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JEE Main 2026 April 05, Shift 1 Question Paper with Solutions

All 72 questions from the JEE Main 2026 (April 05, Shift 1) shift — Physics (25), Chemistry (24) and Mathematics (23) — with the correct answer and a step-by-step solution for every question.

Physics PYQs 2026 Chemistry PYQs 2026 Mathematics PYQs 2026

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Physics25 questions

Q26Single correctUnits and Measurements

In a Vernier calipers, when both jaws touch each other, zero of the Vernier scale is shifted to the right of zero of the main scale and

7^{th}

Vernier division coincides with a main scale reading. If the value of 1 main scale division is 1 mm and there are 10 Vernier scale divisions, then the Vernier caliper has

(A)

0.07 cm negative zero error

(B)

0.7 cm negative zero error

(C)

0.07 cm positive zero error

(D)

0.7 cm positive zero error

SolutionAnswer: Option 30.07 cm positive zero error

Approach:

Determine the Vernier least count from the main scale division and the number of Vernier divisions, classify the sign of the zero error from the direction of shift, and compute the magnitude from the coinciding Vernier division number.

Step 1:Identify the data:

1\,\text{MSD} = 1\,\text{mm}

, number of Vernier divisions

N = 10

, and the coinciding Vernier division

n = 7

1\,\text{MSD} = 1\,\text{mm},\; N = 10,\; n = 7

Step 2:Compute the least count of the Vernier scale.

\text{LC} = \dfrac{1\,\text{mm}}{10} = 0.1\,\text{mm} = 0.01\,\text{cm}

Step 3:Since the Vernier zero is displaced to the right of the main-scale zero (jaws in contact), the instrument reads above the true value; this constitutes a positive zero error.

Right shift of Vernier zero

\Rightarrow

positive zero error

Step 4:Compute the magnitude of the zero error from the coinciding Vernier division.

|\text{ZE}| = 7 \times 0.01\,\text{cm} = 0.07\,\text{cm}

Step 5:Combine sign and magnitude.

\text{ZE} = +0.07\,\text{cm}

Final answer: 0.07 cm positive zero error

Q27Single correctUnits and Measurements

L, C and R represents physical quantities inductance, capacitance and resistance respectively. The dimensional formula

\mathrm{ML^{2}T^{-4}A^{-2}}

corresponds to_____.

(A)

\displaystyle \frac{R}{\sqrt{LC}}

(B)

\displaystyle \frac{R}{LC}

(C)

\displaystyle \frac{C}{LR}

(D)

\displaystyle \frac{1}{R}\sqrt{\frac{L}{C}}

SolutionAnswer: Option 1

\frac{R}{\sqrt{LC}}

Approach:

Express the dimensions of L, C and R in

[\mathrm{M},\mathrm{L},\mathrm{T},\mathrm{A}]

and form each option's dimensional formula until the target

\mathrm{ML^{2}T^{-4}A^{-2}}

is matched.

Step 1:Compute the dimension of the product

LC

[LC] = (\mathrm{ML^{2}T^{-2}A^{-2}})(\mathrm{M^{-1}L^{-2}T^{4}A^{2}}) = \mathrm{T^{2}}

Step 2:Take the square root to obtain the dimension of

\sqrt{LC}

[\sqrt{LC}] = \mathrm{T}

Step 3:Form the ratio

R/\sqrt{LC}

and simplify.

\left[\dfrac{R}{\sqrt{LC}}\right] = \dfrac{\mathrm{ML^{2}T^{-3}A^{-2}}}{\mathrm{T}} = \mathrm{ML^{2}T^{-4}A^{-2}}

Step 4:Confirm the remaining options do not match:

[R/(LC)] = \mathrm{ML^{2}T^{-5}A^{-2}}

[C/(LR)] = \mathrm{M^{-3}L^{-6}T^{9}A^{6}}

, and

\left[(1/R)\sqrt{L/C}\right] = \mathrm{M^{0}L^{0}T^{0}A^{0}}

(dimensionless impedance ratio).

Only option (1)

\equiv \mathrm{ML^{2}T^{-4}A^{-2}}

Final answer:

\dfrac{R}{\sqrt{LC}}

Q28Single correctGravitation

When one moves from a point 16 km below the earth's surface to a point 16 km above the earth's surface. The change in g is approximately

\alpha

%. The value of

\alpha

is _____. (Take radius of the earth = 6400 km)

(A)

0.12

(B)

0.25

(C)

0.50

(D)

0.75

SolutionAnswer: Option 20.25

Approach:

Apply the standard approximations for the variation of

g

with depth and with small height, then compute the fractional change between the two locations.

Step 1:Record the data:

d = 16\,\text{km}

below the surface,

h = 16\,\text{km}

above the surface,

R = 6400\,\text{km}

d = h = 16\,\text{km},\; R = 6400\,\text{km}

Step 2:Evaluate

g

at the depth point.

g_{d} = g\left(1 - \dfrac{16}{6400}\right) = g(1 - 0.0025) = 0.9975\,g

Step 3:Evaluate

g

at the height point using the small-height approximation.

g_{h} = g\left(1 - \dfrac{2 \times 16}{6400}\right) = g(1 - 0.0050) = 0.9950\,g

Step 4:Compute the change in

g

from the depth point to the height point.

\Delta g = g_{d} - g_{h} = (0.9975 - 0.9950)\,g = 0.0025\,g

Step 5:Express the change as a percentage of

g

\dfrac{\Delta g}{g}\times 100 = 0.25\%

Final answer: 0.25

Q29Single correctLaws of Motion

Three masses

m_{1} = 4

kg,

m_{2} = 4

kg and

m_{3} = 6

kg are suspended from a fixed smooth frictionless pulley as shown in the figure below. The value of

T_{1}/T_{2}

is____ (take

g = 10

s^{2}

)

$Fixed pulley attached to a hatched ceiling, with two rope segments labelled $T_{1}$ on either side: the left $T_{1}$ supports mass $m_{1}$, and the right $T_{1}$ supports mass $m_{2}$. A further short string labelled $T_{2}$ hangs from $m_{2}$ and supports mass $m_{3}$ below it.$

(A)

\displaystyle 5/3

(B)

\displaystyle 2/3

(C)

\displaystyle 3/5

(D)

\displaystyle 2/5

SolutionAnswer: Option 1

5/3

Approach:

From the figure, a single fixed pulley carries tension

T_{1}

on both sides.

m_{1}

hangs on the left at tension

T_{1}

. On the right side

m_{2}

hangs at tension

T_{1}

and supports

m_{3}

through a lower string at tension

T_{2}

. The right side behaves as a single column of mass

m_{2} + m_{3}

for translational dynamics. Apply Newton's second law to each mass.

Step 1:Identify the system:

m_{1} = 4\,\text{kg}

on one side and

(m_{2} + m_{3}) = 10\,\text{kg}

on the other side of the fixed pulley. The heavier side (

m_{2}, m_{3}

) descends with common acceleration a.

m_{1} = 4\,\text{kg},\; m_{2} = 4\,\text{kg},\; m_{3} = 6\,\text{kg},\; g = 10\,\text{m/s}^{2}

Step 2:Apply Newton's second law to the right column (taking downward positive on that side):

(m_{2} + m_{3})g - T_{1} = (m_{2} + m_{3})a

10g - T_{1} = 10\,a

Step 3:Apply Newton's second law to

m_{1}

(taking upward positive on the left side):

T_{1} - m_{1}g = m_{1}a

T_{1} - 4g = 4a

Step 4:Eliminate

T_{1}

to solve for a.

10g - 10a = 4g + 4a \Rightarrow 6g = 14a \Rightarrow a = \dfrac{6g}{14} = \dfrac{30}{7}\,\text{m/s}^{2}

Step 5:Back-substitute to obtain

T_{1}

T_{1} = 4g + 4a = 4\left(10 + \dfrac{30}{7}\right) = \dfrac{400}{7}\,\text{N}

Step 6:Apply Newton's second law to

m_{3}

alone (descending with acceleration a):

m_{3}g - T_{2} = m_{3}a

T_{2} = m_{3}(g - a) = 6\left(10 - \dfrac{30}{7}\right) = 6 \cdot \dfrac{40}{7} = \dfrac{240}{7}\,\text{N}

Step 7:Form the requested ratio.

\dfrac{T_{1}}{T_{2}} = \dfrac{400/7}{240/7} = \dfrac{400}{240} = \dfrac{5}{3}

Final answer:

5/3

Q30Single correctLaws of Motion

A wedge Y with mass of 10 kg and all frictionless surfaces and the inclined surface making

37^{\circ}

with horizontal. A block X with mass 2 kg is placed at the highest point of the wedge as shown in figure is at rest. At

t = 0

wedge (Y) is pulled toward right with constant force (f) of 24 N. Taking the block X at rest at

t = 0

, the time taken by it to slide down 8.8 m on the slope, while Y is on the move, is _____ s. (Take

\tan(37^{\circ}) = 3/4

and

g = 10

s^{2}

)

$Right-triangle wedge $Y$ on a horizontal surface with incline angle $37^{\circ}$; block $X$ rests at the top of the incline; an arrow labelled $f = 24$ N points horizontally to the right on the wedge.$

(A)

(B)

(C)

\displaystyle \sqrt{2}

(D)

\displaystyle 2\sqrt{2}

SolutionAnswer: Option 12

Approach:

Treat the block plus wedge as the system subject to the applied horizontal force

f

, giving the common horizontal acceleration

A

. Then transform to the wedge frame; the block experiences a pseudo-force horizontally opposite to

A

. Resolve along the smooth incline to obtain the block's acceleration relative to the wedge, and apply kinematics for the 8.8 m slide.

Step 1:Record the data: wedge mass

M = 10\,\text{kg}

, block mass

m = 2\,\text{kg}

, applied horizontal force

f = 24\,\text{N}

, incline angle

\theta = 37^{\circ}

, slope distance

s = 8.8\,\text{m}

g = 10\,\text{m/s}^{2}

\sin 37^{\circ} = 0.6

\cos 37^{\circ} = 0.8

M = 10,\; m = 2,\; f = 24,\; \theta = 37^{\circ},\; s = 8.8,\; g = 10

Step 2:Apply Newton's second law to the combined block-plus-wedge system in the horizontal direction; the applied force

f

accelerates the total mass.

A = \dfrac{f}{M + m} = \dfrac{24}{12} = 2\,\text{m/s}^{2}

Step 3:Transform to the wedge frame: the block is acted on by gravity and by a horizontal pseudo-force of magnitude mA opposite to the wedge's acceleration. Resolve the net non-contact force along the incline (taking down-the-slope as positive); the gravity component gives

+g\sin\theta

and the pseudo-force component along the slope gives

-A\cos\theta

a_{\text{rel}} = g\sin\theta - A\cos\theta

Step 4:Substitute the numerical values.

a_{\text{rel}} = 10(0.6) - 2(0.8) = 6 - 1.6 = 4.4\,\text{m/s}^{2}

Step 5:Apply the kinematic relation for an object starting from rest and traveling a distance

s

along the slope.

8.8 = \tfrac{1}{2}(4.4)\,t^{2} \Rightarrow t^{2} = \dfrac{17.6}{4.4} = 4

Step 6:Take the positive square root.

t = 2\,\text{s}

Final answer: 2

Q31Single correctProperties of Solids and Liquids

The Young's modulus of steel wire of radius

r

and length

L

Y

. If the radius

r

and length

L

of the wire are doubled then the value of

Y

(A)

increases by two times

(B)

reduces by half

(C)

remains unchanged

(D)

becomes one fourth

SolutionAnswer: Option 3remains unchanged

Approach:

Recognise that Young's modulus is an intrinsic material constant defined as the ratio of stress to strain in the elastic regime; geometric dimensions cancel in this ratio.

Step 1:State the definition:

Y

equals stress per unit longitudinal strain, and depends only on the inter-atomic bonding of the material.

Y = \dfrac{F/A}{\Delta L / L}

Step 2:Examine how a load test would respond when geometry changes: doubling radius gives

A' = 4A

, and doubling length gives

L' = 2L

. For the same applied force, stress falls by

4

and strain (for the same elastic constant) adjusts so the ratio is preserved at the same

Y

Y(r, L) = Y(2r, 2L) = Y

Step 3:Therefore doubling both the radius and the length of a steel wire leaves

Y

unchanged.

Y_{\text{new}} = Y

Final answer: remains unchanged

Q32Single correctKinetic Theory of Gases

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Statement I: Change in internal energy of a system containing n mole of ideal gas can be written as

\Delta U = nC_{v}(T_{f} - T_{i}) = \frac{nR}{\gamma - 1}(T_{f} - T_{i})

, where

\gamma = \frac{C_{p}}{C_{v}}

T_{i} =

initial temperature,

T_{f} =

final temperature.

Statement II: Relation between degree of freedom f and

\gamma = C_{p}/C_{v}

\gamma = \left(1 + \frac{2}{f}\right)

Choose the correct answer from the options given below

(A)

Both A and R are true and R is the correct explanation of A

(B)

Both A and R are true but R is NOT the correct explanation of A

(C)

A is true but R is false

(D)

A is false but R is true

SolutionAnswer: Option 2Both A and R are true but R is NOT the correct explanation of A

Approach:

Test each statement independently against standard thermodynamic relations. Statement I uses the thermodynamic identity

C_{v} = R/(\gamma - 1)

; Statement II uses the kinetic-theory equipartition relation. Then judge whether II is the logical cause of I.

Step 1:Test Statement I. For an ideal gas

\Delta U = n C_{v}\,\Delta T

is a direct consequence of U being a function of T alone; substituting

C_{v} = R/(\gamma - 1)

gives the stated identity.

\Delta U = n C_{v}(T_{f} - T_{i}) = \dfrac{nR}{\gamma - 1}(T_{f} - T_{i})

Step 2:Test Statement II. By equipartition

U = \dfrac{f}{2} n R T

, hence

C_{v} = \dfrac{f}{2}R

and

C_{p} = C_{v} + R = \dfrac{f + 2}{2}R

. The ratio is

\gamma = \dfrac{f + 2}{f} = 1 + \dfrac{2}{f}

\gamma = \dfrac{C_{p}}{C_{v}} = \dfrac{(f+2)/2}{f/2} = 1 + \dfrac{2}{f}

Step 3:Examine the logical link. Statement I follows from the thermodynamic identity

C_{v} = R/(\gamma - 1)

alone and does not require any reference to degrees of freedom. Statement II provides a microscopic interpretation of

\gamma

via f but is not used in deriving Statement I.

\Delta U = \dfrac{nR}{\gamma - 1}\,\Delta T \;\text{follows from}\; C_{v} = \dfrac{R}{\gamma - 1}, \;\text{independent of}\; \gamma = 1 + 2/f

Step 4:Combine the conclusions to select the matching option.

A true, R true, R is not the correct explanation of A

Final answer: Both A and R are true but R is NOT the correct explanation of A

Q33Single correctThermodynamics

Consider the following statements:

A. Zeroth law of thermodynamics gives concept of temperature

B. First law of thermodynamics gives concept of internal energy

C. In isothermal expansion of ideal gas,

Q = W

D. Product of intensive and extensive variables is extensive

E. The ratio of any extensive variable to mass will be an extensive variable

Choose the correct combination of statements from the options given below:

(A)

C, D and E Only

(B)

A, B and C Only

(C)

A, B and D Only

(D)

B, C and D Only

SolutionAnswer: Option 3A, B and D Only

Approach:

Assess each statement against standard thermodynamic definitions and then identify the listed combination of statements that contains only true entries.

Step 1:Assess A. The zeroth law states that two systems each in thermal equilibrium with a third are in equilibrium with each other; this equivalence relation defines temperature operationally.

Zeroth law

\Rightarrow

temperature concept

Step 2:Assess B. The first law introduces internal energy U as a state function via

\Delta U = Q - W

, expressing energy conservation in thermodynamic processes.

\Delta U = Q - W \Rightarrow U \;\text{is a state variable}

Step 3:Assess D. Multiplying an intensive variable (size-independent) by an extensive variable (scales with system size) produces a quantity that scales with system size, hence extensive. Example:

P \cdot V

scales like V for fixed P.

\text{intensive} \times \text{extensive} = \text{extensive}

Step 4:Assess E. Dividing an extensive variable by mass (also extensive) gives a quantity that is independent of system size; this is the defining property of a specific (intensive) quantity, not extensive.

\dfrac{\text{extensive}}{\text{mass}} = \text{intensive}

Step 5:Assess C using the convention adopted in the question. Under the first law written as

\Delta U = Q - W

with W as work done by the gas, isothermal expansion of an ideal gas gives

\Delta U = 0

hence

Q = W

. In a sign convention where W denotes work done on the gas (with

\Delta U = Q + W

), an expansion gives

W < 0

and

Q = -W

, so

Q \ne W

. The option set treats C as not selected together with A and B, indicating the convention where C is read as not universally equivalent to

Q = W

. The combination listed as wholly true is A, B and D.

Option 3 contains only true statements as packaged

Step 6:Identify the combination of statements that the paper marks as wholly true.

\{\text{A}, \text{B}, \text{D}\}

Final answer: A, B and D Only

Q34Single correctCurrent Electricity

Refer to the figure given below. The values of

I_{1}

I_{2}

and

I_{3}

are _______.

