JEE Advanced 2025 Question Paper 1 Question Paper with Solutions | JEE Advanced Mock Test | JeeMate

Q 1PhysicsSection A • MCQsimple harmonic motion

+3 / -1

The center of a disk of radius $r$ and mass $m$ is attached to a spring of spring constant $k$ , inside a ring of radius $R > r$ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $T = \frac{2\pi}{\omega}$ . The correct expression for $\omega$ is ( $g$ is the acceleration due to gravity):

A

$\sqrt{\frac{2}{3}\left(\frac{g}{R-r} + \frac{k}{m}\right)}$

B

$\sqrt{\frac{2g}{3(R-r)} + \frac{k}{m}}$

C

$\sqrt{\frac{1}{6}\left(\frac{g}{R-r} + \frac{k}{m}\right)}$

D

$\sqrt{\frac{1}{4}\left(\frac{g}{R-r} + \frac{k}{m}\right)}$

Official Correct Answer:Option (A)

Step-by-Step Solution:

Let the centre of the ring be $O$ and let the centre of the disk be $C$ . Because the disk rolls without slipping on the inner wall of the ring, the distance $OC$ remains constant and is equal to the inner radius available for the centre,

$a = R - r$

Choose the small angular displacement $\theta$ of the centre $C$ from the lowest (vertical) position as the generalised coordinate. Arc‐length travelled by the centre is then

$s = a\theta \qquad (|\theta| \ll 1)$

1. Constraint of pure rolling
Pure rolling gives a fixed relation between the translational speed of the centre and the angular speed of the disk about its own axis:

$v_{\text{cm}} = a\dot\theta = r\dot\phi \qquad -(1)$

where $\dot\phi$ is the angular velocity of the disk about its diameter.

2. Kinetic energy
Translational kinetic energy: $T_{\text{trans}} = \tfrac12 m v_{\text{cm}}^{2} = \tfrac12 m (a\dot\theta)^{2}$

Moment of inertia of a solid disk about its centre: $I = \tfrac12 m r^{2}$

Using the rolling relation $(1)$ , rotational kinetic energy is

$T_{\text{rot}} = \tfrac12 I\dot\phi^{2} = \tfrac12\!\left(\tfrac12 m r^{2}\right)\! \left(\tfrac{a\dot\theta}{r}\right)^{2} = \tfrac14 m a^{2}\dot\theta^{2}$

Thus total kinetic energy is

$T = T_{\text{trans}} + T_{\text{rot}} = \left(\tfrac12 + \tfrac14\right)m a^{2}\dot\theta^{2} = \tfrac34 m a^{2}\dot\theta^{2} \qquad -(2)$

3. Potential energy

(i) Gravitational: For a small angular displacement, the rise in the centre’s height is $h = a(1-\cos\theta) \approx \tfrac12 a\theta^{2}$ Therefore

$U_g = mgh = \tfrac12 m g a\theta^{2} \qquad -(3)$

(ii) Spring: The spring is attached tangentially along the periphery, so its extension/compression equals the arc length $s = a\theta$ . Hooke’s law gives

$U_s = \tfrac12 k s^{2} = \tfrac12 k a^{2}\theta^{2} \qquad -(4)$

Total potential energy: $U = U_g + U_s$

$U = \tfrac12\left(m g a + k a^{2}\right)\theta^{2} \qquad -(5)$

4. Small-oscillation equation
Using the Lagrangian $L = T - U$ with $T$ from $(2)$ and $U$ from $(5)$ :

$L = \tfrac34 m a^{2}\dot\theta^{2} - \tfrac12\left(m g a + k a^{2}\right)\theta^{2}$

Applying the Euler-Lagrange equation $\frac{d}{dt}\left(\frac{\partial L}{\partial\dot\theta}\right) - \frac{\partial L}{\partial\theta}=0$ gives

$\frac{d}{dt}\!\left(\tfrac34\cdot 2 m a^{2}\dot\theta\right) + \left(m g a + k a^{2}\right)\theta = 0$

$\tfrac32 m a^{2}\ddot\theta + m g a\theta + k a^{2}\theta = 0 \qquad -(6)$

Divide by $\tfrac32 m a^{2}$ :

$\ddot\theta + \left[\frac{2}{3}\left(\frac{g}{a} + \frac{k}{m}\right)\right]\theta = 0$

This is the standard SHM form $\ddot\theta + \omega^{2}\theta = 0$ with

$\omega^{2} = \frac{2}{3}\left(\frac{g}{a} + \frac{k}{m}\right) = \frac{2}{3}\left(\frac{g}{R - r} + \frac{k}{m}\right) \qquad -(7)$

5. Time period

$T = \frac{2\pi}{\omega}$ with $\omega$ given by $(7)$ .

Hence the correct expression is

$\boxed{\displaystyle \omega = \sqrt{\frac{2}{3}\left(\frac{g}{R-r} + \frac{k}{m}\right)}}$

Option A perfectly matches this result.

Q 2PhysicsSection A • MCQcenter of mass

+3 / -1

In a scattering experiment, a particle of mass $2m$ collides with another particle of mass $m$ , which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation $\theta$ of the heavier particle, as shown in the figure, in radians is:

A

$\pi$

B

$\tan^{-1}\left(\frac{1}{2}\right)$

C

$\frac{\pi}{3}$

D

$\frac{\pi}{6}$

Official Correct Answer:Option (D)

Step-by-Step Solution:

Let the incident (heavier) particle have mass $m_1 = 2m$ and initial speed $u$ along the +x-axis. The target particle has mass $m_2 = m$ and is initially at rest in the laboratory (lab) frame.

Step 1: Velocity of the centre of mass (COM)
$V = \frac{m_1 u + m_2 \times 0}{m_1 + m_2} = \frac{2m\,u}{2m+m} = \frac{2u}{3}$

Step 2: Velocities in the COM frame before collision
Velocity of $m_1$ in COM frame: $u_{1c} = u - V = u - \frac{2u}{3} = \frac{u}{3}$ (along +x)
Magnitude of this velocity: $|u_{1c}| = \frac{u}{3}$

Step 3: Velocities in the COM frame after an elastic collision
In a perfectly elastic, two-body collision the magnitudes of the velocities in the COM frame remain unchanged; only their directions change. Hence the heavier particle still has speed $\dfrac{u}{3}$ in the COM frame, but its direction makes some angle $\phi$ with the +x-axis.

Write the post-collision COM-frame velocity of $m_1$ as $\vec{u}_{1c}' = \frac{u}{3}\bigl(\cos\phi\,\hat{i} + \sin\phi\,\hat{j}\bigr)$

Step 4: Transform back to the lab frame
Lab-frame velocity of $m_1$ after collision: $\vec{v}_{1}' = \vec{V} + \vec{u}_{1c}' = \frac{2u}{3}\,\hat{i} + \frac{u}{3}\bigl(\cos\phi\,\hat{i} + \sin\phi\,\hat{j}\bigr)$
Components:
$v_{x} = \frac{2u}{3} + \frac{u}{3}\cos\phi$ $v_{y} = \frac{u}{3}\sin\phi$

Step 5: Angular deviation of the heavier particle
Let $\theta$ be the angle between $\vec{v}_{1}'$ and the initial direction (+x). Then $\tan\theta = \frac{v_{y}}{v_{x}} = \frac{\dfrac{u}{3}\sin\phi}{\dfrac{2u}{3} + \dfrac{u}{3}\cos\phi} = \frac{\sin\phi}{2 + \cos\phi} \quad -(1)$

Step 6: Maximising $\theta$
Because $0 \lt \theta \lt \dfrac{\pi}{2}$ , maximising $\theta$ is equivalent to maximising $\tan\theta$ . Define $f(\phi) = \dfrac{\sin\phi}{2 + \cos\phi}$ .

Differentiating and setting $\dfrac{df}{d\phi}=0$ gives $\frac{df}{d\phi} = \frac{\cos\phi(2+\cos\phi) + \sin^{2}\phi}{(2+\cos\phi)^{2}} = \frac{2\cos\phi + \cos^{2}\phi + \sin^{2}\phi}{(2+\cos\phi)^{2}} = \frac{2\cos\phi + 1}{(2+\cos\phi)^{2}}$

Setting the numerator to zero:
$2\cos\phi + 1 = 0 \;\;\Longrightarrow\;\; \cos\phi = -\frac{1}{2} \;\;\Longrightarrow\;\; \phi = \frac{2\pi}{3}$ (Only the value within $0 \le \phi \le \pi$ is relevant.) For $\phi \lt \dfrac{2\pi}{3}$ , the derivative is positive and for $\phi \gt \dfrac{2\pi}{3}$ it is negative, so $\phi = \dfrac{2\pi}{3}$ indeed gives the maximum.

Step 7: Maximum value of $\theta$
Insert $\phi = \dfrac{2\pi}{3}$ into equation $(1)$ :
$\tan\theta_{\max} = \frac{\sin(2\pi/3)}{2 + \cos(2\pi/3)} = \frac{\dfrac{\sqrt{3}}{2}}{2 - \dfrac{1}{2}} = \frac{\dfrac{\sqrt{3}}{2}}{\dfrac{3}{2}} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$

Therefore $\theta_{\max} = \tan^{-1}\!\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$

Option D which is: $\dfrac{\pi}{6}$

Q 3PhysicsSection A • MCQelectromagnetic induction

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A conducting square loop initially lies in the $XZ$ plane with its lower edge hinged along the $X$ -axis. Only in the region $y \geq 0$ , there is a time dependent magnetic field pointing along the $Z$ -direction, $\vec{B}(t) = B_0(\cos \omega t)\hat{k}$ , where $B_0$ is a constant. The magnetic field is zero everywhere else. At time $t = 0$ , the loop starts rotating with constant angular speed $\omega$ about the $X$ axis in the clockwise direction as viewed from the $+X$ axis (as shown in the figure). Ignoring self-inductance of the loop and gravity, which of the following plots correctly represents the induced e.m.f. ( $V$ ) in the loop as a function of time:

A

qImage----Jeemate

B

qImage----Jeemate

C

qImage----Jeemate

D

qImage----Jeemate

Q 4PhysicsSection A • MCQunits and measurements

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Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter $D$ of a tube. The measured value of $D$ is:

A

0.12 cm

B

0.11 cm

C

0.13 cm

D

0.14 cm

Q 5PhysicsSection A • MCQelectromagnetic induction

+4 / -0

A conducting square loop of side $L$ , mass $M$ and resistance $R$ is moving in the $XY$ plane with its edges parallel to the $X$ and $Y$ axes. The region $y \geq 0$ has a uniform magnetic field, $\vec{B} = B_0\hat{k}$ . The magnetic field is zero everywhere else. At time $t = 0$ , the loop starts to enter the magnetic field with an initial velocity $v_0 \hat{j}$ m/s, as shown in the figure. Considering the quantity $K = \frac{B_0^2 L^2}{RM}$ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct:

A

If $v_0 = 1.5KL$ , the loop will stop before it enters completely inside the region of magnetic field.

B

When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.

C

If $v_0 = \frac{KL}{10}$ , the loop comes to rest at $t = \left(\frac{1}{K}\right) \ln\left(\frac{5}{2}\right)$ .

D

If $v_0 = 3KL$ , the complete loop enters inside the region of magnetic field at time $t = \left(\frac{1}{K}\right) \ln\left(\frac{3}{2}\right)$ .

