridization, see-saw geometry).

SF_{4} : AX_{4} E, see-saw

Step 2:BrF

_{4}^{-}

—Br has 7 valence e

^{-}

+1 (charge) = 8; four bonds + two lone pairs (A

X_{4} E_{2}

) → square planar; not see-saw.

BrF_{4}^{-} : AX_{4} E_{2}, square planar

Step 3:C

H_{4}

—C has 4 bond pairs, no lone pair (A

X_{4}

) → tetrahedral.

CH_{4} : AX_{4}, tetrahedral

Step 4:IF

_{4}^{+}

—I has 7 valence e

^{-}

−1 (charge) = 6; four bonds + one lone pair (A

X_{4}

E) → see-saw.

IF_{4}^{+} : AX_{4} E, see-saw

Step 5:Xe

F_{4}

—Xe has 8 valence e

^{-}

; four bonds + two lone pairs (A

X_{4} E_{2}

) → square planar.

XeF_{4} : AX_{4} E_{2}, square planar

Step 6:Xe

O_{2} F_{2}

—Xe forms two Xe=O and two Xe–F

σ

-bonds with one lone pair (A

X_{4}

E) → see-saw.

XeO_{2} F_{2} : AX_{4} E, see-saw

Step 7:Only IF

_{4}^{+}

and Xe

O_{2} F_{2}

share A

X_{4}

E geometry with S

F_{4}

(see-saw).

Isostructural with SF_{4} : IF_{4}^{+}, XeO_{2} F_{2}

Step 8:The isostructural species are C and E only.

C and E only

Final answer: C and E only

Q54Single correctChemical Thermodynamics

Gas 'A' undergoes change from state 'X' to state 'Y', in this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state 'X' by another process during which 6 J of heat is evolved. In the reverse process of 'Y' to 'X':

(A)

18 J of the work is done by the gas 'A'

(B)

2 J of the work is done by the gas 'A'

(C)

12 J of the work is done on the gas 'A' by the surrounding.

(D)

14 J of the work is done on the gas 'A' by the surrounding

SolutionAnswer: Option 414 J of the work is done on the gas 'A' by the surrounding

Approach:

Apply the first law of thermodynamics over the cyclic process; internal energy is a state function so

Δ U_{X \to Y} + Δ U_{Y \to X} = 0

. Use sign conventions consistently and solve for the work in the reverse step.

Step 1:Process X

\to

Y absorbs heat

q_{X Y}

= +10 J and the gas does work

w_{X Y}

= +18 J on surroundings.

q_{X Y} = + 10 J, w_{X Y, by} = + 18 J

Step 2:Compute change in internal energy for the forward process using

Δ U = q - w_{by}

Δ U_{X Y} = 10 - 18 = - 8 J

Step 3:For the reverse process Y

\to

X, internal energy change is opposite in sign.

Δ U_{Y X} = - Δ U_{X Y} = + 8 J

Step 4:Heat is evolved (released) during Y

\to

X, so

q_{Y X}

-

6 J. Apply first law to solve for w

_{Y X, by}

Δ U_{Y X} = q_{Y X} - w_{Y X, by} \Rightarrow 8 = - 6 - w_{Y X, by}

Step 5:Negative w

_{by}

means work is done on the gas by the surroundings; magnitude = 14 J.

w_{on gas} = + 14 J

Step 6:Sum heats and works over cycle:

Δ U_{cycle} = (10 + (- 6)) - (18 + (- 14)) = 4 - 4 = 0

, consistent with state function.

Δ U_{cycle} = q_{net} - w_{net,by} = 4 - 4 = 0

Step 7:14 J of work is done on the gas by the surroundings, matching option 4.

w_{on gas} = 14 J

Final answer: 14 J of the work is done on the gas 'A' by the surrounding

Q55Single correctSolutions

Solution A is prepared by dissolving 1 g of a protein (molar mass = 50000 g mo

l^{- 1}

) in 0.5 L of water at 300 K. Its osmotic pressure is x bar. Solution B is made by dissolving 2 g of same protein in 1 L of water at 300 K. Osmotic pressure of solution B is y bar. Entire solution of A is mixed with entire solution of B at same temperature. The osmotic pressure of resultant solution is z bar. x, y and z respectively are (R = 0.083 L bar mo

l^{- 1}

K^{- 1}

(A)

9.96 \times 1 0^{- 4}; 9.96 \times 1 0^{- 4}; 9.96 \times 1 0^{- 4}

(B)

9.96 \times 1 0^{- 4}; 9.96 \times 1 0^{- 4}; 19.92 \times 1 0^{- 4}

(C)

4.98 \times 1 0^{- 4}; 4.98 \times 1 0^{- 4}; 9.96 \times 1 0^{- 4}

(D)

4.98 \times 1 0^{- 4}; 4.98 \times 1 0^{- 4}; 4.98 \times 1 0^{- 4}

SolutionAnswer: Option 1

9.96 \times 1 0^{- 4}; 9.96 \times 1 0^{- 4}; 9.96 \times 1 0^{- 4}

Approach:

Use van't Hoff equation

π = CRT

with molar concentration computed from mass/molar-mass per unit volume; for the mixture, compute the total moles divided by total volume.

Step 1:Protein molar mass M = 50000 g mo

l^{- 1}

, T = 300 K, R = 0.083 L bar mo

l^{- 1}

K^{- 1}

M = 50000 g mo l^{- 1}, T = 300 K

Step 2:Solution A—1 g protein in 0.5 L. Moles

n_{A} = 1/50000 = 2 \times 1 0^{- 5}

mol;

C_{A} = 2 \times 1 0^{- 5} /0.5 = 4 \times 1 0^{- 5}

C_{A} = \frac{1/50000}{0.5} = 4 \times 1 0^{- 5} mol L^{- 1}

Step 3:Compute

x = π_{A} = C_{A} RT

x = 4 \times 1 0^{- 5} \times 0.083 \times 300 = 9.96 \times 1 0^{- 4} bar

Step 4:Solution B—2 g protein in 1 L. Moles

n_{B} = 2/50000 = 4 \times 1 0^{- 5}

mol;

C_{B} = 4 \times 1 0^{- 5}

M (same as A).

C_{B} = \frac{2/50000}{1} = 4 \times 1 0^{- 5} mol L^{- 1}

Step 5:

y = π_{B} = C_{B} RT = 9.96 \times 1 0^{- 4}

bar.

y = 9.96 \times 1 0^{- 4} bar

Step 6:Mixing A and B gives total moles

n = n_{A} + n_{B} = 6 \times 1 0^{- 5}

mol in total volume

V = 0.5 + 1 = 1.5

L, so

C_{mix} = 6 \times 1 0^{- 5} /1.5 = 4 \times 1 0^{- 5}

C_{mix} = \frac{6 \times 1 0 ^{- 5}}{1.5} = 4 \times 1 0^{- 5} mol L^{- 1}

Step 7:

z = π_{mix} = C_{mix} RT = 9.96 \times 1 0^{- 4}

bar.

z = 9.96 \times 1 0^{- 4} bar

Step 8:Since both A and B have identical molar concentrations, the mixture concentration equals either parent; osmotic pressure stays unchanged.

x = y = z = 9.96 \times 1 0^{- 4} bar

Step 9:

x, y, z = 9.96 \times 1 0^{- 4}

bar each, matching option 1.

x = y = z = 9.96 \times 1 0^{- 4} bar

Final answer:

9.96 \times 1 0^{- 4}; 9.96 \times 1 0^{- 4}; 9.96 \times 1 0^{- 4}

Q56Single correctEquilibrium

At 2

5^{\circ}

C, 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is added respectively are: Given:

K_{a} = 5 \times 1 0^{- 4}

p K_{a} = 3.3

(A)

0.7, 2.0

(B)

2.0, 3.3

(C)

1.1, 2.2

(D)

3.0, 2.2

SolutionAnswer: Option 22.0, 3.3

Approach:

Use weak-acid approximation

[H^{+}] = K_{a} C

for the starting pH, and the Henderson–Hasselbalch equation at the half-neutralization point where [HX] = [X

^{-}

], giving pH = p

K_{a}

Step 1:20.0 mL of 0.2 M HX (

n_{H X} = 4

mmol),

K_{a} = 5 \times 1 0^{- 4}

p K_{a} = 3.3

; titrant 0.2 M NaOH.

n_{HX} = 0.020 \times 0.2 = 4 \times 1 0^{- 3} mol

Step 2:No NaOH added. Apply weak-acid approximation with C = 0.2 M.

[H^{+}] = (5 \times 1 0^{- 4}) (0.2) = 1 \times 1 0^{- 4} = 1 0^{- 2} M

Step 3:PH at start.

pH = - lo g (1 0^{- 2}) = 2.0

Step 4:10 mL of 0.2 M NaOH = 2 mmol OH

^{-}

added, which converts half of HX into NaX (half-equivalence point).

n_{OH^{-}} = 0.010 \times 0.2 = 2 \times 1 0^{- 3} mol = \frac{1}{2} n_{HX}

Step 5:At half-equivalence, [HX] = [X

^{-}

], so Henderson–Hasselbalch reduces to pH = p

K_{a}

pH = p K_{a} + lo g (1) = 3.3

Step 6:Approximation

[H^{+}] ≪ C

holds since 1

0^{- 2}

≪ 0.2 (α = 5

, validating the weak-acid formula.

α = \frac{1 0 ^{- 2}}{0.2} = 0.05

Step 7:(a) pH = 2.0, (b) pH = 3.3 — option 2.

pH_{a} = 2.0, pH_{b} = 3.3

Final answer: 2.0, 3.3

Q57Single correctChemical Kinetics

Consider the reaction

a X \to bY

, for which the rate constant at 30

0^{\circ}

C is

1 \times 1 0^{- 3} mol^{- 1} L s^{- 1}

. Which of the following statements are true?

(A)

A and B only

(B)

A, B and C only

(C)

A, B, D and E only

(D)

C and D only

SolutionAnswer: Option 1A and B only

Approach:

Identify reaction order from the units of the rate constant; test each statement against the second-order rate law, its integrated form, half-life expression, and known kinetic behaviour of

N_{2} O_{5}

decomposition.

Step 1:Rate constant

k = 1 \times 1 0^{- 3} mol^{- 1} L s^{- 1}

. Units of k for an n-th order reaction are

mol^{1 - n} L^{n - 1} s^{- 1}

;

mol^{- 1} L s^{- 1}

corresponds to

n = 2

units (k) = mo l^{1 - n} L^{n - 1} s^{- 1} \Rightarrow n = 2

Step 2:For r = k[X

]^{2}

, if [X] becomes 4[X], rate becomes (4

)^{2}

= 16 times original.

\frac{r _{new}}{r _{old}} = (\frac{4 [ X ]}{[ X ]})^{2} = 16

Step 3:Established in step 1 that the reaction is second order.

n = 2

Step 4:For second order,

t_{1/2}

= 1/(k[X

]_{0}

), which depends on initial concentration.

t_{1/2} \propto \frac{1}{[ X ] _{0}}

Step 5:Gas-phase decomposition of

N_{2}

O_{5}

is a classical first-order reaction (r = k[

N_{2}

O_{5}

]), not second order.

2 N_{2} O_{5} \to 4N O_{2} + O_{2}, first order in [N_{2} O_{5}]

Step 6:Ln([

R_{0}

]/[R]) vs t is the integrated rate law for first order; second order requires 1/[X] vs t.

ln \frac{[ R ] _{0}}{[ R ]} = k t \Rightarrow first order

Step 7:Only A and B remain true; C, D, E violate second-order kinetics.

True = {A, B}

Step 8:Correct option is A and B only, matching option 1.

A and B only

Final answer: A and B only

Q58Single correctClassification of Elements and Periodicity in Properties

The correct set that contain all kinds (basic, acids, amphoteric and neutral) of Oxides is:

(A)

a_{2}

K_{2}

O, A

l_{2}

O_{3}

and A

s_{2}

O_{3}

(B)

l_{2}

O_{3}

, A

s_{2}

O_{3}

, CO and NO

(C)

K_{2}

O, C

l_{2}

O_{7}

, A

s_{2}

O_{3}

and NO

(D)

a_{2}

N_{2}

O, A

l_{2} O_{3}

and CO

SolutionAnswer: Option 3

K_{2}

O, C

l_{2}

O_{7}

, A

s_{2}

O_{3}

and NO

Approach:

Classify each oxide in every option as basic, acidic, amphoteric, or neutral and pick the option that contains exactly one representative from each of the four categories.

Step 1:Identify standard classifications. Basic: N

a_{2}

K_{2}

O. Acidic: C

l_{2}

O_{7}

. Amphoteric: A

l_{2}

O_{3}

, A

s_{2} O_{3}

. Neutral: CO, NO,

N_{2}

Categories tabulated

Step 2:N

a_{2}

O (basic),

K_{2}

O (basic), A

l_{2}

O_{3}

(amphoteric), A

s_{2}

O_{3}

(amphoteric). Missing acidic and neutral.

{basic, basic, amph, amph}

Step 3:A

l_{2}

O_{3}

(amph), A

s_{2}

O_{3}

(amph), CO (neutral), NO (neutral). Missing basic and acidic.

{amph, amph, neutral, neutral}

Step 4:

K_{2}

O (basic), C

l_{2}

O_{7}

(acidic), A

s_{2}

O_{3}

(amphoteric), NO (neutral). All four categories present.

{basic, acidic, amph, neutral}

Step 5:N

a_{2}

O (basic),

N_{2}

O (neutral), A

l_{2} O_{3}

(amph), CO (neutral). Missing acidic.

{basic, neutral, amph, neutral}

Step 6:C

l_{2}

O_{7}

is the anhydride of HCl

O_{4}

(strongly acidic); NO is a well-known neutral oxide (does not react with acids or bases under normal conditions); A

s_{2}

O_{3}

dissolves in both HCl and NaOH (amphoteric);

K_{2}

O reacts with water to give KOH (basic).