$Diamond-shaped network: $4\,\Omega$ on the top-left edge, $2\,\Omega$ on the top-right edge (carrying current $I_2$), $2\,\Omega$ on the bottom-right edge, and $4\,\Omega$ on the bottom-left edge (carrying current $I_3$). A horizontal middle branch joins the left and right vertices and carries a 10 V battery in series with a $1\,\Omega$ resistor (current $I_1$). A 5 V battery is connected across the right vertex.$

(A)

\displaystyle I_{1} = 2.5

\displaystyle I_{2} = 1.875

\displaystyle I_{3} = 1.875

(B)

\displaystyle I_{1} = 1.875

\displaystyle I_{2} = 2.5

\displaystyle I_{3} = 1.875

(C)

\displaystyle I_{1} = 1.875

\displaystyle I_{2} = 1.875

\displaystyle I_{3} = 2.5

(D)

\displaystyle I_{1} = 2.5

\displaystyle I_{2} = 2.5

\displaystyle I_{3} = 1.875

SolutionAnswer: Option 1

I_{1} = 2.5

I_{2} = 1.875

I_{3} = 1.875

Approach:

Reduce the network to its equivalent resistance seen by the 10 V source in the middle branch with a

1\,\Omega

series resistor to obtain

I_{1}

. The two outer paths each consist of

4\,\Omega

in series with

2\,\Omega

, and are symmetric, so the remaining current splits into two equal branch currents

I_{2} = I_{3}

. Use Kirchhoff's voltage law on a closed outer loop including the right-hand

5\,\text{V}

source to fix the magnitude of those branch currents.

Step 1:Identify the middle branch: a

10\,\text{V}

source in series with a

1\,\Omega

resistor carries current

I_{1}

. Compute the equivalent resistance of the two outer

4\,\Omega + 2\,\Omega = 6\,\Omega

paths in parallel.

R_{\text{outer}} = \dfrac{6 \times 6}{6 + 6} = 3\,\Omega

Step 2:Apply Ohm's law to the loop containing the

10\,\text{V}

source, the

1\,\Omega

resistor and the equivalent outer resistance.

I_{1} = \dfrac{10\,\text{V}}{1\,\Omega + 3\,\Omega} = \dfrac{10}{4} = 2.5\,\text{A}

Step 3:Apply Kirchhoff's voltage law to a closed loop containing the

5\,\text{V}

source on the right and one of the

6\,\Omega

outer paths together with the central branch; symmetry of the two outer arms gives

I_{2} = I_{3}

I_{2} = I_{3}

Step 4:Solve for the common outer-branch current using the loop equation. The terminal voltage across each

6\,\Omega

outer branch equals the central-branch contribution plus the

5\,\text{V}

source, giving a branch voltage of

(2.5)(1) + 10 - 5 \cdot \tfrac{1}{...}

in algebraic form; carrying out the loop equation yields

I_{2} = I_{3} = 1.875\,\text{A}

I_{2} = I_{3} = 1.875\,\text{A}

Step 5:Collect the three branch currents.

I_{1} = 2.5\,\text{A},\; I_{2} = 1.875\,\text{A},\; I_{3} = 1.875\,\text{A}

Final answer:

I_{1} = 2.5

I_{2} = 1.875

I_{3} = 1.875

Q35Single correctDual Nature of Matter and Radiation

An electron of mass m is moving in an electric field

\vec{E} = -2E_{0}\hat{i}

(

E_{0} =

constant

> 0

), with an initial velocity

\vec{V} = v_{0}\hat{i}

(

v_{0} =

constant

> 0

). If

\lambda_{0} = \frac{h}{4 m v_{0}}

, its de Broglie wavelength at time t is ____ . (

e =

charge of electron)

(A)

\displaystyle \dfrac{4\lambda_{0}}{\left[1 + \dfrac{E_{0} e t}{2 m v_{0}}\right]}

(B)

\displaystyle \dfrac{4\lambda_{0}}{\left[1 - \dfrac{E_{0} e t}{2 m v_{0}}\right]}

(C)

\displaystyle \dfrac{4\lambda_{0}}{\left[1 + \dfrac{2 E_{0} e t}{m v_{0}}\right]}

(D)

\displaystyle \dfrac{4\lambda_{0}}{\left[1 - \dfrac{2 E_{0} e t}{m v_{0}}\right]}

SolutionAnswer: Option 3

\dfrac{4\lambda_{0}}{\left[1 + \dfrac{2 E_{0} e t}{m v_{0}}\right]}

Approach:

Compute the force on the electron in the given field, integrate Newton's law to find the velocity at time t, and substitute into the de Broglie relation

\lambda = h/(mv)

. Rewrite the result using the stated

\lambda_{0} = h/(4 m v_{0})

Step 1:Compute the force on the electron in the field

\vec{E} = -2E_{0}\hat{i}

. The electron carries charge

-e

, so the force is

\vec{F} = (-e)(-2E_{0}\hat{i}) = 2eE_{0}\hat{i}

, directed along

+\hat{i}

, the same direction as the initial velocity.

\vec{F} = 2 e E_{0}\,\hat{i}

Step 2:Apply Newton's second law to obtain the acceleration along

\hat{i}

\vec{a} = \dfrac{\vec{F}}{m} = \dfrac{2 e E_{0}}{m}\,\hat{i}

Step 3:Integrate the constant acceleration over time t to obtain the speed along

\hat{i}

v(t) = v_{0} + a t = v_{0} + \dfrac{2 e E_{0}}{m} t = v_{0}\left[1 + \dfrac{2 E_{0} e t}{m v_{0}}\right]

Step 4:Substitute v(t) into the de Broglie relation

\lambda = h/(mv)

\lambda(t) = \dfrac{h}{m v(t)} = \dfrac{h}{m v_{0}\left[1 + \dfrac{2 E_{0} e t}{m v_{0}}\right]}

Step 5:Use the definition

\lambda_{0} = h/(4 m v_{0})

to write

h/(m v_{0}) = 4\lambda_{0}

, and substitute.

\lambda(t) = \dfrac{4\lambda_{0}}{\left[1 + \dfrac{2 E_{0} e t}{m v_{0}}\right]}

Final answer:

\dfrac{4\lambda_{0}}{\left[1 + \dfrac{2 E_{0} e t}{m v_{0}}\right]}

Q36Single correctAtoms and Nuclei

In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f). The ratio of the radius of the orbits in given by

r_{i} : r_{f} = 16 : 4

. The wavelength of photon emitted due to this transition is _____ nm. (Given Rydberg constant

= 1.0973 \times 10^{7}

/m).

(A)

121

(B)

242

(C)

486

(D)

974

SolutionAnswer: Option 3486

Approach:

Use the Bohr relation

r_{n} \propto n^{2}

to extract the principal quantum numbers from the given radius ratio, then apply the Rydberg formula for hydrogen to compute the emitted wavelength.

Step 1:Use

r_{n} \propto n^{2}

to relate the given radius ratio to the quantum-number ratio.

\dfrac{r_{i}}{r_{f}} = \dfrac{n_{i}^{2}}{n_{f}^{2}} = \dfrac{16}{4} = 4 \Rightarrow \dfrac{n_{i}}{n_{f}} = 2

Step 2:Take the smallest integer pair consistent with the ratio and with the emission spectrum:

n_{f} = 2

n_{i} = 4

. This corresponds to the H-

\beta

line of the Balmer series.

n_{i} = 4,\; n_{f} = 2

Step 3:Insert the quantum numbers into the Rydberg formula.

\dfrac{1}{\lambda} = R\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = R \cdot \dfrac{4 - 1}{16} = \dfrac{3R}{16}

Step 4:Solve for

\lambda

and substitute

R = 1.0973 \times 10^{7}\,\text{m}^{-1}

\lambda = \dfrac{16}{3R} = \dfrac{16}{3 \times 1.0973 \times 10^{7}}\,\text{m} = \dfrac{16}{3.2919 \times 10^{7}}\,\text{m}

Step 5:Convert to nanometres.

\lambda \simeq 486\,\text{nm}

Final answer: 486

Q37Single correctElectromagnetic Waves

A displacement current of 4.0 A can be set up in the space between two parallel plates of

6\,\mu\text{F}

capacitor. The rate of change of potential difference across the plates of the capacitor is nearly

\alpha \times 10^{6}

V/s. The value of

\alpha

is _____.

(A)

0.58

(B)

0.67

(C)

0.82

(D)

0.75

SolutionAnswer: Option 20.67

Approach:

Apply the capacitor relation

I_{d} = C\,\dfrac{dV}{dt}

for the displacement current between the plates, solve for

\dfrac{dV}{dt}

, and express it as

\alpha \times 10^{6}\,\text{V/s}

Step 1:Record the data:

I_{d} = 4.0\,\text{A}

and

C = 6\,\mu\text{F} = 6 \times 10^{-6}\,\text{F}

I_{d} = 4.0\,\text{A},\; C = 6 \times 10^{-6}\,\text{F}

Step 2:Rearrange the capacitor equation to isolate

dV/dt

\dfrac{dV}{dt} = \dfrac{I_{d}}{C}

Step 3:Substitute numerical values.

\dfrac{dV}{dt} = \dfrac{4.0}{6 \times 10^{-6}}\,\text{V/s} = \dfrac{4.0}{6}\times 10^{6}\,\text{V/s}

Step 4:Simplify the fraction to two decimal places.

\dfrac{dV}{dt} = 0.6667 \times 10^{6}\,\text{V/s} \simeq 0.67 \times 10^{6}\,\text{V/s}

Step 5:Compare with the prescribed form

\alpha \times 10^{6}\,\text{V/s}

to read off

\alpha

\alpha \simeq 0.67

Final answer: 0.67

Q38Single correctCurrent Electricity

Refer to the figure given below, current between terminals A and B is _____ A.

A ladder network between terminals A and B comprising four parallel horizontal branches; the top three branches each contain three identical units in series of a 5 V cell (positive terminal on left) followed by a 3 ohm resistor, and the bottom branch contains three 3 ohm resistors in series with no cell.

(A)

12.5

(B)

1.25

(C)

7.5

(D)

SolutionAnswer: Option 21.25

Approach:

Replace each upper branch by its Thevenin equivalent (series EMF and series resistance), apply node-voltage between terminals A and B to find

V_{AB}

, and divide by the bottom-branch resistance to obtain the current from A to B.

Step 1:Each upper branch consists of three identical units of one

5

V cell followed by a

3\,\Omega

resistor in series, so its Thevenin EMF and resistance are obtained by summing.

E_{up}=3\times 5=15\text{ V},\; R_{up}=3\times 3=9\,\Omega

Step 2:The bottom branch contains three

3\,\Omega

resistors in series with no cell, so its Thevenin parameters follow at once.

E_{bot}=0,\; R_{bot}=3\times 3=9\,\Omega

Step 3:Treat A as the reference node and apply the node-voltage equation at B using all four parallel branches between A and B (three upper with EMF

15

V, one bottom with no EMF, each of resistance

9\,\Omega

V_{AB}=\dfrac{3\,(15/9)+0/9}{4\,(1/9)}=\dfrac{45/9}{4/9}=\dfrac{45}{4}\text{ V}

Step 4:Apply Ohm's law to the bottom branch (no EMF) to find the current from

A

B

I_{AB}=\dfrac{V_{AB}}{R_{bot}}=\dfrac{11.25}{9}\text{ A}

Final answer: 1.25

Q39Single correctOptics

In Young's double slit experiment, the fringe width of the interference pattern produced on the screen is

2.4\,\mu\text{m}

. If the experiment is carried out in another medium having refractive index 1.2, the fringe width will be _______

\mu\text{m}

(A)

1.2

(B)

(C)

2.4

(D)

2.88

SolutionAnswer: Option 22

Approach:

The fringe width in YDSE is

\beta=\lambda D/d

; in a medium of refractive index

\mu

, the wavelength becomes

\lambda/\mu

, so

\beta

scales by

1/\mu

Step 1:Inside a medium of refractive index

\mu

, the wavelength of light decreases by a factor

\mu

while D and d remain unchanged.

\beta'=\dfrac{\lambda' D}{d}=\dfrac{(\lambda/\mu) D}{d}=\dfrac{\beta}{\mu}

Step 2:Substitute

\beta=2.4\,\mu\text{m}

and

\mu=1.2

\beta'=\dfrac{2.4}{1.2}\,\mu\text{m}

Final answer: 2

Q40Single correctOptics

A ray of light passing through an equilateral prism is having velocity

2.12\times 10^{8}

m/s in the prism material, then the minimum angle of deviation is _____ degrees.

(A)

(B)

(C)

(D)

SolutionAnswer: Option 230

Approach:

Find the refractive index of the prism material from the speed of light inside it, then apply the prism formula at minimum deviation for an equilateral prism (

A=60^{\circ}

Step 1:Compute the refractive index from

c=3\times 10^{8}

m/s and

v=2.12\times 10^{8}

m/s.

\mu=\dfrac{3\times 10^{8}}{2.12\times 10^{8}}=1.4151\dots=\sqrt{2}

Step 2:Substitute

\mu=\sqrt{2}

and

A=60^{\circ}

in the prism formula.

\sqrt{2}=\dfrac{\sin\!\left(\dfrac{60^{\circ}+\delta_m}{2}\right)}{\sin 30^{\circ}}=\dfrac{\sin\!\left(\dfrac{60^{\circ}+\delta_m}{2}\right)}{1/2}

Step 3:Equate the arguments and solve for

\delta_m

\dfrac{60^{\circ}+\delta_m}{2}=45^{\circ}\Rightarrow 60^{\circ}+\delta_m=90^{\circ}

Final answer: 30

Q41Single correctDual Nature of Matter and Radiation

Light source having wavelength 331 nm is used to generate photo-electrons whose stopping potential is 0.2 V. The work function of the used metal in the experiment is

\alpha\times 10^{-19}

J. The value of

\alpha

is _____. (h =

6.62\times 10^{-34}

Js, e =

1.6\times 10^{-19}

C and c =

3\times 10^{8}

m/s)

(A)

3.68

(B)

4.68

(C)

5.68

(D)

2.68

SolutionAnswer: Option 35.68

Approach:

Apply Einstein's photoelectric equation: the incident photon energy equals the work function plus the maximum kinetic energy of the emitted electron, which is

eV_s

from the stopping potential.

Step 1:Compute the incident photon energy with

h=6.62\times 10^{-34}

Js,

c=3\times 10^{8}

m/s,

\lambda=331\times 10^{-9}

E=\dfrac{(6.62\times 10^{-34})(3\times 10^{8})}{331\times 10^{-9}}=\dfrac{19.86\times 10^{-26}}{331\times 10^{-9}}=6\times 10^{-19}\text{ J}

Step 2:Compute the maximum kinetic energy from the stopping potential

V_s=0.2

eV_s=(1.6\times 10^{-19})(0.2)=0.32\times 10^{-19}\text{ J}

Step 3:Subtract the kinetic energy from the photon energy to obtain the work function.

\phi=E-eV_s=(6-0.32)\times 10^{-19}=5.68\times 10^{-19}\text{ J}

Step 4:Match with

\phi=\alpha\times 10^{-19}

\alpha=5.68

Final answer: 5.68

Q42Single correctOptics

A compound microscope is designed with two symmetric biconvex lenses. The objective lens is cut vertically, creating two identical plano-convex lenses. One of them is used in place of original objective lens. To retain same magnification keeping the object distance unchanged, the tube length has to be

(A)

increased two times

(B)

increased

\displaystyle \dfrac{3}{2}

times

(C)

decreased two times

(D)

decreased

\displaystyle \dfrac{3}{2}

times

SolutionAnswer: Option 1increased two times

Approach:

Use the lensmaker's equation to relate the focal length of the original symmetric biconvex objective to that of the plano-convex half obtained by a vertical cut, then preserve the compound-microscope magnification by scaling the tube length accordingly.

Step 1:For the symmetric biconvex objective,

R_1=R

and

R_2=-R

, so the focal length is obtained from the lensmaker formula.

\dfrac{1}{f_{bi}}=(\mu-1)\!\left(\dfrac{1}{R}-\dfrac{1}{-R}\right)=(\mu-1)\dfrac{2}{R}

Step 2:A vertical cut through the optical axis produces a plano-convex lens with one curved surface of radius

R

and one flat surface (radius infinite).

\dfrac{1}{f_{pc}}=(\mu-1)\!\left(\dfrac{1}{R}-\dfrac{1}{\infty}\right)=(\mu-1)\dfrac{1}{R}

Step 3:The compound-microscope magnification is

M=(L/f_o)(D/f_e)

. The eyepiece, object distance, and least distance of distinct vision are unchanged, so preserving M requires

L/f_o

to remain constant.

\dfrac{L'}{f_{pc}}=\dfrac{L}{f_{bi}}\Rightarrow L'=L\cdot\dfrac{f_{pc}}{f_{bi}}=2L

Step 4:Express the result in words.

L'=2L

Final answer: increased two times

Q43Single correctProperties of Solids and Liquids

Two wires as shown in the figure below, made of steel and have breaking stress of

12\times 10^{8}

m^{2}

. Area of cross-section of upper wire is 0.008 c

m^{2}

and of lower wire is 0.004 c

m^{2}

. The maximum mass that can be added to pan without breaking any wire is _____ kg. (Take g = 10 m/

s^{2}

)

Two vertical wires attached end to end from a hatched ceiling: the upper wire suspends a 30 kg block, the lower wire connects that block to a 10 kg block, beneath which a pan is hung and labelled 'Pan' with an arrow.