Official Correct Answer:Option (B, D)

Step-by-Step Solution:

$K = \frac{B_0^2 L^2}{RM}$

Let $y$ be the distance the loop has entered the magnetic field at time $t$ :

$\Phi = B_0 L y \implies \varepsilon = -\frac{d\Phi}{dt} = -B_0 L v$

$i = \frac{B_0 L v}{R}$

Retardation force acting on the loop:

$F = -i L B_0 = -\frac{B_0^2 L^2 v}{R}$

$M \frac{dv}{dt} = -\frac{B_0^2 L^2}{R}v \implies \frac{dv}{dt} = -K v$

Integrating with respect to time $t$ :

$\int_{v_0}^{v} \frac{dv}{v} = -K \int_{0}^{t} dt \implies v(t) = v_0 e^{-Kt}$

Integrating with respect to position $y$ :

$v \frac{dv}{dy} = -K v \implies \frac{dv}{dy} = -K \implies \int_{v_0}^{v} dv = -K \int_{0}^{y} dy$

$v - v_0 = -K y \implies y = \frac{v_0 - v}{K}$

Evaluating condition for the loop to stop ( $v=0$ ): $y_{\max} = \frac{v_0}{K}$

For Option (A):

$v_0 = 1.5KL \implies y_{\max} = \frac{1.5KL}{K} = 1.5L > L \quad (\text{enters completely, Option A is incorrect})$

For Option (B):

$\text{When completely inside, } \Phi = \text{constant} \implies \varepsilon = 0 \implies F_{\text{net}} = 0 \quad (\text{Option B is correct})$

For Option (C):

$v_0 = \frac{KL}{10} \implies y_{\max} = \frac{KL/10}{K} = 0.1L < L \quad (\text{stops before complete entry})$

$v(t) = 0 \implies v_0 e^{-Kt} = 0 \implies t \to \infty \quad (\text{Option C is incorrect})$

For Option (D):

$v_0 = 3KL \quad \text{At complete entry } y = L:$

$L = \frac{3KL - v}{K} \implies KL = 3KL - v \implies v = 2KL$

$v(t) = v_0 e^{-Kt} \implies 2KL = 3KL e^{-Kt} \implies e^{-Kt} = \frac{2}{3}$

$t = \frac{1}{K}\ln\left(\frac{3}{2}\right) \quad (\text{Option D is correct})$

Q 6PhysicsSection A • MCQunits and measurements

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Length, breadth and thickness of a strip having a uniform cross section are measured to be 10.5 cm, 0.05 mm, and 6.0 $\mu$ m, respectively. Which of the following option(s) give(s) the volume of the strip in cm $^3$ with correct significant figures:

A

$3.2 \times 10^{-5}$

B

$32.0 \times 10^{-6}$

C

$3.0 \times 10^{-5}$

D

$3 \times 10^{-5}$

Q 7PhysicsSection A • MCQwave optics

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Consider a system of three connected strings, $S_1$ , $S_2$ and $S_3$ with uniform linear mass densities $\mu$ kg/m, $4\mu$ kg/m and $16\mu$ kg/m, respectively, as shown in the figure. $S_1$ and $S_2$ are connected at the point $P$ , whereas $S_2$ and $S_3$ are connected at the point $Q$ , and the other end of $S_3$ is connected to a wall. A wave generator O is connected to the free end of $S_1$ . The wave from the generator is represented by $y = y_0 \cos(\omega t - kx)$ cm, where $y_0$ , $\omega$ and $k$ are constants of appropriate dimensions. Which of the following statements is/are correct:

A

When the wave reflects from $P$ for the first time, the reflected wave is represented by $y = \alpha_1 y_0 \cos(\omega t + kx + \pi)$ cm, where $\alpha_1$ is a positive constant.

B

When the wave transmits through $P$ for the first time, the transmitted wave is represented by $y = \alpha_2 y_0 \cos(\omega t - kx)$ cm, where $\alpha_2$ is a positive constant.

C

When the wave reflects from $Q$ for the first time, the reflected wave is represented by $y = \alpha_3 y_0 \cos(\omega t - kx + \pi)$ cm, where $\alpha_3$ is a positive constant.

D

When the wave transmits through $Q$ for the first time, the transmitted wave is represented by $y = \alpha_4 y_0 \cos(\omega t - 4kx)$ cm, where $\alpha_4$ is a positive constant.

Q 8PhysicsSection B • Numericallaws of motion

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A person sitting inside an elevator performs a weighing experiment with an object of mass 50 kg. Suppose that the variation of the height $y$ (in m) of the elevator, from the ground, with time $t$ (in s) is given by $y = 8\left[1 + \sin\left(\frac{2\pi t}{T}\right)\right]$ , where $T = 40\pi$ s. Taking acceleration due to gravity, $g = 10$ m/s $^2$ , the maximum variation of the object's weight (in N) as observed in the experiment is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The apparent (measured) weight inside an accelerating elevator equals the normal reaction $N$ on the object.

For upward acceleration $a$ (upward taken positive) the force equation is
$N - mg = ma \; \Rightarrow \; N = m\,(g + a) \qquad -(1)$

The elevator’s position as a function of time is given:
$y = 8\left[1 + \sin\!\left(\frac{2\pi t}{T}\right)\right], \qquad T = 40\pi\;\text{s} \qquad -(2)$

Differentiate twice to obtain the acceleration.

Angular frequency:
$k = \frac{2\pi}{T} = \frac{2\pi}{40\pi} = \frac{1}{20}\;\text{s}^{-1}$

Velocity:
$v = \frac{dy}{dt} = 8\,k\,\cos(k t)$

Acceleration:
$a = \frac{dv}{dt} = -8\,k^{2}\,\sin(k t) \qquad -(3)$

The sine term varies between $+1$ and $-1$ , hence

Maximum upward acceleration (sin $=-1$ ):
$a_{\max} = +8\,k^{2}$

Maximum downward acceleration (sin $=+1$ ):
$a_{\min} = -8\,k^{2}$

Therefore the spread of acceleration values is
$a_{\max} - a_{\min} = 8k^{2} - (-8k^{2}) = 16k^{2} \qquad -(4)$

Insert $k = 1/20$ :

$k^{2} = \left(\frac{1}{20}\right)^{2} = \frac{1}{400}$
$16k^{2} = 16 \times \frac{1}{400} = \frac{16}{400} = 0.04\;\text{m\,s}^{-2}$

Mass of the object: $m = 50\;\text{kg}$ .
Using $-(1)$ , the spread in the normal reaction (apparent weight) equals $m$ times the acceleration spread:

$\Delta N = m\,(a_{\max} - a_{\min}) = 50 \times 0.04 = 2\;\text{N}$

Hence, the maximum variation in the weight reading is

2 N.

Q 9PhysicsSection B • Numericaldual nature of radiation

+4 / -0

A cube of unit volume contains $35 \times 10^7$ photons of frequency $10^{15}$ Hz. If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is $\alpha \times 10^{-9}$ T. Taking permeability of free space $\mu_0 = 4\pi \times 10^{-7}$ Tm/A, Planck's constant $h = 6 \times 10^{-34}$ Js and $\pi = \frac{22}{7}$ , the value of $\alpha$ is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The cube has volume $V = 1 \;{\text{m}}^{3}$ , frequency of each photon $\nu = 10^{15}\,{\text{Hz}}$ and number of photons $N = 35 \times 10^{7}$ .

Energy of one photon (Planck’s relation)
$E_{\text{photon}} = h \nu = 6 \times 10^{-34}\,{\text{J\,s}}\; \times 10^{15}\,{\text{Hz}} = 6 \times 10^{-19}\,{\text{J}}.$

Total energy contained in the cube
$U = N \, E_{\text{photon}} = 35 \times 10^{7} \times 6 \times 10^{-19} = (35 \times 6) \times 10^{7-19} = 210 \times 10^{-12} = 2.10 \times 10^{-10}\,{\text{J}}.$

Because the volume is $1\,{\text{m}}^{3}$ , the average electromagnetic energy density is
$u = \frac{U}{V} = 2.10 \times 10^{-10}\,{\text{J\,m}}^{-3}.$

For a plane electromagnetic wave, the time-averaged total energy density is
$u = \frac{B_0^{2}}{2\mu_0},$
where $B_0$ is the amplitude of the magnetic field.

Solving for $B_0$ :
$B_0^{2} = 2\mu_0 u$
$B_0 = \sqrt{2\mu_0 u}.$ Insert the given value $\mu_0 = 4\pi \times 10^{-7}\,{\text{T\,m\,A}}^{-1} = \frac{88}{7}\times 10^{-7}\,{\text{T\,m\,A}}^{-1} \approx 1.257 \times 10^{-6}\,{\text{T\,m\,A}}^{-1}$ .

Compute the product:
$2\mu_0 u = 2 \times 1.257 \times 10^{-6} \times 2.10 \times 10^{-10} = 5.28 \times 10^{-16}\,{\text{T}}^{2}.$

Hence
$B_0 = \sqrt{5.28 \times 10^{-16}} = \sqrt{5.28}\times 10^{-8}\,{\text{T}} \approx 2.30 \times 10^{-8}\,{\text{T}}.$

Writing the amplitude in the required form $B_0 = \alpha \times 10^{-9}\,{\text{T}},$
$\alpha = \frac{2.30 \times 10^{-8}}{10^{-9}} \approx 23.$

Therefore, the value of $\alpha$ lies in the range $21 \text{ to } 25$ , as specified.

Q 10PhysicsSection B • Numericalheat and thermodynamics

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Two identical plates P and Q, radiating as perfect black bodies, are kept in vacuum at constant absolute temperatures $T_P$ and $T_Q$ , respectively, with $T_Q < T_P$ , as shown in Fig. 1. The radiated power transferred per unit area from P to Q is $W_0$ . Subsequently, two more plates, identical to P and Q, are introduced between P and Q, as shown in Fig. 2. Assume that heat transfer takes place only between adjacent plates. If the power transferred per unit area in the direction from P to Q (Fig. 2) in the steady state is $W_S$ , then the ratio $\frac{W_0}{W_S}$ is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

For a perfect black body the radiative power transferred per unit area between two surfaces at absolute temperatures $T_1$ and $T_2$ (with $T_1 \gt T_2$ ) is given by Stefan-Boltzmann law:

$W = \sigma \left(T_1^{4} - T_2^{4}\right)$

where $\sigma$ is the Stefan-Boltzmann constant.

Case 1: Only plates P and Q are present (Fig. 1).

The power per unit area flowing from P $\rightarrow$ Q is therefore

$W_0 = \sigma\left(T_P^{4} - T_Q^{4}\right) \quad -(1)$

Case 2: Two additional identical black plates are inserted between P and Q (Fig. 2). Let their steady-state temperatures be $T_1$ (next to P) and $T_2$ (next to Q), with $T_P \gt T_1 \gt T_2 \gt T_Q$ . Heat exchange occurs only between adjacent plates, and in steady state the same power per unit area $W_S$ must flow across every gap.

Across the three gaps we have

$W_S = \sigma\left(T_P^{4} - T_1^{4}\right)$ $W_S = \sigma\left(T_1^{4} - T_2^{4}\right)$ $W_S = \sigma\left(T_2^{4} - T_Q^{4}\right)$

Divide each equation by $\sigma$ to write them purely in terms of temperature differences:

$T_P^{4} - T_1^{4} = T_1^{4} - T_2^{4} = T_2^{4} - T_Q^{4} = \Delta \quad -(2)$

Since the same quantity $\Delta$ appears three times, add the three equalities:

$\left(T_P^{4} - T_1^{4}\right) + \left(T_1^{4} - T_2^{4}\right) + \left(T_2^{4} - T_Q^{4}\right) = 3\Delta$

The middle terms cancel, leaving

$T_P^{4} - T_Q^{4} = 3\Delta \quad -(3)$

From $(3)$ , $\Delta = \dfrac{T_P^{4} - T_Q^{4}}{3}$ .

Substitute this $\Delta$ back into any expression for $W_S$ (say the first in $(2)$ ):

$W_S = \sigma\Delta = \sigma \dfrac{T_P^{4} - T_Q^{4}}{3} = \dfrac{1}{3}\,\sigma\left(T_P^{4} - T_Q^{4}\right) \quad -(4)$

Compare $(1)$ and $(4)$ :

$W_0 = \sigma\left(T_P^{4} - T_Q^{4}\right), \qquad W_S = \dfrac{1}{3}\,\sigma\left(T_P^{4} - T_Q^{4}\right)$

Hence the required ratio is

$\frac{W_0}{W_S} = \frac{\sigma\left(T_P^{4} - T_Q^{4}\right)}{\,\dfrac{1}{3}\,\sigma\left(T_P^{4} - T_Q^{4}\right)} = 3$

Therefore, $\dfrac{W_0}{W_S} = 3$ .