C l_{2} O_{7} + H_{2} O \to 2 HCl O_{4}; K_{2} O + H_{2} O \to 2 KOH

Step 7:Only option 3 contains all four kinds of oxides.

K_{2} O, C l_{2} O_{7}, A s_{2} O_{3}, NO

Final answer:

K_{2}

O, C

l_{2}

O_{7}

, A

s_{2}

O_{3}

and NO

Q59Single correctClassification of Elements and Periodicity in Properties

Given below are two statements:
Statement I: The second ionization enthalpy of B, Al and Ga is in order of B > Al > Ga.
Statement II: The correct order in terms of first ionization enthalpy is Si < Ge < Pb < Sn.
In the light of the above statement, choose the correct answer from the option given below:

(A)

Both statement I and statement II are true

(B)

Both statement I and statement II are false

(C)

Statement I is true but statement II is false

(D)

Statement I is false but statement II is true

SolutionAnswer: Option 2Both statement I and statement II are false

Approach:

Evaluate the two statements against standard second-IE data for Group 13 (B, Al, Ga) and first-IE data for Group 14 (Si, Ge, Sn, Pb), accounting for poor d-shielding in Ga and inert-pair stabilisation in Pb.

Step 1:List I

E_{2}

(kJ mo

l^{- 1}

) for Group 13 elements B, Al, Ga from standard data.

I E_{2} (B) = 2427, I E_{2} (G a) = 1979, I E_{2} (A l) = 1816

Step 2:Rank I

E_{2}

from the values; B (small size, high

Z_{e ff}

) tops, Ga sits above Al because the intervening 3

d^{10}

shell shields poorly and raises

Z_{e ff}

on the 4s electron of Ga

^{+}

I E_{2} : B > G a > A l

Step 3:Compare with the proposed Statement I order B > Al > Ga.

Proposed: B > A l > G a; Actual: B > G a > A l

Step 4:List I

E_{1}

(kJ mo

l^{- 1}

) for Group 14 elements Si, Ge, Sn, Pb from standard data.

I E_{1} (S i) = 786, I E_{1} (G e) = 761, I E_{1} (P b) = 715, I E_{1} (S n) = 708

Step 5:Rank I

E_{1}

; Si > Ge owing to smaller size; Pb > Sn because 6

s^{2}

inert pair is poorly shielded by 4

f^{14} 5 d^{10}

raising

Z_{e ff}

I E_{1} : S i > G e > P b > S n

Step 6:Compare with the proposed Statement II order Si < Ge < Pb < Sn (the reverse direction).

Proposed: S i < G e < P b < S n

Step 7:Both statements contradict the standard IE data, so both are false.

S_{I} = False, S_{II} = False

Final answer: Both Statement I and Statement II are false

Q60Single correctd- and f-Block Elements

Given below are two statements:
Statement I: Among Zn, Mn, Sc and Cu, the energy required to remove the third valence electron is highest for Zn and lowest for Sc.
Statement II: The correct order of the following complexes in terms of CFSE is

[C o (H_{2} O)_{6}]^{2 +} < [C o (H_{2} O)_{6}]^{3 +} < [C o (e n)_{3}]^{3 +}

.
In the light of the above statement, choose the correct answer from the option given below:

(A)

Both statement I and statement II are true

(B)

Both statement I and statement II are false

(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 true

Approach:

Compare I

E_{3}

values across Sc, Mn, Cu, Zn using the configurations of

M^{2 +}

ions, then rank CFSE of the three Co complexes using the dependence of

Δ_{o}

on oxidation state and ligand field strength.

Step 1:Write the

M^{2 +}

configurations for Sc, Mn, Cu, Zn.

S c^{2 +} : [A r] 3 d^{1} M n^{2 +} : [A r] 3 d^{5} C u^{2 +} : [A r] 3 d^{9} Z n^{2 +} : [A r] 3 d^{10}

Step 2:I

E_{3}

removes the third electron from

M^{2 +}

. Z

n^{2 +}

has the very stable 3

d^{10}

shell, so I

E_{3}

is highest. S

c^{2 +}

has a single 3d electron easily removed leaving a noble-gas [Ar] core, so I

E_{3}

is lowest.

I E_{3} (Z n) \approx 3837, I E_{3} (C u) \approx 3554, I E_{3} (M n) \approx 3248, I E_{3} (S c) \approx 2393 kJ mol^{- 1}

Step 3:Order shows highest I

E_{3}

for Zn and lowest for Sc, matching Statement I.

I E_{3} : Z n > C u > M n > S c

Step 4:For Statement II, list the three Co complexes and note oxidation state and ligand type.

[C o (H_{2} O)_{6}]^{2 +}

: Co(II), weak-field

H_{2}

[C o (H_{2} O)_{6}]^{3 +}

: Co(III), weak-field

H_{2}

[C o (e n)_{3}]^{3 +}

: Co(III), stronger-field ethylenediamine.

[C o (H_{2} O)_{6}]^{2 +}, [C o (H_{2} O)_{6}]^{3 +}, [C o (e n)_{3}]^{3 +}

Step 5:For the same metal and same ligand, raising the charge from +2 to +3 contracts orbitals and increases

Δ_{o}

, hence

[C o (H_{2} O)_{6}]^{3 +}

has larger

Δ_{o}

than

[C o (H_{2} O)_{6}]^{2 +}

. Among Co(III) complexes, en is higher in the spectrochemical series than

H_{2}

O, so

[C o (e n)_{3}]^{3 +}

has the largest

Δ_{o}

Δ_{o} ([C o (H_{2} O)_{6}]^{2 +}) < Δ_{o} ([C o (H_{2} O)_{6}]^{3 +}) < Δ_{o} ([C o (e n)_{3}]^{3 +})

Step 6:Order of CFSE matches the proposed inequality in Statement II.

CFSE : [C o (H_{2} O)_{6}]^{2 +} < [C o (H_{2} O)_{6}]^{3 +} < [C o (e n)_{3}]^{3 +}

Step 7:Both statements are consistent with standard data, so the correct option is Option 1.

S_{I} = True, S_{II} = True

Final answer: Both Statement I and Statement II are true

Q61Single correctCoordination Compounds

Which of the following complexes will show coordination isomerism?
A.

[A g (N H_{3})_{2}] [A g (CN)_{2}]

[C o (N H_{3})_{6}] [C r (CN)_{6}]

[C o (N H_{3})_{6}] [C o (CN)_{6}]

[F e (N H_{3})_{6}] [C o (CN)_{6}]

[C o (N H_{3})_{6}] [F e (CN)_{6}]

(A)

B, C and D only

(B)

B, D and E only

(C)

A, C and D only

(D)

C, D and E only

SolutionAnswer: Option 2B, D and E only

Approach:

Coordination isomerism arises in salts with both cationic and anionic complex spheres when the two metal centres differ, allowing ligand exchange between the spheres to give a distinct isomer; screen each complex for this condition.

Step 1:Identify the metal in the cation and in the anion of each salt.

A : A g / A g B : C o / C r C : C o / C o D : F e / C o E : C o / F e

Step 2:Apply the criterion that coordination isomerism requires two different metals in the two spheres so that ligand interchange yields a structurally distinct salt.

M_{c a t i o n} \neq = M_{ani o n} \Rightarrow coordination isomerism

Step 3:

[A g (N H_{3})_{2}] [A g (CN)_{2}]

has Ag in both spheres; interchange gives the same compound back.

A : M_{c a t i o n} = M_{ani o n} = A g

Step 4:

[C o (N H_{3})_{6}] [C r (CN)_{6}]

has Co and Cr; interchange gives

[C r (N H_{3})_{6}] [C o (CN)_{6}]

B : C o \neq = C r

Step 5:

[C o (N H_{3})_{6}] [C o (CN)_{6}]

has Co in both spheres (same oxidation state implied), no new isomer on interchange.

C : M_{c a t i o n} = M_{ani o n} = C o

Step 6:

[F e (N H_{3})_{6}] [C o (CN)_{6}]

has Fe and Co; interchange gives

[C o (N H_{3})_{6}] [F e (CN)_{6}]

, a distinct salt.

D : F e \neq = C o

Step 7:

[C o (N H_{3})_{6}] [F e (CN)_{6}]

has Co and Fe; interchange gives

[F e (N H_{3})_{6}] [C o (CN)_{6}]

, distinct from E.

E : C o \neq = F e

Step 8:Coordination isomerism is observed in B, D and E only, matching Option 2.

Answer set = {B, D, E}

Final answer: B, D and E only

Q62Single correctPurification and Characterisation of Organic Compounds

Complete combustion of X g of an organic compound gave 0.25 g of C

O_{2}

and 0.12 g of

H_{2}

O. If the % of carbon is 25% and of hydrogen is 4.89%, then

X = \dots \times 1 0^{- 3}

g (Nearest integer). (Molar mass of C, H and O are 12, 1 and 16 g mo

l^{- 1}

respectively)

(A)

273

(B)

(C)

2730

(D)

227

SolutionAnswer: Option 1273

Approach:

Convert the masses of C

O_{2}

and

H_{2}

O into masses of C and H, then use the given percentage of each element to back-calculate the sample mass X, retaining the value rounded to the nearest integer of

1 0^{- 3}

Step 1:Note the given quantities

m_{C O_{2}} = 0.25

m_{H_{2} O} = 0.12

g, %C

= 25

, %H

= 4.89

, with M(C)

= 12

, M(H)

= 1

, M(O)

= 16

m_{C O_{2}} = 0.25 g, m_{H_{2} O} = 0.12 g

Step 2:Compute mass of carbon obtained in combustion.

m_{C} = \frac{12}{44} \times 0.25 = 0.0 \overline{681} \approx 0.06 g

Step 3:Compute mass of hydrogen obtained in combustion.

m_{H} = \frac{2}{18} \times 0.12 = 0.01 \overline{3} \approx 0.01 g

Step 4:Use the carbon percentage to solve for X.

25 = \frac{0.06}{X} \times 100 \Rightarrow X = \frac{0.06 \times 100}{25} \times \frac{1}{1}

Step 5:Use the hydrogen percentage to solve for X.

4.89 = \frac{0.01}{X} \times 100 \Rightarrow X = \frac{1}{4.89} \approx 0.2045 g (from rounded m_{H})

Step 6:With the exact carbon mass 0.0682 g and %C

= 25

X = \frac{0.0682 \times 100}{25} = 0.2727

g, consistent with the hydrogen channel.

X \approx 0.2727 g

Step 7:Convert X to the requested units of

1 0^{- 3}

X = 0.2727 g = 272.7 \times 1 0^{- 3} g \approx 273 \times 1 0^{- 3} g

Final answer:

X = 273 \times 1 0^{- 3}

Q63Single correctSome Basic Principles of Organic Chemistry

Given below are two statements:

(A)

Both statement I and statement II are true

(B)

Both statement I and statement II are false

(C)

Statement I and statement II is false

(D)

Statement I is false but statement II is true

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

Approach:

Apply the standard rules that carbocations are stabilised by electron-releasing groups via

+ R

and carbanions by electron-withdrawing groups via

- R

, examining each substituent through the para-aryl ring directly conjugated with the charged centre.

Step 1:Identify the para substituents on the two aryl rings and the charged benzylic centre in each statement.

- OCH_{3}

donates a lone pair to the ring (

+ R

);

- N O_{2}

withdraws via its

π

system (

- R

- OCH_{3} : + R (ERG) - N O_{2} : - R (EWG)

Step 2:For Statement I, the lone pair on

- OCH_{3}

delocalises through the para-methoxyphenyl ring into the empty p-orbital of the adjacent C

^{+}

, distributing the positive charge over the methoxy oxygen.

MeO - A r - C + H_{2} \leftrightarrow MeO + = A r = C H_{2}

Step 3:The presence of a para

- OCH_{3}

on a benzylic cation is the classic resonance stabilisation case; hence Statement I is TRUE.

+ R of - OC H_{3} \Rightarrow stabilises C^{+}

Step 4:For Statement II, the carbanion lone pair delocalises through the para-nitrophenyl ring into the

π^{*}

- N O_{2}

, placing the negative charge on the strongly electronegative oxygens of nitro.

O_{2} N - A r - C - H \leftrightarrow O - - N + (= O) - A r = C H

Step 5:

- R

delocalisation of a benzylic carbanion onto para

- N O_{2}

is the standard textbook example; hence Statement II is TRUE.

- R of - N O_{2} \Rightarrow stabilises C^{-}

Step 6:Both statements correctly identify the dominant resonance stabilisation in each species, matching Option 1.

S_{I} = True, S_{II} = True

Final answer: Both Statement I and Statement II are true

Q64Single correctHydrocarbons

The compound (X) on
(i) heating in the presence of anhydrous

AlCl_{3}

and HCl gas gives 2,4-dimethyl pentane,
(ii) aromatization gives toluene, and
(iii) cyclisation gives methyl cyclohexane.
The correct name of compound (X) is:

(A)

Hept-2-ene

(B)

Hepta-1,3,5-triene

(C)

Heptane

(D)

Hepta-2,4,6-triene

SolutionAnswer: Option 3Heptane

Approach:

All three transformations (Lewis-acid catalysed isomerisation, dehydrocyclisation to toluene, cyclisation to methylcyclohexane) are diagnostic of a straight-chain

C_{7}

alkane; deduce that X is n-heptane and reject the alkene/triene options because they would not undergo

AlCl_{3} / HCl

skeletal isomerisation to 2,4-dimethyl pentane.