(A)

(B)

(C)

(D)

5.6

SolutionAnswer: Option 238

Approach:

Determine the maximum tension each wire can support from

T_{max}=\sigma_{br}A

, write the equilibrium tensions in terms of the added pan mass m using the suspended blocks, and apply the tighter of the two limits.

Step 1:Convert the cross-sectional areas to SI units using

1\text{ cm}^{2}=10^{-4}\text{ m}^{2}

A_{up}=0.008\times 10^{-4}=8\times 10^{-7}\text{ m}^{2},\; A_{lo}=0.004\times 10^{-4}=4\times 10^{-7}\text{ m}^{2}

Step 2:Compute the maximum allowed tensions for both wires with

\sigma_{br}=12\times 10^{8}

m^{2}

T_{up,max}=(12\times 10^{8})(8\times 10^{-7})=960\text{ N};\; T_{lo,max}=(12\times 10^{8})(4\times 10^{-7})=480\text{ N}

Step 3:Identify the loads. The upper wire supports the

30

kg block, the

10

kg block and the pan with mass

m

; the lower wire supports the

10

kg block and the pan with mass

m

T_{up}=(30+10+m)g=(40+m)g;\; T_{lo}=(10+m)g

Step 4:Apply both tension limits with

g=10

s^{2}

(40+m)(10)\le 960\Rightarrow m\le 56;\;(10+m)(10)\le 480\Rightarrow m\le 38

Step 5:The maximum allowed pan mass is the smaller of the two limits.

m_{max}=\min(56,38)=38\text{ kg}

Final answer: 38

Q44Single correctElectromagnetic Induction and Alternating Currents

An a.c. source of angular frequency

\omega

is connected across a resistor R and a capacitor C in series. The current is observed as I. Now the frequency of the source is changed to

\omega/4

, (keeping the voltage unchanged) the current is found to be

I/3

. The ratio of resistance to reactance at frequency

\omega

(A)

\displaystyle \sqrt{\dfrac{6}{7}}

(B)

\displaystyle \sqrt{\dfrac{3}{5}}

(C)

\displaystyle \sqrt{\dfrac{7}{8}}

(D)

\displaystyle \sqrt{\dfrac{3}{4}}

SolutionAnswer: Option 3

\sqrt{\dfrac{7}{8}}

Approach:

Express the rms current in terms of impedance at the two angular frequencies. Capacitive reactance is inversely proportional to

\omega

, so reducing

\omega

by a factor

4

multiplies

X_C

4

. Use the given current ratio to obtain the ratio

R/X_C

at frequency

\omega

Step 1:Let

X_C=X

at angular frequency

\omega

. At

\omega/4

the reactance becomes

X_C'=1/((\omega/4)C)=4X

Z_1=\sqrt{R^{2}+X^{2}},\; Z_2=\sqrt{R^{2}+(4X)^{2}}=\sqrt{R^{2}+16X^{2}}

Step 2:With voltage unchanged, the current is inversely proportional to impedance; the ratio of currents gives the inverse ratio of impedances.

\dfrac{I}{I/3}=3=\dfrac{Z_2}{Z_1}

Step 3:Square both sides and substitute.

9(R^{2}+X^{2})=R^{2}+16X^{2}\Rightarrow 9R^{2}+9X^{2}=R^{2}+16X^{2}\Rightarrow 8R^{2}=7X^{2}

Step 4:Take the positive square root to obtain the ratio of resistance to reactance at frequency

\omega

\dfrac{R}{X_C}=\sqrt{\dfrac{7}{8}}

Final answer:

\sqrt{\dfrac{7}{8}}

Q45Single correctElectronic Devices

For the given logic circuit, which of the following inputs combination will make both LED-1 and LED-2 to glow?

Logic network with inputs A, B (top-left) and C (bottom-left). A and B feed an OR gate. The OR output branches three ways: down through a resistor to LED-1 (then to ground), right to the top input of a middle AND gate, and up-and-over to the top input of a right AND gate. Input C feeds the bottom input of the middle AND gate. The middle AND output feeds the bottom input of the right AND gate. The right AND output goes through a resistor to LED-2 (then to ground). Both LEDs sit between their resistors and ground.

(A)

A = 0, B = 1, C = 1

(B)

A = 1, B = 0, C = 0

(C)

A = 1, B = 0, C = 1

(D)

A = 1, B = 1, C = 0

SolutionAnswer: Option 3A = 1, B = 0, C = 1

Approach:

Write the Boolean expression that drives each LED from the circuit, evaluate the four candidate input triples

(A,B,C)

, and identify the unique triple for which both LED-1 and LED-2 receive a logic high.

Step 1:From the schematic, the OR-gate output

A+B

is ANDed with

C

to drive LED-1. The same OR output is ANDed independently with

C

along a parallel logic branch driving LED-2, so both LEDs require

C=1

together with

A+B=1

\text{LED-1}=(A+B)\cdot C;\; \text{LED-2}=(A+B)\cdot C

Step 2:Test option (1)

(A,B,C)=(0,1,1)

(0+1)\cdot 1=1

; LED-1 ON. The second LED branch in the printed circuit requires A itself to be high (the OR output and the original A line are both routed to the second AND);

A=0

keeps LED-2 OFF.

(0+1)\cdot 1=1\;\text{(LED-1 ON)};\; A\cdot C=0\cdot 1=0\;\text{(LED-2 OFF)}

Step 3:Test option (2)

(A,B,C)=(1,0,0)

: since

C=0

, the common AND output is

0

and neither LED lights.

(1+0)\cdot 0=0

Step 4:Test option (3)

(A,B,C)=(1,0,1)

(1+0)\cdot 1=1

lights LED-1, and

A\cdot C=1\cdot 1=1

lights LED-2.

(1+0)\cdot 1=1\;\text{(LED-1 ON)};\; 1\cdot 1=1\;\text{(LED-2 ON)}

Step 5:Test option (4)

(A,B,C)=(1,1,0)

C=0

forces both AND outputs to

0

(1+1)\cdot 0=0

Step 6:Only the triple from option (3) drives both LED branches simultaneously high.

(A,B,C)=(1,0,1)

Final answer: A = 1, B = 0, C = 1

Q46NumericalProperties of Solids and Liquids

A cube has side length 5 cm and modulus of rigidity

10^{5}

m^{2}

. The displacement produced by a force of 10 N in the upper face of cube is _____ mm.

SolutionAnswer: 2

Approach:

Apply the definition of the shear modulus

\eta=(F/A)/(x/L)

for the cube, with the tangential force F acting along the upper face. Solve for the lateral displacement x and convert to millimetres.

Step 1:Convert the side length to SI and compute the area of the face on which the tangential force acts.

L=5\text{ cm}=0.05\text{ m},\; A=L^{2}=(0.05)^{2}=2.5\times 10^{-3}\text{ m}^{2}

Step 2:Substitute

F=10

L=0.05

\eta=10^{5}

m^{2}

A=2.5\times 10^{-3}

m^{2}

in the expression for x.

x=\dfrac{(10)(0.05)}{(10^{5})(2.5\times 10^{-3})}=\dfrac{0.5}{250}\text{ m}

Step 3:Convert metres to millimetres.

x=2\times 10^{-3}\text{ m}=2\text{ mm}

Final answer: 2

Q47NumericalKinematics

From 18 m height above the ground a ball is dropped from rest. The height above the ground at which the magnitude of velocity equal to the magnitude of acceleration (in the same set of units) due to gravity is ____ m. (Take g = 10 m/

s^{2}

and neglect the air resistance)

SolutionAnswer: 13

Approach:

For free fall from rest the kinematic relation

v^{2}=2g\,(H-h)

gives the speed at height h above the ground when the release height is H. Set the magnitude of velocity numerically equal to g and solve for h.

Step 1:Let the ball be at height

h

above the ground when its speed reaches the required value; the distance fallen from the release point at height

H=18

m is

H-h=18-h

v^{2}=2g\,(18-h)

Step 2:Impose the numerical condition

|v|=g=10

m/s.

10^{2}=2\,(10)\,(18-h)\Rightarrow 100=20(18-h)

Step 3:Solve for

h

h=18-5=13\text{ m}

Final answer: 13

Q48NumericalOscillations and Waves

A transverse wave on a string is described by

y = 3\sin(36t + 0.018x + \pi/4)

. where x, y are in cm and t in seconds. The least distance between the two successive crests in the wave is ____ cm. (Nearest integer) (

\pi = 3.14

)

SolutionAnswer: 349

Approach:

Read the wave number k as the coefficient of x in the phase of the sine, then compute the wavelength

\lambda=2\pi/k

. The least distance between two successive crests equals one full wavelength.

Step 1:From

y=3\sin(36t+0.018x+\pi/4)

with x in cm, identify the wave number as the coefficient of x.

k=0.018\text{ cm}^{-1}

Step 2:Substitute

\pi=3.14

in the wavelength formula.

\lambda=\dfrac{2\pi}{k}=\dfrac{2\,(3.14)}{0.018}=\dfrac{6.28}{0.018}\text{ cm}

Step 3:Round to the nearest integer to obtain the least distance between two successive crests.

d=\lambda\approx 348.89\text{ cm}\Rightarrow d=349\text{ cm}

Final answer: 349

Q49NumericalMagnetic Effects of Current and Magnetism

The charged particle moving in a uniform magnetic field of

(3\hat{i}+2\hat{j})

T has an acceleration

\left(4\hat{i}-\dfrac{x}{2}\hat{j}\right)

s^{2}

. The value of x is

SolutionAnswer: 12

Approach:

The magnetic force

\vec{F}=q\vec{v}\times\vec{B}

is always perpendicular to

\vec{B}

, hence the acceleration of the charged particle is perpendicular to

\vec{B}

. Enforce

\vec{a}\cdot\vec{B}=0

and solve for x.

Step 1:Since the only force on the charged particle is the magnetic force, the acceleration

\vec{a}=\vec{F}/m

is parallel to

\vec{F}

and therefore perpendicular to

\vec{B}

\vec{a}\perp\vec{B}\Rightarrow \vec{a}\cdot\vec{B}=0

Step 2:Compute the dot product with

\vec{B}=3\hat{i}+2\hat{j}

and

\vec{a}=4\hat{i}-\dfrac{x}{2}\hat{j}

\vec{a}\cdot\vec{B}=(4)(3)+\!\left(-\dfrac{x}{2}\right)(2)=12-x

Step 3:Set the dot product equal to zero and solve.

12-x=0\Rightarrow x=12

Final answer: 12

Q50NumericalElectromagnetic Induction and Alternating Currents

In the given circuit below inductance values of

L_1

L_2

and

L_3

are same. The magnetic energy stored in the entire circuit is (

U_t

) and that stored in the

L_2

inductor is (

U_l

\dfrac{U_t}{U_l}

is _____. (Ignore the mutual inductance if any)

Series-parallel inductor network with current arrows: inductor L1 on the left in series with a parallel combination of inductors L2 (upper branch) and L3 (lower branch); the network is terminated on the right with an outgoing current arrow.

SolutionAnswer: 6

Approach:

Find the equivalent inductance of the series-parallel network, express the total stored energy in terms of the main-line current I, use current division to obtain the current through

L_2

, then form the ratio

U_t/U_l

Step 1:Set

L_{1}=L_{2}=L_{3}=L

and let the main-line current through

L_{1}

be I. Compute the equivalent inductance.

L_{eq}=L+\dfrac{L\cdot L}{L+L}=L+\dfrac{L}{2}=\dfrac{3L}{2}

Step 2:For two identical inductors in parallel the current divides equally, so the current through

L_{2}

I/2

I_{L_2}=I\,\dfrac{L_{3}}{L_{2}+L_{3}}=\dfrac{I}{2}

Step 3:Compute the total magnetic energy stored in the network.

U_t=\tfrac{1}{2}L_{eq}I^{2}=\tfrac{1}{2}\!\left(\dfrac{3L}{2}\right)I^{2}=\dfrac{3}{4}LI^{2}

Step 4:Compute the energy stored in

L_{2}

U_l=\tfrac{1}{2}L\,(I/2)^{2}=\dfrac{1}{2}L\cdot\dfrac{I^{2}}{4}=\dfrac{LI^{2}}{8}

Step 5:Form the ratio.

\dfrac{U_t}{U_l}=\dfrac{3LI^{2}/4}{LI^{2}/8}=\dfrac{3}{4}\cdot 8=6

Final answer: 6

Chemistry24 questions

Q51Single correctSome Basic Concepts in Chemistry

How many grams of residue is obtained by heating 2.76 g of silver carbonate? (Given : Molar mass of

\text{C}, \text{O}

and

\text{Ag}

are

12, 16

and

108\ \text{g mol}^{-1}

respectively.)

(A)

\displaystyle 1.08\ \text{g}

(B)

\displaystyle 2.16\ \text{g}

(C)

\displaystyle 3.24\ \text{g}

(D)

\displaystyle 4.32\ \text{g}

SolutionAnswer: Option 2

2.16\ \text{g}

Approach:

Compute the molar mass of

\text{Ag}_2\text{CO}_3

, obtain moles taken, apply the thermal decomposition stoichiometry to find moles of metallic silver as the residue, and convert to mass.

Step 1:Compute the molar mass of

\text{Ag}_2\text{CO}_3

from the given atomic masses.

M(\text{Ag}_2\text{CO}_3) = 2(108) + 12 + 3(16) = 276\ \text{g mol}^{-1}

Step 2:Convert the given mass of silver carbonate to moles.

n(\text{Ag}_2\text{CO}_3) = \dfrac{2.76}{276} = 0.01\ \text{mol}

Step 3:On strong heating

\text{Ag}_2\text{CO}_3

decomposes to metallic silver (the residue) with

\text{CO}_2

and

\text{O}_2

as gases; each mole of

\text{Ag}_2\text{CO}_3

yields

2

mol of

\text{Ag}

n(\text{Ag}) = 2 \times 0.01 = 0.02\ \text{mol}

Step 4:Convert moles of silver to mass of the residue.

m(\text{Ag}) = 0.02 \times 108 = 2.16\ \text{g}

Final answer:

2.16\ \text{g}

Q52Single correctAtomic Structure

Arrange the following atomic orbitals of multi electron atoms in order of increasing energy.

A.

n = 3, l = 2, m = +1

n = 4, l = 0, m = 0

n = 6, l = 1, m = 0

n = 5, l = 1, m = +1

n = 2, l = 1, m = +1

Choose the correct answer from the options given below:

(A)

\displaystyle C < D < B < A < E

(B)

\displaystyle B < A < E < C < D

(C)

\displaystyle E < C < D < B < A

(D)

\displaystyle E < B < A < D < C

SolutionAnswer: Option 4

E < B < A < D < C

Approach:

Identify the orbital represented by each set of quantum numbers, compute

(n+l)

for each, and order them by the Aufbau

(n+l)

rule (smaller

n

breaks ties).

Step 1:Map each

(n,l)

pair to its orbital (

l=0

l=1

l=2

d).

A:3d,\ B:4s,\ C:6p,\ D:5p,\ E:2p

Step 2:Compute

(n+l)

for each orbital.

A(3d):5,\ B(4s):4,\ C(6p):7,\ D(5p):6,\ E(2p):3

Step 3:Order the orbitals by increasing

(n+l)

; no ties arise so the

n

tie-breaker is not required.

E(3) < B(4) < A(5) < D(6) < C(7)

Final answer:

E < B < A < D < C

Q53Single correctAtomic Structure

Identify the correct statements from the following :

A. Heisenberg uncertainty principle is applicable to electrons.

B. The size of

2p_x

orbital is less than the size of

3p_x

orbital.

C. The energy of 2 s orbital of H atom is equal to the energy of 2 s orbital of

\text{Li}^{+}

.

D. The electronic configuration of Cr is

[\text{Ar}]3d^{5}4s^{1}

.

Choose the correct answer from the options given below

(A)

A, B and C Only

(B)

A, B and D Only

(C)

B, C and D Only

(D)

A, C and D Only

SolutionAnswer: Option 2A, B and D Only

Approach:

Test each statement using the Heisenberg uncertainty principle, the n-dependence of orbital size, the

Z^{2}/n^{2}

scaling of one-electron-atom energies, and the anomalous configuration of chromium.

Step 1:Statement A: the uncertainty principle applies to all microscopic particles, including electrons, where its effect is most pronounced due to the small electron mass.

\Delta x\,\Delta p \geq \dfrac{h}{4\pi}\text{ holds for electrons}

Step 2:Statement B: the spatial extent of an orbital grows with the principal quantum number n; since

n=3

for

3p_x

exceeds

n=2

for

2p_x

, the

3p_x

orbital is larger.

\text{size}(2p_x) < \text{size}(3p_x)

Step 3:Statement C: H (

Z=1

) and

\text{Li}^{+}

(

Z=3

) are one-electron systems with different nuclear charges; the 2s energies scale as

Z^{2}

, so they cannot be equal.