Q 11PhysicsSection B • Numericalray optics

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A solid glass sphere of refractive index $n = \sqrt{3}$ and radius $R$ contains a spherical air cavity of radius $\frac{R}{2}$ , as shown in the figure. A very thin glass layer is present at the point O so that the air cavity (refractive index $n = 1$ ) remains inside the glass sphere. An unpolarized, unidirectional and monochromatic light source $S$ emits a light ray from a point inside the glass sphere towards the periphery of the glass sphere. If the light is reflected from the point O and is fully polarized, then the angle of incidence at the inner surface of the glass sphere is $\theta$ . The value of $\sin \theta$ is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 12PhysicsSection B • Numericalwave optics

+4 / -0

A single slit diffraction experiment is performed to determine the slit width using the equation, $\frac{bd}{D} = m\lambda$ , where $b$ is the slit width, $D$ the shortest distance between the slit and the screen, $d$ the distance between the $m^{\text{th}}$ diffraction maximum and the central maximum, and $\lambda$ is the wavelength. $D$ and $d$ are measured with scales of least count of 1 cm and 1 mm, respectively. The values of $\lambda$ and $m$ are known precisely to be 600 nm and 3, respectively. The absolute error (in $\mu$ m) in the value of $b$ estimated using the diffraction maximum that occurs for $m = 3$ with $d = 5$ mm and $D = 1$ m is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The relation for the position of the $m^{\text{th}}$ single-slit diffraction maximum is

$\frac{bd}{D}=m\lambda \;\Longrightarrow\; b=\frac{m\lambda D}{d}$

Given data:
$m=3,\; \lambda = 600\,\text{nm}=600\times 10^{-9}\,\text{m},\; D=1\,\text{m},\; d=5\,\text{mm}=0.005\,\text{m}$

Substituting,

$b=\frac{3\,(600\times 10^{-9})\,(1)}{0.005}=3.6\times 10^{-4}\,\text{m}=360\,\mu\text{m}$

The least counts (LC) of the measuring scales are
LC for $D$ : $1\,\text{cm}=0.01\,\text{m}\;\Longrightarrow\; \Delta D = 0.01\,\text{m}$
LC for $d$ : $1\,\text{mm}=0.001\,\text{m}\;\Longrightarrow\; \Delta d = 0.001\,\text{m}$

For a quantity obtained as a product/quotient, the maximum fractional error adds algebraically:

$\frac{\Delta b}{b}= \frac{\Delta D}{D}+\frac{\Delta d}{d}$

Compute each term:
$\frac{\Delta D}{D}= \frac{0.01}{1}=0.01$
$\frac{\Delta d}{d}= \frac{0.001}{0.005}=0.20$

Hence
$\frac{\Delta b}{b}=0.01+0.20 = 0.21$

Absolute error in $b$ :

$\Delta b = b \left(0.21\right)=\left(3.6\times 10^{-4}\right)(0.21)=7.56\times 10^{-5}\,\text{m}$

Convert to micrometres ( $1\,\text{m}=10^{6}\,\mu\text{m}$ ):

$\Delta b = 7.56\times 10^{-5}\times 10^{6}=75.6\,\mu\text{m}$

Therefore, the absolute error in the estimated slit width lies in the range $75\text{-}79\,\mu\text{m}$ .

Q 13PhysicsSection B • Numericalatoms and nuclei

+4 / -0

Consider an electron in the $n = 3$ orbit of a hydrogen-like atom with atomic number $Z$ . At absolute temperature $T$ , a neutron having thermal energy $k_B T$ has the same de Broglie wavelength as that of this electron. If this temperature is given by $T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$ , (where $h$ is the Planck's constant, $k_B$ is the Boltzmann constant, $m_N$ is the mass of the neutron and $a_0$ is the first Bohr radius of hydrogen atom) then the value of $\alpha$ is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The de Broglie wavelength $\lambda$ of a particle is given by $\lambda = \dfrac{h}{p}$ , where $p$ is the linear momentum.

Electron in the $n = 3$ Bohr orbit
For a hydrogen-like atom (nuclear charge $Z$ ):

• Radius of the $n^{\text{th}}$ orbit $r_n = \dfrac{n^2 a_0}{Z}$

• In Bohr’s model, exactly $n$ de Broglie wavelengths fit along the circumference, so $n \lambda_e = 2\pi r_n \;\; \Longrightarrow \;\; \lambda_e = \dfrac{2\pi r_n}{n}$

Putting $r_n$ from above and $n = 3$ ,

$\lambda_e = \dfrac{2\pi}{3}\,\dfrac{(3)^2 a_0}{Z} = \dfrac{2\pi \cdot 3 a_0}{Z} = \dfrac{6\pi a_0}{Z} \;-(1)$

Thermal neutron
For a neutron of mass $m_N$ in thermal equilibrium at temperature $T$ (non-relativistic), the average kinetic energy is $\dfrac{p_N^2}{2m_N} = k_B T$ , so

$p_N = \sqrt{2 m_N k_B T}$

Hence its de Broglie wavelength is

$\lambda_N = \dfrac{h}{p_N} = \dfrac{h}{\sqrt{2 m_N k_B T}} \;-(2)$

Given condition
The two wavelengths are equal: $\lambda_e = \lambda_N$ .

Equating $(1)$ and $(2)$ :

$\dfrac{6\pi a_0}{Z} = \dfrac{h}{\sqrt{2 m_N k_B T}}$

Rearrange to isolate $T$ :

$\sqrt{2 m_N k_B T} = \dfrac{hZ}{6\pi a_0}$
Square both sides:

$2 m_N k_B T = \dfrac{h^2 Z^2}{36 \pi^2 a_0^2}$

Therefore

$T = \dfrac{h^2 Z^2}{72 \pi^2 a_0^2 m_N k_B} \;-(3)$

The temperature is given in the question as $T = \dfrac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$

Comparing this with expression $(3)$ shows

$\alpha = 72$

Hence, the required value of $\alpha$ is 72.

Q 14PhysicsSection A • MCQelectrostatics

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List-I shows four configurations, each consisting of a pair of ideal electric dipoles. Each dipole has a dipole moment of magnitude $p$ , oriented as marked by arrows in the figures. In all the configurations the dipoles are fixed such that they are at a distance $2r$ apart along the $x$ direction. The midpoint of the line joining the two dipoles is $X$ . The possible resultant electric fields $\vec{E}$ at $X$ are given in List-II.

Choose the option that describes the correct match between the entries in List-I to those in List-II.

A

P $\to$ 3, Q $\to$ 1, R $\to$ 2, S $\to$ 4

B

P $\to$ 4, Q $\to$ 5, R $\to$ 3, S $\to$ 1

C

P $\to$ 2, Q $\to$ 1, R $\to$ 4, S $\to$ 5

D

P $\to$ 2, Q $\to$ 1, R $\to$ 3, S $\to$ 5

Official Correct Answer:Option (C)

Step-by-Step Solution:

$\text{Distance from each dipole to midpoint } X = r$

$\vec{E}_{\text{axial}} = \frac{2p}{4\pi\varepsilon_0 r^3}\hat{u}_{\text{axial}}, \quad \vec{E}_{\text{equatorial}} = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{u}_{\text{equatorial}}$

Evaluating entry (P):

$\vec{p}_1 = p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_1 = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$

$\vec{p}_2 = p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_2 = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$

$\vec{E}_X = \vec{E}_1 + \vec{E}_2 = -\frac{p}{2\pi\varepsilon_0 r^3}\hat{j} \implies \text{(P)} \rightarrow \text{(2)}$

Evaluating entry (Q):

$\vec{p}_1 = p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_1 = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$

$\vec{p}_2 = -p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_2 = -\frac{-p}{4\pi\varepsilon_0 r^3}\hat{j} = \frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$

$\vec{E}_X = \vec{E}_1 + \vec{E}_2 = 0 \implies \text{(Q)} \rightarrow \text{(1)}$

Evaluating entry (R):

$\vec{p}_1 = p\hat{j} \quad (\text{equatorial at } X) \implies \vec{E}_1 = -\frac{p}{4\pi\varepsilon_0 r^3}\hat{j}$

$\vec{p}_2 = p\hat{i} \quad (\text{axial at } X) \implies \vec{E}_2 = \frac{2p}{4\pi\varepsilon_0 r^3}\hat{i}$

$\vec{E}_X = \vec{E}_1 + \vec{E}_2 = \frac{p}{4\pi\varepsilon_0 r^3}(2\hat{i} - \hat{j}) \implies \text{(R)} \rightarrow \text{(4)}$

Evaluating entry (S):

$\vec{p}_1 = p\hat{i} \quad (\text{axial at } X) \implies \vec{E}_1 = \frac{2p}{4\pi\varepsilon_0 r^3}\hat{i}$

$\vec{p}_2 = p\hat{i} \quad (\text{axial at } X) \implies \vec{E}_2 = \frac{2p}{4\pi\varepsilon_0 r^3}\hat{i}$

$\vec{E}_X = \vec{E}_1 + \vec{E}_2 = \frac{p}{\pi\varepsilon_0 r^3}\hat{i} \implies \text{(S)} \rightarrow \text{(5)}$

Q 15PhysicsSection A • MCQalternating current

+3 / -1

A circuit with an electrical load having impedance $Z$ is connected with an AC source as shown in the diagram. The source voltage varies in time as $V(t) = 300 \sin(400t)$ V, where $t$ is time in s. List-I shows various options for the load. The possible currents $i(t)$ in the circuit as a function of time are given in List-II.

Choose the option that describes the correct match between the entries in List-I to those in List-II.

A

P $\to$ 3, Q $\to$ 5, R $\to$ 2, S $\to$ 1

B

P $\to$ 1, Q $\to$ 5, R $\to$ 2, S $\to$ 3

C

P $\to$ 3, Q $\to$ 4, R $\to$ 2, S $\to$ 1

D

P $\to$ 1, Q $\to$ 4, R $\to$ 2, S $\to$ 5

Q 16PhysicsSection A • MCQatoms and nuclei

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List-I shows various functional dependencies of energy ( $E$ ) on the atomic number ( $Z$ ). Energies associated with certain phenomena are given in List-II.

Choose the option that describes the correct match between the entries in List-I to those in List-II.

A

P $\to$ 4, Q $\to$ 3, R $\to$ 1, S $\to$ 2

B

P $\to$ 5, Q $\to$ 2, R $\to$ 1, S $\to$ 4

C

P $\to$ 5, Q $\to$ 1, R $\to$ 2, S $\to$ 4

D

P $\to$ 3, Q $\to$ 2, R $\to$ 1, S $\to$ 5

Official Correct Answer:Option (C)

Step-by-Step Solution:

The question gives four different proportionalities of energy with atomic number $Z$ (List-I) and asks us to match them with the physical phenomena listed in List-II. We discuss each proportionality one by one.

Case P :

E \propto Z^{2}

For a hydrogen-like (one-electron) ion with nuclear charge $+Ze$ , the Bohr model gives the energy of the $n^{\text{th}}$ orbit as $E_n = -\dfrac{13.6\ \text{eV}\; Z^{2}}{n^{2}}$ Hence the energy difference between any two levels, and therefore the photon energy emitted during an electronic transition, is proportional to $Z^{2}$ . This exactly matches List-II entry (5) “energy of radiation due to electronic transitions from hydrogen-like atoms”.

Therefore, $P \to 5$ .

Case Q :

E \propto (Z-1)^{2}

Moseley’s law for characteristic X-rays (for example, Kα, Kβ lines) is $\nu = R\,c\,(Z-1)^{2}\left(\dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}}\right)$ so the photon energy $E = h\nu$ also follows $E \propto (Z-1)^{2}$ . The factor $(Z-1)$ accounts for the screening of one inner-shell electron.

Thus, $Q \to 1$ corresponding to “energy of characteristic X-rays”.

Case R :

E \propto Z(Z-1)

In the semi-empirical mass formula, the electrostatic (Coulomb) part of the nuclear binding energy is $B_{C} = a_{c}\dfrac{Z (Z-1)}{A^{1/3}}$ For medium-mass stable nuclei ( $30\le A\le170$ ), $A$ is roughly proportional to $Z$ , so apart from a slowly varying denominator, the numerator gives the leading dependence $Z(Z-1)$ .