Step 1:Count the carbons in the products. Toluene (

C_{6} H_{5} CH_{3}

) and methylcyclohexane (

C_{7} H_{14}

) both carry 7 C atoms; 2,4-dimethylpentane is also

C_{7}

Toluene: C_{7} H_{8}, Methylcyclohexane: C_{7} H_{14}, 2, 4 -DMP: C_{7} H_{16}

Step 2:Anhydrous

AlCl_{3}

/HCl is a textbook Lewis-acid catalyst for skeletal isomerisation of saturated alkanes; alkenes/polyenes would polymerise or undergo addition under these conditions, not branch to 2,4-DMP.

n - C_{7} H_{16} A lC l_{3} / H Cl branched C_{7} H_{16} (2, 4 -DMP)

Step 3:Aromatisation (dehydrocyclisation) of n-heptane over

C r_{2} O_{3} / A l_{2} O_{3}

\sim

773 K loses 4

H_{2}

to give toluene.

C H_{3} (C H_{2})_{5} C H_{3} C r_{2} O_{3}, 773 K C_{6} H_{5} C H_{3} + 4 H_{2}

Step 4:Catalytic cyclisation of n-heptane closes the chain to methylcyclohexane with loss of one

H_{2}

C H_{3} (C H_{2})_{5} C H_{3} \to C_{6} H_{11} C H_{3} + H_{2}

Step 5:Hept-2-ene, hepta-1,3,5-triene and hepta-2,4,6-triene are unsaturated and would not survive

AlCl_{3} / HCl

to give 2,4-DMP; only n-heptane satisfies all three transformations.

Alkene/triene options eliminated by test (i)

Step 6:Compound X is n-heptane (Option 3).

X = n -heptane

Final answer: Heptane

Q65Single correctOrganic Compounds Containing Halogens

Correct statement regarding alkyl halides (R-X) among the following are:
A. Alcohol being less polar solvent as compared to water, alcoholic KOH favours elimination reaction with R-X.
B. Order of reactivity towards

S_{N} 1

mechanism:

C_{6} H_{5} - C H_{2} - Cl > C_{6} H_{5} - C H Cl - C_{6} H_{5}

.
C. Non substituted aryl halides exhibit properties similar to alkyl halides.
D. Vinyl chloride is an example of haloalkene and allyl chloride is an example of haloalkyne.
E. R-Cl can be prepared by reaction of R-OH with

SOCl_{2}

, but Ar-Cl cannot be prepared by reacting Ar-OH with

SOCl_{2}

(A)

A, B and C only

(B)

A and D only

(C)

A and E only

(D)

D and E only

SolutionAnswer: Option 3A and E only

Approach:

Test each statement against standard alkyl/aryl halide chemistry: solvent polarity controls

E 2

S_{N} 2

; benzylic-cation stability governs

S_{N} 1

; aryl halide resonance differs from alkyl halides; allyl chloride is a haloalkane; Darzen's procedure works for alcohols but not phenols.

Step 1:List each statement A-E and the diagnostic chemistry needed to test it.

A, B, C, D, E

as given

Step 2:Ethanol is less polar than water; in the less polar medium

O H^{-}

is less solvated but more basic, favouring deprotonation (

E 2

) over nucleophilic substitution. So alcoholic KOH causes elimination of R-X.

Alcoholic KOH \Rightarrow E 2 (alkene)

Step 3:The intermediate cation in

C_{6} H_{5} - C H Cl - C_{6} H_{5}

is the diphenyl methyl cation (stabilised by two aryl rings), whereas benzyl chloride gives only the benzyl cation (one aryl ring). The diphenyl cation is more stable, so

C_{6} H_{5} CHClC_{6} H_{5}

reacts faster, not slower. The stated order is reversed.

Stability: (C_{6} H_{5})_{2} C + H > C_{6} H_{5} C + H_{2} \Rightarrow Rate : (C_{6} H_{5})_{2} C H Cl > C_{6} H_{5} C H_{2} Cl

Step 4:Aryl halides have partial C-X double-bond character from resonance of the halogen lone pair into the ring, making them much less reactive in nucleophilic substitution than alkyl halides; properties are NOT similar.

A r - X : partial C = X character

Step 5:Vinyl chloride

C H_{2} = CHCl

is correctly a haloalkene, but allyl chloride

C H_{2} = C H - C H_{2} - Cl

has Cl on an s

p^{3}

carbon and is a haloalkane (not haloalkyne).

Allyl chloride : C H_{2} = C H - C H_{2} - Cl \Rightarrow haloalkane

Step 6:

SOCl_{2}

converts R-OH to R-Cl smoothly with gaseous S

O_{2}

and HCl by-products. With Ar-OH the C-O bond has partial double-bond character (resonance into the ring), so the OH cannot be displaced and Ar-Cl is not obtained.

R - O H + SOC l_{2} \to R - Cl + S O_{2} + H Cl; A r - O H + SOC l_{2} \neq \to A r - Cl

Step 7:Cross-check the true/false assignments — A true (alcohol vs water medium for

O H^{-}

), E true (

SOCl_{2}

with R-OH); B, C, D shown false above. Two correct statements confirms option 3.

Correct statements = {A, E}

Step 8:Only A and E are correct, matching Option 3.

Correct set = {A, E}

Final answer: A and E only

Q66Single correctAldehydes, Ketones and Carboxylic Acids

An organic compound x, where the molar ratio of C, O and H are equal, on treatment with 50% KOH under reflux followed by acidification produced y. The most likely structure of y is: [Molar mass of x is 58 g mo

l^{- 1}

]

(A)

Acrylic acid: CH_2=CH-C(=O)-OH; ethenyl group attached to a carboxylic acid carbon, with the carbonyl oxygen drawn above the C and a hydroxyl to the right.

(B)

C H_{3} - C H = C = C H - O

(C)

Glycolic acid drawn as O=C-OH stacked above CH_2 with -OH attached: hydroxyacetic acid (HOOC-CH_2-OH).

(D)

Acetic acid: CH_3-C(=O)-OH; methyl group attached to a carboxylic acid carbon, with the carbonyl oxygen drawn above the C and a hydroxyl to the right.

SolutionAnswer: Option 3

H OOC - C H_{2} - O H

(hydroxy-acetic acid / glycolic acid)

Approach:

Use the equal C:O:H molar ratio and the molar mass 58 g mo

l^{- 1}

to fix the molecular formula as

C_{2} H_{2} O_{2}

; assign x as glyoxal (OHC-CHO), apply the intramolecular Cannizzaro reaction with 50% KOH that simultaneously oxidises one CHO to COO

^{-}

and reduces the other to C

H_{2}

OH, and acidify to obtain glycolic acid as y.

Step 1:Let the molecular formula be

C_{n} O_{n} H_{n}

with the equal-ratio condition; the molar mass is

29 n

M = 12 n + 16 n + n = 29 n

Step 2:Set

29 n = 58

to find n.

29 n = 58 \Rightarrow n = 2

Step 3:

C_{2} H_{2} O_{2}

with two oxygens and only two hydrogens fits OHC-CHO (glyoxal) - a dialdehyde with no

α

-H, exactly the condition for the Cannizzaro reaction with concentrated KOH.

x = O H C - C H O (g l yo x a l)

Step 4:Under 50% KOH reflux, an intramolecular Cannizzaro occurs - hydroxide attacks one CHO and a hydride migrates from that tetrahedral intermediate to the other CHO; one carbonyl is oxidised to carboxylate and the other reduced to primary alcohol on the same molecule.

O H C - C H O 50% K O H H OC H_{2} - CO O^{-} K^{+}

Step 5:Acidification protonates the carboxylate to give the free carboxylic acid.

H OC H_{2} - CO O^{-} K^{+} H^{+} H OC H_{2} - COO H

Step 6:Glycolic acid has formula

C_{2} H_{4} O_{3}

, which is balanced versus the reactants (

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

); only option 3 shows both -OH on C

H_{2}

and -COOH on the same

C_{2}

skeleton.

C_{2} H_{2} O_{2} + H_{2} O \to HOCH_{2} COOH

Step 7:Y is glycolic acid, HOOC-C

H_{2}

-OH, matching Option 3.

y = H OC H_{2} - COO H

Final answer:

y = H OOC - C H_{2} - O H

(glycolic acid)

Q67Single correctAldehydes, Ketones and Carboxylic Acids

A molecule (x) with the following structure, under mild acid condition, is hydrolysed to produce (Y) and (Z). Identify the correct statement about (Y) and (Z).

(A)

A, B and C only

(B)

B and C only

(C)

C and D only

(D)

A and D only

SolutionAnswer: Option 4A and D only

Approach:

Hydrolyse the unsymmetrical bis-enol ether to two enols, tautomerise to a propanal (Y) and an acetone (Z), then test each statement A-D against carbonyl chemistry of an aldehyde versus a ketone.

Step 1:Ether (x) = (E)-C

H_{3}

CH=CH-O-C(C

H_{3}

)=C

H_{2}

(1-propenyl isopropenyl ether); identify (Y), (Z) and test statements A-D about identity, NaHC

O_{3}

, HCN and 2,4-DNP.

(x) = C H_{3} C H = C H - O - C (C H_{3}) = C H_{2}

Step 2:Protonate and hydrolyse the enol ether to give two enols.

(x) H_{3} O^{+} C H_{3} C H = CHOH + C H_{2} = C (O H) C H_{3}

Step 3:Tautomerise each enol to its carbonyl form.

C H_{3} C H = CHOH ⇌ C H_{3} C H_{2} CHO (Y); C H_{2} = C (O H) C H_{3} ⇌ C H_{3} COCH_{3} (Z)

Step 4:Evaluate statement A (molar mass) by computing M for Y and Z.

M (C_{3} H_{6} O) = 3 (12) + 6 (1) + 16 = 58 g mol^{- 1}

Step 5:Evaluate statement B (NaHC

O_{3}

). Neither aldehyde nor ketone bears -COOH; both fail to react with NaHC

O_{3}

RCHO + NaHCO_{3} \neq \to; R_{2} CO + NaHCO_{3} \neq \to

Step 6:Evaluate statement C (HCN rate). Aldehydes (less steric, more

δ^{+}

) react with HCN faster than ketones; propanal

>

acetone.

k_{H CN} (C H_{3} C H_{2} CHO) > k_{H CN} (C H_{3} COCH_{3})

Step 7:Evaluate statement D (2,4-DNP). Both aldehydes and ketones undergo nucleophilic addition with 2,4-DNP to give hydrazones.

RCHO + H_{2} NNHAr \to RCH = NNHAr; R_{2} CO + H_{2} NNHAr \to R_{2} C = NNHAr

Step 8:Collect true statements and back-substitute. True

= {A, D}

; this matches option 4.

True statements = {A, D}

Step 9:The correct statements are A and D only.

A and D only

Final answer: Option 4 - A and D only

Q68Single correctOrganic Compounds Containing Halogens

Identify compounds A and E in the following reaction sequence:

Reaction scheme: para-nitroethylbenzene (benzene ring with -C_2H_5 at C1 and -NO_2 at C4) treated successively with (i) Br_2/AlBr_3 to give A, (ii) Sn/HCl to give B, (iii) NaNO_2/HCl at 273-278 K to give C, (iv) C_2H_5OH to give D, (v) (i) KMnO_4/KOH and then (ii) H_3O^+ to give E.

(A)

Option 1 - A: benzene ring with -C_2H_5 at C1, -Br at C3 and -NO_2 at C4; E: benzene ring with -COOH at C1 and -Br at C3 (meta-bromobenzoic acid).

(B)

Option 2 - A: benzene ring with -C_2H_5 at C1, -Br at C2 and -NO_2 at C4; E: benzene ring with -COOH at C1 and -Br at C2 (ortho-bromobenzoic acid).

(C)

Option 3 - A: benzene ring with -CH_2CH_2Br side chain at C1 and -NO_2 at C4 (side-chain bromination); E: benzoic acid (benzene ring with -COOH at C1, no halogen).

(D)

Option 4 - A: benzene ring with -C_2H_5 at C1, -Br at C2 and -NO_2 at C4; E: benzene ring with -COOH at C1, -Br at C2 and -OH at C4 (2-bromo-4-hydroxybenzoic acid).

SolutionAnswer: Option 2A = 2-bromo-1-ethyl-4-nitrobenzene (Br ortho to -

C_{2}

H_{5}

, N

O_{2}

para to -

C_{2}

H_{5}

); E = 2-bromobenzoic acid (o-bromobenzoic acid)

Approach:

March through the five steps using directing effects, functional-group inter-conversion (Sn/HCl, diazonium, deamination by ethanol) and side-chain oxidation by KMn

O_{4}

to identify each intermediate; identify A after Step-1 and E after Step-5.

Step 1:Substrate is p-nitroethylbenzene (

C_{2}

H_{5}

C_{1}

, N

O_{2}

C_{4}

); target A = product of B

r_{2}

/AlB

r_{3}

; target E = product after Sn/HCl, NaN

O_{2}

/HCl(273-278 K),

C_{2}

H_{5}

OH, KMn

O_{4}

/KOH then

H_{3} O^{+}

p - O_{2} N - C_{6} H_{4} - C_{2} H_{5}

Step 2:In B

r_{2}

/AlB

r_{3}

, -

C_{2}

H_{5}

is o/p activator and -N

O_{2}

is m-deactivator. The para position to -

C_{2}

H_{5}

is blocked by -N

O_{2}

; both ortho positions to -

C_{2}

H_{5}

are also meta to -N

O_{2}

(mutually reinforcing). Br enters one ortho position to -

C_{2}

H_{5}

A = 2 -Br- 1 -ethyl- 4 -nitrobenzene

Step 3:Sn/HCl reduces the aromatic -N

O_{2}

to -N

H_{2}

(other groups unaffected).