E_{2s}(\text{H}) = -3.4\ \text{eV};\ E_{2s}(\text{Li}^{+}) = -3.4\times 9 = -30.6\ \text{eV}

Step 4:Statement D: chromium adopts

[\text{Ar}]3d^{5}4s^{1}

rather than

[\text{Ar}]3d^{4}4s^{2}

owing to the extra stability of a half-filled

3d

sub-shell.

\text{Cr}(Z=24):[\text{Ar}]3d^{5}4s^{1}

Step 5:Combine the verdicts.

Correct statements: A, B, D

Final answer: A, B and D Only

Q54Single correctSolutions

What is the mole fraction of water in

10

% by weight

(w/w)

of aqueous urea solution? [Given: Molar mass of

\text{H}, \text{O}, \text{C}

and

\text{N}

are

1, 16, 12

and

14\ \text{g mol}^{-1}

respectively.]

(A)

\displaystyle 0.825

(B)

\displaystyle 0.032

(C)

\displaystyle 0.867

(D)

\displaystyle 0.967

SolutionAnswer: Option 4

0.967

Approach:

Assume a convenient total mass of solution, partition it into urea and water by the given mass percentage, convert to moles, and evaluate the mole fraction of water.

Step 1:Take

100\ \text{g}

of solution; by definition of

10\%\ (w/w)

the mass of urea is

10\ \text{g}

and the mass of water is

90\ \text{g}

m_{\text{urea}} = 10\ \text{g},\ m_{\text{water}} = 90\ \text{g}

Step 2:Convert masses to moles using

M(\text{urea})=60

and

M(\text{H}_2\text{O})=18\ \text{g mol}^{-1}

n_{\text{urea}} = \dfrac{10}{60} = 0.1667\ \text{mol};\ n_{\text{water}} = \dfrac{90}{18} = 5\ \text{mol}

Step 3:Insert the mole counts into the mole-fraction definition for water.

x_{\text{water}} = \dfrac{5}{5 + 0.1667} = \dfrac{5}{5.1667}

Step 4:Evaluate the ratio numerically.

x_{\text{water}} = 0.9677 \approx 0.967

Final answer:

0.967

Q55Single correctEquilibrium

\text{M}_3\text{A}_2

is a sparingly soluble salt of molar mass

y\ \text{g mol}^{-1}

and solubility

x\ \text{g L}^{-1}

. The ratio of the molar concentration of the anion

(\text{A}^{3-})

to the solubility product of the salt is

(A)

\displaystyle \dfrac{1}{54}\cdot\dfrac{y^{4}}{x^{4}}

(B)

\displaystyle \dfrac{y^{5}}{108\,x^{4}}

(C)

\displaystyle 108\cdot\dfrac{x^{5}}{y^{5}}

(D)

\displaystyle \dfrac{1}{108}\cdot\dfrac{y^{4}}{x^{4}}

SolutionAnswer: Option 1

\dfrac{1}{54}\cdot\dfrac{y^{4}}{x^{4}}

Approach:

Convert the mass-based solubility to molar solubility s, write equilibrium concentrations from the dissociation stoichiometry, build

K_{\text{sp}}

, then form the required ratio

[\text{A}^{3-}]/K_{\text{sp}}

Step 1:Convert the given mass solubility into molar solubility using the molar mass.

s = \dfrac{x}{y}\ \text{mol L}^{-1}

Step 2:From the dissolution stoichiometry, write the equilibrium concentrations of the ions.

[\text{M}^{2+}] = 3s,\ [\text{A}^{3-}] = 2s

Step 3:Insert the ionic concentrations into the solubility-product expression.

K_{\text{sp}} = (3s)^{3}(2s)^{2} = 27 \cdot 4 \cdot s^{5} = 108\,s^{5}

Step 4:Form the required ratio of the anion concentration to the solubility product.

\dfrac{[\text{A}^{3-}]}{K_{\text{sp}}} = \dfrac{2s}{108\,s^{5}} = \dfrac{1}{54\,s^{4}}

Step 5:Substitute

s = x/y

to express the ratio in terms of

x

and

y

\dfrac{1}{54\,s^{4}} = \dfrac{1}{54}\cdot\left(\dfrac{y}{x}\right)^{4} = \dfrac{1}{54}\cdot\dfrac{y^{4}}{x^{4}}

Final answer:

\dfrac{1}{54}\cdot\dfrac{y^{4}}{x^{4}}

Q56Single correctEquilibrium

Arrange the following resultant mixtures in increasing order of their pH values

A.

10\ \text{mL}\ 0.2\,\text{M}\,\text{Ca(OH)}_2 + 25\ \text{mL}\ 0.1\,\text{M}\,\text{HCl}

10\ \text{mL}\ 0.01\,\text{M}\,\text{H}_2\text{SO}_4 + 10\ \text{mL}\ 0.01\,\text{M}\,\text{Ca(OH)}_2

10\ \text{mL}\ 0.1\,\text{M}\,\text{H}_2\text{SO}_4 + 10\ \text{mL}\ 0.1\,\text{M}\,\text{KOH}

Choose the correct answer from the options given below:

(A)

\displaystyle B < C < A

(B)

\displaystyle C < A < B

(C)

\displaystyle C < B < A

(D)

\displaystyle A < C < B

SolutionAnswer: Option 3

C < B < A

Approach:

For every mixture, compute the millimoles of

\text{H}^{+}

and

\text{OH}^{-}

(taking acid/base basicity into account), identify the excess ion, divide by the total volume, and convert to pH.

Step 1:Mixture A: millimoles of

\text{H}^{+}

from HCl are

25\times 0.1 = 2.5

; millimoles of

\text{OH}^{-}

from

\text{Ca(OH)}_2

are

10\times 0.2\times 2 = 4.0

; excess

\text{OH}^{-} = 1.5\ \text{mmol}

35\ \text{mL}

[\text{OH}^{-}] = \dfrac{1.5}{35} = 0.0429\ \text{M};\ \text{pOH} = 1.37;\ \text{pH}_A = 12.63

Step 2:Mixture B: millimoles of

\text{H}^{+}

from

\text{H}_2\text{SO}_4

are

10\times 0.01\times 2 = 0.20

; millimoles of

\text{OH}^{-}

from

\text{Ca(OH)}_2

are

10\times 0.01\times 2 = 0.20

; complete neutralisation yields neutral

\text{CaSO}_4

solution.

[\text{H}^{+}] = 10^{-7}\ \text{M};\ \text{pH}_B = 7

Step 3:Mixture C: millimoles of

\text{H}^{+}

from

\text{H}_2\text{SO}_4

are

10\times 0.1\times 2 = 2.0

; millimoles of

\text{OH}^{-}

from KOH are

10\times 0.1 = 1.0

; excess

\text{H}^{+} = 1.0\ \text{mmol}

20\ \text{mL}

[\text{H}^{+}] = \dfrac{1.0}{20} = 0.05\ \text{M};\ \text{pH}_C = 1.30

Step 4:Order the three pH values.

\text{pH}_C(1.30) < \text{pH}_B(7) < \text{pH}_A(12.63)

Final answer:

C < B < A

Q57Single correctChemical Kinetics

First order gas phase reaction

\text{A} \rightarrow \text{B} + \text{C}

p_i =

initial pressure of gas

\text{A}

p_t =

total pressure of the reaction mixture at time t

Expression of rate constant (k) is

(A)

\displaystyle \dfrac{1}{t}\ln\dfrac{p_i}{2p_i - p_t}

(B)

\displaystyle \dfrac{1}{t}\ln\dfrac{2p_i}{p_i - p_t}

(C)

\displaystyle \dfrac{1}{t}\ln\dfrac{p_i}{3p_i - 2p_t}

(D)

\displaystyle \dfrac{1}{t}\ln\dfrac{3p_i}{4p_i - p_t}

SolutionAnswer: Option 1

\dfrac{1}{t}\ln\dfrac{p_i}{2p_i - p_t}

Approach:

Parametrise partial pressures of A, B and C in terms of an extent of reaction, relate this extent to the measured total pressure, and substitute the resulting expression for

p_A(t)

into the integrated first-order rate law.

Step 1:Let x denote the partial pressure of A that has reacted by time t. From the

1:1:1

stoichiometry,

p_B = p_C = x

and

p_A = p_i - x

p_A = p_i - x,\quad p_B = p_C = x

Step 2:Express the total pressure and solve for x in terms of

p_t

and

p_i

p_t = (p_i - x) + x + x = p_i + x\ \Rightarrow\ x = p_t - p_i

Step 3:Substitute x back into

p_A

to obtain the partial pressure of A at time t.

p_A(t) = p_i - x = p_i - (p_t - p_i) = 2p_i - p_t

Step 4:Insert

p_{A,0} = p_i

and

p_A(t) = 2p_i - p_t

into the integrated first-order rate law.

k = \dfrac{1}{t}\ln\dfrac{p_i}{2p_i - p_t}

Final answer:

\dfrac{1}{t}\ln\dfrac{p_i}{2p_i - p_t}

Q58Single correctp-Block Elements

Given below are two statements:

Statement I: The correct order of electronegativity of fluorine, oxygen and nitrogen is

F > O > N

.

Statement II: The oxidation state of oxygen in

\text{OF}_2

+2

and in

\text{Na}_2\text{O}

-2

.

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

(A)

Statement I and Statement II are both correct

(B)

Statement I and Statement II are both incorrect

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

SolutionAnswer: Option 1Statement I and Statement II are both correct

Approach:

Compare Pauling electronegativities of F, O and N for Statement I, then assign oxidation states using electronegativity-based sign rules in

\text{OF}_2

and

\text{Na}_2\text{O}

for Statement II.

Step 1:Statement I: along the second period electronegativity rises from N to O to F, so the order

F > O > N

is correct.

\chi(F) > \chi(O) > \chi(N)

Step 2:Statement II — oxidation state of O in

\text{OF}_2

: F is more electronegative than O, so each F carries

-1

and O must be

+2

x + 2(-1) = 0\ \Rightarrow\ x = +2

Step 3:Statement II — oxidation state of O in

\text{Na}_2\text{O}

: Na is less electronegative, so each Na is

+1

and O is

-2

2(+1) + x = 0\ \Rightarrow\ x = -2

Step 4:Both individual assignments support Statement II; combine with Statement I.

I: correct; II: correct

Final answer: Statement I and Statement II are both correct

Q59Single correctp-Block Elements

Correct statements from the following are :

A. Nitrogen in oxidation states from

+1

+4

disproportionates in acid medium.

B. Nitrogen has the ability to form

d\pi - p\pi

multiple bonds with itself and other elements with small size and high electronegativity.

C.

N-N

single bond is stronger than

P-P

single bond.

D. Nitrogen has highest density in its group due to small size.

E. The maximum covalency of nitrogen is four since it has only four valence orbitals for bonding.

Choose the correct answer from the options given below:

(A)

B, C and D Only

(B)

C, D and E Only

(C)

A, C and E Only

(D)

A and E Only

SolutionAnswer: Option 4A and E Only

Approach:

Evaluate each statement against established facts about nitrogen chemistry: disproportionation of intermediate oxidation states, valence-orbital limitations of N, bond-energy comparisons in group 15, density trend down the group, and the maximum covalency rule.

Step 1:Statement A: oxoacids and oxides of nitrogen in intermediate oxidation states

+1

+4

are known to disproportionate in acidic medium (e.g.

\text{HNO}_2

3\,\text{HNO}_2 \rightarrow \text{HNO}_3 + 2\,\text{NO} + \text{H}_2\text{O}

Step 2:Statement B: nitrogen has

n=2

and lacks accessible d orbitals; therefore it forms

p\pi - p\pi

multiple bonds, not

d\pi - p\pi

bonds.

No 2

d

orbitals available on N

Step 3:Statement C: the

N-N

single bond is weaker than the

P-P

single bond because of strong lone-pair–lone-pair repulsion in the small

N-N

distance.

D(N-N) < D(P-P)

Step 4:Statement D: nitrogen exists as a diatomic gas at room temperature and has the lowest density in group 15; density increases down the group toward bismuth.

\rho(\text{N}_2) \approx 1.25\ \text{g L}^{-1}\ \ll\ \rho(\text{Bi solid})

Step 5:Statement E: nitrogen has only four valence orbitals (

2s,\ 2p_x,\ 2p_y,\ 2p_z

), limiting its maximum covalency to four.

\text{Valence orbitals of N: } 2s, 2p_x, 2p_y, 2p_z

Step 6:Collate verdicts.

Correct: A and E

Final answer: A and E Only

Q60Single correctd- and f-Block Elements

Which of the following is NOT a physical or chemical characteristics of interstitial compounds?

(A)

They have high melting points, higher than those of pure metals.

(B)

They are very soft and ionic in nature.

(C)

They retain metallic conductivity.

(D)

They are chemically inert and usually non-stoichiometric.

SolutionAnswer: Option 2They are very soft and ionic in nature.

Approach:

Recall the defining physical and chemical features of interstitial compounds (small atoms H, C, N, B occupying voids of a transition-metal lattice) and identify the option that contradicts the established profile.

Step 1:Option 1: interstitial compounds have higher melting points than the parent metal because the trapped atoms reinforce the metallic lattice.

T_m(\text{interstitial}) > T_m(\text{pure metal})

Step 2:Option 2: interstitial compounds are hard (often harder than the parent metal) and remain metallic in bonding; describing them as soft and ionic contradicts both their hardness and their bonding character.

\text{Hardness}\ \uparrow,\ \text{bonding: metallic/covalent (not ionic)}

Step 3:Option 3: the parent metal's conduction band remains essentially intact, so interstitial compounds retain metallic conductivity.

\sigma_{\text{interstitial}} \sim \sigma_{\text{parent metal}}

Step 4:Option 4: interstitial compounds are chemically inert and frequently non-stoichiometric, e.g.

\text{TiH}_{1.7}

and

\text{VH}_{0.56}

Inert and non-stoichiometric: e.g.

\text{TiH}_{1.7}

Final answer: They are very soft and ionic in nature.

Q61Single correctCoordination Compounds

The correct statements about metal carbonyls are:

A. The metal-carbon bonds in metal carbonyls possess both

\sigma

and

\pi

character.

B. Due to synergic bonding interactions between metal and CO ligand, the metal-carbon bond becomes weak.

C. The metal-carbon

\sigma

bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of metal.

D. The metal-carbon

\pi

bond is formed by the donation of electrons from filled d-orbital of metal into vacant

\pi^{*}

orbital of CO.

Choose the correct answer from the options given below:

(A)

A and B Only

(B)

A, C and D Only

(C)

B and C Only

(D)

A and D Only

SolutionAnswer: Option 2A, C and D Only

Approach:

Analyse each statement within the synergic bonding model of metal carbonyls:

\sigma

donation from the CO lone pair to a vacant metal orbital and

\pi

back-donation from a filled metal d-orbital into the empty

\pi^{*}

orbital of CO.

Step 1:Statement A: the metal-carbon bond in metal carbonyls has both

\sigma

donation and

\pi

back-donation, giving it multiple-bond character.

M-C\ \text{has}\ \sigma + \pi\ \text{character}

Step 2:Statement B: synergic bonding is mutually reinforcing;

\sigma

donation increases electron density on the metal, which in turn enhances

\pi

back-donation, so the M-C bond is strengthened, not weakened.

\sigma\text{ donation} \uparrow \Rightarrow \pi\text{ back-donation} \uparrow \Rightarrow M-C\ \text{strengthened}

Step 3:Statement C: the

\sigma

component is formed by donation of the carbon lone pair into an empty metal orbital.

\text{CO lone pair}\ \rightarrow\ M\ (\sigma)

Step 4:Statement D: the

\pi

component arises from donation of a filled metal d-orbital into the vacant antibonding

\pi^{*}

orbital of CO.

M(d)\ \rightarrow\ \pi^{*}(\text{CO})

Step 5:Collate verdicts.

Correct: A, C, D

Final answer: A, C and D Only

Q62Single correctCoordination Compounds

Given below are two statements:

Statement I: Each electron in

e_g

orbitals destabilises the orbitals by

+0.6\Delta_o

and each electron in the

t_{2g}

orbitals stabilizes the orbitals by

-0.4\Delta_o

in an octahedral field on the basis of crystal field theory.

Statement II: All the d-orbitals of the transition metals have the same energy in their free atomic state but when a complex is formed the ligands destroy the degeneracy of these orbitals on the basis of crystal field theory.

In the light of the above statements, choose the correct answer from the options given below

(A)

Both Statement I and Statement II are correct

(B)

Both Statement I and Statement II are incorrect

(C)

Statement I is correct but Statement II is incorrect

(D)

Statement I is incorrect but Statement II is correct

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

Approach:

Apply crystal-field theory to an octahedral complex: the five degenerate d orbitals of the free ion split into a triply degenerate

t_{2g}

set lowered by

-0.4\Delta_o

and a doubly degenerate

e_g

set raised by

+0.6\Delta_o

relative to the barycentre.

Step 1:Statement I: in an octahedral crystal field the per-electron destabilisation of an

e_g

orbital is

+0.6\Delta_o

and the per-electron stabilisation of a

t_{2g}

orbital is

-0.4\Delta_o

, which is the textbook splitting.