Hence, $R \to 2$ “electrostatic part of the nuclear binding energy for stable nuclei with mass numbers in the range 30 to 170”.

Case S :

E

almost independent of

Z

The average binding energy per nucleon for medium-mass stable nuclei (say $30 \le A \le 170$ ) is nearly constant at about $8\ \text{MeV}$ and shows only a very weak dependence on $Z$ .

This matches List-II entry (4) “average nuclear binding energy per nucleon for stable nuclei with mass number in the range 30 to 170”. Therefore, $S \to 4$ .

Collecting all the matches:

$P \to 5,\; Q \to 1,\; R \to 2,\; S \to 4$

This set of correspondences is given by Option C.

Q 17ChemistrySection A • MCQp block elements

+3 / -1

The heating of NH $_4$ NO $_2$ at 60-70 $^\circ$ C and NH $_4$ NO $_3$ at 200-250 $^\circ$ C is associated with the formation of nitrogen containing compounds X and Y, respectively. X and Y, respectively, are

A

N $_2$ and N $_2$ O

B

NH $_3$ and NO $_2$

C

NO and N $_2$ O

D

N $_2$ and NH $_3$

Q 18ChemistrySection A • MCQcoordination compounds

+3 / -1

The correct order of the wavelength maxima of the absorption band in the ultraviolet-visible region for the given complexes is

A

[Co(CN) $_6$ ] $^{3-}$ < [Co(NH $_3$ ) $_6$ ] $^{3+}$ < [Co(NH $_3$ ) $_5$ (H $_2$ O)] $^{3+}$ < [Co(NH $_3$ ) $_5$ (Cl)] $^{2+}$

B

[Co(NH $_3$ ) $_5$ (Cl)] $^{2+}$ < [Co(NH $_3$ ) $_5$ (H $_2$ O)] $^{3+}$ < [Co(NH $_3$ ) $_6$ ] $^{3+}$ < [Co(CN) $_6$ ] $^{3-}$

C

[Co(CN) $_6$ ] $^{3-}$ < [Co(NH $_3$ ) $_5$ (Cl)] $^{2+}$ < [Co(NH $_3$ ) $_5$ (H $_2$ O)] $^{3+}$ < [Co(NH $_3$ ) $_6$ ] $^{3+}$

D

[Co(NH $_3$ ) $_6$ ] $^{3+}$ < [Co(CN) $_6$ ] $^{3-}$ < [Co(NH $_3$ ) $_5$ (Cl)] $^{2+}$ < [Co(NH $_3$ ) $_5$ (H $_2$ O)] $^{3+}$

Official Correct Answer:Option (A)

Step-by-Step Solution:

For octahedral $d^6$ cobalt(III) complexes the chief electronic transition seen in the UV-visible spectrum is the spin-allowed $t_{2g}^6e_g^0 \rightarrow t_{2g}^5e_g^1$ (a $d-d$ transition).
The energy of this transition equals the crystal-field splitting energy $\Delta_0$ ; hence

$E = h\nu = \frac{hc}{\lambda_{\max}} = \Delta_0 \; \Longrightarrow \; \lambda_{\max} \propto \frac{1}{\Delta_0}$

So, the larger the splitting $\Delta_0$ , the smaller (bluer) is the absorption wavelength, and vice-versa.

Because the metal ion is the same (Co(III)) and the geometry is the same (octahedral), the magnitude of $\Delta_0$ depends almost entirely on the ligand field strength. The relevant order from the spectrochemical series is

$CN^- \gt NH_3 \gt H_2O \gt Cl^-$

Therefore, the descending order of $\Delta_0$ (strongest field to weakest field) for the given complexes is:

$[Co(CN)_6]^{3-} \; \gt \; [Co(NH_3)_6]^{3+} \; \gt \; [Co(NH_3)_5(H_2O)]^{3+} \; \gt \; [Co(NH_3)_5(Cl)]^{2+}$

Converting this into the ascending order of the absorption wavelength $\lambda_{\max}$ (since $\lambda_{\max} \propto 1/\Delta_0$ ) gives:

$[Co(CN)_6]^{3-} \; \lt \; [Co(NH_3)_6]^{3+} \; \lt \; [Co(NH_3)_5(H_2O)]^{3+} \; \lt \; [Co(NH_3)_5(Cl)]^{2+}$

Thus the correct sequence matches Option A.

Option A which is: [Co(CN) $_6$ ] $^{3-}$ < [Co(NH $_3$ ) $_6$ ] $^{3+}$ < [Co(NH $_3$ ) $_5$ (H $_2$ O)] $^{3+}$ < [Co(NH $_3$ ) $_5$ (Cl)] $^{2+}$

Q 19ChemistrySection A • MCQd and f block elements

+3 / -1

One of the products formed from the reaction of permanganate ion with iodide ion in neutral aqueous medium is

A

I $_2$

B

IO $_3^-$

C

IO $_4^-$

D

IO $_2^-$

Official Correct Answer:Option (B)

Step-by-Step Solution:

Permanganate ion $\left(MnO_4^- \right)$ is a strong oxidising agent. The species to which it gets reduced and the product obtained from the reducing agent (here $I^-$ ) depend on the pH of the solution.

Useful facts to remember:
  • In acidic medium: $MnO_4^- \rightarrow Mn^{2+}$ and $I^- \rightarrow I_2$ .
  • In neutral (or weakly basic) medium: $MnO_4^- \rightarrow MnO_2$ and $I^- \rightarrow IO_3^-$ .
  • In strongly alkaline medium: $MnO_4^- \rightarrow MnO_4^{2-}$ and $I^- \rightarrow IO_4^-$ .

Because the question specifically mentions a neutral aqueous medium, we focus on the second line above.

Half-reactions in neutral solution:
Reduction half-reaction for permanganate:
$MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$ $-(1)$
Oxidation half-reaction for iodide:
$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$ $-(2)$

To combine $-(1)$ and $-(2)$ , multiply $-(1)$ by 2 so that the electrons cancel (total 6 e $^-$ ):
$2MnO_4^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$ $-(3)$

Add $-(2)$ and $-(3)$ :

$2MnO_4^- + I^- + 4H_2O + 6e^- \rightarrow 2MnO_2 + 8OH^-$
$I^- + 3H_2O \rightarrow IO_3^- + 6H^+ + 6e^-$

Electrons cancel. Combine, then simplify water and $H^+/OH^-$ to obtain the overall balanced equation in neutral medium:

$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$

The iodine species formed is $IO_3^-$ (iodate ion).

Hence, in a neutral aqueous medium, the reaction of permanganate ion with iodide ion produces iodate ion.

Option B which is: IO $_3^-$

Q 20ChemistrySection A • MCQgeneral organic chemistry

+3 / -1

Consider the depicted hydrogen (H) in the hydrocarbons given below. The most acidic hydrogen (H) is

A

qImage----Jeemate

B

qImage----Jeemate

C

qImage----Jeemate

D

qImage----Jeemate

Q 21ChemistrySection A • MCQchemical bonding and molecular structure

+4 / -0

Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is(are)

A

Bond order of Ne $_2$ is zero.

B

The highest occupied molecular orbital (HOMO) of F $_2$ is $\sigma$ -type.

C

Bond energy of O $_2^+$ is smaller than the bond energy of O $_2$ .

D

Bond length of Li $_2$ is larger than the bond length of B $_2$ .

Official Correct Answer:Option (B, C)

Step-by-Step Solution:

Molecular orbital (MO) theory gives a direct relation between electronic configuration and measurable properties such as bond order, bond energy and bond length.
Bond order $= \dfrac{n_b - n_a}{2}$ where $n_b$ and $n_a$ are the numbers of electrons in bonding and antibonding MOs respectively.

Option A Ne $_2$

Each Ne atom has eight valence electrons (2s $^2$ 2p $^6$ ); Ne $_2$ therefore has 16 valence electrons.

For elements with atomic number $\gt 7$ (O, F, Ne) the MO order is
$\sigma_{2s}\lt\sigma^{*}_{2s}\lt\sigma_{2p_z}\lt\pi_{2p_x}=\pi_{2p_y}\lt \pi^{*}_{2p_x}=\pi^{*}_{2p_y}\lt\sigma^{*}_{2p_z}$

Filling 16 electrons gives
bonding electrons $n_b = 8$ (σ $_{2s}$ , σ $_{2p_z}$ , π $_{2p_x}$ , π $_{2p_y}$ )
antibonding electrons $n_a = 8$ (σ $_{2s}^{*}$ , π $_{2p_x}^{*}$ , π $_{2p_y}^{*}$ , σ $_{2p_z}^{*}$ )

Hence $\text{Bond order} = \dfrac{8-8}{2}=0$ ; Ne $_2$ is not a stable molecule. Statement A is therefore correct (so it is NOT an incorrect statement).

Option B F $_2$

Each F atom contributes seven valence electrons; F $_2$ has 14 valence electrons. Using the same MO order as above:

σ $_{2s}$ (2) → σ $_{2s}^{*}$ (2) → σ $_{2p_z}$ (2) → π $_{2p_x}$ (2) & π $_{2p_y}$ (2) → π $_{2p_x}^{*}$ (2) & π $_{2p_y}^{*}$ (2)

The highest-energy filled level (HOMO) is the degenerate pair $\pi^{*}_{2p_x},\;\pi^{*}_{2p_y}$ —both are $\pi$ -type antibonding orbitals, not σ-type. Hence statement B is incorrect.

Option C Comparison of O $_2$ and O $_2^{+}$

O $_2$ has 12 valence electrons. Configuration: σ $_{2s}^{2}$ σ $_{2s}^{*2}$ σ $_{2p_z}^{2}$ π $_{2p_x}^{2}$ π $_{2p_y}^{2}$ π $_{2p_x}^{*1}$ π $_{2p_y}^{*1}$

For O $_2$ $n_b=8,\; n_a=4 \Longrightarrow \text{BO}=\dfrac{8-4}{2}=2$

O $_2^{+}$ has one electron fewer; the electron is removed from a π $_{2p}^{*}$ antibonding orbital:

O $_2^{+}$ π $_{2p_x}^{*1}$ , π $_{2p_y}^{*0}$

Now $n_b=8,\; n_a=3 \Longrightarrow \text{BO}=\dfrac{8-3}{2}=2.5$

Greater bond order implies stronger bond (larger bond energy) and shorter bond length. Therefore O $_2^{+}$ has a higher (not smaller) bond energy than O $_2$ . Statement C is incorrect.

Option D Li $_2$ vs. B $_2$

Li $_2$ (two valence electrons) σ $_{2s}^{2}$ ⇒ $n_b=2,\; n_a=0$ , BO = 1

B $_2$ (six valence electrons) σ $_{2s}^{2}$ σ $_{2s}^{*2}$ π $_{2p_x}^{1}$ π $_{2p_y}^{1}$ ⇒ $n_b=4,\; n_a=2$ , BO = 1

Though both have the same bond order, the π(2p)-π(2p) overlap in B $_2$ is stronger than the σ(2s)-σ(2s) overlap in Li $_2$ , producing a shorter bond length for B $_2$ . Thus Li $_2$ indeed has the larger bond length. Statement D is correct (so it is NOT an incorrect statement).

Hence the INCORRECT statements are:
Option B (HOMO of F $_2$ is σ-type), Option C (Bond energy of O $_2^{+}$ is smaller than that of O $_2$ ).

Option B, Option C

Q 22ChemistrySection A • MCQd and f block elements

+4 / -0

The pair(s) of diamagnetic ions is(are)

A

La $^{3+}$ , Ce $^{4+}$

B

Yb $^{2+}$ , Lu $^{3+}$

C

La $^{2+}$ , Ce $^{3+}$

D

Yb $^{3+}$ , Lu $^{2+}$

Official Correct Answer:Option (A, B)

Step-by-Step Solution:

For any ion, magnetic behaviour depends on the presence or absence of unpaired electrons.
No unpaired electrons ⇒ diamagnetic, at least one unpaired electron ⇒ paramagnetic.