A S n / H Cl B = 2 -Br- 1 -ethyl- 4 -aminobenzene

Step 4:NaN

O_{2}

/HCl at 273-278 K converts -N

H_{2}

to a stable diazonium salt -N

_{2}^{+} C l^{-}

B N a N O_{2} / H Cl, 273 - 278 K C = 2 -Br- 1 -ethyl- 4 -diazoniumchloride benzene

Step 5:

C_{2} H_{5}

OH reduces the diazonium group, replacing -N

_{2}^{+}

by -H (deamination), so the N

H_{2}

position becomes H.

C C_{2} H_{5} O H D = 1 -ethyl- 2 -bromobenzene (o -bromoethylbenzene)

Step 6:KMn

O_{4}

/KOH then

H_{3} O^{+}

oxidises the entire aromatic side-chain (-

C_{2} H_{5}

) to -COOH; ring -Br is unchanged.

D (i) K M n O_{4} / K O H, (ii) H_{3} O^{+} E = 2 -bromobenzoic acid

Step 7:A = 2-bromo-1-ethyl-4-nitrobenzene and E = o-bromobenzoic acid. Match to options A and E; only option 2 has both.

A, E match option 2

Step 8:The correct pair is option 2.

A = 2 -bromo- 1 -ethyl- 4 -nitrobenzene; E = o -bromobenzoic acid

Final answer: Option 2 - A = 2-bromo-1-ethyl-4-nitrobenzene; E = o-bromobenzoic acid

Q69Single correctBiomolecules

Identify the correct pair having amino acid (A) and the hormone (B) that is an iodinated derivative of the amino acid (A). (T and Y represent one letter code for amino acids)

(A)

A = T; B = Insulin

(B)

A = T; B = Thyroxine

(C)

A = Y; B = Thyroxine

(D)

A = Y; B = Insulin

SolutionAnswer: Option 3A = Y; B = Thyroxine

Approach:

Use the NCERT statement that thyroxine is the iodinated derivative of the amino acid tyrosine (one-letter code Y); insulin is a 51-residue polypeptide and is not an iodinated single amino acid.

Step 1:Pair (A, B) where B is the hormone formed by iodination of amino acid A; one-letter codes T (Threonine) and Y (Tyrosine) are used.

A \in {T, Y}, B \in {Insulin, Thyroxine}

Step 2:Identify the hormone that is an iodinated amino-acid derivative. Insulin is a 51-residue protein hormone (no iodine); thyroxine (

T_{4}

) is 3,5,3',5'-tetraiodo derivative of tyrosine.

Thyroxine = 3,5,3’,5’-tetraiodothyronine (iodinated tyrosine)

Step 3:Tyrosine carries one-letter code Y, so A = Y.

Tyrosine \equiv Y

Step 4:Only option 3 carries A = Y and B = Thyroxine simultaneously.

Option 3 : A = Y, B = Thyroxine

Step 5:The correct pair is A = Y (tyrosine), B = Thyroxine.

A = Y; B = Thyroxine

Final answer: Option 3 - A = Y (tyrosine); B = Thyroxine

Q70Single correctd- and f-Block Elements

Among

F e^{2 +}

F e^{3 +}

C r^{2 +}

and

Z n^{2 +}

, the ion that shows a positive borax bead test and has the highest ionization enthalpy is:

(A)

F e^{2 +}

(B)

Z n^{2 +}

(C)

C r^{2 +}

(D)

F e^{3 +}

SolutionAnswer: Option 4

F e^{3 +}

Approach:

Apply two criteria sequentially: (i) the borax bead test demands a coloured transition-metal ion with partly filled d-orbitals; (ii) among the surviving ions, the one whose formation needs the highest cumulative ionization enthalpy is the answer.

Step 1:Choose ion from

{F e^{2 +}, F e^{3 +}, C r^{2 +}, Z n^{2 +}}

that gives a positive borax bead test and has the highest ionization enthalpy.

Targets: borax-bead-positive AND highest IE

Step 2:Write electronic configurations.

F e^{2 +} : 3 d^{6}

;

F e^{3 +} : 3 d^{5}

;

C r^{2 +} : 3 d^{4}

;

Z n^{2 +} : 3 d^{10}

F e^{2 +} : 3 d^{6}, F e^{3 +} : 3 d^{5}, C r^{2 +} : 3 d^{4}, Z n^{2 +} : 3 d^{10}

Step 3:

Z n^{2 +}

3 d^{10}

(filled), so it is colourless and fails the borax bead test; eliminate it.

Z n^{2 +} : 3 d^{10} \Rightarrow no borax bead

Step 4:

F e^{2 +}

F e^{3 +}

and

C r^{2 +}

all have partly filled d-orbitals and give coloured borax beads (F

e^{2 +}

e^{3 +}

yellow/brown, C

r^{2 +}

green).

F e^{2 +}, F e^{3 +}, C r^{2 +} \Rightarrow positive borax bead

Step 5:Compare cumulative IE. To form F

e^{3 +}

requires I

E_{1}

E_{2}

E_{3}

of Fe; F

e^{2 +}

and C

r^{2 +}

require only I

E_{1}

E_{2}

. The third IE of Fe is large (removal from stable 3

d^{6}

to 3

d^{5}

shell), making

Σ IE (F e^{3 +})

the highest in the set.

Σ IE (F e^{3 +}) > Σ IE (F e^{2 +}), Σ IE (C r^{2 +})

Step 6:F

e^{3 +}

satisfies both filters (positive borax bead + highest IE), confirming option 4.

F e^{3 +} : positive borax bead, highest IE

Step 7:The required ion is F

e^{3 +}

F e^{3 +}

Final answer:

F e^{3 +}

- Option 4

Q71NumericalDual Nature of Matter and Radiation

The surface of sodium metal is irradiated with radiation of wavelength x nm. The kinetic energy of ejected electrons is

2.8 \times 1 0^{- 20}

J. The work function of sodium is

2.3

eV. The value of x is

\dots\dots \times 1 0^{2}

nm (Nearest integer). (Given:

h = 6.6 \times 1 0^{- 34}

Js;

1 eV = 1.6 \times 1 0^{- 19}

c = 3.0 \times 1 0^{8} ms^{- 1}

)

SolutionAnswer: 5

Approach:

Apply Einstein's photoelectric equation

h c / λ = ϕ + K E

, solve for

λ

in metres, convert to nm and then express in the requested form

x \times 1 0^{2}

nm.

Step 1:Given

ϕ = 2.3

eV,

K E = 2.8 \times 1 0^{- 20}

h = 6.6 \times 1 0^{- 34}

Js,

c = 3.0 \times 1 0^{8}

s^{- 1}

1 eV = 1.6 \times 1 0^{- 19}

J; find

λ

in the form

x \times 1 0^{2}

nm.

ϕ = 2.3 eV, K E = 2.8 \times 1 0^{- 20} J

Step 2:Convert the work function to joules.

ϕ = 2.3 \times 1.6 \times 1 0^{- 19} = 3.68 \times 1 0^{- 19} J

Step 3:Substitute in Einstein's equation.

\frac{( 6.6 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{λ} = 3.68 \times 1 0^{- 19} + 2.8 \times 1 0^{- 20}

Step 4:Add the right-hand-side energies on a common

1 0^{- 20}

basis.

3.68 \times 1 0^{- 19} + 2.8 \times 1 0^{- 20} = 36.8 \times 1 0^{- 20} + 2.8 \times 1 0^{- 20} = 39.6 \times 1 0^{- 20} J

Step 5:Solve for

λ

λ = \frac{( 6.6 \times 1 0 ^{- 34} ) ( 3.0 \times 1 0 ^{8} )}{39.6 \times 1 0 ^{- 20}} = \frac{19.8 \times 1 0 ^{- 26}}{39.6 \times 1 0 ^{- 20}} = 0.5 \times 1 0^{- 6} m

Step 6:Convert to nanometres.

λ = 5 \times 1 0^{- 7} m = 500 \times 1 0^{- 9} m = 500 nm

Step 7:Dimensional and back-substitution.

h c / λ = (1.98 \times 1 0^{- 25}) / (5 \times 1 0^{- 7}) = 3.96 \times 1 0^{- 19}

J, matching

ϕ + K E = 3.68 \times 1 0^{- 19} + 0.28 \times 1 0^{- 19} = 3.96 \times 1 0^{- 19}

\frac{h c}{λ} = 3.96 \times 1 0^{- 19} J = ϕ + K E

Step 8:The stem requests

λ

written as

x \times 1 0^{2}

nm. With

λ = 500

= 5 \times 1 0^{2}

nm,

x = 5

500 nm = 5 \times 1 0^{2} nm \Rightarrow x = 5

Final answer:

x = 5

(i.e.

λ = 5 \times 1 0^{2}

= 500

nm)

Q72NumericalChemical Kinetics

Consider the following gas phase reaction being carried out in a closed vessel at

2 5^{\circ}

2 A (g) ⟶ 4 B (g) + C (g)

time (min) | total pressure of the system (mm Hg) |
30 | 300 |

\infty

| 600 |
The pressure of C(g) at

30

minutes time interval would be

\dots\dots

mm Hg (nearest integer).

SolutionAnswer: 20

Approach:

Set up an ICE-like table of partial pressures; use the

t = \infty

condition to obtain the initial pressure

P_{0}

of A, then the

t = 30

min condition to obtain the extent x, hence

P_{C} (30) = x

Step 1:At

t = 0

only A is present with pressure

P_{0}

; let

2 x

be the pressure of A consumed at

t = 30

min. From stoichiometry,

4 x

pressure of B and x of C form. At

t = \infty

all A is consumed.

t = 0 t = 30 t = \infty 2 A P_{0} P_{0} - 2 x 0 4 B 0 4 x 2 P_{0} C 0 x P_{0} /2

Step 2:Apply total-pressure condition at

t = \infty

P_{t o t} = 0 + 2 P_{0} + P_{0} /2 = 5 P_{0} /2 = 600

\frac{5 P _{0}}{2} = 600 \Rightarrow P_{0} = 240 mm Hg

Step 3:Apply total-pressure condition at

t = 30

min:

(P_{0} - 2 x) + 4 x + x = P_{0} + 3 x = 300

240 + 3 x = 300 \Rightarrow 3 x = 60

Step 4:Identify

P_{C}

t = 30

min:

P_{C} = x = 20

mm Hg.

P_{C} (30) = x = 20 mm Hg

Step 5:Substitute back at

t = 30

P_{A} = 240 - 40 = 200

;

P_{B} = 80

;

P_{C} = 20

; sum

= 300

mm Hg, matching the data.

200 + 80 + 20 = 300 mm Hg

Step 6:The partial pressure of

C (g)

t = 30

min is

20

mm Hg.

P_{C} = 20 mm Hg

Final answer: 20 mm Hg

Q73NumericalElectrochemistry

Consider the following two half-cell reactions along with the standard reduction potential given:

SolutionAnswer: 560

Approach:

Compute

E_{ce ll}^{\circ}

, then maximum electrical work

∣Δ G^{\circ} ∣ = nFE_{ce ll}^{\circ}

for

n = 6

electrons per mole of C

H_{3}

OH; apply

80%

efficiency; equate available work to

∣ P_{e x t} Δ V ∣

for isothermal compression to find

Δ V

Step 1:Cathode =

O_{2}

reduction (higher

E_{re d}^{\circ} = 1.23

V); anode = C

O_{2}

H_{3}

OH (lower

E_{re d}^{\circ} = 0.02

V, written here as reduction but operating in oxidation direction);

n = 6

per mole of C

H_{3}

OH; efficiency

η = 0.80

;

P_{e x t} = 1

kPa

= 1 0^{3}

Pa;

F = 96500

C mo

l^{- 1}

E_{C}^{\circ} = 1.23 V, E_{A}^{\circ} = 0.02 V, n = 6, η = 0.80, P_{e x t} = 1 0^{3} Pa

Step 2:Compute the standard cell EMF.

E_{ce ll}^{\circ} = 1.23 - 0.02 = 1.21 V

Step 3:Compute

∣Δ G^{\circ} ∣

for 1 mol C

H_{3}

OH using

n = 6

∣Δ G^{\circ} ∣ = n F E_{ce ll}^{\circ} = 6 \times 96500 \times 1.21 = 700590 J

Step 4:Apply

80

% efficiency to get available work.

W_{a v} = 0.80 \times 700590 = 560472 J

Step 5:Equate available work to

∣ P_{e x t} Δ V ∣

and solve for

Δ V

560472 = (1 0^{3} Pa) ∣Δ V ∣ \Rightarrow ∣Δ V ∣ = 560.472 m^{3}

Step 6:Dimensional check

Pa \cdot m^{3} = J

1 0^{3} Pa \times 560.47 m^{3} = 5.6047 \times 1 0^{5}

= 560472

= W_{a v}

P_{e x t} Δ V = 1 0^{3} \times 560.47 = 5.6047 \times 1 0^{5} J

Step 7:Nearest-integer change in volume.

∣Δ V ∣ \approx 560 m^{3}

Final answer: 560

m^{3}

Q74Numericald- and f-Block Elements

Number of paramagnetic ions among the following d- and f-block metal ions is

\dots\dots

M n^{2 +}, C u^{2 +}, Z n^{2 +}, Y b^{2 +}, S c^{3 +}, L a^{3 +}, G d^{3 +}, L u^{3 +}, T i^{4 +}, C e^{4 +}

(Atomic number of Mn = 25, Cu = 29, Yb = 70, Sc = 21, La = 57, Gd = 64, Lu = 71, Ti = 22, Ce = 58)

SolutionAnswer: 3

Approach:

Write the d-/f-electronic configuration of each ion and count those with at least one unpaired electron; such ions are paramagnetic.

Step 1:Classify each ion of the list using its valence-shell d- or f-occupancy and count paramagnetic species.

Set = {M n^{2 +}, C u^{2 +}, Z n^{2 +}, Y b^{2 +}, S c^{3 +}, L a^{3 +}, G d^{3 +}, L u^{3 +}, T i^{4 +}, C e^{4 +}}

Step 2:D-block configurations.