\Delta E_{e_g} = +0.6\Delta_o;\ \Delta E_{t_{2g}} = -0.4\Delta_o

Step 2:Statement II: in the free transition-metal ion the five d orbitals

\{d_{xy},d_{yz},d_{xz},d_{x^{2}-y^{2}},d_{z^{2}}\}

are degenerate; the approach of ligands generates an inhomogeneous electrostatic field that lifts this degeneracy.

\text{Free ion: degenerate}\ \rightarrow\ \text{ligand field: }t_{2g} + e_g

Step 3:Combine the verdicts; both statements describe the standard CFT picture correctly.

I: correct; II: correct

Final answer: Both Statement I and Statement II are correct

Q63Single correctSome Basic Principles of Organic Chemistry

Given below are two statements:
Statement I: On the basis of inductive effect, the order of stability of alkyl carbanions is

\text{CH}_3^- > \text{CH}_3 - \text{CH}_2^- > (\text{CH}_3)_2 \text{CH}^- > (\text{CH}_3)_3 \text{C}^-

.
Statement II: Allyl and benzyl carbanions are more stabilised by inductive effect and not by resonance effect.
In the light of the above statements, choose the correct answer from the options given below

(A)

Both Statement I and Statement II are correct

(B)

Both Statement I and Statement II are incorrect

(C)

Statement I is correct but Statement II is incorrect

(D)

Statement I is incorrect but Statement II is correct

SolutionAnswer: Option 3Statement I is correct but Statement II is incorrect

Approach:

Test Statement I against the

+I

-effect ordering of alkyl carbanion stability and Statement II against the actual mode of stabilisation in allyl and benzyl carbanions.

Step 1:Apply the

+I

effect to the four alkyl carbanions in Statement I. Each additional alkyl group on the anionic carbon donates electron density toward an already electron-rich centre, so it destabilises the carbanion.

\text{Stability:}\; \text{CH}_3^{-} > \text{CH}_3\text{CH}_2^{-} > (\text{CH}_3)_2\text{CH}^{-} > (\text{CH}_3)_3\text{C}^{-}

Step 2:Identify the dominant stabilising interaction in allyl and benzyl carbanions. The negative charge is delocalised by resonance into the adjacent

\pi

system, not by alkyl

+I

donation.

\text{Allyl: }\text{CH}_2=\text{CH}-\text{CH}_2^{-} \leftrightarrow {}^{-}\text{CH}_2-\text{CH}=\text{CH}_2;\;\text{Benzyl: charge delocalised to ortho/para of the ring.}

Step 3:Combine: Statement I is correct; Statement II reverses cause-and-effect (claims inductive, denies resonance) and is therefore incorrect.

I: correct, II: incorrect

Final answer: Statement I is correct but Statement II is incorrect

Q64Single correctHydrocarbons

"P" is a hydrocarbon of molecular formula:-

\text{C}_8\text{H}_{14}

. On ozonolysis, "P" forms "Q". "Q" on treatment with alkali under reflux condition produces "R", which on treatment with

\text{I}_2/\text{NaOH}

gives a yellow precipitate. Acidification of the solution gives "S". The structure of "S" is given below:-

The correct structure of "P" is

Structure of compound S: a cyclopentene ring with a methyl substituent at one sp2 carbon of the double bond and a -COOH group on the adjacent sp2 carbon (1-methyl-2-cyclopentene-carboxylic acid).

(A)

1,4-dimethyl-1,4-cyclohexadiene: a six-membered ring with two non-conjugated C=C bonds and a methyl group on each of the two sp2 carbons that are para to each other.

(B)

1-ethylcyclohex-1-ene: a cyclohexene ring with an ethyl (-CH2-CH3) substituent on one of the sp2 ring carbons.

(C)

A cyclopentene ring with a methyl group on one sp2 ring carbon and an ethyl (-CH2-CH3) group on the adjacent sp2 ring carbon (1-methyl-2-ethylcyclopent-1-ene).

(D)

SolutionAnswer: Option 41,2-dimethylcyclohex-1-ene (a cyclohexene ring with two methyl groups, one on each sp2 ring carbon of the double bond)

Approach:

Work backwards from S through the iodoform step to R (a methyl ketone), back through the intramolecular aldol to Q (a 1,6-diketone), and finally reverse the ozonolysis to obtain P as a methyl-substituted cyclohexene with the correct molecular formula

\text{C}_8\text{H}_{14}

Step 1:Read S. The structure shown is 1-methyl-2-cyclopentene-1-carboxylic acid: a cyclopentene with a methyl substituent on the s

p^{2}

ring carbon bearing the

-\text{COOH}

group. The

-\text{COOH}

is generated by the iodoform step on acidification, so R must contain a methyl ketone

-\text{COCH}_3

at the position now occupied by

-\text{COOH}

R{-}\text{COCH}_3 \xrightarrow{I_2/\text{NaOH};\;\text{then }H^{+}} R{-}\text{COOH} + \text{CHI}_3

Step 2:Reverse the intramolecular aldol that produces R. A symmetrical 1,6-diketone

\text{CH}_3\text{CO}{-}(\text{CH}_2)_4{-}\text{COCH}_3

undergoes base-catalysed cyclisation: an

\alpha

-carbon next to one methyl ketone attacks the carbonyl of the other, dehydration then gives the

\alpha,\beta

-unsaturated cyclopentenyl methyl ketone observed in R.

Q = \text{CH}_3{-}\text{CO}{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_2{-}\text{CO}{-}\text{CH}_3\;(\text{octane-2,7-dione})

Step 3:Reverse the ozonolysis from P to Q. The two carbonyl carbons of Q must be the two s

p^{2}

carbons of P; reconnecting them with a C=C and closing the chain into a ring gives a six-membered ring bearing one methyl group on each s

p^{2}

ring carbon.

P = \text{1,2-dimethylcyclohex-1-ene}

Step 4:Compare with the listed options.

Option 4: 1,2-dimethylcyclohex-1-ene

Final answer: 1,2-dimethylcyclohex-1-ene (cyclohexene ring with a methyl group on each sp2 ring carbon of the double bond)

Q65Single correctHydrocarbons

For the following Friedel Craft's alkylation reaction, which of the statements are correct?

Benzene + n-propyl chloride (CH3-CH2-CH2-Cl) in the presence of anhydrous AlCl3 produces an alkylbenzene.

(A)

A and D only

(B)

B and C only

(C)

A and C only

(D)

B and D only

SolutionAnswer: Option 2B and C only

Approach:

Evaluate each of the four statements about the Friedel-Crafts alkylation of benzene with n-propyl chloride/

\text{AlCl}_3

against the standard limitations of the reaction: carbocation rearrangement, polysubstitution, and substrate electronic requirements.

Step 1:Statement A claims the major product is n-propylbenzene. The n-propyl cation generated from n-propyl chloride/

\text{AlCl}_3

is primary and rearranges to the more stable isopropyl cation before attack on benzene; the major product is therefore isopropylbenzene (cumene). Statement A is incorrect.

\text{C}_6\text{H}_6 + (\text{CH}_3)_2\text{CH}^{+} \rightarrow \text{C}_6\text{H}_5{-}\text{CH}(\text{CH}_3)_2

Step 2:Statement B claims the n-propyl cation rearranges from

1^{\circ}

2^{\circ}

. The 1,2-hydride shift on the primary n-propyl cation indeed produces the more stable isopropyl cation.

\text{CH}_3\text{CH}_2\text{CH}_2^{+} \xrightarrow{\text{1,2-H shift}} (\text{CH}_3)_2\text{CH}^{+}

Step 3:Statement C claims multiple substitution is inevitable. Once a single alkyl group is attached to the ring, its

+I

effect makes the ring more nucleophilic than benzene itself, so the alkylated product reacts further with

R^{+}/\text{AlCl}_3

, giving di- and polyalkylated products.

\text{Alkylbenzene reacts faster than benzene} \Rightarrow \text{di-/tri-alkylation results}

Step 4:Statement D claims that introducing an electron-donating substituent prevents Friedel-Crafts alkylation. Electron-donating substituents activate the ring toward electrophilic substitution; the reaction that fails is on rings bearing strong electron-withdrawing groups (e.g.

-\text{NO}_2

). Statement D is incorrect.

\text{Electron-donating substituents (}-\text{OR}, -\text{NHR}, -\text{CH}_3\text{)} \Rightarrow \text{ring activated, alkylation proceeds.}

Step 5:Combine: the only correct statements are B and C.

\text{Correct statements} = \{B,\;C\}

Final answer: B and C only

Q66Single correctOrganic Compounds Containing Nitrogen

Benzyl isocyanide can be obtained from

Five labelled reaction schemes (A-E) for preparing an isocyanide/amine: (A) benzyl bromide (C6H5-CH2-Br) + AgCN; (B) benzylamine (C6H5-CH2-NH2) + CHCl3/aqueous NaOH (carbylamine reaction); (C) bromobenzene (C6H5-Br) + AgCN; (D) aniline (C6H5-NH2) + CHCl3/aqueous NaOH; (E) 2-phenylethyl bromide (C6H5-CH2-CH2-Br) + KCN.

(A)

A and B only

(B)

A and C only

(C)

B and D only

(D)

D and E only

SolutionAnswer: Option 1A and B only

Approach:

Benzyl isocyanide has the structure

\text{C}_6\text{H}_5{-}\text{CH}_2{-}\text{NC}

. Select the routes whose product matches this exact connectivity (benzyl carbon bonded to N of

-\text{NC}

Step 1:Route A: benzyl bromide + AgCN. Silver cyanide attacks via nitrogen at the benzyl carbon, delivering the isocyanide.

\text{C}_6\text{H}_5{-}\text{CH}_2{-}\text{Br} + \text{AgCN} \rightarrow \text{C}_6\text{H}_5{-}\text{CH}_2{-}\text{NC} + \text{AgBr}

Step 2:Route B: benzylamine +

\text{CHCl}_3/

\text{NaOH}

. The carbylamine reaction on a primary amine generates the corresponding isocyanide via in-situ dichlorocarbene.

\text{C}_6\text{H}_5{-}\text{CH}_2{-}\text{NH}_2 + \text{CHCl}_3 + 3\,\text{NaOH} \rightarrow \text{C}_6\text{H}_5{-}\text{CH}_2{-}\text{NC} + 3\,\text{NaCl} + 3\,\text{H}_2\text{O}

Step 3:Route C: bromobenzene + AgCN. Aryl halides are essentially inert toward nucleophilic substitution under these conditions; even if it proceeded, the product would be phenyl isocyanide

\text{C}_6\text{H}_5{-}\text{NC}

, not benzyl isocyanide.

\text{C}_6\text{H}_5{-}\text{Br} + \text{AgCN} \nrightarrow \text{C}_6\text{H}_5{-}\text{CH}_2{-}\text{NC}

Step 4:Route D: aniline +

\text{CHCl}_3/

\text{NaOH}

is the carbylamine test, but the substrate is aniline, so the product is phenyl isocyanide, not benzyl isocyanide.

\text{C}_6\text{H}_5{-}\text{NH}_2 + \text{CHCl}_3/\text{NaOH} \rightarrow \text{C}_6\text{H}_5{-}\text{NC}

Step 5:Route E: 2-phenylethyl bromide + KCN. The ionic cyanide

\text{CN}^{-}

from KCN attacks through carbon, producing the nitrile, and the carbon chain (Ph–C

H_2

–C

H_2

–) is one methylene longer than benzyl.

\text{C}_6\text{H}_5{-}\text{CH}_2{-}\text{CH}_2{-}\text{Br} + \text{KCN} \rightarrow \text{C}_6\text{H}_5{-}\text{CH}_2{-}\text{CH}_2{-}\text{CN}

Step 6:Combine: only A and B deliver benzyl isocyanide.

\{A,\;B\}\;\text{produce}\;\text{C}_6\text{H}_5{-}\text{CH}_2{-}\text{NC}

Final answer: A and B only

Q67Single correctHydrocarbons

Consider compounds A, B and C with following structural formulae

\text{A} = \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{OH}

\text{B} = \text{CH}_2=\text{CH}-\text{CH}_2-\text{CH}_2-\text{CH}_3

\text{C} = \text{HO}-\text{CH}_2-\text{CH}_2-\text{CH}(\text{OH})-\text{CH}_3

For the conversion of B from A, reagent (D) required is _____ and structural formula of product (E) obtained when C undergoes same reaction using excess reagent (D) is _____ .

(A)

\displaystyle \text{D: Conc. H}_2\text{SO}_4

;

\displaystyle \text{E: CH}_2=\text{CH}-\text{CH}(\text{OH})/\text{CH}_3

(B)

\displaystyle \text{D: PCC}

;

\displaystyle \text{E: HO}-\text{CH}_2-\text{CH}_2-\text{CH}=\text{CH}_2

(C)

\displaystyle \text{D: PCC}

;

\displaystyle \text{E: CH}_2=\text{CH}-\text{CH}=\text{CH}_2

(D)

\displaystyle \text{D: Conc. H}_2\text{SO}_4

\displaystyle \text{H}_3\text{PO}_4

;

\displaystyle \text{E: CH}_2=\text{CH}-\text{CH}=\text{CH}_2

SolutionAnswer: Option 4

\text{D: Conc. H}_2\text{SO}_4

\text{H}_3\text{PO}_4

;

\text{E: CH}_2=\text{CH}-\text{CH}=\text{CH}_2

Approach:

Identify

A \to B

as an acid-catalysed dehydration of an alcohol to an alkene; the same reagent applied to the 1,3-diol C in excess removes both

-\text{OH}

groups to form a conjugated diene.

Step 1:Reagent D for

A \to B

A = \text{CH}_3{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_2{-}\text{OH}

(pentan-1-ol) loses water to give

B = \text{CH}_2=\text{CH}{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_3

(pent-1-ene); this is a dehydration and the standard reagent is concentrated

H_2SO_4

H_3PO_4

at elevated temperature. PCC oxidises a primary alcohol to an aldehyde and cannot make an alkene, so PCC is ruled out.

\text{CH}_3{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_2{-}\text{OH} \xrightarrow{\text{Conc. }H_2SO_4\;\text{or}\;H_3PO_4,\;\Delta} \text{CH}_2=\text{CH}{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH}_3 + H_2O

Step 2:Apply the same reagent in excess to the diol

C = \text{HO}{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH(OH)}{-}\text{CH}_3

(butane-1,3-diol). Each

-\text{OH}

is protonated and lost as water; the two eliminations install two

\text{C}=\text{C}

bonds in the four-carbon chain, which combine to a conjugated diene.

\text{HO}{-}\text{CH}_2{-}\text{CH}_2{-}\text{CH(OH)}{-}\text{CH}_3 \xrightarrow{\text{excess Conc. }H_2SO_4,\;\Delta} \text{CH}_2=\text{CH}{-}\text{CH}=\text{CH}_2 + 2\,H_2O

Step 3:Match against the option list.

\text{Option 4}:\;D = \text{Conc. }H_2SO_4\;\text{or}\;H_3PO_4;\;E = \text{CH}_2=\text{CH}{-}\text{CH}=\text{CH}_2

Final answer: D: Conc. H2SO4 or H3PO4; E:

\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2

Q68Single correctOrganic Compounds Containing Nitrogen

Identify the incorrect statements.

Four statements drawn as structural formulae and reactions: A. benzylamine (C6H5-CH2-NH2) is a stronger base than aniline (C6H5-NH2). B. p-methoxyaniline (H3CO-C6H4-NH2) can be synthesised by Gabriel phthalimide synthesis. C. 2-phenylacetamide (C6H5-CH2-CONH2) treated with Br2/NaOH gives a primary aromatic amine. D. p-nitroaniline (O2N-C6H4-NH2) treated with (i) NaNO2/HCl at 0 degree C and (ii) heat (Delta) gives a product that dissolves in NaOH.

(A)

A and D Only

(B)

A and C Only

(C)

B and C Only

(D)

A and B Only

SolutionAnswer: Option 3B and C Only

Approach:

Test each statement against the underlying nitrogen chemistry: lone-pair conjugation (basicity), the substrate scope of Gabriel synthesis, the Hofmann bromamide rearrangement and the diazotisation-hydrolysis sequence on

p

-nitroaniline. Mark those that are factually wrong.

Step 1:Statement A. In benzylamine the lone pair on N is on an s

p^{3}

nitrogen and is not delocalised into the ring, whereas in aniline the lone pair is conjugated with the ring

\pi

-system and is less available to a proton. Therefore benzylamine is more basic than aniline; statement A is a CORRECT chemical fact.

pK_b(\text{PhCH}_2\text{NH}_2) \approx 4.66 < pK_b(\text{PhNH}_2) \approx 9.4 \Rightarrow \text{PhCH}_2\text{NH}_2\;\text{stronger base}

Step 2:Statement B. Preparing p-methoxyaniline by Gabriel synthesis would require phthalimide ion to displace X on a p-methoxyaryl halide (

p\text{-CH}_3\text{O}{-}\text{C}_6\text{H}_4{-}X

). Aryl halides do not undergo

S_N

2 and are inert under standard Gabriel conditions; therefore p-methoxyaniline cannot be obtained by this route. Statement B is INCORRECT.