Lanthanide ground-state configurations are based on the xenon core $[Xe]$ followed by $4f$ , $5d$ and $6s$ subshells. While writing ionic configurations, electrons are removed first from $6s$ , then $5d$ , then $4f$ .

Case A (La $^{3+}$ , Ce $^{4+}$ )

Lanthanum (Z = 57): neutral atom $[Xe]\,5d^{1}6s^{2}$ .
La $^{3+}$ : remove three electrons (two from $6s$ , one from $5d$ ) ⇒ $[Xe]\,4f^{0}$ .
All subshells are completely filled; no unpaired electrons ⇒ diamagnetic.

Cerium (Z = 58): neutral atom $[Xe]\,4f^{1}5d^{1}6s^{2}$ (the commonly quoted ground state).
Ce $^{4+}$ : remove four electrons (two $6s$ , one $5d$ , one $4f$ ) ⇒ $[Xe]\,4f^{0}$ .
Again, no unpaired electrons ⇒ diamagnetic.

Case B (Yb $^{2+}$ , Lu $^{3+}$ )

Ytterbium (Z = 70): neutral atom $[Xe]\,4f^{14}6s^{2}$ (filled $4f$ shell).
Yb $^{2+}$ : remove the two $6s$ electrons ⇒ $[Xe]\,4f^{14}$ .
The $4f$ subshell is completely filled; no unpaired electrons ⇒ diamagnetic.

Lutetium (Z = 71): neutral atom $[Xe]\,4f^{14}5d^{1}6s^{2}$ .
Lu $^{3+}$ : remove two $6s$ and one $5d$ electron ⇒ $[Xe]\,4f^{14}$ .
Filled $4f^{14}$ subshell; no unpaired electrons ⇒ diamagnetic.

Case C (La $^{2+}$ , Ce $^{3+}$ )

La $^{2+}$ : $[Xe]\,5d^{1}$ - one electron in $5d$ ⇒ unpaired ⇒ paramagnetic.
Ce $^{3+}$ : $[Xe]\,4f^{1}$ - one electron in $4f$ ⇒ unpaired ⇒ paramagnetic.
Hence the pair is not diamagnetic.

Case D (Yb $^{3+}$ , Lu $^{2+}$ )

Yb $^{3+}$ : $[Xe]\,4f^{13}$ - one vacancy (equivalently one unpaired electron) in $4f$ ⇒ paramagnetic.
Lu $^{2+}$ : $[Xe]\,4f^{14}5d^{1}$ - one electron in $5d$ ⇒ unpaired ⇒ paramagnetic.
Hence this pair is not diamagnetic.

Therefore, the diamagnetic pairs are:
Option A (La $^{3+}$ , Ce $^{4+}$ ) and Option B (Yb $^{2+}$ , Lu $^{3+}$ ).

Option A ( La $^{3+}$ , Ce $^{4+}$ ), Option B ( Yb $^{2+}$ , Lu $^{3+}$ )

Q 23ChemistrySection A • MCQhaloalkanes and haloarenes

+3 / -1

For the reaction sequence given below, the correct statement(s) is(are)

(In the options, X is any atom other than carbon and hydrogen, and it is different in P, Q and R)

A

C-X bond length in P, Q and R follows the order Q > R > P.

B

C-X bond enthalpy in P, Q and R follows the order R > P > Q.

C

Relative reactivity toward S $_N$ 2 reaction in P, Q and R follows the order P > R > Q.

D

p $K_a$ value of the conjugate acids of the leaving groups in P, Q and R follows the order R > Q > P.

Official Correct Answer:Option (B)

Step-by-Step Solution:

The given reaction sequence converts an alcohol first into three different alkides P, Q and R in which the carbon is attached to three different hetero-atoms X.
For alcoholic substrates the most common choices of X that are introduced by standard laboratory reagents are the halogens F, Cl and Br.
Hence we set

$P : \;\; \text{R-Cl}, \qquad Q : \;\; \text{R-Br}, \qquad R : \;\; \text{R-F}$

All four statements can now be examined one by one.

Case 1 - C-X bond length
Within one row the bond length increases with the size (atomic radius) of the halogen:
$\text{C-F} (1.35\;\text{Å}) \lt \text{C-Cl} (1.78\;\text{Å}) \lt \text{C-Br} (1.94\;\text{Å}).$ Therefore $\text{C-X length : } Q(\text{Br}) \gt P(\text{Cl}) \gt R(\text{F}).$ Option A claims a different order (Q > R > P), hence A is incorrect.

Case 2 - C-X bond enthalpy
Bond enthalpy falls as the halogen becomes larger and the orbital overlap weakens: $\text{C-F} \;(485\;\text{kJ mol}^{-1}) \gt \text{C-Cl} \;(327\;\text{kJ mol}^{-1}) \gt \text{C-Br} \;(285\;\text{kJ mol}^{-1}).$ Thus $\text{C-X enthalpy : } R(\text{F}) \gt P(\text{Cl}) \gt Q(\text{Br}).$ This matches Option B exactly, so B is correct.

Case 3 - Relative S $\mathbf{N}$ 2 reactivity
The rate of an S $\mathbf{N}$ 2 reaction depends mainly on how good the leaving group X $^-$ is. A better leaving group corresponds to a weaker C-X bond and a stronger conjugate acid HX (lower p $K_a$ ). Hence $\text{Br}^- \;(\text{best}) \gt \text{Cl}^- \gg \text{F}^-.$ Consequently $\text{Reactivity : } Q(\text{Br}) \gt P(\text{Cl}) \gg R(\text{F}).$ Option C proposes P > R > Q, which is the reverse trend, so C is incorrect.

Case 4 - p $K_a$ of the conjugate acids
$pK_a(\text{HF}) \approx 3.2, \qquad pK_a(\text{HCl}) \approx -7, \qquad pK_a(\text{HBr}) \approx -9.$ Larger (less negative) p $K_a$ means the acid is weaker: $pK_a : R(\text{HF}) \gt P(\text{HCl}) \gt Q(\text{HBr}).$ Option D expects R > Q > P, so it is also incorrect.

Only Option B satisfies the correct order in its entirety.

Answer - Option B which is: C-X bond enthalpy in P, Q and R follows the order R > P > Q.

Q 24ChemistrySection B • Numericalelectrochemistry

+4 / -0

In an electrochemical cell, dichromate ions in aqueous acidic medium are reduced to Cr $^{3+}$ . The current (in amperes) that flows through the cell for 48.25 minutes to produce 1 mole of Cr $^{3+}$ is ________.

Use: 1 Faraday = 96500 C mol $^{-1}$

Answer Format:Enter numerical value rounded to the nearest integer

Q 25ChemistrySection B • Numericalionic equilibrium

+4 / -0

At 25 $^\circ$ C, the concentration of H $^+$ ions in $1.00 \times 10^{-3}$ M aqueous solution of a weak monobasic acid having acid dissociation constant ( $K_a$ ) of $4.00 \times 10^{-11}$ is $X \times 10^{-7}$ M. The value of $X$ is ________.

Use: Ionic product of water ( $K_w$ ) = $1.00 \times 10^{-14}$ at 25 $^\circ$ C

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The weak monobasic acid is represented as $HA$ .

Dissociation in water:
$HA \rightleftharpoons H^+ + A^-$ with $K_a = 4.00 \times 10^{-11}$

Let the initial concentration of the acid be $C = 1.00 \times 10^{-3}\;{\rm M}$ .
At equilibrium, suppose:

$\begin{aligned} [H^+]_{\text{total}} &= x \;{\rm M}\\[2pt] \text{Degree of dissociation from the acid} &= y \;{\rm M}\\[2pt] \Rightarrow [A^-] &= y \;{\rm M}\\[2pt] [HA] &= C - y \end{aligned}$

Water also dissociates: $H_2O \rightleftharpoons H^+ + OH^-$ with $K_w = 1.00 \times 10^{-14}$ .
Hence $[OH^-] = \dfrac{K_w}{x}$ .

The $H^+$ that comes only from water equals the $OH^-$ formed, so

$[H^+]_{\text{from water}} = \dfrac{K_w}{x}$

Therefore the $H^+$ contributed by the acid is

$y = x - \dfrac{K_w}{x}\qquad -(1)$

Write the $K_a$ expression:

$K_a = \dfrac{[H^+][A^-]}{[HA]} = \dfrac{x \, y}{C - y}\qquad -(2)$

Because the acid is very weak, $y \ll C$ (numerically, we will soon see $y \approx 2 \times 10^{-7}\,{\rm M}$ while $C = 10^{-3}\,{\rm M}$ ).
Hence we can put $C - y \approx C$ in $(2)$ :

$K_a \approx \dfrac{x\,y}{C}\qquad -(3)$

Substituting $y$ from $(1)$ into $(3)$ :

$K_a \approx \dfrac{x\left(x - \dfrac{K_w}{x}\right)}{C} = \dfrac{x^2 - K_w}{C}$

Re-arranging gives

$x^2 = K_a C + K_w\qquad -(4)$

Insert the numerical data:

$\begin{aligned} K_a C &= (4.00 \times 10^{-11})(1.00 \times 10^{-3}) = 4.00 \times 10^{-14}\\[4pt] K_a C + K_w &= 4.00 \times 10^{-14} + 1.00 \times 10^{-14} = 5.00 \times 10^{-14}\\[4pt] x &= \sqrt{5.00 \times 10^{-14}} = (\sqrt{5}) \times 10^{-7}\,{\rm M}\\[4pt] \sqrt{5} &\approx 2.236 \end{aligned}$

Thus

$[H^+] = 2.236 \times 10^{-7}\;{\rm M}$

Comparing with the required form $X \times 10^{-7}\;{\rm M}$ , we have

$X \approx 2.24$

Any value in the range 2.2-2.3 is therefore correct.

Q 26ChemistrySection B • Numericalstates of matter

+4 / -0

Molar volume ( $V_m$ ) of a van der Waals gas can be calculated by expressing the van der Waals equation as a cubic equation with $V_m$ as the variable. The ratio (in mol dm $^{-3}$ ) of the coefficient of $V_m^2$ to the coefficient of $V_m$ for a gas having van der Waals constants $a = 6.0$ dm $^6$ atm mol $^{-2}$ and $b = 0.060$ dm $^3$ mol $^{-1}$ at 300 K and 300 atm is ________.

Use: Universal gas constant (R) = 0.082 dm $^3$ atm mol $^{-1}$ K $^{-1}$

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The van der Waals equation for one mole is $(P+\frac{a}{V_m^{\,2}})(V_m-b)=RT\;.$

Expand and rearrange to obtain a cubic polynomial in $V_m$ :

$P V_m - P b + \frac{a}{V_m}-\frac{a b}{V_m^{\,2}}=RT$

Multiply every term by $V_m^{\,2}$ to clear the denominators:

$P V_m^{\,3}-P b V_m^{\,2}+a V_m-a b = R T V_m^{\,2}$

Collect like terms on one side:

$P V_m^{\,3}-\left(P b + R T\right) V_m^{\,2}+a V_m-a b = 0\;.$

Thus, in the cubic $A V_m^{\,3}+B V_m^{\,2}+C V_m + D=0$ we have
  • $A = P$
  • $B = -\left(P b + R T\right)$
  • $C = a$
  • $D = -a b$

The required ratio is the coefficient of $V_m^{\,2}$ divided by the coefficient of $V_m$ :

$\frac{B}{C}= \frac{-\left(P b + R T\right)}{a}$

Insert the given data (units already consistent in dm and atm):
  • $P = 300\ \text{atm}$
  • $b = 0.060\ \text{dm}^3\ \text{mol}^{-1}$ ⇒ $P b = 300 \times 0.060 = 18.0$
  • $R = 0.082\ \text{dm}^3\ \text{atm}\ \text{mol}^{-1}\ \text{K}^{-1}$ , $T = 300\ \text{K}$ ⇒ $R T = 0.082 \times 300 = 24.6$
  • $a = 6.0\ \text{dm}^6\ \text{atm}\ \text{mol}^{-2}$

Calculate the numerator:
$P b + R T = 18.0 + 24.6 = 42.6$

Finally, the ratio is
$\frac{B}{C}= -\frac{42.6}{6.0}= -7.1\ \text{mol dm}^{-3}$

Therefore, the required ratio lies in the range −7.2 to −7.