M n^{2 +} (Z = 25 - 2) : 3 d^{5}

(5 unpaired, para);

C u^{2 +} (Z = 29 - 2) : 3 d^{9}

(1 unpaired, para);

Z n^{2 +} (Z = 30 - 2) : 3 d^{10}

(0 unpaired, dia);

S c^{3 +} (Z = 21 - 3) : 3 d^{0}

(dia);

T i^{4 +} (Z = 22 - 4) : 3 d^{0}

(dia).

M n^{2 +} : 3 d^{5} (p), C u^{2 +} : 3 d^{9} (p), Z n^{2 +} : 3 d^{10} (d), S c^{3 +} : 3 d^{0} (d), T i^{4 +} : 3 d^{0} (d)

Step 3:F-block configurations.

Y b^{2 +} (Z = 70 - 2) : 4 f^{14}

(dia);

L a^{3 +} (Z = 57 - 3) : 4 f^{0}

(dia);

G d^{3 +} (Z = 64 - 3) : 4 f^{7}

(7 unpaired, para);

L u^{3 +} (Z = 71 - 3) : 4 f^{14}

(dia);

C e^{4 +} (Z = 58 - 4) : 4 f^{0}

(dia).

Y b^{2 +} : 4 f^{14} (d), L a^{3 +} : 4 f^{0} (d), G d^{3 +} : 4 f^{7} (p), L u^{3 +} : 4 f^{14} (d), C e^{4 +} : 4 f^{0} (d)

Step 4:Collect all paramagnetic ions.

Para set = {M n^{2 +}, C u^{2 +}, G d^{3 +}}

Step 5:Cross-check unpaired-electron counts. M

n^{2 +}

^{-}

, C

u^{2 +}

^{-}

, G

d^{3 +}

^{-}

; all others 0 unpaired electrons. Count of paramagnetic ions

= 3

n_{p a r a} = 3

Step 6:Number of paramagnetic ions

= 3

3

Final answer: 3

Q75NumericalOrganic Compounds Containing Nitrogen

Consider the following reactions sequence

SolutionAnswer: 1

Approach:

Trace the four steps to identify product P, write its molecular formula, then use the Carius relation (one Br per molecule) to compute the mass of AgBr obtained from 1.0 g of P via the molar-mass ratio

M (AgBr) / M (P)

Step 1:Substrate p-nitrotoluene (-C

H_{3}

C_{1}

, -N

O_{2}

C_{4}

); four-step sequence (i) Sn/HCl, OH

^{-}

(ii) (C

H_{3} CO)_{2}

O (iii) B

r_{2}

/AlB

r_{3}

(iv)

H_{3} O^{+}

; target = mass of AgBr from 1.0 g of P.

p - O_{2} N - C_{6} H_{4} - C H_{3}

Step 2:Sn/HCl reduces -N

O_{2}

to -NH

_{3}^{+} C l^{-}

and OH

^{-}

liberates -N

H_{2}

, giving p-toluidine.

p - O_{2} N - C_{6} H_{4} - C H_{3} S n / H Cl; O H^{-} p - H_{2} N - C_{6} H_{4} - C H_{3}

Step 3:Acetic anhydride acetylates -N

H_{2}

to -NHCOC

H_{3}

, lowering its activating power and protecting it during electrophilic bromination.

p - H_{2} N - C_{6} H_{4} - C H_{3} (C H_{3} CO)_{2} O p - C H_{3} CON H - C_{6} H_{4} - C H_{3}

Step 4:With -NHCOC

H_{3}

(strong o/p-director) at

C_{1}

and -C

H_{3}

(weaker o/p-director) at

C_{4}

, the para position to -NHCOC

H_{3}

is blocked. Bromination occurs at the ortho position to the dominant director (-NHCOC

H_{3}

), giving the 2-bromo derivative.

Br enters C2 (ortho to -NHCOCH_{3})

Step 5:

H_{3} O^{+}

hydrolyses the amide back to the free amine, leaving the C-Br intact.

2 -Br- 4 - C H_{3} - C_{6} H_{3} - N H COC H_{3} H_{3} O^{+} P = 2 -bromo- 4 -methylaniline

Step 6:Compute molar masses.

M (P) = 7 (12) + 8 (1) + 80 + 14 = 84 + 8 + 80 + 14 = 186 g mol^{- 1}; M (AgBr) = 108 + 80 = 188 g mol^{- 1}

Step 7:Apply the Carius mass relation (one Br per molecule, so

n (AgBr) = n (P)

mass AgBr = \frac{188}{186} \times 1.0 = 1.0108 g

Step 8:Dimensional check via moles.

n (P) = 1.0/186 = 5.376 \times 1 0^{- 3}

mol

= n (AgBr)

; mass AgBr

= 5.376 \times 1 0^{- 3} \times 188 = 1.0108

g, matches.

n_{P} = n_{A g B r} = 5.376 \times 1 0^{- 3} mol \Rightarrow m_{A g B r} = 1.011 g

Step 9:Rounding

1.0108

g to the nearest integer gives

1

g of AgBr.

1 g

Final answer: 1 g

Mathematics25 questions

Q1Single correctComplex Numbers and Quadratic Equations

Let

α, β

be the roots of the equation

x^{2} - 3 x + r = 0

and

\frac{α}{2}, 2 β

be the roots of the equation

x^{2} + 3 x + r = 0

. If the roots of the equation

x^{2} + 6 x = m

are

2 α + β + 2 r

and

α - 2 β - \frac{r}{2}

, then m is equal to

(A)

- 135

(B)

- 567

(C)

135

(D)

567

SolutionAnswer: Option 4

567

Approach:

Apply Vieta's formulas to both quadratics to determine

α, β, r

, then construct the new quadratic from the prescribed roots and read off m.

Step 1:Given

α, β

are roots of

x^{2} - 3 x + r = 0

and

α /2, 2 β

are roots of

x^{2} + 3 x + r = 0

. The roots of

x^{2} + 6 x - m = 0

are

2 α + β + 2 r

and

α - 2 β - r /2

. Target: value of m.

α + β = 3, α β = r; \frac{α}{2} + 2 β = - 3, α β = r

Step 2:Solve the two linear equations

α + β = 3

and

α /2 + 2 β = - 3

. Multiply the second by

2

to get

α + 4 β = - 6

; subtract the first to obtain

3 β = - 9

3 β = - 9 \Rightarrow β = - 3, α = 6

Step 3:Compute

r

using the product relation.

r = α β = (6) (- 3) = - 18

Step 4:Evaluate the prescribed roots of

x^{2} + 6 x - m = 0

2 α + β + 2 r = 12 - 3 - 36 = - 27; α - 2 β - \frac{r}{2} = 6 + 6 + 9 = 21

Step 5:Apply product of roots to

x^{2} + 6 x - m = 0

where product equals

- m

(- 27) (21) = - m \Rightarrow - 567 = - m \Rightarrow m = 567

Step 6:Sum of roots gives

- 27 + 21 = - 6

which equals the coefficient

- 6

x^{2} + 6 x - m = 0

, confirming the pair are correct roots.

- 27 + 21 = - 6 ✓

Step 7:Hence

m = 567

, matching option (4).

m = 567

Final answer:

m = 567

Q2Single correctComplex Numbers and Quadratic Equations

Let the circles

C_{1} : ∣ z ∣ = r

and

C_{2} : ∣ z - 3 - 4 i ∣ = 5, z \in C

, be such that

C_{2}

lies within

C_{1}

. If

z_{1}

moves on

C_{1}

z_{2}

moves on

C_{2}

and

min ∣ z_{1} - z_{2} ∣ = 2

, then

max ∣ z_{1} - z_{2} ∣

is equal to

(A)

12

(B)

17

(C)

22

(D)

24

SolutionAnswer: Option 3

22

Approach:

Use that for circle

C_{2}

lying inside

C_{1}

, the extreme values of

∣ z_{1} - z_{2} ∣

occur along the line joining the centres, giving

min = r - d - R

and

max = r + d + R

where d is the distance between centres and R the radius of

C_{2}

Step 1:

C_{1}

is centred at

O_{1} = (0, 0)

with radius r;

C_{2}

is centred at

O_{2} = (3, 4)

with radius

R = 5

. Given

C_{2}

lies within

C_{1}

and

min ∣ z_{1} - z_{2} ∣ = 2

. Target:

max ∣ z_{1} - z_{2} ∣

O_{1} = (0, 0), O_{2} = (3, 4), R = 5

Step 2:Compute the distance between centres.

d = 3^{2} + 4^{2} = 25 = 5

Step 3:Apply the minimum-distance condition along the line joining the centres:

min ∣ z_{1} - z_{2} ∣ = r - (d + R)

r - (5 + 5) = 2 \Rightarrow r = 12

Step 4:Compute the maximum distance along the same line on the opposite side.

max ∣ z_{1} - z_{2} ∣ = r + d + R = 12 + 5 + 5 = 22

Step 5:

min + max = 2 r

for two such circles, giving

2 + 22 = 24 = 2 (12)

, which is consistent.

2 + 22 = 24 = 2 r

Step 6:The maximum distance equals

22

, matching option (3).

max ∣ z_{1} - z_{2} ∣ = 22

Final answer:

max ∣ z_{1} - z_{2} ∣ = 22

Q3Single correctMatrices and Determinants

If the system of equations

x + 5 y + 6 z = 4, 2 x + 3 y + 4 z = 7, x + 6 y + a z = b

has infinitely many solutions, then the point (a, b) lies on the line

(A)

y - x = 3

(B)

x - y = 3

(C)

x + y = 11

(D)

x + y = 12

SolutionAnswer: Option 2

x - y = 3

Approach:

Impose the consistency conditions for infinitely many solutions: the coefficient determinant and any one substitution determinant must vanish; solve them to obtain

a

and

b

, then check the candidate lines.

Step 1:Coefficient matrix

A = 121536 64 a

with RHS

(4, 7, b)

. Target: (a,b).

A x = (4, 7, b)^{T}

Step 2:Compute

Δ = det A

along the first row.

Δ = 1 \cdot (3 a - 24) - 5 \cdot (2 a - 4) + 6 \cdot (12 - 3) = 3 a - 24 - 10 a + 20 + 54 = - 7 a + 50

Step 3:Set

Δ = 0

to obtain a.

50 - 7 a = 0 \Rightarrow a = \frac{50}{7}

Step 4:Replace the third column by the RHS to form

Δ_{z} = det 121536 47 b

and expand.

Δ_{z} = 1 \cdot (3 b - 42) - 5 \cdot (2 b - 7) + 4 \cdot (12 - 3) = 3 b - 42 - 10 b + 35 + 36 = - 7 b + 29

Step 5:Set

Δ_{z} = 0

29 - 7 b = 0 \Rightarrow b = \frac{29}{7}

Step 6:Compute

a - b

to test the candidate lines.

a - b = \frac{50}{7} - \frac{29}{7} = \frac{21}{7} = 3

Step 7:Substitute

(a, b) = (50/7, 29/7)

into

x - y = 3

50/7 - 29/7 = 3

, identity holds. The other candidates fail.

\frac{50}{7} - \frac{29}{7} = 3 ✓

Step 8:The point

(a, b)

lies on

x - y = 3

, matching option (2).

x - y = 3

Final answer:

(a, b)

lies on

x - y = 3

Q4Single correctSequence and Series

Let

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

be an A.P. and

g_{1} = a_{1}, g_{2}, g_{3}, \dots

be an increasing G.P. If

a_{1} = a_{2} + g_{2} = 1

and

a_{3} + g_{3} = 4

, then

a_{10} + g_{5}

is equal to

(A)

81

(B)

76

(C)

62

(D)

55

SolutionAnswer: Option 4

55

Approach:

Parametrise the A.P. and G.P. with

a_{1} = g_{1} = 1

, common difference d, common ratio r; reduce the two given conditions to a quadratic in r, pick the increasing-GP root, then evaluate

a_{10} + g_{5}

Step 1:A.P. has

a_{1} = 1

and common difference d, so

a_{n} = 1 + (n - 1) d

. G.P. has

g_{1} = 1

and common ratio

r > 0

with

r > 1

for an increasing G.P., so

g_{n} = r^{n - 1}

. Target:

a_{10} + g_{5}

a_{n} = 1 + (n - 1) d, g_{n} = r^{n - 1}

Step 2:Use

a_{2} + g_{2} = 1

with

a_{2} = 1 + d

and

g_{2} = r

(1 + d) + r = 1 \Rightarrow d = - r

Step 3:Use

a_{3} + g_{3} = 4

with

a_{3} = 1 + 2 d

and

g_{3} = r^{2}

, and substitute

d = - r

1 + 2 d + r^{2} = 4 \Rightarrow 1 - 2 r + r^{2} = 4 \Rightarrow r^{2} - 2 r - 3 = 0

Step 4:Factorise the quadratic.