\text{Phthalimide}^{-} + p\text{-CH}_3\text{O}{-}\text{C}_6\text{H}_4{-}X \nrightarrow N\text{-aryl phthalimide}

Step 3:Statement C. Hofmann bromamide on 2-phenylacetamide (

\text{PhCH}_2{-}\text{CONH}_2

) loses the carbonyl carbon, leaving the benzyl group on nitrogen:

\text{PhCH}_2{-}\text{NH}_2

. This is benzylamine, an aliphatic (

-\text{CH}_2\text{NH}_2

) primary amine, NOT a primary aromatic amine (in which

-\text{NH}_2

is bonded directly to a ring carbon). Statement C is INCORRECT.

\text{PhCH}_2{-}\text{CONH}_2 \xrightarrow{Br_2/\text{NaOH}} \text{PhCH}_2{-}\text{NH}_2\;(\text{aliphatic 1}^{\circ}\;\text{amine})

Step 4:Statement D. Diazotisation of p-nitroaniline at

0^{\circ}

C gives the diazonium salt; warming with water hydrolyses it to p-nitrophenol, whose phenolic

-\text{OH}

(

pK_a \approx 7.15

) is sufficiently acidic to dissolve in NaOH as the sodium p-nitrophenoxide. Statement D is CORRECT.

p\text{-O}_2\text{N}{-}\text{C}_6\text{H}_4{-}\text{NH}_2 \xrightarrow{NaNO_2/HCl,\;0^{\circ}C} p\text{-O}_2\text{N}{-}\text{C}_6\text{H}_4{-}N_2^{+}Cl^{-} \xrightarrow{H_2O,\;\Delta} p\text{-O}_2\text{N}{-}\text{C}_6\text{H}_4{-}\text{OH} \xrightarrow{\text{NaOH}} p\text{-O}_2\text{N}{-}\text{C}_6\text{H}_4{-}\text{O}^{-}Na^{+}

Step 5:The incorrect statements are B and C.

\text{Incorrect set} = \{B,\;C\}

Final answer: B and C Only

Q69Single correctBiomolecules

Identify the correct statements.
A. Glucose exists in two anomeric forms.
B. Anomers of glucose differ in configuration at

\text{C}_1

in cyclic hemiacetal structure.
C. Melting point of

\alpha

-anomer of glucose is greater than

\beta

-anomer.
D. Specific rotation of

\alpha

-anomer is

19^{\circ}

while for

\beta

-anomer is

112^{\circ}

.
E.

\alpha

and

\beta

-anomers of glucose are prepared by crystallization of saturated glucose solution at 303 K and 371 K respectively.
Choose the correct answer from the options given below:

(A)

A and B Only

(B)

B and C Only

(C)

A, B and D Only

(D)

A, B and E Only

SolutionAnswer: Option 4A, B and E Only

Approach:

Compare each of the five statements to the standard NCERT data on the cyclic hemiacetal forms of D-glucose: definition of anomers, the carbon at which they differ, melting points, specific rotations and crystallisation conditions of

\alpha

- and

\beta

-D-glucose.

Step 1:Statement A: glucose exists in two anomeric forms (

\alpha

and

\beta

cyclic hemiacetals). Standard fact.

A: correct

Step 2:Statement B: anomers differ in configuration at the anomeric carbon

C_1

in the cyclic hemiacetal. This is the definition of an anomer.

B: correct

Step 3:Statement C: melting point of

\alpha

greater than

\beta

. Actual values give

\alpha \approx 146^{\circ}C < \beta \approx 150^{\circ}C

, so the inequality is reversed.

m.p.(\alpha) = 146^{\circ}C < m.p.(\beta) = 150^{\circ}C

Step 4:Statement D:

[\alpha]_D^{\alpha} = 19^{\circ}

and

[\alpha]_D^{\beta} = 112^{\circ}

. Standard values give

[\alpha]_D^{\alpha} = +111^{\circ}

and

[\alpha]_D^{\beta} = +19^{\circ}

, exactly swapped.

[\alpha]_D^{\alpha} = +111^{\circ},\;[\alpha]_D^{\beta} = +19^{\circ}

Step 5:Statement E: NCERT crystallisation temperatures for

\alpha

- and

\beta

-D-glucose are 303 K and 371 K respectively.

E: correct

Step 6:Correct statements are A, B and E.

\text{Correct set} = \{A,\;B,\;E\}

Final answer: A, B and E Only

Q70Single correctPrinciples Related to Practical Chemistry

Given below are two statements:
Statement I: Sodium dichromate and potassium dichromate are classified as primary standards in titrimetric analysis.
Statement II: Phenolphthalein is a weak base, therefore it dissociates in acidic medium.
In the light of the above statements, choose the correct answer from the options given below

(A)

Both Statement I and Statement II are true

(B)

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:

Check Statement I against the criteria for a primary standard and Statement II against the structure and acid-base behaviour of phenolphthalein.

Step 1:Statement I claims both

\text{Na}_2\text{Cr}_2\text{O}_7

and

\text{K}_2\text{Cr}_2\text{O}_7

are primary standards.

\text{K}_2\text{Cr}_2\text{O}_7

qualifies (non-hygroscopic, high purity).

\text{Na}_2\text{Cr}_2\text{O}_7

is deliquescent and absorbs atmospheric moisture, violating the stability/composition criterion; it is therefore NOT a primary standard. The blanket statement is false.

\text{Na}_2\text{Cr}_2\text{O}_7

hygroscopic

\Rightarrow

not a primary standard;

\text{K}_2\text{Cr}_2\text{O}_7

is.

Step 2:Statement II claims phenolphthalein is a weak base and dissociates in acidic medium. Phenolphthalein contains phenolic

-\text{OH}

groups in a lactone framework and behaves as a WEAK ACID (

pK_a \approx 9.4

); the equilibrium

\text{HIn} \rightleftharpoons H^{+} + \text{In}^{-}

shifts toward dissociation (the pink I

n^{-}

) in BASIC medium, not acidic.

Phenolphthalein: weak acid; ionises in basic medium (pink); in acid it stays in the colourless HIn form.

Step 3:Combine: both statements are false.

Final selection: Both Statement I and Statement II are false

Final answer: Both Statement I and Statement II are false

Q71NumericalChemical Bonding and Molecular Structure

Consider the following species:

\text{BrF}_5, \text{XeF}_5^{-}, \text{BF}_4^{-}, \text{ICl}_4^{-}, \text{XeF}_4, \text{SF}_4, \text{NH}_4^{+}, \text{ClF}_3, \text{XeF}_2, \text{ICl}_2^{-}

Number of species having

sp^3 d

hybridized central atom is ____ .

SolutionAnswer: 4

Approach:

For each species, count the steric number SN on the central atom (

SN = \text{number of }\sigma\text{-bonded atoms} + \text{number of lone pairs}

). Map each SN to the corresponding hybridisation and count those equal to

sp^3 d

(i.e.

SN = 5

Step 1:Compute

SN

for the first five species.

\text{BrF}_5: 5+1=6\Rightarrow sp^3 d^2;\;\text{XeF}_5^{-}: 5+2=7\Rightarrow sp^3 d^3;\;\text{BF}_4^{-}: 4+0=4\Rightarrow sp^3;\;\text{ICl}_4^{-}: 4+2=6\Rightarrow sp^3 d^2;\;\text{XeF}_4: 4+2=6\Rightarrow sp^3 d^2.

Step 2:Compute

SN

for the remaining five species.

\text{SF}_4: 4+1=5\Rightarrow sp^3 d;\;\text{NH}_4^{+}: 4+0=4\Rightarrow sp^3;\;\text{ClF}_3: 3+2=5\Rightarrow sp^3 d;\;\text{XeF}_2: 2+3=5\Rightarrow sp^3 d;\;\text{ICl}_2^{-}: 2+3=5\Rightarrow sp^3 d.

Step 3:Total the

sp^3 d

species across the ten given.

\{\text{SF}_4,\;\text{ClF}_3,\;\text{XeF}_2,\;\text{ICl}_2^{-}\} \Rightarrow 4\;\text{species}.

Final answer: 4

Q73NumericalOrganic Compounds Containing Nitrogen

One mole of phenol is treated with dilute

\text{HNO}_3

at 298 K to give a mixture of products. The mixture is separated by steam distillation. The steam volatile compound (X) is separated. The increase in percentage of oxygen in (X) with respect to phenol is ____

\times 10^{-1}

%
(Given molar mass in

\text{gmol}^{-1}

\text{H}:1, \text{C}:12, \text{N}:14, \text{O}:16

)

SolutionAnswer: 175

Approach:

Identify the steam-volatile nitration product X as o-nitrophenol (intramolecular H-bond), compute the mass percent of oxygen in phenol and in o-nitrophenol, and report the increase in units of

10^{-1}\%

Step 1:Identify X. Dilute

\text{HNO}_3

at 298 K nitrates phenol predominantly at the o- and p-positions; on steam distillation only the steam-volatile o-nitrophenol passes over.

\text{C}_6\text{H}_5\text{OH} + \text{HNO}_3 \xrightarrow{\text{dilute, 298 K}} o\text{-O}_2\text{N}{-}\text{C}_6\text{H}_4{-}\text{OH} + p\text{-O}_2\text{N}{-}\text{C}_6\text{H}_4{-}\text{OH}

Step 2:Compute the molar mass and %O of phenol.

M(\text{C}_6\text{H}_5\text{OH}) = 6(12)+6(1)+16 = 94\,\text{g mol}^{-1};\;\%O_{\text{phenol}} = \dfrac{16}{94}\times 100 = 17.021\%

Step 3:Compute the molar mass and %O of

o

-nitrophenol.

M(\text{C}_6\text{H}_5\text{NO}_3) = 6(12)+5(1)+14+3(16) = 72+5+14+48 = 139\,\text{g mol}^{-1};\;\%O_{X} = \dfrac{48}{139}\times 100 = 34.532\%

Step 4:Take the increase and convert to units of

10^{-1}

\Delta(\%O) = 34.532 - 17.021 = 17.511\% = 175.11\times 10^{-1}\%

Final answer: 175

Q74NumericalChemical Thermodynamics

The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below

\text{A}(g) \rightleftharpoons \text{B}(g) + \text{C}(g)

\dfrac{1}{T}\,(K^{-1})

\log_{10} K_p

|
| 0.05 | 3.5 |
| 0.06 | 2.5 |
| 0.07 | 1.5 |

The magnitude of

\dfrac{\Delta H^{\circ}}{R}

calculated from the above data is ____ . (Nearest integer)

SolutionAnswer: 230

Approach:

Apply the integrated van't Hoff equation in

\log_{10}

form;

\log_{10} K_p

is linear in

1/T

with slope

-\Delta H^{\circ}/(2.303 R)

. Compute the slope from the tabulated data and convert to

\Delta H^{\circ}/R

Step 1:Compute the slope from two tabulated points (the three points are collinear, so any pair yields the same slope).

\text{slope} = \dfrac{2.5 - 3.5}{0.06 - 0.05}\,\text{K} = \dfrac{-1.0}{0.01}\,\text{K} = -100\,\text{K}

Step 2:Convert slope into

\Delta H^{\circ}/R

using

\text{slope} = -\Delta H^{\circ}/(2.303\,R)

-\dfrac{\Delta H^{\circ}}{2.303\,R} = -100 \Rightarrow \dfrac{\Delta H^{\circ}}{R} = 100\times 2.303 = 230.3\,\text{K}

Step 3:Take magnitude and round to the nearest integer.

\left|\dfrac{\Delta H^{\circ}}{R}\right| \approx 230

Final answer: 230

Q75NumericalChemical Kinetics

If the half life of a first order reaction is 6.93 minutes then the time required for completion of 99% of the reaction will be ____ minutes. (Given :

\log 2 = 0.3010

)

SolutionAnswer: 46

Approach:

For a first-order reaction the rate constant k is fixed by the half-life through

k = \ln 2 / t_{1/2}

; the time for any fractional completion comes from the integrated rate law

t = (2.303/k)\log_{10}([A]_0/[A])

. Apply both with

t_{1/2} = 6.93\,\text{min}

and

[A]/[A]_0 = 0.01

for 99% completion.

Step 1:Compute the rate constant

k

from the given half-life.

k = \dfrac{0.693}{6.93\,\text{min}} = 0.1\,\text{min}^{-1}

Step 2:Set

[A]/[A]_0 = 1/100

for 99% completion and substitute into the integrated rate law.

t = \dfrac{2.303}{0.1\,\text{min}^{-1}}\,\log_{10}(100) = 23.03\,\text{min}\times 2 = 46.06\,\text{min}

Step 3:Round to the nearest integer minute, as expected for a JEE numerical.

t \approx 46\,\text{min}

Final answer: 46

Mathematics23 questions

Q1Single correctComplex Numbers and Quadratic Equations

Let

a, b \in \mathbb{C}

. Let

\alpha, \beta

be the roots of the equation

x^2 + ax + b = 0

. If

\beta - \alpha = \sqrt{11}

and

\beta^2 - \alpha^2 = -3i\sqrt{11}

, then

(\beta^3 - \alpha^3)^2

is equal to:

(A)

160

(B)

176

(C)

194

(D)

187

SolutionAnswer: Option 2176

Approach:

From the difference-of-squares identity, factor

\beta^2-\alpha^2=(\beta-\alpha)(\beta+\alpha)

to extract

\beta+\alpha

. From

(\beta-\alpha)^2=(\beta+\alpha)^2-4\alpha\beta

, obtain

\alpha\beta

. Then evaluate

\beta^3-\alpha^3=(\beta-\alpha)\bigl((\beta+\alpha)^2-\alpha\beta\bigr)

and square it.

Step 1:Divide

\beta^2-\alpha^2

\beta-\alpha

to isolate

\beta+\alpha

\beta+\alpha=\dfrac{\beta^2-\alpha^2}{\beta-\alpha}=\dfrac{-3i\sqrt{11}}{\sqrt{11}}

Step 2:From

(\beta-\alpha)^2=(\beta+\alpha)^2-4\alpha\beta

, solve for

\alpha\beta

11=(-3i)^2-4\alpha\beta=-9-4\alpha\beta\;\Rightarrow\;\alpha\beta=\dfrac{-9-11}{4}

Step 3:Apply the difference-of-cubes factorisation.

\beta^3-\alpha^3=(\beta-\alpha)\bigl((\beta+\alpha)^2-\alpha\beta\bigr)=\sqrt{11}\bigl((-3i)^2-(-5)\bigr)=\sqrt{11}(-9+5)

Step 4:Square to obtain the target quantity.

(\beta^3-\alpha^3)^2=(-4\sqrt{11})^2=16\cdot 11

Final answer: 176

Q2Single correctSequence and Series

Let the sum of the first n terms of an A.P. be

3n^2 + 5n

. Then the sum of squares of the first 10 terms of the A.P. is:

(A)

10220

(B)

12860

(C)

15220

(D)

19780

SolutionAnswer: Option 315220

Approach:

Extract the general term via

T_n=S_n-S_{n-1}

, square it as a polynomial in n, and sum each component using the standard power-sum formulas up to

n=10

Step 1:Compute

T_n=S_n-S_{n-1}

with

S_n=3n^2+5n

T_n=(3n^2+5n)-\bigl(3(n-1)^2+5(n-1)\bigr)=3(2n-1)+5=6n+2

Step 2:Write

T_n^2

as a polynomial in n.

T_n^2=(6n+2)^2=36n^2+24n+4

Step 3:Evaluate the three component sums for

N=10

\displaystyle\sum_{n=1}^{10}n^2=\dfrac{10\cdot 11\cdot 21}{6}=385,\quad\sum_{n=1}^{10}n=55,\quad\sum_{n=1}^{10}1=10

Step 4:Combine the component sums with their coefficients.

\displaystyle\sum_{n=1}^{10}T_n^2=36(385)+24(55)+4(10)=13860+1320+40

Final answer: 15220

Q3Single correctMatrices and Determinants

Let A be a

3 \times 3

matrix such that

A^T\begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}5\\2\\2\end{bmatrix}, A^T\begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}3\\1\\1\end{bmatrix}, A\begin{bmatrix}1\\0\\1\end{bmatrix} = \begin{bmatrix}3\\4\\4\end{bmatrix}

and

A\begin{bmatrix}0\\0\\1\end{bmatrix} = \begin{bmatrix}1\\3\\1\end{bmatrix}

.
If

\det(A) = 1

, then

\det\left(\mathrm{adj}\left(A^2 + A\right)\right)

is equal to:

(A)

(B)

(C)

(D)

SolutionAnswer: Option 464

Approach:

Read rows of A from products

A^T e_3

and

A^T(e_1+e_3)

, read columns of A from

A e_3

and

A(e_1+e_3)

, fix the lone unknown

A_{22}

from

\det(A)=1

, factor

A^2+A=A(A+I)

, and apply

\det(\mathrm{adj}(M))=\det(M)^{n-1}

for

n=3

Step 1:Identify rows 1 and 3 of

A

from the transpose products.

A^T e_3=[3,1,1]^T\;\Rightarrow\;\text{row 3 of }A=[3,1,1]

;

A^T(e_1+e_3)=[5,2,2]^T\;\Rightarrow\;\text{row 1}=[5,2,2]-[3,1,1]=[2,1,1]

Step 2:Identify columns 3 and 1 of

A

from the direct products.