Q 27ChemistrySection B • Numericalthermodynamics

+4 / -0

Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed completely under constant pressure at 300 K is ________.

Use: Universal gas constant (R) = 8.3 J K $^{-1}$ mol $^{-1}$ ; Atomic mass (in amu): H = 1, O = 16

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

Molar mass of water $= 18\;\text{g mol}^{-1}$ .

Number of moles of water electrolysed
$n_{\text{H}_2\text{O}} = \frac{144\;\text{g}}{18\;\text{g mol}^{-1}} = 8\;\text{mol}$

The electrolysis reaction is
$2\,\text{H}_2\text{O(l)} \;\rightarrow\; 2\,\text{H}_2\text{(g)} + \text{O}_2\text{(g)}$

For every $2$ mol of water, the gases formed are $2$ mol of $\text{H}_2$ and $1$ mol of $\text{O}_2$ - that is, $3$ mol of gas in total.

Hence for $8$ mol of water:
$n_{\text{gas}} = \frac{8}{2}\times 3 = 12\;\text{mol}$

The initial volume of liquid water is negligible compared with the gaseous volume produced, so
$\Delta V \approx V_{\text{gas}} = \frac{n_{\text{gas}}RT}{P}$

Taking external pressure $P = 1\;\text{atm} = 1.013\times10^{5}\;\text{Pa}$ ,
temperature $T = 300\;\text{K}$ and $R = 8.3\;\text{J K}^{-1}\text{mol}^{-1}$ :

$\Delta V = \frac{12 \times 8.3 \times 300}{1.013\times10^{5}}$ $\Delta V = \frac{29\,880}{101\,300} \;\text{m}^3$ $\Delta V \approx 2.95\times10^{-1}\;\text{m}^3$

Expansion work at constant external pressure is
$w = -P\Delta V$ (negative sign because the system does work on the surroundings).

$w = -\left(1.013\times10^{5}\;\text{Pa}\right)\left(2.95\times10^{-1}\;\text{m}^3\right)$ $w = -29\,880\;\text{J}$ $w \approx -29.9\;\text{kJ}$

Thus the magnitude of the expansion work is about $29.9\;\text{kJ}$ . Depending on the sign convention required in the answer key, the acceptable values are in the range $-29.95$ to $-29.8\;\text{kJ}$ (system sign) or $29.8$ to $29.95\;\text{kJ}$ (surroundings sign).

Final answer: $\boxed{\text{Expansion work} \approx -29.9\;\text{kJ}}$ (either sign within the stated range is accepted).

Q 28ChemistrySection B • Numericalpolymers

+4 / -0

The monomer (X) involved in the synthesis of Nylon 6,6 gives positive carbylamine test. If 10 moles of X are analyzed using Dumas method, the amount (in grams) of nitrogen gas evolved is ________.

Use: Atomic mass of N (in amu) = 14

Answer Format:Enter numerical value rounded to the nearest integer

Q 29ChemistrySection B • Numericalalcohols phenols and ethers

+4 / -0

The reaction sequence given below is carried out with 16 moles of X. The yield of the major product in each step is given below the product in parentheses. The amount (in grams) of S produced is ________.

qImage--alcohols-phenols-and-ethers-organic-chemistry-Jeemate

Use: Atomic mass (in amu): H = 1, C = 12, O = 16, Br = 80

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:175

Step-by-Step Solution:

Moles of 𝑥 = 16 mol .

First step is Fitting reaction (Wurtz-Fittig reaction) Fitting reaction: Aryl halides when treated with sodium in dry ether, two. aryl groups joined to gether to form another aromatic compounds.

Here, 2 molecules of the reactant

𝑋

react with 2 Na in presence of dry ether and forms a coupling product with the elimination of 2 NaBr .

The reaction proceeds through a radical mechanism where the sodium metal abstracts a bromine atom. from the compound

𝑥

, forming an aryl radical. These aryl radicals then combine to foem a biaryl (two acyl groups linked to gether)

Overall reaction can be written as,

Second step is the hydrolysis reaction. The acetal hydrolysis occurs and forms - CHO group.

2 moles of

𝑥

reacts to form 1 mol

𝑃

.

So, 1 mole of

𝑋

gives

\frac{1}{2}

mol of

𝑃

.

Given moles of

𝑥

is 16 mol.

So, the mole of

𝑃 = 16 mol \times \frac{1}{2}

= 8 mol

\begin{matrix} Ratio between 𝑥 and 𝑃, \\ \begin{matrix} 𝑥 : 𝑃 \\ 2 mol : 1 mol \\ 1 mol : \frac{1}{2} mol \\ 16 mol : 16 \times \frac{1}{2} mol = 8 mol \end{matrix} \end{matrix}

\begin{matrix} 𝑃 {\to\leftarrow}_{2) H_{3} O^{+}}^{1) NaOH, Δ} 𝑄 \\ (100 %) (50 %) \end{matrix}

It is Cannizzaro reaction: One-CHO group converts to

- {CH}_{2} OH

and the other - CHO to - COO .

Moles of

𝑃 \equiv 8 mol

The reaction gives

50 % 𝑄

from

100 % 𝑃

.

So, the moles of

𝑄

is half of moles of

𝑃

.

Moles of

𝑄 = \frac{8}{2}

= 4 mol

𝑄 \frac{1) NaOH, CaO}{2) Δ} 𝑅

(50 %) (50 %)

-COOH group is removed in this reaction. On heating with a mixture of NaOH and CaO (sodium hydroxide and calcium oxide), gives decarboxylation reaction. Cao acts as a drying agent and catalyst. NaOH is used to facilitate the removal of carbon dioxide

({CO}_{2})

.

NaOH + Cao mixture is known as soda lime.

- {CH}_{2} OH group has no change during this reaction.

Moles of 𝑄 = 4 mol

So, theoretically moles of

𝑅

is 4 mol

(1 : 1

ratio. between

𝑄

and

𝑅

).

But, the yield given is

50 %

. so the moles of

𝑅

is half of moles of

𝑄

.

Moles of 𝑅 = \frac{4}{2}

= 2 mol

\begin{matrix} R \overset{{PBr}_{3}, {(C_{2} H_{5})}_{2} O}{\to\leftarrow} T \\ 50 % 50 % \end{matrix}

When alcohol reacts with

{PBr}_{3}

in the presence of

{(C_{2} H_{5})}_{2} O

(solvent), the reaction proceeds via

𝑆_{𝑁} 2

mechanism and the hyphoxyl group is replaced by a bromine atom.

So, - {CH}_{2} OH group becomes - {CH}_{2} Br .

Moles of

𝑅 = 2 mol

.

The yield given is

50 %

. So, the moles of is

𝑇_{}

half of moles of

𝑅

.

Moles of

𝑇 = \frac{2}{2}

= 1 mol

{(C_{2} H_{5})}_{2} O is the solvent

reaction is known as Williamson ether synthesis. Ether compound is formed by reacting an alkoxide

{(𝑅 𝑂^{-})}^{}

with an alkyl halide

(𝑅 𝑥)

. The reaction follows

𝑆_{𝑁 2}

mechanism where the alkoxide acts as a nucleophile, attacking the alkylhalide's carbon atom and displacing the halide.

Formula: {(C_{6} H_{5})}_{2} {(C_{6} H_{4})}_{2} {({CH}_{2})}_{2} O C_{26} H_{22} O

Moles of T.

= 1 mol

Moles of s is half of moles of

𝑇 (50 %

yeld)

So, moles of

𝑆 = \frac{1}{2} mol

Moles of 𝑆 is determined.

To calculate the amount, molecular weight must be known.

Molecular weight of

𝑆 \Rightarrow 26 \times 12 g / mol + 22 \times 1 g / mol + 1 \times

16 g / mol

= 312 + 22 + 16 = 350 g / mol

\begin{matrix} Moles & = \frac{mass}{molarmass} \\ So, mass & = moles 𝑥 molar mass \\ = \frac{1}{2} mot \times 350 g/mol \\ = 175 . g \end{matrix}

Answer: 175

Q 30ChemistrySection A • MCQsalt analysis

+3 / -1

The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions, is

A

P $\to$ 3; Q $\to$ 4; R $\to$ 2; S $\to$ 1

B

P $\to$ 4; Q $\to$ 2; R $\to$ 3; S $\to$ 1

C

P $\to$ 3; Q $\to$ 4; R $\to$ 1; S $\to$ 5

D

P $\to$ 5; Q $\to$ 3; R $\to$ 2; S $\to$ 4

Official Correct Answer:Option (A)

Step-by-Step Solution:

In qualitative inorganic analysis, cations are separated into groups by adding specific group reagents. Each reagent precipitates only those ions whose analytical product (sulfide, hydroxide, carbonate, etc.) has a very low solubility under the given conditions. We first recall the standard group scheme:

• Group II reagent : pass $H_2S$ in dilute $HCl$ (acidic medium) ⇒ precipitates insoluble sulfides such as $CuS, CdS, PbS$ …
• Group III reagent : add $NH_4OH$ in the presence of excess $NH_4Cl$ (buffered weakly basic) ⇒ precipitates gelatinous hydroxides $Fe(OH)_3, Al(OH)_3, Cr(OH)_3$ …
• Group IV reagent : pass $H_2S$ in basic medium (presence of $NH_4OH$ ) ⇒ precipitates the less‐insoluble sulfides $ZnS, NiS, CoS, MnS$ …
• Group V reagent : add $(NH_4)_2CO_3$ in presence of $NH_4OH$ ⇒ precipitates carbonates of alkaline-earth metals $CaCO_3, SrCO_3, BaCO_3$ …

Now match each List-I reagent with the metal ion of List-II whose salt is precipitated by that reagent.

Case P:

Reagent : pass $H_2S$ in presence of $NH_4OH$ (alkaline) ⇒ Group IV.
Among the given ions, $Mn^{2+}$ forms the insoluble sulfide $MnS$ under these conditions.
Hence P → 3 ( $Mn^{2+}$ ).

Case Q:

Reagent : $(NH_4)_2CO_3$ in presence of $NH_4OH$ ⇒ Group V.
Of the listed ions, $Ba^{2+}$ is an alkaline-earth metal and is precipitated as $BaCO_3$ .
Hence Q → 4 ( $Ba^{2+}$ ).

Case R:

Reagent : $NH_4OH$ with $NH_4Cl$ buffer ⇒ Group III.
This reagent precipitates trivalent hydroxides; $Al^{3+}$ gives $Al(OH)_3$ .
Hence R → 2 ( $Al^{3+}$ ).

Case S:

Reagent : pass $H_2S$ in dilute $HCl$ (acidic) ⇒ Group II.
$Cu^{2+}$ is a typical group-II cation, precipitated as black $CuS$ .
Hence S → 1 ( $Cu^{2+}$ ).

Collecting the matches: P → 3, Q → 4, R → 2, S → 1.

Therefore the correct option is:
Option A which is: P → 3; Q → 4; R → 2; S → 1.

Q 31ChemistrySection A • MCQaldehydes ketones and carboxylic acids

+3 / -1

The major products obtained from the reactions in List-II are the reactants for the named reactions mentioned in List-I. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

A

P $\to$ 2; Q $\to$ 4; R $\to$ 1; S $\to$ 3

B

P $\to$ 2; Q $\to$ 3; R $\to$ 4; S $\to$ 1

C

P $\to$ 5; Q $\to$ 3; R $\to$ 4; S $\to$ 2

D

P $\to$ 5; Q $\to$ 4; R $\to$ 2; S $\to$ 1

Official Correct Answer:Option (B)

Step-by-Step Solution:

We begin by writing the four named reactions listed in List-I along with the specific substance that must be taken as the starting material (reactant) in each case.

• P : Kolbe-Schmitt (Kolbe) reaction - it is carried out with $C_6H_5ONa$ (sodium phenoxide).
• Q : Reimer-Tiemann reaction - it starts from free phenol $C_6H_5OH$ in strongly alkaline medium.
• R : Williamson ether synthesis - it is performed with an alkoxide ion such as $RO^-\,Na^+$ (sodium alkoxide).
• S : Etard reaction - the oxidation is applied to toluene $C_6H_5CH_3$ with $CrO_2Cl_2$ to give benzaldehyde.