(r - 3) (r + 1) = 0 \Rightarrow r = 3 or r = - 1

Step 5:For an increasing G.P. starting at

1

require

r > 1

, so

r = 3

and hence

d = - 3

r = 3, d = - 3

Step 6:Compute

a_{10}

and

g_{5}

and add them.

a_{10} = 1 + 9 (- 3) = - 26; g_{5} = 3^{4} = 81; a_{10} + g_{5} = - 26 + 81 = 55

Step 7:Substitute back into the given condition

a_{3} + g_{3} = (1 + 2 (- 3)) + 3^{2} = - 5 + 9 = 4

, matching the problem statement.

a_{3} + g_{3} = - 5 + 9 = 4 ✓

Step 8:Hence

a_{10} + g_{5} = 55

, matching option (4).

a_{10} + g_{5} = 55

Final answer:

a_{10} + g_{5} = 55

Q5Single correctSequence and Series

The sum

\frac{1 ^{3}}{1} + \frac{1 ^{3} + 2 ^{3}}{1 + 3} + \frac{1 ^{3} + 2 ^{3} + 3 ^{3}}{1 + 3 + 5} + \dots

up to 8 terms, is:

(A)

70

(B)

71

(C)

72

(D)

73

SolutionAnswer: Option 2

71

Approach:

Simplify the general term using the closed forms for the sum of cubes and the sum of the first

n

odd numbers, then sum to

n = 8

Step 1:The

n^{t h}

term is

t_{n} = \frac{1 ^{3} + 2 ^{3} + \dots + n ^{3}}{1 + 3 + 5 + \dots + ( 2 n - 1 )}

. Target:

S_{8} = \sum_{n = 1}^{8} t_{n}

t_{n} = \frac{\sum _{k = 1}^{n} k ^{3}}{\sum _{k = 1}^{n} ( 2 k - 1 )}

Step 2:Apply the two closed forms.

t_{n} = \frac{[ n ( n + 1 ) /2 ] ^{2}}{n ^{2}} = \frac{( n + 1 ) ^{2}}{4} = \frac{n ^{2} + 2 n + 1}{4}

Step 3:Sum to

n

terms by linearity.

S_{n} = \frac{1}{4} (\sum_{k = 1}^{n} k^{2} + 2 \sum_{k = 1}^{n} k + \sum_{k = 1}^{n} 1) = \frac{1}{4} (\frac{n ( n + 1 ) ( 2 n + 1 )}{6} + n (n + 1) + n)

Step 4:Substitute

n = 8

. Here

\sum_{k = 1}^{8} k^{2} = 204, \sum_{k = 1}^{8} k = 36

S_{8} = \frac{1}{4} (204 + 2 \cdot 36 + 8) = \frac{1}{4} (204 + 72 + 8) = \frac{284}{4} = 71

Step 5:Direct check of

t_{n}

values

1, 9/4, 4, 25/4, 9, 49/4, 16, 81/4

sums to

(4 + 9 + 16 + 25 + 36 + 49 + 64 + 81) /4 = 284/4 = 71

\sum t_{n} = \frac{284}{4} = 71 ✓

Step 6:

S_{8} = 71

, matching option (2).

S_{8} = 71

Final answer:

S_{8} = 71

Q6Single correctBinomial Theorem

If for

3 \leq r \leq 30

(30 - r 30) + 3 (31 - r 30) + 3 (32 - r 30) + (33 - r 30) = (r m)

, then m equals:

(A)

31

(B)

32

(C)

33

(D)

34

SolutionAnswer: Option 3

33

Approach:

Use the symmetry

(30 - k 30) = (k 30)

to rewrite each term as

(r - j 30)

for

j = 0, 1, 2, 3

; the coefficients

1, 3, 3, 1

identify the sum as the Vandermonde convolution of

(1 + x)^{3}

with

(1 + x)^{30}

Step 1:Given

L = (30 - r 30) + 3 (31 - r 30) + 3 (32 - r 30) + (33 - r 30)

for

3 \leq r \leq 30

. Target: integer m with

L = (r m)

L = \sum_{j = 0}^{3} (j 3) (30 + j - r 30)

Step 2:Apply symmetry

(30 + j - r 30) = (r - j 30)

to each term.

(30 - r 30) = (r 30), (31 - r 30) = (r - 1 30), (32 - r 30) = (r - 2 30), (33 - r 30) = (r - 3 30)

Step 3:Combine with coefficients

1, 3, 3, 1 = (0 3), (1 3), (2 3), (3 3)

L = \sum_{j = 0}^{3} (j 3) (r - j 30)

Step 4:Apply Vandermonde's identity with

p = 3, q = 30

\sum_{j = 0}^{3} (j 3) (r - j 30) = (r 33)

Step 5:Check at

r = 3

: LHS

= (27 30) + 3 (28 30) + 3 (29 30) + (30 30) = 4060 + 3 \cdot 435 + 3 \cdot 30 + 1 = 4060 + 1305 + 90 + 1 = 5456 = (3 33)

, since

(3 33) = 33 \cdot 32 \cdot 31/6 = 5456

5456 = (3 33) ✓

Step 6:Matching

(r m) = (r 33)

gives

m = 33

, option (3).

m = 33

Final answer:

m = 33

Q7Single correctPermutations and Combinations

Let

P_{n}

denote the total number of triangles formed by joining the vertices of an n-sided regular polygon. If

P_{n + 1} - P_{n} = 66

, then the sum of all distinct prime divisors of n is

(A)

7

(B)

8

(C)

5

(D)

6

SolutionAnswer: Option 3

5

Approach:

Express

P_{n} = (3 n)

and use Pascal's identity

(3 n + 1) - (3 n) = (2 n)

to solve for n, then list the distinct prime divisors.

Step 1:

P_{n} = (3 n)

counts triangles from n polygon vertices. Given

P_{n + 1} - P_{n} = 66

. Target: sum of distinct prime divisors of n.

P_{n} = (3 n)

Step 2:Apply Pascal's identity with

k = 3

(3 n + 1) - (3 n) = (2 n)

Step 3:Set the difference equal to

66

and solve.

\frac{n ( n - 1 )}{2} = 66 \Rightarrow n (n - 1) = 132 \Rightarrow n^{2} - n - 132 = 0 \Rightarrow n = 12

(positive root)

Step 4:Factorise

n = 12 = 2^{2} \cdot 3

; distinct primes are

{2, 3}

12 = 2^{2} \cdot 3

Step 5:Sum the distinct primes.

2 + 3 = 5

Step 6:Check

P_{13} - P_{12} = (3 13) - (3 12) = 286 - 220 = 66

, matching the given condition.

286 - 220 = 66 ✓

Step 7:The sum of all distinct prime divisors of

n = 12

5

, matching option (3).

2 + 3 = 5

Final answer: Sum of distinct prime divisors of

n

5

Q8Single correctStatistics and Probability

A man throws a fair coin repeatedly. He gets 10 points for each head he throws and 5 points for each tail he throws. If the probability that he gets exactly 30 points is

\frac{m}{n}, g cd (m, n) = 1

, then

m + n

is equal to

(A)

53

(B)

55

(C)

107

(D)

105

SolutionAnswer: Option 3

107

Approach:

Find all (h,t) pairs with

10 h + 5 t = 30

, count the orderings of each composition, weight by

(1/2)^{h + t}

, and add. Since each ordering corresponds to a distinct sample path that achieves exactly

30

as a partial sum, this gives the required probability.

Step 1:Let h = number of heads, t = number of tails so far. Each head adds

10

, each tail adds 5, so a total of

30

requires

10 h + 5 t = 30

, i.e.

2 h + t = 6

. Target:

P (total reaches 30)

2 h + t = 6

Step 2:Enumerate non-negative integer solutions

(h, t)

(h, t) \in {(0, 6), (1, 4), (2, 2), (3, 0)}

Step 3:For each (h,t) count the orderings ending with the toss that brings the total to

30

. Since every ordering of h heads and t tails attains the total exactly at the final toss of that block, the probability contribution is

(h h + t) (1/2)^{h + t}

P_{(h, t)} = (h h + t) (1/2)^{h + t}

Step 4:Sum over the four pairs.

P = (0 6) \frac{1}{2 ^{6}} + (1 5) \frac{1}{2 ^{5}} + (2 4) \frac{1}{2 ^{4}} + (3 3) \frac{1}{2 ^{3}} = \frac{1}{64} + \frac{5}{32} + \frac{6}{16} + \frac{1}{8}

Step 5:Common denominator

64

gives

1 + 10 + 24 + 8 = 43

P = \frac{1 + 10 + 24 + 8}{64} = \frac{43}{64}

Step 6:Since

43

is prime and does not divide

64 = 2^{6}

g cd (43, 64) = 1

, so

m = 43, n = 64

m = 43, n = 64

Step 7:Cross-check via recursion

q_{k} = \frac{1}{2} q_{k - 5} + \frac{1}{2} q_{k - 10}

with

q_{0} = 1

q_{k} = 0

for

k < 0

q_{5} = 1/2, q_{10} = 3/4, q_{15} = 5/8, q_{20} = 11/16, q_{25} = 21/32, q_{30} = \frac{1}{2} (21/32) + \frac{1}{2} (11/16) = 21/64 + 22/64 = 43/64

, matching.

q_{30} = \frac{43}{64} ✓

Step 8:Hence

m + n = 43 + 64 = 107

, matching option (3).

m + n = 107

Final answer:

m + n = 107

Q9Single correctStatistics and Probability

The mean and variance of

n

observations are 8 and 16, respectively. If the sum of the first

(n - 1)

observations is 48 and the sum of squares of the first

(n - 1)

observations is 496, then the value of

n

(A)

21

(B)

16

(C)

13

(D)

7

SolutionAnswer: Option 4

7

Approach:

Translate mean and variance into sums, isolate

x_{n}

, then reduce to a quadratic in n.

Step 1:Let the n observations be

x_{1}, x_{2}, \dots, x_{n}

with

\sum_{i = 1}^{n - 1} x_{i} = 48

and

\sum_{i = 1}^{n - 1} x_{i}^{2} = 496

\overset{x}{ˉ} = 8, σ^{2} = 16

Step 2:Mean gives

\sum x_{i} = 8 n

, so

x_{n} = 8 n - 48

x_{n} = 8 n - 48 = 8 (n - 6)

Step 3:Variance gives

\sum x_{i}^{2} = n (σ^{2} + \overset{x}{ˉ}^{2}) = n (16 + 64) = 80 n

and

\sum x_{i}^{2} = 496 + x_{n}^{2}

80 n = 496 + 64 (n - 6)^{2}

Step 4:Multiply out and simplify the quadratic.

64 n^{2} - 848 n + 2800 = 0 \Rightarrow 4 n^{2} - 53 n + 175 = 0

Step 5:Solve using the quadratic formula; discriminant

= 5 3^{2} - 4 \cdot 4 \cdot 175 = 2809 - 2800 = 9

n = \frac{53 \pm 3}{8} = 7 or \frac{25}{4}

Step 6:The valid number of observations is

n = 7

n = 7

Final answer:

n = 7

Q10Single correctCoordinate Geometry

Let a circle pass through the origin and its centre be the point of intersection of two mutually perpendicular lines

x + (k - 1) y + 3 = 0

and

2 x + k^{2} y - 4 = 0

. If the line

x - y + 2 = 0

intersects the circle at the points A and B, then

(A B)^{2}

is equal to

(A)

10

(B)

27

(C)

18

(D)

34

SolutionAnswer: Option 3

18

Approach:

Use perpendicularity of the two lines to fix

k

, find their intersection (the centre), compute the radius from the origin, and apply the chord-length formula.

Step 1:Slope of

x + (k - 1) y + 3 = 0

m_{1} = - \frac{1}{k - 1}

and slope of

2 x + k^{2} y - 4 = 0

m_{2} = - \frac{2}{k ^{2}}

m_{1} m_{2} = \frac{2}{k ^{2} ( k - 1 )}

Step 2:Set

m_{1} m_{2} = - 1

, giving

k^{2} (k - 1) = - 2

; testing

k = - 1

yields

1 \cdot (- 2) = - 2

, so

k = - 1

k = - 1

Step 3:Substitute

k = - 1

: the lines become

x - 2 y + 3 = 0

and

2 x + y - 4 = 0

. Solve to get the centre.

y = 4 - 2 x, x - 2 (4 - 2 x) + 3 = 0 \Rightarrow x = 1, y = 2

Step 4:Circle passes through origin, so radius equals

∣ OC ∣

r = 1^{2} + 2^{2} = 5

Step 5:Perpendicular distance from

C (1, 2)

to chord

x - y + 2 = 0

d = \frac{∣1 - 2 + 2∣}{2} = \frac{1}{2}

Step 6:Apply chord-length formula.

A B^{2} = 4 (r^{2} - d^{2}) = 4 (5 - \frac{1}{2}) = 18

Step 7:The square of chord length is

18

(A B)^{2} = 18

Final answer:

(A B)^{2} = 18

Q11Single correctCo-ordinate Geometry

Let O be the origin, and P and Q be two points on the rectangular hyperbola

x y = 12

such that the mid point of the line segment PQ is

(\frac{1}{2}, - \frac{1}{2})

. Then the area of the triangle OPQ equals:

(A)

\frac{3}{2}

(B)

\frac{5}{2}

(C)

\frac{7}{2}

(D)

\frac{9}{2}

SolutionAnswer: Option 3

\frac{7}{2}

Approach:

Use the mid-point chord identity

T = S_{1}

to obtain the chord line, find its intersections with

x y = 12

, and compute the triangle area.

Step 1:Midpoint is

(h, k) = (\frac{1}{2}, - \frac{1}{2})

on the hyperbola

x y = 12

x k + y h = 2 hk

Step 2:Plug in to get

- \frac{x}{2} + \frac{y}{2} = 2 \cdot \frac{1}{2} \cdot (- \frac{1}{2}) = - \frac{1}{2}

, i.e.

x - y = 1

x - y = 1

Step 3:Substitute

x = y + 1

into

x y = 12

(y + 1) y = 12 \Rightarrow y^{2} + y - 12 = 0

Step 4:Compute intersection points

P

and

Q

P = (4, 3), Q = (- 3, - 4)

Step 5:Midpoint of P and Q is

(\frac{4 - 3}{2}, \frac{3 - 4}{2}) = (\frac{1}{2}, - \frac{1}{2})

M = (\frac{1}{2}, - \frac{1}{2})

Step 6:Area of

△ OPQ

using the origin-vertex formula.