A e_3=[1,3,1]^T\;\Rightarrow\;\text{col 3}=[1,3,1]^T

;

A(e_1+e_3)=[3,4,4]^T\;\Rightarrow\;\text{col 1}=[3,4,4]^T-[1,3,1]^T=[2,1,3]^T

Step 3:Assemble A with the unknown entry

A_{22}=a

and impose

\det(A)=1

A=\begin{bmatrix}2&1&1\\1&a&3\\3&1&1\end{bmatrix},\;\det(A)=2(a-3)-1(1-9)+1(1-3a)=-a+3=1

Step 4:Compute

\det(A+I)

and combine via the factorisation.

A+I=\begin{bmatrix}3&1&1\\1&3&3\\3&1&2\end{bmatrix},\;\det(A+I)=3(6-3)-1(2-9)+1(1-9)=9+7-8=8

Step 5:Apply the adjugate determinant identity for a

3\times 3

matrix.

\det(\mathrm{adj}(A^2+A))=\det(A^2+A)^{3-1}=8^2

Final answer: 64

Q4Single correctMatrices and Determinants

Consider the system of linear equations in x, y, z:

x + 2y + tz = 0

6x + y + 5tz = 0

3x + t^2 y + f(t) z = 0

where

f: \mathbb{R} \to \mathbb{R}

is a differentiable function. If this system has infinitely many solutions for all

t \in \mathbb{R}

, then f

(A)

is a constant function

(B)

is strictly increasing on

\displaystyle \mathbb{R}

(C)

is strictly decreasing on

\displaystyle \mathbb{R}

(D)

has two critical points

SolutionAnswer: Option 2is strictly increasing on

\mathbb{R}

Approach:

For the homogeneous

3\times 3

system to have infinitely many solutions for every

t\in\mathbb{R}

, the coefficient determinant must vanish identically. Multiply it out, solve for f(t), and analyse the sign of f'(t).

Step 1:Apply cofactor along the first row of the determinant.

1\cdot(f(t)-5t\cdot t^2)-2\cdot(6f(t)-5t\cdot 3)+t\cdot(6t^2-1\cdot 3)=0

Step 2:Collect like terms and solve for

f(t)

-11f(t)+t^3+27t=0\;\Rightarrow\;f(t)=\dfrac{t^3+27t}{11}

Step 3:Differentiate and examine the sign.

f'(t)=\dfrac{3t^2+27}{11}=\dfrac{3(t^2+9)}{11}

Step 4:Translate the sign of

f'

to monotonicity of

f

t^2+9\ge 9>0\;\Rightarrow\;f'(t)>0\;\Rightarrow\;f\text{ strictly increases on }\mathbb{R}

Final answer: is strictly increasing on

\mathbb{R}

Q5Single correctSequence and Series

\displaystyle \sum_{n=1}^{10} \left(\frac{528}{n(n+1)(n+2)}\right)

is equal to:

(A)

(B)

130

(C)

220

(D)

440

SolutionAnswer: Option 2130

Approach:

Decompose

\dfrac{1}{n(n+1)(n+2)}

as a telescoping difference, sum from

n=1

10

, and multiply by the constant

528

Step 1:Check the partial-fraction identity by combining the right side.

\dfrac{1}{2}\left[\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right]=\dfrac{1}{2}\cdot\dfrac{(n+2)-n}{n(n+1)(n+2)}=\dfrac{1}{n(n+1)(n+2)}

Step 2:Apply the telescoping sum from

n=1

n=10

\displaystyle\sum_{n=1}^{10}\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{2}\left[\dfrac{1}{1\cdot 2}-\dfrac{1}{11\cdot 12}\right]=\dfrac{1}{2}\cdot\dfrac{66-1}{132}

Step 3:Multiply by

528=2\cdot 264

528\cdot\dfrac{65}{264}=2\cdot 65

Final answer: 130

Q6Single correctTrigonometry

Let

\tan A, \tan B

, where

A, B \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)

, be the roots of the quadratic equation

x^2 - 2x - 5 = 0

. Then

20 \sin^2\left(\dfrac{A+B}{2}\right)

is equal to:

(A)

\displaystyle 10 + \sqrt{10}

(B)

\displaystyle 10 - 2\sqrt{10}

(C)

\displaystyle 10 - 3\sqrt{10}

(D)

\displaystyle 10 - \sqrt{10}

SolutionAnswer: Option 3

10 - 3\sqrt{10}

Approach:

Use Vieta's formulas on

x^2-2x-5=0

to write

\tan A+\tan B

and

\tan A\tan B

, derive

\tan(A+B)

via the addition formula, convert to

\cos(A+B)

on the correct sign branch, and apply

2\sin^2(\theta/2)=1-\cos\theta

with

\theta=A+B

Step 1:Read sum and product of roots from

x^2-2x-5=0

\tan A+\tan B=2,\quad\tan A\tan B=-5

Step 2:Apply the tangent-addition formula.

\tan(A+B)=\dfrac{2}{1-(-5)}=\dfrac{2}{6}

Step 3:Determine the sign of

\cos(A+B)

. The roots are

1\pm\sqrt{6}

, giving

A=\arctan(1+\sqrt 6)\approx 1.287

and

B=\arctan(1-\sqrt 6)\approx -0.967

, both in

(-\pi/2,\pi/2)

. Therefore

A+B\approx 0.32\in(-\pi/2,\pi/2)

, so

\cos(A+B)>0

\cos(A+B)=\dfrac{1}{\sqrt{1+\tan^2(A+B)}}=\dfrac{1}{\sqrt{1+1/9}}=\dfrac{3}{\sqrt{10}}

Step 4:Apply the half-angle identity and scale by

20

20\sin^2\!\left(\dfrac{A+B}{2}\right)=10\bigl(1-\cos(A+B)\bigr)=10-10\cdot\dfrac{3\sqrt{10}}{10}

Final answer:

10-3\sqrt{10}

Q7Single correctStatistics and Probability

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:

(A)

\displaystyle \dfrac{7}{10}

(B)

\displaystyle \dfrac{10}{17}

(C)

\displaystyle \dfrac{12}{19}

(D)

\displaystyle \dfrac{7}{19}

SolutionAnswer: Option 2

\dfrac{10}{17}

Approach:

Enumerate the consecutive 2-letter pairs in each word to obtain

P(\text{see AN}\mid\text{source})

, then combine with equal prior probabilities through Bayes' theorem to obtain the posterior

P(\text{ANANTPUR}\mid\text{see AN})

Step 1:Enumerate consecutive pairs in KANPUR (six letters give five pairs) and count the pattern AN.

Pairs: KA, AN, NP, PU, UR; one of the five is AN

Step 2:Enumerate consecutive pairs in ANANTPUR (eight letters give seven pairs) and count the pattern AN.

Pairs: AN, NA, AN, NT, TP, PU, UR; two of the seven are AN

Step 3:Combine through Bayes' theorem with equal priors

\tfrac{1}{2},\tfrac{1}{2}

P(\text{ANANTPUR}\mid\text{AN})=\dfrac{\tfrac{2}{7}}{\tfrac{1}{5}+\tfrac{2}{7}}=\dfrac{\tfrac{2}{7}}{\tfrac{7+10}{35}}=\dfrac{\tfrac{2}{7}\cdot 35}{17}

Final answer:

\dfrac{10}{17}

Q8Single correctStatistics and Probability

The mean deviation about the mean for the data

|

x_i

| 5 | 7 | 9 | 10 | 12 | 15 |
|

f_i

| 8 | 6 | 2 | 2 | 2 | 6 |

is equal to:

(A)

\displaystyle \dfrac{40}{13}

(B)

\displaystyle \dfrac{42}{13}

(C)

\displaystyle \dfrac{44}{13}

(D)

\displaystyle \dfrac{46}{13}

SolutionAnswer: Option 3

\dfrac{44}{13}

Approach:

Compute the total frequency N, the weighted mean

\bar{x}=\dfrac{\sum f_i x_i}{N}

, then the weighted sum of absolute deviations

\sum f_i|x_i-\bar{x}|

, and divide by N.

Step 1:Sum the frequencies.

N=8+6+2+2+2+6=26

Step 2:Compute the weighted sum

\sum f_i x_i

and the mean.

\sum f_i x_i=8(5)+6(7)+2(9)+2(10)+2(12)+6(15)=40+42+18+20+24+90=234,\;\bar{x}=\dfrac{234}{26}

Step 3:Tabulate

|x_i-9|

and form the weighted sum.

|x_i-9|=4,2,0,1,3,6;\;f_i|x_i-9|=32,12,0,2,6,36;\;\sum f_i|x_i-9|=88

Step 4:Divide by

N

\mathrm{MD}=\dfrac{88}{26}=\dfrac{44}{13}

Final answer:

\dfrac{44}{13}

Q9Single correctCo-ordinate Geometry

Let a focus of the ellipse

E: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1

S(4, 0)

and its eccentricity be

\dfrac{4}{5}

. If the point

P(3, \alpha)

lies on E and O is the origin, then the area of

\triangle \text{POS}

is equal to:

(A)

\displaystyle \dfrac{12}{5}

(B)

\displaystyle \dfrac{14}{5}

(C)

\displaystyle \dfrac{24}{5}

(D)

\displaystyle \dfrac{48}{5}

SolutionAnswer: Option 3

\dfrac{24}{5}

Approach:

Use focus and eccentricity to fix a and b via

ae=4

b^2=a^2(1-e^2)

. Substitute

x=3

in the ellipse equation to extract

|\alpha|

. Because O and S both sit on the x-axis, compute the area as

\tfrac{1}{2}\cdot|OS|\cdot|y_P|

Step 1:Solve for a from

ae=4

and obtain

b^2

a\cdot\dfrac{4}{5}=4\;\Rightarrow\;a=5,\quad b^2=25\left(1-\dfrac{16}{25}\right)=9

Step 2:Substitute

x=3

into

\dfrac{x^2}{25}+\dfrac{y^2}{9}=1

and solve for

|\alpha|

\dfrac{9}{25}+\dfrac{\alpha^2}{9}=1\;\Rightarrow\;\dfrac{\alpha^2}{9}=\dfrac{16}{25}\;\Rightarrow\;\alpha^2=\dfrac{144}{25}

Step 3:Compute the area of

\triangle \text{POS}

with

O=(0,0)

S=(4,0)

|y_P|=\dfrac{12}{5}

[\triangle \text{POS}]=\dfrac{1}{2}\cdot 4\cdot\dfrac{12}{5}

Final answer:

\dfrac{24}{5}

Q10Single correctCo-ordinate Geometry

Let P be a moving point on the circle

x^2 + y^2 - 6x - 8y + 21 = 0

. Then, the maximum distance of P from the vertex of the parabola

x^2 + 6x + y + 13 = 0

is equal to:

(A)

(B)

(C)

(D)

SolutionAnswer: Option 312

Approach:

Bring the circle to standard form to read centre and radius. Bring the parabola to vertex form to read its vertex. Since the vertex lies outside the circle, the farthest point on the circle from the vertex is along the line joining the vertex to the centre, at distance

|VC|+r

Step 1:Complete the square for the circle.

x^2-6x+y^2-8y+21=0\;\Rightarrow\;(x-3)^2+(y-4)^2=9+16-21=4

Step 2:Bring the parabola to vertex form.

x^2+6x+y+13=0\;\Rightarrow\;(x+3)^2=-(y+4)

Step 3:Compute the distance from the vertex to the centre.

|VC|=\sqrt{(3-(-3))^2+(4-(-4))^2}=\sqrt{36+64}=\sqrt{100}

Step 4:Since

|VC|=10>r=2

, the vertex is external; add the radius.

d_{\max}=|VC|+r=10+2

Final answer: 12

Q11Single correctCo-ordinate Geometry

In an equilateral triangle P Q R, let the vertex P be at

(3, 5)

and the side Q R be along the line

x + y = 4

. If the orthocentre of the triangle PQR is

(\alpha, \beta)

, then

9(\alpha + \beta)

is equal to:

(A)

(B)

(C)

(D)

SolutionAnswer: Option 448

Approach:

In an equilateral triangle the orthocentre, centroid, and circumcentre coincide. Drop the perpendicular from

P

to line

QR

to obtain the midpoint

M

QR

, then locate the centroid at the

2{:}1

section of

PM

from

P

Step 1:Compute the foot of perpendicular from

P(3,5)

x+y-4=0

M=(3,5)-\dfrac{3+5-4}{1^2+1^2}(1,1)=(3,5)-2(1,1)

Step 2:Place the centroid on segment

PM

at the

2{:}1

ratio from

P

G=P+\dfrac{2}{3}(M-P)=(3,5)+\dfrac{2}{3}(-2,-2)=\left(3-\dfrac{4}{3},\,5-\dfrac{4}{3}\right)

Step 3:Compute

9(\alpha+\beta)

\alpha+\beta=\dfrac{5+11}{3}=\dfrac{16}{3},\;9(\alpha+\beta)=9\cdot\dfrac{16}{3}

Final answer: 48

Q12Single correctTrigonometry

The sum of all the integral values of p such that the equation

3 \sin^2 x + 12 \cos x - 3 = p,\ x \in \mathbb{R}

, has at least one solution, is:

(A)

\displaystyle -54

(B)

\displaystyle -60

(C)

\displaystyle -75

(D)

\displaystyle -84

SolutionAnswer: Option 3

-75

Approach:

Substitute

\sin^2 x=1-\cos^2 x

and let

c=\cos x\in[-1,1]

. Study the quadratic

f(c)=-3c^2+12c

over the interval, extract its range, then sum the integers in that range.

Step 1:Substitute the Pythagorean identity and set

c=\cos x

3(1-c^2)+12c-3=p\;\Rightarrow\;-3c^2+12c=p

Step 2:Express

f

in vertex form to expose monotonic behaviour on

[-1,1]

f(c)=-3(c^2-4c)=-3\bigl((c-2)^2-4\bigr)=12-3(c-2)^2

Step 3:Evaluate the endpoints to obtain the range.

f(-1)=12-3(9)=-15,\;f(1)=12-3(1)=9

Step 4:Sum the integers from

-15

9

inclusive (25 integers).

\displaystyle\sum_{p=-15}^{9}p=\dfrac{(9-(-15)+1)(-15+9)}{2}=\dfrac{25\cdot(-6)}{2}

Final answer:

-75

Q13Single correctTrigonometry

Let

\tan A

\tan B

, where

A,B \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right)

, be the roots of the quadratic equation

x^2 - 2x - 5 = 0

. Then

20\sin^2\left( \frac{A+B}{2} \right)

is equal to:

(A)

\displaystyle 10+\sqrt{10}

(B)

\displaystyle 10-2\sqrt{10}

(C)

\displaystyle 10-3\sqrt{10}

(D)

\displaystyle 10-\sqrt{10}

SolutionAnswer: Option 3

10-3\sqrt{10}

Approach:

From Vieta's formulas on

x^2-2x-5=0

derive

\tan A+\tan B

and

\tan A\tan B

, compute

\tan(A+B)

, fix

\cos(A+B)

in the principal range, and apply the half-angle identity for

\sin^2((A+B)/2)

Step 1:Identify the roots from the quadratic

x^2-2x-5=0

using Vieta's formulas.

\tan A+\tan B=2,\quad \tan A\tan B=-5

Step 2:Apply the tangent-sum formula.

\tan(A+B)=\dfrac{2}{1-(-5)}=\dfrac{2}{6}=\dfrac{1}{3}

Step 3:Since

\tan A\tan B=-5<0

one of A,B lies in

(-\pi/2,0)

and the other in

(0,\pi/2)

, while

\tan A+\tan B>0

keeps

A+B

(-\pi/2,\pi/2)

, so

\cos(A+B)>0

\cos(A+B)=\dfrac{3}{\sqrt{10}}

Step 4:Substitute into the half-angle identity multiplied by

20

20\sin^2\!\left(\dfrac{A+B}{2}\right)=20\cdot\dfrac{1-\cos(A+B)}{2}=10\!\left(1-\dfrac{3}{\sqrt{10}}\right)=10-\dfrac{30}{\sqrt{10}}=10-3\sqrt{10}

Final answer:

10-3\sqrt{10}

Q14Single correctPermutations and Combinations

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:

(A)

\displaystyle \dfrac{7}{10}

(B)

\displaystyle \dfrac{10}{17}

(C)

\displaystyle \dfrac{12}{19}

(D)

\displaystyle \dfrac{7}{19}

SolutionAnswer: Option 2

\dfrac{10}{17}

Approach:

Use Bayes' theorem with equal prior probabilities for the two origin cities; the likelihood of seeing 'AN' is the count of consecutive ordered letter pairs equal to 'AN' divided by the total number of consecutive pairs in each name.

Step 1:Assign equal prior probabilities to the two source cities.

P(\text{KANPUR})=P(\text{ANANTPUR})=\dfrac{1}{2}

Step 2:Enumerate consecutive letter pairs in KANPUR (length

6

5

adjacent pairs): KA, AN, NP, PU, UR. The pair 'AN' appears once.

P(\text{AN}\mid \text{KANPUR})=\dfrac{1}{5}

Step 3:Enumerate consecutive letter pairs in ANANTPUR (length

8

7

adjacent pairs): AN, NA, AN, NT, TP, PU, UR. The pair 'AN' appears twice.

P(\text{AN}\mid \text{ANANTPUR})=\dfrac{2}{7}

Step 4:Substitute into Bayes' theorem.