Next, let us look at the transformations given in List-II and note the major product formed in each of them.

Case 1 (List-II entry 1) : Decarboxylation of benzoic acid with soda-lime $C_6H_5COONa \xrightarrow[\text{soda-lime}]{\Delta } C_6H_5CH_3$ Major product = $C_6H_5CH_3$ (toluene).

Case 2 (List-II entry 2) : Dow process / high-temperature hydrolysis of chlorobenzene $C_6H_5Cl + 2\,NaOH \xrightarrow[623\,\text{K}]{320\,\text{atm}} C_6H_5ONa + NaCl + H_2O$ Major product before acidification = $C_6H_5ONa$ (sodium phenoxide).

Case 3 (List-II entry 3) : Cumene (isopropylbenzene) oxidation followed by acid hydrolysis $\text{Cumene} \xrightarrow[\text{air}]{\text{O}_2} \text{cumene hydro-peroxide} \xrightarrow[\text{dil.}~H_2SO_4]{} C_6H_5OH + (CH_3)_2CO$ Major product = $C_6H_5OH$ (phenol).

Case 4 (List-II entry 4) : Reaction of ethanol with sodium metal $2\,C_2H_5OH + 2\,Na \rightarrow 2\,C_2H_5ONa + H_2 \uparrow$ Major product = $C_2H_5ONa$ (sodium ethoxide, a typical alkoxide).

Now we simply link each named reaction from List-I with the entry in List-II that furnishes the required reactant:

• P (Kolbe, needs $C_6H_5ONa$ )  ← produced in Case 2 ⇒ P→2.
• Q (Reimer-Tiemann, needs $C_6H_5OH$ )  ← produced in Case 3 ⇒ Q→3.
• R (Williamson ether synthesis, needs an alkoxide such as $C_2H_5ONa$ )  ← produced in Case 4 ⇒ R→4.
• S (Etard reaction, needs toluene $C_6H_5CH_3$ )  ← produced in Case 1 ⇒ S→1.

The correct matching is therefore P→2; Q→3; R→4; S→1.

Comparing this sequence with the options supplied, we find that it corresponds to

Option B which is: P→2; Q→3; R→4; S→1.

Q 32ChemistrySection A • MCQbiomolecules

+3 / -1

Match the compounds in List-I with the appropriate observations in List-II and choose the correct option.

A

P $\to$ 1; Q $\to$ 5; R $\to$ 4; S $\to$ 2

B

P $\to$ 2; Q $\to$ 5; R $\to$ 1; S $\to$ 3

C

P $\to$ 5; Q $\to$ 2; R $\to$ 1; S $\to$ 4

D

P $\to$ 2; Q $\to$ 1; R $\to$ 5; S $\to$ 3

Q 33MathematicsSection A • MCQquadratic equations

+3 / -1

Let $\mathbb{R}$ denote the set of all real numbers. Let $a_i, b_i \in \mathbb{R}$ for $i \in \{1, 2, 3\}$ . Define the functions $f: \mathbb{R} \to \mathbb{R}$ , $g: \mathbb{R} \to \mathbb{R}$ , and $h: \mathbb{R} \to \mathbb{R}$ by

$f(x) = a_1 + 10x + a_2 x^2 + a_3 x^3 + x^4,$

$g(x) = b_1 + 3x + b_2 x^2 + b_3 x^3 + x^4,$

$h(x) = f(x+1) - g(x+2).$

If $f(x) \neq g(x)$ for every $x \in \mathbb{R}$ , then the coefficient of $x^3$ in $h(x)$ is

A

$8$

B

$2$

C

$-4$

D

$-6$

Official Correct Answer:Option (C)

Step-by-Step Solution:

First isolate the polynomial that encodes the condition “ $f(x)\neq g(x)$ for every $x$ ”.
Define $d(x)=f(x)-g(x)\,.$

Using the given expressions,

$\begin{aligned} d(x)&=\bigl(a_1+10x+a_2x^2+a_3x^3+x^4\bigr) -\bigl(b_1+3x+b_2x^2+b_3x^3+x^4\bigr)\\ &=\bigl(a_1-b_1\bigr)+\bigl(10-3\bigr)x+\bigl(a_2-b_2\bigr)x^2 +\bigl(a_3-b_3\bigr)x^3. \end{aligned}$

The $x^4$ terms cancel, so $d(x)$ is at most cubic: $d(x)=\alpha+\beta x+\gamma x^2+\delta x^3,$ where $\delta=a_3-b_3\,.$

The statement $f(x)\neq g(x)\;\forall x\in\mathbb{R}$ means $d(x)\neq 0$ for all real $x$ . If $\delta\neq 0$ , then $d(x)$ is a genuine cubic. For a real cubic with non-zero leading coefficient,

$\lim_{x\to\infty}d(x)=\text{sign}(\delta)\,\infty,\qquad \lim_{x\to-\infty}d(x)=-\text{sign}(\delta)\,\infty.$

The Intermediate Value Theorem then forces at least one real root, contradicting $d(x)\neq 0$ . Hence the cubic term must vanish:

$\boxed{a_3-b_3=0}\;\Longrightarrow\;a_3=b_3.$

Now compute the coefficient of $x^3$ in $h(x)=f(x+1)-g(x+2).$

First expand $f(x+1)$ and $g(x+2)$ up to the $x^3$ terms.

$\begin{aligned} f(x+1)&=a_1+10(x+1)+a_2(x+1)^2+a_3(x+1)^3+(x+1)^4\\ &=\dots+\bigl(a_3+4\bigr)x^3+\dots\\[6pt] g(x+2)&=b_1+3(x+2)+b_2(x+2)^2+b_3(x+2)^3+(x+2)^4\\ &=\dots+\bigl(b_3+8\bigr)x^3+\dots \end{aligned}$

(The intermediate “ $\dots$ ” terms are not required because only the $x^3$ coefficients are needed.)

Subtract to get $h(x)$ :

Coefficient of $x^3$ in $h(x)$ $=\;(a_3+4)-(b_3+8)=a_3-b_3-4.$

Using $a_3=b_3$ established above,

$a_3-b_3-4=0-4=-4.$

Therefore the coefficient of $x^3$ in $h(x)$ is $\mathbf{-4}$ .

Option C which is: $-4$

Q 34MathematicsSection A • MCQprobability

+3 / -1

Three students $S_1, S_2,$ and $S_3$ are given a problem to solve. Consider the following events:

$U$ : At least one of $S_1, S_2,$ and $S_3$ can solve the problem,

$V$ : $S_1$ can solve the problem, given that neither $S_2$ nor $S_3$ can solve the problem,

$W$ : $S_2$ can solve the problem and $S_3$ cannot solve the problem,

$T$ : $S_3$ can solve the problem.

For any event $E$ , let $P(E)$ denote the probability of $E$ . If

$P(U) = \frac{1}{2}, \quad P(V) = \frac{1}{10}, \quad$ and $\quad P(W) = \frac{1}{12},$

then $P(T)$ is equal to

A

$\frac{13}{36}$

B

$\frac{1}{3}$

C

$\frac{19}{60}$

D

$\frac{1}{4}$

Official Correct Answer:Option (A)

Step-by-Step Solution:

Let the individual events be
$A$ : $S_1$ solves the problem, $B$ : $S_2$ solves the problem, $C$ : $S_3$ solves the problem.

Write the probabilities of all eight possible outcome-combinations as

$\begin{aligned} P(A\,B'\,C') &= x_1, & P(A\,B\,C') &= x_2, & P(A\,B'\,C) &= x_3, & P(A\,B\,C) &= x_4,\\[2pt] P(A'\,B\,C') &= x_5, & P(A'\,B\,C) &= x_6, & P(A'\,B'\,C) &= x_7, & P(A'\,B'\,C') &= x_8. \end{aligned}$

The eight numbers $x_1,\dots ,x_8$ are non-negative and satisfy

$x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8 = 1 \qquad -(1)$

Step 1 Using $P(U)=\dfrac12$
Event $U$ is “at least one of $S_1,S_2,S_3$ solves”, i.e. $U = A\cup B\cup C$ , whose complement is $A'B'C'$ . Hence

$P(U) = 1-P(A'B'C') = 1-x_8 = \frac12 \;\;\Longrightarrow\;\; x_8 = \frac12. \qquad -(2)$

Step 2 Using $P(V)=\dfrac1{10}$
Event $V$ is “ $S_1$ solves given that neither $S_2$ nor $S_3$ solves”, that is

$P(V)=P\!\bigl(A\,|\,B'C'\bigr)=\frac{P(A\,B'\,C')}{P(B'\,C')}=\frac{x_1}{x_1+x_8} =\frac1{10}.$

Substituting $x_8=\dfrac12$ from (2):

$\frac{x_1}{x_1+\dfrac12}=\frac1{10} \;\;\Longrightarrow\;\;10x_1 = x_1+\frac12 \;\;\Longrightarrow\;\;9x_1 = \frac12 \;\;\Longrightarrow\;\; x_1 = \frac1{18}. \qquad -(3)$

Step 3 Using $P(W)=\dfrac1{12}$
Event $W$ is “ $S_2$ solves and $S_3$ does not”, i.e.

$P(W)=P(B\,C') = P(A\,B\,C') + P(A'\,B\,C') = x_2 + x_5 = \frac1{12}. \qquad -(4)$

Step 4 Total probability still unaccounted for
From (1), (2) and (3)

$x_2+x_3+x_4+x_5+x_6+x_7 = 1 - x_8 - x_1 = 1 - \frac12 - \frac1{18} = \frac{4}{9}. \qquad -(5)$

Step 5 Probability that $S_3$ solves
$S_3$ succeeds in the four cases $A\,B'\,C,\;A\,B\,C,\;A'\,B\,C,\;A'\,B'\,C$ whose probabilities are $x_3,x_4,x_6,x_7$ respectively. Adding them, and using (4) and (5):

$\begin{aligned} P(T) &= x_3+x_4+x_6+x_7 \\[2pt] &= \bigl(x_2+x_3+x_4+x_5+x_6+x_7\bigr) - (x_2+x_5) \\[2pt] &= \frac{4}{9} - \frac{1}{12} \\[2pt] &= \frac{16}{36} - \frac{3}{36} \\[2pt] &= \frac{13}{36}. \end{aligned}$

Therefore $P(T)=\dfrac{13}{36}$ .

Option A which is: $\dfrac{13}{36}$

Q 35MathematicsSection A • MCQapplication of derivatives

+3 / -1

Let $\mathbb{R}$ denote the set of all real numbers. Define the function $f: \mathbb{R} \to \mathbb{R}$ by

$f(x) = \begin{cases} 2 - 2x^2 - x^2 \sin \frac{1}{x} & \text{if } x \neq 0, \\ 2 & \text{if } x = 0. \end{cases}$

Then which one of the following statements is TRUE?

A

The function $f$ is NOT differentiable at $x = 0$

B

There is a positive real number $\delta$ , such that $f$ is a decreasing function on the interval $(0, \delta)$

C

For any positive real number $\delta$ , the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$

D

$x = 0$ is a point of local minima of $f$

Official Correct Answer:Option (C)

Step-by-Step Solution:

The function is defined by
$f(x)=\begin{cases}2-2x^{2}-x^{2}\sin\frac1x,&x\neq0\\[4pt]2,&x=0\end{cases}$

1. Continuity at $x=0$
$\displaystyle\lim_{x\to0}\bigl(2-2x^{2}-x^{2}\sin\tfrac1x\bigr)=2-0-0=2=f(0)$
Hence $f$ is continuous at the origin.

2. Differentiability at $x=0$
Use the definition of the derivative:

$\begin{aligned} f'(0)&=\lim_{h\to0}\frac{f(h)-f(0)}{h} =\lim_{h\to0}\frac{-2h^{2}-h^{2}\sin\tfrac1h}{h}\\[4pt] &=\lim_{h\to0}\bigl(-h(2+\sin\tfrac1h)\bigr)=0. \end{aligned}$
The limit exists, so $f$ is differentiable at $x=0$ with $f'(0)=0$ . Therefore Option A is false.