Area = \frac{1}{2} ∣4 (- 4) - (- 3) (3) ∣ = \frac{1}{2} ∣ - 16 + 9∣ = \frac{7}{2}

Final answer:

\frac{7}{2}

Q12Single correctCo-ordinate Geometry

Let the parabola

y = x^{2} + p x + q

passing through the point

(1, - 1)

be such that the distance between its vertex and the x-axis is minimum. Then the value of

p^{2} + q^{2}

is:

(A)

2

(B)

4

(C)

5

(D)

8

SolutionAnswer: Option 2

4

Approach:

Use the point on the parabola to relate

p

and

q

, reduce vertex distance to a single-variable function in

p

, and minimise.

Step 1:Parabola

y = x^{2} + p x + q

passes through

(1, - 1)

- 1 = 1 + p + q

Step 2:For

a = 1

b = p

c = q

, vertex y-coordinate is

\frac{4 q - p ^{2}}{4}

. Substitute

q = - 2 - p

y_{v} = \frac{4 ( - 2 - p ) - p ^{2}}{4} = - \frac{p ^{2} + 4 p + 8}{4}

Step 3:Complete the square:

p^{2} + 4 p + 8 = (p + 2)^{2} + 4

y_{v} = - \frac{( p + 2 ) ^{2} + 4}{4}

Step 4:Distance from x-axis is

∣ y_{v} ∣ = \frac{( p + 2 ) ^{2} + 4}{4}

d (p) = \frac{( p + 2 ) ^{2} + 4}{4}

Step 5:Minimise d(p):

(p + 2)^{2} \geq 0

with equality at

p = - 2

, giving

d_{m i n} = 1

and

q = - 2 - p = 0

p = - 2, q = 0

Step 6:Compute

p^{2} + q^{2}

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

Final answer:

4

Q13Single correctTrigonometry

Let

P = {θ \in [0, 4 π] : tan^{2} θ \neq = 1}

and

S = {a \in Z : 2 (cos^{8} θ - sin^{8} θ) sec 2 θ = a^{2}, θ \in P}

. Then n(S) is:

(A)

0

(B)

1

(C)

2

(D)

3

SolutionAnswer: Option 1

0

Approach:

Simplify

2 (cos^{8} θ - sin^{8} θ) sec 2 θ

to a function of

sin 2 θ

, then determine which integers are attainable subject to the domain restriction.

Step 1:Set

E (θ) = 2 (cos^{8} θ - sin^{8} θ) sec 2 θ

E (θ) = 2 (cos^{8} θ - sin^{8} θ) sec 2 θ

Step 2:Factor

cos^{8} θ - sin^{8} θ = (cos^{4} θ + sin^{4} θ) (cos^{2} θ + sin^{2} θ) (cos^{2} θ - sin^{2} θ) = (cos^{4} θ + sin^{4} θ) cos 2 θ

cos^{8} θ - sin^{8} θ = (cos^{4} θ + sin^{4} θ) cos 2 θ

Step 3:Multiply by

2 sec 2 θ

to cancel

cos 2 θ

E (θ) = 2 (cos^{4} θ + sin^{4} θ) = 2 - sin^{2} 2 θ

Step 4:Domain restriction

tan^{2} θ \neq = 1

excludes

θ = \frac{π}{4} + \frac{nπ}{2}

, where

sin^{2} 2 θ = 1

. Together with

sec 2 θ

defined (i.e.

cos 2 θ \neq = 0

, also

sin^{2} 2 θ \neq = 1

), the achievable range is

sin^{2} 2 θ \in [0, 1)

sin^{2} 2 θ \in [0, 1)

Step 5:Therefore

E (θ) = a^{2} \in (1, 2]

1 < a^{2} \leq 2

Step 6:The only integer satisfying

1 < a^{2} \leq 2

would need

a^{2} = 2

, but

2 \in / Z

, so no integer a works.

a^{2} = 2 \Rightarrow a \in / Z

Step 7:The set

S

is empty, so

n (S) = 0

n (S) = 0

Final answer:

0

Q14Single correctVector Algebra

Let the vectors

a = - \hat{i} + \hat{j} + 3 \hat{k}

and

b = \hat{i} + 3 \hat{j} + \hat{k}

. For some

λ, μ \in R

, let

c = λ a + μ b

. If

c \cdot (3 \hat{i} - 6 \hat{j} + 2 \hat{k}) = 10

and

c \cdot (\hat{i} + \hat{j} + \hat{k}) = - 2

, then

∣ c ∣^{2}

is equal to:

(A)

8

(B)

12

(C)

14

(D)

15

SolutionAnswer: Option 2

12

Approach:

Convert the two dot-product conditions into a linear system in

λ, μ

via linearity, solve, then compute

∣ c ∣^{2}

Step 1:Take

u = 3 \hat{i} - 6 \hat{j} + 2 \hat{k}

and

v = \hat{i} + \hat{j} + \hat{k}

c \cdot u = 10, c \cdot v = - 2

Step 2:Compute

a \cdot u = (- 1) (3) + (1) (- 6) + (3) (2) = - 3

and

b \cdot u = (1) (3) + (3) (- 6) + (1) (2) = - 13

- 3 λ - 13 μ = 10

Step 3:Compute

a \cdot v = - 1 + 1 + 3 = 3

and

b \cdot v = 1 + 3 + 1 = 5

3 λ + 5 μ = - 2

Step 4:Add both equations to eliminate

λ

- 8 μ = 8 \Rightarrow μ = - 1

, then

3 λ - 5 = - 2 \Rightarrow λ = 1

λ = 1, μ = - 1

Step 5:Form

c = a - b

c = (- 1 - 1, 1 - 3, 3 - 1) = - 2 \hat{i} - 2 \hat{j} + 2 \hat{k}

Step 6:Check

c \cdot u = - 6 + 12 + 4 = 10

and

c \cdot v = - 2 - 2 + 2 = - 2

, both consistent.

c \cdot u = 10, c \cdot v = - 2

Step 7:Magnitude squared.

∣ c ∣^{2} = 4 + 4 + 4 = 12

Final answer:

12

Q15Single correctThree Dimensional Geometry

Let the point A be the foot of perpendicular drawn from the point

P (a, b, 0)

on the line

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

. If the mid point of the line segment PA is

(0, \frac{3}{4}, - \frac{1}{4})

, then the value of

a^{2} + b^{2} + α^{2}

is equal to:

(A)

1

(B)

2

(C)

6

(D)

9

SolutionAnswer: Option 1

1

Approach:

Parameterise A on the line, apply the midpoint formula to express

a, b, α

in terms of the parameter r, then enforce

P A \cdot d = 0

to solve.

Step 1:Let

A = (1 + 2 r, 2 + r, α + 3 r)

on the line; direction

d = (2, 1, 3)

and

P = (a, b, 0)

A = (1 + 2 r, 2 + r, α + 3 r)

Step 2:Midpoint conditions give

a + 1 + 2 r = 0

b + 2 + r = \frac{3}{2}

α + 3 r = - \frac{1}{2}

a = - 1 - 2 r, b = - \frac{1}{2} - r, α = - \frac{1}{2} - 3 r

Step 3:Compute

P A = (1 + 2 r - a, 2 + r - b, α + 3 r)

. Substitute the relations:

1 + 2 r - a = 2 + 4 r

2 + r - b = \frac{5}{2} + 2 r

α + 3 r = - \frac{1}{2}

P A = (2 + 4 r, \frac{5}{2} + 2 r, - \frac{1}{2})

Step 4:Enforce

P A \cdot (2, 1, 3) = 0

2 (2 + 4 r) + (\frac{5}{2} + 2 r) + 3 (- \frac{1}{2}) = 0

4 + 8 r + \frac{5}{2} + 2 r - \frac{3}{2} = 0 \Rightarrow 10 r + 5 = 0

Step 5:Substitute

r = - \frac{1}{2}

a = - 1 + 1 = 0

b = - \frac{1}{2} + \frac{1}{2} = 0

α = - \frac{1}{2} + \frac{3}{2} = 1

a = 0, b = 0, α = 1

Step 6:

A = (1 - 1, 2 - \frac{1}{2}, 1 - \frac{3}{2}) = (0, \frac{3}{2}, - \frac{1}{2})

; midpoint of

P (0, 0, 0)

and A is

(0, \frac{3}{4}, - \frac{1}{4})

, matching.

M = (0, \frac{3}{4}, - \frac{1}{4})

Step 7:Compute the required sum.

a^{2} + b^{2} + α^{2} = 0 + 0 + 1 = 1

Final answer:

1

Q16Single correctVector Algebra

Two adjacent sides of a parallelogram PQRS are given by

PQ = \hat{j} + \hat{k}

and

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

. If the side PS is rotated about the point P by an acute angle

α

in the plane of the parallelogram so that it becomes perpendicular to the side PQ, then

sin^{2} (\frac{5 α}{2}) - sin^{2} (\frac{α}{2})

is equal to:

(A)

\frac{1}{2}

(B)

\frac{3}{2}

(C)

\frac{3}{4}

(D)

\frac{2 3}{5}

SolutionAnswer: Option 2

\frac{3}{2}

Approach:

Find the angle between

PQ

and

PS

using the dot product, determine the rotation

α

that makes them perpendicular, then evaluate the trig expression via

sin^{2} A - sin^{2} B = sin (A + B) sin (A - B)

Step 1:

PQ = (0, 1, 1), PS = (1, - 1, 0)

; magnitudes

∣ PQ ∣ = ∣ PS ∣ = 2

u \cdot v = 0 \cdot 1 + 1 \cdot (- 1) + 1 \cdot 0 = - 1

Step 2:Cosine of angle

θ

between PQ and PS.

cos θ = \frac{- 1}{2 \cdot 2} = - \frac{1}{2}

Step 3:Rotate PS by acute

α

to make angle

9 0^{\circ}

; reducing the obtuse angle of

12 0^{\circ}

9 0^{\circ}

requires

α = 3 0^{\circ}

α = 12 0^{\circ} - 9 0^{\circ} = 3 0^{\circ}

Step 4:Apply

sin^{2} A - sin^{2} B = sin (A + B) sin (A - B)

with

A = \frac{5 α}{2}

B = \frac{α}{2}

, so

A + B = 3 α

and

A - B = 2 α

sin^{2} \frac{5 α}{2} - sin^{2} \frac{α}{2} = sin (3 α) sin (2 α)

Step 5:Evaluate at

α = 3 0^{\circ}

sin 9 0^{\circ} \cdot sin 6 0^{\circ} = 1 \cdot \frac{3}{2}

sin 9 0^{\circ} \cdot sin 6 0^{\circ} = \frac{3}{2}

Step 6:The required value is

\frac{3}{2}

\frac{3}{2}

Final answer:

\frac{3}{2}

Q17Single correctIntegral Calculus

The value of

\int_{0}^{20 π} (sin^{4} x + cos^{4} x) d x

is equal to:

(A)

\frac{15 π}{2}

(B)

25 π

(C)

15 π

(D)

\frac{25 π}{2}

SolutionAnswer: Option 3

15 π

Approach:

Use the period of

sin^{4} x + cos^{4} x

(which is

π /2

) to reduce the long integral, then simplify the integrand via a power-reduction identity and integrate.

Step 1:Let

f (x) = sin^{4} x + cos^{4} x

. It has period

\frac{π}{2}

, and

20 π = 40 \cdot \frac{π}{2}

I = \int_{0}^{20 π} f (x) d x = 40 \int_{0}^{π /2} f (x) d x

Step 2:Simplify

f (x) = (sin^{2} x + cos^{2} x)^{2} - 2 sin^{2} x cos^{2} x = 1 - \frac{sin ^{2} 2 x}{2}

f (x) = 1 - \frac{sin ^{2} 2 x}{2}

Step 3:Evaluate

\int_{0}^{π /2} 1 d x = \frac{π}{2}

and

\int_{0}^{π /2} sin^{2} 2 x d x = \frac{π}{4}

\int_{0}^{π /2} f (x) d x = \frac{π}{2} - \frac{1}{2} \cdot \frac{π}{4} = \frac{3 π}{8}

Step 4:Multiply by the number of periods.

I = 40 \cdot \frac{3 π}{8} = 15 π

Step 5:Total integral equals

15 π

\int_{0}^{20 π} (sin^{4} x + cos^{4} x) d x = 15 π

Final answer:

15 π

Q18Single correctLimit, Continuity and Differentiability

Let f(x) be a polynomial of degree

5

and have extrema at

x = 1

and

x = - 1

. If

x \to 0 lim (\frac{f ( x )}{x ^{3}}) = - 5

, then

f (2) - f (- 2)

is equal to:

(A)

0

(B)

50

(C)

92

(D)

112

SolutionAnswer: Option 4

112

Approach:

Use the finite-limit condition to fix the low-order coefficients of the degree-5 polynomial, then impose the extrema conditions to solve for the remaining coefficients and compute the required difference.

Step 1:From

x \to 0 lim \frac{f ( x )}{x ^{3}} = - 5

the coefficients of

x^{0}, x^{1}, x^{2}

in f vanish and the coefficient of

x^{3}

equals

- 5

, so the degree-5 polynomial has the form below.

f (x) = a x^{5} + b x^{4} - 5 x^{3}

Step 2:Differentiate and impose extrema conditions

f^{'} (1) = 0

and

f^{'} (- 1) = 0

f^{'} (x) = 5 a x^{4} + 4 b x^{3} - 15 x^{2} \Rightarrow 5 a + 4 b - 15 = 0, 5 a - 4 b - 15 = 0

Step 3:Solve the simultaneous equations.