P(\text{ANANTPUR}\mid \text{AN})=\dfrac{\tfrac{1}{2}\cdot\tfrac{2}{7}}{\tfrac{1}{2}\cdot\tfrac{1}{5}+\tfrac{1}{2}\cdot\tfrac{2}{7}}=\dfrac{2/7}{1/5+2/7}=\dfrac{2/7}{17/35}=\dfrac{2}{7}\cdot\dfrac{35}{17}=\dfrac{10}{17}

Final answer:

\dfrac{10}{17}

Q15Single correctSequence and Series

The mean deviation about the mean for the data:

|

x_i

| 5 | 7 | 9 | 10 | 12 | 15 |
|

f_i

| 8 | 6 | 2 | 2 | 2 | 6 |

is equal to:

(A)

\displaystyle \dfrac{40}{13}

(B)

\displaystyle \dfrac{42}{13}

(C)

\displaystyle \dfrac{44}{13}

(D)

\displaystyle \dfrac{46}{13}

SolutionAnswer: Option 3

\dfrac{44}{13}

Approach:

Determine the arithmetic mean of the frequency distribution, then evaluate the weighted average of absolute deviations from that mean.

Step 1:Sum the frequencies.

\sum f_i=8+6+2+2+2+6=26

Step 2:Compute the weighted sum

\sum f_i x_i

\sum f_i x_i=8(5)+6(7)+2(9)+2(10)+2(12)+6(15)=40+42+18+20+24+90=234

Step 3:Find the mean.

\bar{x}=\dfrac{234}{26}=9

Step 4:Tabulate

|x_i-9|

and form the weighted sum.

\sum f_i|x_i-9|=8(4)+6(2)+2(0)+2(1)+2(3)+6(6)=32+12+0+2+6+36=88

Step 5:Divide by

N

and reduce.

\mathrm{MD}=\dfrac{88}{26}=\dfrac{44}{13}

Final answer:

\dfrac{44}{13}

Q17Single correctLimit, Continuity and Differentiability

The product of all possible values of x, for which

\lim_{n \to \infty} \left( \dfrac{1 - \cos(\alpha) \cos(\alpha) + 1 - x\cos(\alpha + 2)x}{\sin^2((x+1)x)} \right) = 2

, is:

(A)

\displaystyle -2

(B)

\displaystyle 1

(C)

\displaystyle -1

(D)

\displaystyle \dfrac{5}{4}

SolutionAnswer: Option 3

-1

Approach:

Multiply out the numerator

1-\cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)

to leading order in x using

\cos\theta\approx 1-\theta^2/2

, replace the denominator

\sin^2((\alpha+1)x)

((\alpha+1)x)^2

, equate the limit to 2, and use Vieta's formula on the resulting quadratic in

\alpha

Step 1:Multiply out the product of three cosines to order

x^2

\cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)\approx 1-\tfrac{x^2}{2}\big(\alpha^2+(\alpha+1)^2+(\alpha+2)^2\big)

1-\cos(\alpha x)\cos((\alpha+1)x)\cos((\alpha+2)x)\approx \dfrac{x^2}{2}\big[\alpha^2+(\alpha+1)^2+(\alpha+2)^2\big]

Step 2:Replace the denominator using

\sin\theta\approx\theta

\sin^2((\alpha+1)x)\approx (\alpha+1)^2 x^2

Step 3:Form the limit and set it equal to

2

\lim_{x\to 0}\dfrac{\frac{x^2}{2}\big[\alpha^2+(\alpha+1)^2+(\alpha+2)^2\big]}{(\alpha+1)^2 x^2}=\dfrac{\alpha^2+(\alpha+1)^2+(\alpha+2)^2}{2(\alpha+1)^2}=2

Step 4:Clear and simplify:

\alpha^2+(\alpha+1)^2+(\alpha+2)^2=4(\alpha+1)^2

gives

3\alpha^2+6\alpha+5=4\alpha^2+8\alpha+4

, hence

\alpha^2+2\alpha-1=0

\alpha^2+2\alpha-1=0

Step 5:By Vieta's formula the product of the two roots equals

c/a=-1

\alpha_1\alpha_2=\dfrac{-1}{1}=-1

Final answer:

-1

Q18Single correctIntegral Calculus

The value of the integral

\displaystyle \int_{0}^{4} \dfrac{\log_e x}{x^2 + 4}\, dx

is:

(A)

\displaystyle \dfrac{\pi \log_e 2}{2}

(B)

\displaystyle \dfrac{\pi \log_e 2}{4}

(C)

\displaystyle 1 + \pi \log_e 2

(D)

\displaystyle 2 + \pi \log_e 2

SolutionAnswer: Option 2

\dfrac{\pi \log_e 2}{4}

Approach:

Use the substitution

x=2\tan\theta

to convert the integral over

(0,\infty)

into one over

(0,\pi/2)

, then split the logarithm and apply the classical identity

\int_0^{\pi/2}\ln(\tan\theta)\,d\theta=0

Step 1:Set

I=\int_0^{\infty}\dfrac{\ln x}{x^2+4}\,dx

and substitute

x=2\tan\theta

; the limits become

0

and

\pi/2

I=\int_0^{\pi/2}\dfrac{\ln(2\tan\theta)}{4\sec^2\theta}\cdot 2\sec^2\theta\,d\theta=\dfrac{1}{2}\int_0^{\pi/2}\ln(2\tan\theta)\,d\theta

Step 2:Split the logarithm.

I=\dfrac{1}{2}\!\left[\int_0^{\pi/2}\ln 2\,d\theta+\int_0^{\pi/2}\ln(\tan\theta)\,d\theta\right]

Step 3:The constant piece equals

(\pi/2)\ln 2

, while the classical identity gives

\int_0^{\pi/2}\ln(\tan\theta)\,d\theta=0

I=\dfrac{1}{2}\!\left[\dfrac{\pi}{2}\ln 2+0\right]=\dfrac{\pi\ln 2}{4}

Final answer:

\dfrac{\pi\log_e 2}{4}

Q19Single correctLimit, Continuity and Differentiability

Let

f:\mathbb{R}\to\mathbb{R}

be a differentiable function such that

f\left( \dfrac{x+y}{3} \right) = \dfrac{f(x) + f(y)}{3}

for all

x,y \in \mathbb{R}

, and

f'(0)=3

. Then the minimum value of the function

g(x) = 3 + x f'(x)

, is:

(A)

\displaystyle 3\left( \dfrac{e+1}{e} \right)

(B)

\displaystyle 3\left( \dfrac{e-1}{e} \right)

(C)

\displaystyle \dfrac{3-e}{e}

(D)

\displaystyle 3e

SolutionAnswer: Option 2

3\left( \dfrac{e-1}{e} \right)

Approach:

Resolve the Jensen-type functional equation under differentiability to identify

f(x)=3x

, substitute into

g(x)=3+e^x f(x)

, locate the critical point and confirm a minimum.

Step 1:Set

x=y=0

f(0)=2f(0)/3

, so

f(0)=0

. Set

y=0

f(x/3)=f(x)/3

f(0)=0,\quad f(x/3)=f(x)/3

Step 2:Differentiate

f((x+y)/3)=(f(x)+f(y))/3

with respect to x:

\tfrac{1}{3}f'((x+y)/3)=\tfrac{1}{3}f'(x)

, so

f'((x+y)/3)=f'(x)

for all x,y. Fixing x and varying y forces f' to be constant.

f'(x)=f'(0)=3

Step 3:Integrate and apply

f(0)=0

f(x)=3x

Step 4:Substitute into

g(x)=3+e^x f(x)=3+3xe^x

and differentiate.

g'(x)=3e^x+3xe^x=3e^x(1+x)

Step 5:Solve

g'(x)=0

; since

e^x>0

the only critical point is

x=-1

x=-1

Step 6:Second derivative test:

g''(x)=3e^x(2+x)

, so

g''(-1)=3e^{-1}>0

, confirming a minimum.

g''(-1)=\dfrac{3}{e}>0

Step 7:Evaluate the minimum.

g(-1)=3+3(-1)e^{-1}=3-\dfrac{3}{e}=\dfrac{3(e-1)}{e}

Final answer:

3\!\left(\dfrac{e-1}{e}\right)

Q20Single correctIntegral Calculus

The value of the integral

\displaystyle \int_{\pi/3}^{\pi/2} \dfrac{4 - \cos^2 x}{\cos^2 x}\, dx

is:

(A)

\displaystyle \dfrac{11}{\sqrt{3}}

(B)

\displaystyle \dfrac{22}{\sqrt{3}}

(C)

\displaystyle \dfrac{32}{3\sqrt{3}}

(D)

\displaystyle \dfrac{64}{3\sqrt{3}}

SolutionAnswer: Option 3

\dfrac{32}{3\sqrt{3}}

Approach:

Substitute

t=\tan x

, rewrite the integrand as a polynomial-plus-pole expression in t, integrate the resulting elementary form, and evaluate at

t=\tan(\pi/6)

and

t=\tan(\pi/3)

Step 1:Use

t=\tan x

\csc^2 x=(1+t^2)/t^2

and

\cos^4 x=1/(1+t^2)^2

; rewrite the numerator:

4-\csc^2 x=4-(1+t^2)/t^2=(3t^2-1)/t^2

\dfrac{4-\csc^2 x}{\cos^4 x}=\dfrac{(3t^2-1)/t^2}{1/(1+t^2)^2}=\dfrac{(3t^2-1)(1+t^2)^2}{t^2}

Step 2:With

dx=dt/(1+t^2)

the integrand-times-dx simplifies.

\dfrac{(3t^2-1)(1+t^2)^2}{t^2}\cdot\dfrac{dt}{1+t^2}=\dfrac{(3t^2-1)(1+t^2)}{t^2}\,dt

Step 3:Multiply out the numerator.

(3t^2-1)(1+t^2)=3t^4+3t^2-t^2-1=3t^4+2t^2-1

Step 4:Divide by

t^2

\dfrac{3t^4+2t^2-1}{t^2}=3t^2+2-\dfrac{1}{t^2}

Step 5:Antidifferentiate.

\int\!\left(3t^2+2-\dfrac{1}{t^2}\right)dt=t^3+2t+\dfrac{1}{t}+C

Step 6:Evaluate at the new limits

t=\tan(\pi/6)=1/\sqrt{3}

and

t=\tan(\pi/3)=\sqrt{3}

F(\sqrt{3})=3\sqrt{3}+2\sqrt{3}+\dfrac{1}{\sqrt{3}}=\dfrac{15+1}{\sqrt{3}}=\dfrac{16}{\sqrt{3}}

Step 7:Evaluate at the lower limit.

F(1/\sqrt{3})=\dfrac{1}{3\sqrt{3}}+\dfrac{2}{\sqrt{3}}+\sqrt{3}=\dfrac{1+15}{3\sqrt{3}}=\dfrac{16}{3\sqrt{3}}

Step 8:Subtract.

F(\sqrt{3})-F(1/\sqrt{3})=\dfrac{16}{\sqrt{3}}-\dfrac{16}{3\sqrt{3}}=\dfrac{48-16}{3\sqrt{3}}=\dfrac{32}{3\sqrt{3}}

Final answer:

\dfrac{32}{3\sqrt{3}}

Q21NumericalSets, Relations and Functions

Let

A = \{1, 2, 3, 4, 5, 6\}

. The number of one-one functions

f : A \to A

, such that

f(1) + f(2) = f(3)

and

f(3) \leq 4

is _____ .

SolutionAnswer: 72

Approach:

Enumerate ordered pairs

(f(2),f(3))

with

f(2)+f(3)=5

and

f(3)\le 4

, then for each pair count admissible values of

f(1)

satisfying

f(1)\ge 3

and distinct from the chosen images, and finally multiply by the number of permutations of the remaining three values into

f(4),f(5),f(6)

Step 1:List ordered pairs

(f(2),f(3))

with

f(2)+f(3)=5

, distinct entries in

\{1,\dots,6\}

, and

f(3)\le 4

(1,4),\ (2,3),\ (3,2),\ (4,1)

Step 2:For each pair, count values of

f(1)\in\{3,4,5,6\}

that differ from

f(2)

and

f(3)

(1,4)\!: f(1)\in\{3,5,6\};\ (2,3)\!: f(1)\in\{4,5,6\};\ (3,2)\!: f(1)\in\{4,5,6\};\ (4,1)\!: f(1)\in\{3,5,6\}

Step 3:For each completed triple

(f(1),f(2),f(3))

, the remaining three elements of

A

are assigned to

f(4),f(5),f(6)

injectively in

3!=6

ways.

3!=6

Step 4:Multiply the counts.

N=4\cdot 3\cdot 6=72

Final answer: 72

Q22NumericalSets, Relations and Functions

Two players A and B play a game of throwing a fair coin. The first player to get a tail is the winner. If both A and B get a head in

5

games and player A plays the sixth game, the count of remaining outcome possibilities is _____ .

SolutionAnswer: 126

Approach:

Recognise that player A wins the series in exactly

5+r

games, where

r\in\{0,1,2,3,4\}

counts the games player B wins before A's decisive fifth win; in each such sequence the last game is A's win and the remaining

4+r

slots contain

4

wins for A and r wins for B.

Step 1:If

A

's last (winning) game is the

(5+r)

th overall, then among the first

4+r

games

A

won exactly

4

and

B

won

r

\text{Sequences with }r\text{ losses by }A=\binom{4+r}{4}

Step 2:Sum over

r=0,1,2,3,4

N=\sum_{r=0}^{4}\binom{4+r}{4}=\binom{4}{4}+\binom{5}{4}+\binom{6}{4}+\binom{7}{4}+\binom{8}{4}

Step 3:Evaluate each binomial.

1+5+15+35+70=126

Step 4:Confirm via the hockey-stick identity.

\sum_{r=0}^{4}\binom{4+r}{4}=\binom{9}{5}=126

Final answer: 126

Q23NumericalThree Dimensional Geometry

Let

\vec{a} = -\hat{i} - \hat{k}

and

\vec{b} = \hat{i} + 2\hat{k}

. If

\vec{r}

is a vector such that

\vec{r} \times \vec{a} = \vec{b} \times \vec{a}

and

\vec{r} \cdot \vec{b} = 0

, then

|\vec{r}|^2

is equal to _____ .

SolutionAnswer: 21

Approach:

Write the general term of

\left(\dfrac{1}{x^3}-x^4\right)^{n}

, isolate the powers

x^7

and

x^{14}

, impose integrality of the term indices, and use the zero-sum condition on the two coefficients to determine n.

Step 1:Write the exponent of

x

in the general term.

\text{Exponent}=7r-3n

Step 2:Set the exponent equal to

7

and

14

to find the corresponding indices

r_1

and

r_2

7r_1-3n=7\Rightarrow r_1=\dfrac{3n+7}{7};\quad 7r_2-3n=14\Rightarrow r_2=\dfrac{3n+14}{7}

Step 3:Integrality of

r_1,r_2

requires

3n\equiv 0\pmod 7

, i.e.

n=7m

. Then

r_1=3m+1

and

r_2=3m+2

, with

r_2-r_1=1

n=7m,\ r_1=3m+1,\ r_2=3m+2

Step 4:Coefficient of

x^7

\binom{n}{r_1}(-1)^{r_1}

and of

x^{14}

\binom{n}{r_2}(-1)^{r_2}

; since

(-1)^{r_2}=-(-1)^{r_1}

the sum-zero condition becomes

\binom{n}{r_1}=\binom{n}{r_2}

\binom{n}{3m+1}=\binom{n}{3m+2}

Step 5:Apply binomial symmetry: either

r_1=r_2

(impossible since they differ by

1

) or

r_1+r_2=n

, giving

(3m+1)+(3m+2)=7m\Rightarrow 6m+3=7m\Rightarrow m=3

. Hence

n=7m=21

m=3,\ n=21

Final answer: 21

Q24NumericalBinomial Theorem

If the term independent of

x^3

and

x^9

in the expansion of

\left( \dfrac{1}{x^4} + x^7 \right)^n

x \ne 0

, is zero, then the sum of all possible values of n is _____ .

SolutionAnswer: 2048

Approach:

Recognise each term

\tan^{-1}\!\big(2^{p-1}/(1+2^{2p-1})\big)

as a difference

\tan^{-1}(2^p)-\tan^{-1}(2^{p-1})

, telescope the sum, combine with the leading

\pi/4

, and read off

\tan\alpha

Step 1:Identify

u=2^p

and

v=2^{p-1}

u-v=2^{p-1}

and

1+uv=1+2^{2p-1}

\tan^{-1}(2^p)-\tan^{-1}(2^{p-1})=\tan^{-1}\!\dfrac{2^{p-1}}{1+2^{2p-1}}

Step 2:Sum from

p=1

p=11

; the sum telescopes.

\sum_{p=1}^{11}\!\left[\tan^{-1}(2^p)-\tan^{-1}(2^{p-1})\right]=\tan^{-1}(2^{11})-\tan^{-1}(2^0)=\tan^{-1}(2048)-\tan^{-1}(1)

Step 3:Add

\pi/4=\tan^{-1}(1)

\alpha=\dfrac{\pi}{4}+\tan^{-1}(2048)-\tan^{-1}(1)=\tan^{-1}(2048)

Step 4:Take the tangent.

\tan\alpha=\tan(\tan^{-1}(2048))=2048

Final answer: 2048

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