Adding: 10 a = 30 \Rightarrow a = 3; Subtracting: 8 b = 0 \Rightarrow b = 0

Step 4:Evaluate f at

x = \pm 2

using the fact that f is an odd function.

f (2) = 3 (32) - 5 (8) = 96 - 40 = 56; f (- 2) = - f (2) = - 56

Step 5:Compute the required difference.

f (2) - f (- 2) = 56 - (- 56) = 112

Final answer:

112

Q19Single correctIntegral Calculus

Let

f (x) = \int (\frac{16 x + 24}{x ^{2} + 2 x - 15}) d x

. If

f (4) = 14 lo g_{e} (3)

and

f (7) = lo g_{e} (2^{α} \cdot 3^{β}), α, β \in N

, then

α + β

is equal to:

(A)

31

(B)

37

(C)

39

(D)

41

SolutionAnswer: Option 3

39

Approach:

Split the numerator into the derivative of the denominator plus a constant, integrate via a log term and partial fractions, fix the constant from

f (4)

, then evaluate

f (7)

and identify

α, β

Step 1:Write the numerator as

A (2 x + 2) + B

where

2 x + 2

is the derivative of

x^{2} + 2 x - 15

. Matching coefficients gives

A = 8, B = 8

16 x + 24 = 8 (2 x + 2) + 8

Step 2:Split the integral into the log-derivative piece and the constant-numerator piece, then use partial fractions on the second piece.

f (x) = 8 ln ∣ x^{2} + 2 x - 15∣ + 8 \cdot \frac{1}{8} ln \frac{x - 3}{x + 5} + C = 8 ln ∣ x^{2} + 2 x - 15∣ + ln \frac{x - 3}{x + 5} + C

Step 3:Use

f (4) = 14 ln 3

. At

x = 4

x^{2} + 2 x - 15 = 9

and

(x - 3) / (x + 5) = 1/9

, giving

C = 0

8 ln 9 + ln (\frac{1}{9}) + C = 16 ln 3 - 2 ln 3 + C = 14 ln 3 + C = 14 ln 3 \Rightarrow C = 0

Step 4:Evaluate

f (7)

. At

x = 7

x^{2} + 2 x - 15 = 48 = 2^{4} \cdot 3

and

(x - 3) / (x + 5) = 4/12 = 1/3

f (7) = 8 ln (2^{4} \cdot 3) + ln (\frac{1}{3}) = 32 ln 2 + 8 ln 3 - ln 3 = 32 ln 2 + 7 ln 3 = ln (2^{32} \cdot 3^{7})

Step 5:Add

α

and

β

α + β = 32 + 7 = 39

Final answer:

39

Q20Single correctDifferential Equations

Let

x = x (y)

be the solution of the differential equation

2 y^{2} \frac{d x}{d y} - 2 x y + x^{2} = 0, y > 1, x (e) = e

. Then

x (e^{2})

is equal to:

(A)

\frac{3}{2} e^{2}

(B)

\frac{2}{3} e^{2}

(C)

e^{2}

(D)

2 e^{2}

SolutionAnswer: Option 2

\frac{2}{3} e^{2}

Approach:

Recognise the Bernoulli form in x(y), substitute

t = 1/ x

to linearise, solve via integrating factor, fix the constant from

x (e) = e

, then evaluate at

y = e^{2}

Step 1:Divide the given ODE by

2 y^{2}

to put it in standard Bernoulli form.

\frac{d x}{d y} - \frac{x}{y} + \frac{x ^{2}}{2 y ^{2}} = 0

Step 2:Divide by

x^{2}

and substitute

t = 1/ x

(so

d t / d y = - x^{- 2} d x / d y

- \frac{d t}{d y} - \frac{t}{y} + \frac{1}{2 y ^{2}} = 0 \Rightarrow \frac{d t}{d y} + \frac{t}{y} = \frac{1}{2 y ^{2}}

Step 3:Integrating factor is

e^{\int d y / y} = y

. Multiply through.

\frac{d}{d y} (t y) = \frac{1}{2 y}

Step 4:Integrate both sides.

t y = \frac{1}{2} ln y + C \Rightarrow \frac{y}{x} = \frac{1}{2} ln y + C

Step 5:Apply

x (e) = e

to find

C

\frac{e}{e} = \frac{1}{2} ln e + C = \frac{1}{2} + C \Rightarrow C = \frac{1}{2}

Step 6:Solve for x and substitute

y = e^{2}

\frac{y}{x} = \frac{1}{2} (ln y + 1) = \frac{1}{2} ln (ey) \Rightarrow x = \frac{2 y}{ln ( ey )}; x (e^{2}) = \frac{2 e ^{2}}{ln ( e ^{3} )} = \frac{2 e ^{2}}{3}

Final answer:

\frac{2}{3} e^{2}

Q21NumericalSets, Relations and Functions

Let

A = {2, 3, 4, 5, 6}

. Let R be a relation on the set

A \times A

given by (x, y) R (z, w) if and only if x divides z and

y \leq w

. Then the number of elements in R is

SolutionAnswer: 120

Approach:

Decouple the two independent constraints: count ordered pairs (x,z) with

x ∣ z

in A, and ordered pairs (y,w) with

y \leq w

in A, then multiply.

Step 1:The conditions

x ∣ z

(in the first coordinate) and

y \leq w

(in the second coordinate) are independent, so

∣ R ∣

factorises.

∣ R ∣ = N_{x ∣ z} \cdot N_{y \leq w}

Step 2:For each

x \in A

, count multiples of x lying in

A = {2, 3, 4, 5, 6}

2 \to {2, 4, 6}

gives

3

;

3 \to {3, 6}

gives

2

;

4 \to {4}

5 \to {5}

6 \to {6}

each give

1

N_{x ∣ z} = 3 + 2 + 1 + 1 + 1 = 8

Step 3:With

n = ∣ A ∣ = 5

, the number of ordered pairs (y,w) satisfying

y \leq w

equals

(2 5) + 5 = 15

N_{y \leq w} = \frac{5 \cdot 6}{2} = 15

Step 4:Multiply the independent counts.

∣ R ∣ = 8 \times 15 = 120

Step 5:The relation has

120

ordered pairs.

∣ R ∣ = 120

Final answer: 120

Q22NumericalMatrices and Determinants

Consider the matrices

A = [24 - 2 - 2]

and

B = [3193]

. If matrices P and Q are such that

P A = B

and

A Q = B

, then the absolute value of the sum of the diagonal elements of

2 (P + Q)

is ..........

SolutionAnswer: 34

Approach:

Invert A, set

P = B A^{- 1}

and

Q = A^{- 1} B

, compute each matrix, then form

2 (P + Q)

and take the absolute value of its trace.

Step 1:Compute

det A

and

A^{- 1}

det A = (2) (- 2) - (- 2) (4) = - 4 + 8 = 4 \Rightarrow A^{- 1} = \frac{1}{4} [- 2 - 4 22] = [- \frac{1}{2} - 1 \frac{1}{2} \frac{1}{2}]

Step 2:Compute

P = B A^{- 1}

P = [3193] [- \frac{1}{2} - 1 \frac{1}{2} \frac{1}{2}] = [- \frac{3}{2} - 9 - \frac{1}{2} - 3 \frac{3}{2} + \frac{9}{2} \frac{1}{2} + \frac{3}{2}] = [- \frac{21}{2} - \frac{7}{2} 62]

Step 3:Compute

Q = A^{- 1} B

Q = [- \frac{1}{2} - 1 \frac{1}{2} \frac{1}{2}] [3193] = [- \frac{3}{2} + \frac{1}{2} - 3 + \frac{1}{2} - \frac{9}{2} + \frac{3}{2} - 9 + \frac{3}{2}] = [- 1 - \frac{5}{2} - 3 - \frac{15}{2}]

Step 4:Use linearity of trace.

tr (2 (P + Q)) = 2 (tr (P) + tr (Q)) = 2 (- \frac{17}{2} - \frac{17}{2}) = - 34

Step 5:Take the absolute value.

∣ - 34 ∣ = 34

Final answer: 34

Q23NumericalCo-ordinate Geometry

Let A be the point

(3, 0)

and circles with variable diameter AB touch the circle

x^{2} + y^{2} = 36

internally. Let the curve C be the locus of the point B. If the eccentricity of C is e, then

72 e^{2}

is equal to ..........

SolutionAnswer: 18

Approach:

Translate internal tangency between the variable circle (diameter

A B

) and the fixed circle of radius

6

into a focal-distance relation; the resulting locus is an ellipse from which

e

is read off.

Step 1:Let

B = (x, y)

. The variable circle has centre M at the midpoint of AB and radius

∣ A B ∣/2

. Internal tangency to

x^{2} + y^{2} = 36

(centre O, radius

6

) requires

∣ OM ∣ = 6 - ∣ A B ∣/2

(\frac{x + 3}{2})^{2} + (\frac{y}{2})^{2} = 6 - \frac{1}{2} (x - 3)^{2} + y^{2}

Step 2:Multiply both sides by

2

(x + 3)^{2} + y^{2} + (x - 3)^{2} + y^{2} = 12

Step 3:This is the locus definition of an ellipse with foci

S^{'} = (- 3, 0)

and

S = (3, 0)

and major-axis length

2 a = 12

2 a = 12 \Rightarrow a = 6; ∣ S S^{'} ∣ = 6 \Rightarrow 2 a e = 6 \Rightarrow a e = 3

Step 4:Compute the eccentricity.

e = \frac{a e}{a} = \frac{3}{6} = \frac{1}{2}

Step 5:Compute

72 e^{2}

72 e^{2} = 72 \cdot \frac{1}{4} = 18

Final answer: 18

Q24NumericalIntegral Calculus

If the area of the region bounded by

16 x^{2} - 9 y^{2} = 144

and

8 x - 3 y = 24

is A, then

3 (A + 6 lo g_{e} (3))

is equal to ..........

SolutionAnswer: 24

Approach:

Convert the hyperbola to standard form, find intersection of the line with the right branch, integrate the difference between the upper hyperbola arc and the line from

x = 3

x = 5

, then substitute into the requested expression.

Step 1:Divide

16 x^{2} - 9 y^{2} = 144

144

to obtain the standard form.

\frac{x ^{2}}{9} - \frac{y ^{2}}{16} = 1

Step 2:Find intersection of the line

8 x - 3 y = 24

with the hyperbola. From the line,

3 y = 8 x - 24

, so

9 y^{2} = (8 x - 24)^{2}

, while from the hyperbola

9 y^{2} = 16 x^{2} - 144

. Equating gives

(8 x - 24)^{2} = 16 x^{2} - 144

64 x^{2} - 384 x + 576 = 16 x^{2} - 144 \Rightarrow 48 x^{2} - 384 x + 720 = 0 \Rightarrow x^{2} - 8 x + 15 = 0 \Rightarrow x = 3, 5

Step 3:For

x \in [3, 5]

on the right branch, the upper arc is

y_{H} = \frac{4}{3} x^{2} - 9

and the line gives

y_{L} = \frac{8 x - 24}{3}

. Comparing at

x = 4

shows

y_{H} = \frac{4 7}{3} \approx 3.53 > \frac{8}{3} \approx 2.67 = y_{L}

, so the hyperbola is the upper boundary.

A = \int_{3}^{5} (\frac{4}{3} x^{2} - 9 - \frac{8 x - 24}{3}) d x

Step 4:Evaluate the linear part.

\int_{3}^{5} \frac{8 x - 24}{3} d x = \frac{1}{3} [4 x^{2} - 24 x]_{3}^{5} = \frac{1}{3} ((100 - 120) - (36 - 72)) = \frac{1}{3} (16) = \frac{16}{3}

Step 5:Evaluate the radical part using the standard integral with

a = 3

\int_{3}^{5} x^{2} - 9 d x = [\frac{x}{2} x^{2} - 9 - \frac{9}{2} ln x + x^{2} - 9]_{3}^{5} = (\frac{5}{2} \cdot 4 - \frac{9}{2} ln 9) - (0 - \frac{9}{2} ln 3) = 10 - \frac{9}{2} ln 3

Step 6:Combine to obtain

A

A = \frac{4}{3} (10 - \frac{9}{2} ln 3) - \frac{16}{3} = \frac{40}{3} - 6 ln 3 - \frac{16}{3} = 8 - 6 ln 3

Step 7:Evaluate the requested expression

3 (A + 6 ln 3)

3 (A + 6 ln 3) = 3 ((8 - 6 ln 3) + 6 ln 3) = 3 \cdot 8 = 24

Final answer: 24

Q25NumericalLimit, Continuity and Differentiability

The number of points in the interval

[2, 4]

at which the function

f (x) = [x^{2} - x - \frac{1}{2}]

, where

[\cdot]

denotes the greatest integer function, is discontinuous, is ..........

SolutionAnswer: 10

Approach:

Let

g (x) = x^{2} - x - \frac{1}{2}

. Since g is strictly increasing on

[2, 4]

f = [g]

is discontinuous exactly at the unique preimages of integers lying strictly inside

g ([2, 4])

; count those integers.

Step 1:Set

g (x) = x^{2} - x - \frac{1}{2}

and compute the endpoint values on

[2, 4]

g (2) = 4 - 2 - \frac{1}{2} = \frac{3}{2}; g (4) = 16 - 4 - \frac{1}{2} = \frac{23}{2}

Step 2:

g^{'} (x) = 2 x - 1 > 0

[2, 4]

, so

g

is strictly increasing and continuous; therefore

g

takes each integer in its range at exactly one interior point.

g^{'} (x) = 2 x - 1 \geq 3 > 0 \forall x \in [2, 4]

Step 3:Count integers n with

\frac{3}{2} < n < \frac{23}{2}

(endpoints

g (2), g (4)

are non-integer so the boundary

x \in {2, 4}

contributes nothing). The integers are

2, 3, 4, 5, 6, 7, 8, 9, 10, 11

n \in {2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

Step 4:At each such level n there is a unique

x_{n} \in (2, 4)

with

g (x_{n}) = n

; at

x_{n}

the function [g] jumps from

n - 1

to n, so f is discontinuous there.

# {discontinuities of f} = # {n \in Z : \frac{3}{2} < n < \frac{23}{2}} = 10

Step 5:The function is discontinuous at

10

points of

[2, 4]

Answer = 10

Final answer: 10

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