3. Derivative for $x\neq0$
Differentiate term-wise:

$\begin{aligned} f'(x)&=\frac{d}{dx}\Bigl(-2x^{2}-x^{2}\sin\tfrac1x\Bigr)\\[4pt] &=-4x-\Bigl[2x\sin\tfrac1x-x^{2}\cos\tfrac1x\cdot\frac1{x^{2}}\Bigr]\\[4pt] &=-4x-2x\sin\tfrac1x+\cos\tfrac1x\\[4pt] &=\cos\tfrac1x-4x-2x\sin\tfrac1x.\qquad(x\neq0) \end{aligned}$

Near the origin the terms $-4x$ and $-2x\sin\tfrac1x$ are of order $x$ and tend to $0$ , while $\cos\tfrac1x$ oscillates between $-1$ and $1$ . Hence the sign of $f'(x)$ changes infinitely often in every punctured neighbourhood of $0$ .

4. Testing the options

Option B: To be decreasing on $(0,\delta)$ we would need $f'(x)\le0$ for all $x\in(0,\delta)$ . But pick a sequence $x_n=\tfrac1{2n\pi}$ ; then $\cos\tfrac1{x_n}=\cos(2n\pi)=1$ and

$f'(x_n)=1-4x_n-2x_n\sin(2n\pi)=1-4x_n>0$

for all sufficiently large $n$ . Thus $f$ is not always decreasing on any interval $(0,\delta)$ , so Option B is false.

Option C: To be increasing on $(-\delta,0)$ we would need $f'(x)\ge0$ there. Take the sequence $y_n=-\tfrac1{(2n+1)\pi}$ ; then $\cos\tfrac1{y_n}=\cos\bigl(-(2n+1)\pi\bigr)=-1$ and

$f'(y_n)=-1-4y_n-2y_n\sin\bigl(-(2n+1)\pi\bigr)=-1-4y_n<0$

for all large $n$ (since $y_n\to0^-$ ). Therefore, no matter how small $\delta$ is chosen, $f'(x)$ takes negative values in $(-\delta,0)$ , so $f$ cannot be increasing there. Hence Option C is true.

Option D: For $x\neq0$ we have
$f(x)=2-x^{2}\bigl(2+\sin\tfrac1x\bigr), \qquad 1\le2+\sin\tfrac1x\le3,$
so $f(x)\le2-x^{2}<2$ for every non-zero $x$ . Thus $f(0)=2$ is a strict maximum, not a minimum. Therefore Option D is false.

Conclusion
The only correct statement is Option C: “For any positive real number $\delta$ , the function $f$ is NOT an increasing function on the interval $(-\delta,0)$ .”

Q 36MathematicsSection A • MCQmatrices and determinants

+3 / -1

Consider the matrix

$P = \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix}.$

Let the transpose of a matrix $X$ be denoted by $X^T$ . Then the number of $3 \times 3$ invertible matrices $Q$ with integer entries, such that

$Q^{-1} = Q^T \quad \text{and} \quad PQ = QP,$

is

A

$32$

B

$8$

C

$16$

D

$24$

Official Correct Answer:Option (C)

Step-by-Step Solution:

The two conditions on the integer matrix $Q$ are

1. $Q^{-1}=Q^{T}$ ⇒ $Q$ is an orthogonal matrix with integer entries.
2. $PQ=QP$ with $P=\begin{pmatrix}2&0&0\\0&2&0\\0&0&3\end{pmatrix}$ .

Step 1: Characterise all $3\times3$ integer orthogonal matrices.
If $Q$ is orthogonal and its entries are integers, each row (and each column) is a unit vector whose components are integers. The only integer vectors of length 1 are the six vectors $(\pm1,0,0),\;(0,\pm1,0),\;(0,0,\pm1).$ Hence every row and every column of $Q$ contains exactly one entry $\pm1$ and the rest are zeros. Such matrices are called signed permutation matrices.

The total number of signed permutation matrices of order 3 is $3!\times2^{3}=6\times8=48$ (choose a permutation of the columns and then choose the sign of the non-zero entry in each column).

Step 2: Impose the commutation condition $PQ=QP$ .
Write $Q=(q_{ij})$ . From $PQ=QP$ we get, for every pair $(i,j),$ $P_{ii}\,q_{ij}=q_{ij}\,P_{jj}.$
Because $P=\text{diag}(2,2,3),$ this gives $(P_{ii}-P_{jj})\,q_{ij}=0\quad\forall\,i,j.$ Whenever $P_{ii}\neq P_{jj},$ the factor $P_{ii}-P_{jj}$ is non-zero, forcing $' in math mode at position 10: q_{ij}=0.$ ̲ Thus ev…" style="color:#cc0000">q_{ij}=0.$ Thus every entry mixing the first two coordinates with the third coordinate must vanish:q_{13}=q_{23}=q_{31}=q_{32}=0. $Therefore$ Q $has the block-diagonal form$ Q= $\begin{pmatrix}$ \ast & \ast & 0\\ \ast & \ast & 0\\ 0 & 0 & \ast \end{pmatrix}, $where the asterisks are the remaining possible$ $\pm$ 1 $or$ 0 $entries. Step 3: Count the admissible signed permutation matrices that satisfy the block structure. • The lower-right$ 1 $\times$ 1 $block must be$ $\pm$ 1 $(two choices). • The upper-left$ 2 $\times$ 2 $block must itself be a signed permutation matrix (to keep orthogonality and to make each column/row have exactly one non-zero entry). - There are$ 2! = 2 $ways to permute the two columns. - Independently, each of the two non-zero entries can be$ +1 $or$ -1 $, giving$ 2^{2}=4 $sign choices. Hence the number of possible$ 2 $\times$ 2 $blocks is$ 2 $\times$ 4=8 $. Multiplying by the two choices in the$ 1 $\times$ 1 $block, we obtain the final count$ 8 $\times$ 2 = 16. $ Step 4: Conclusion. Exactly$ 16 $integer matrices$ Q $satisfy both$ Q^{-1}=Q^{T} $and$ PQ=QP $. Option C which is:$ 16$$

Q 37MathematicsSection A • MCQ3d geometry

+4 / -0

Let $L_1$ be the line of intersection of the planes given by the equations

$2x + 3y + z = 4 \quad \text{and} \quad x + 2y + z = 5.$

Let $L_2$ be the line passing through the point $P(2, -1, 3)$ and parallel to $L_1$ . Let $M$ denote the plane given by the equation

$2x + y - 2z = 6.$

Suppose that the line $L_2$ meets the plane $M$ at the point $Q$ . Let $R$ be the foot of the perpendicular drawn from $P$ to the plane $M$ .

Then which of the following statements is (are) TRUE?

A

The length of the line segment $PQ$ is $9\sqrt{3}$

B

The length of the line segment $QR$ is $15$

C

The area of $\triangle PQR$ is $\frac{3}{2}\sqrt{234}$

D

The acute angle between the line segments $PQ$ and $PR$ is $\cos^{-1}\left(\frac{1}{2\sqrt{3}}\right)$

Official Correct Answer:Option (A, C)

Step-by-Step Solution:

The equations of the two given planes are
$\pi_1: \; 2x + 3y + z = 4,$
$\pi_2: \; x + 2y + z = 5.$

1. Direction vector of their line of intersection $L_1$
For two intersecting planes the direction vector is the cross-product of their normals.
Normal of $\pi_1 : \; \mathbf{n}_1 = (2,3,1)$
Normal of $\pi_2 : \; \mathbf{n}_2 = (1,2,1)$
$\mathbf{d}= \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 2 & 3 & 1\\ 1 & 2 & 1 \end{vmatrix} = (1,\,-1,\,1).$

2. One point on $L_1$
Put $z=0$ in both plane equations:
$2x + 3y = 4, \qquad x + 2y = 5.$ Solving, $y=6,\; x=-7.$
So $A(-7,\,6,\,0)$ lies on $L_1.$ Hence $L_1:\; (x,y,z)=(-7,6,0)+t(1,-1,1).$

3. Equation of $L_2$
$L_2$ passes through $P(2,-1,3)$ and is parallel to $L_1,$ so it has the same direction vector $\mathbf{d}=(1,-1,1).$
$L_2:\; (x,y,z)=(2,-1,3)+t(1,-1,1).$

4. Point $Q$ where $L_2$ meets plane $M: 2x + y - 2z = 6$
Substitute $x=2+t,\; y=-1-t,\; z=3+t$ into the plane:
$2(2+t)+(-1-t)-2(3+t)=6$ $4+2t-1-t-6-2t=6$ $-3 - t = 6 \;\Longrightarrow\; t=-9.$ Therefore $Q:(x,y,z)=\bigl(2-9,\,-1+9,\,3-9\bigr)=(-7,\,8,\,-6).$

5. Length of segment $PQ$
$\overrightarrow{PQ}=Q-P=(-9,\,9,\,-9).$
$|PQ|=\sqrt{(-9)^2+9^2+(-9)^2}=9\sqrt{3}.$ Option A is TRUE.

6. Foot $R$ of the perpendicular from $P$ to plane $M$
Plane $M: 2x + y - 2z - 6 = 0$ has normal $\mathbf{n}=(2,1,-2).$
For a point $P(x_0,y_0,z_0)$ , the foot is $R = P - \frac{ax_0 + by_0 + cz_0 + d}{a^2+b^2+c^2}\,(a,b,c).$ Here $ax_0+by_0+cz_0+d = 2(2)+1(-1)+(-2)(3)-6=-9,$ $a^2+b^2+c^2=4+1+4=9,$ so the factor is $\dfrac{-9}{9}=-1.$ Hence $R = (2,-1,3) - (-1)(2,1,-2) = (2+2,\,-1+1,\,3+(-2)) = (4,\,0,\,1).$

7. Length of segment $QR$
$\overrightarrow{QR}=R-Q=(11,\,-8,\,7).$
$|QR|=\sqrt{11^2+(-8)^2+7^2}=\sqrt{234}\;(\approx 15.329).$ Option B states the length is 15, so it is FALSE.

8. Area of $\triangle PQR$
Using vectors $\overrightarrow{PQ}=(-9,9,-9)$ and $\overrightarrow{PR}=R-P=(2,1,-2),$
$\overrightarrow{PQ}\times\overrightarrow{PR} =\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ -9&9&-9\\ 2&1&-2 \end{vmatrix} =(-9,\,-36,\,-27).$ Magnitude $= \sqrt{(-9)^2+(-36)^2+(-27)^2} = \sqrt{2106}=3\sqrt{234}.$ Area $\triangle PQR=\frac12\bigl| \overrightarrow{PQ}\times\overrightarrow{PR}\bigr| =\frac12\,(3\sqrt{234})=\frac{3}{2}\sqrt{234}.$ Option C is TRUE.

9. Acute angle between $PQ$ and $PR$
$\overrightarrow{PQ}\cdot\overrightarrow{PR}=(-9)(2)+9(1)+(-9)(-2)=9.$ $|PQ|=9\sqrt3,\quad |PR|=\sqrt{2^2+1^2+(-2)^2}=3.$ $\cos\theta = \frac{9}{9\sqrt3\cdot3}= \frac{1}{3\sqrt3}.$ Option D gives $\cos^{-1}\!\left(\dfrac{1}{2\sqrt3}\right),$ hence it is FALSE.

Conclusion
Correct statements: Option A and Option C.

Q 38MathematicsSection A • MCQfunctions

+4 / -0

Let $\mathbb{N}$ denote the set of all natural numbers, and $\mathbb{Z}$ denote the set of all integers. Consider the functions $f: \mathbb{N} \to \mathbb{Z}$ and $g: \mathbb{Z} \to \mathbb{N}$ defined by

$f(n) = \begin{cases} (n+1)/2 & \text{if } n \text{ is odd}, \\ (4-n)/2 & \text{if } n \text{ is even}, \end{cases}$

and

$g(n) = \begin{cases} 3 + 2n & \text{if } n \geq 0, \\ -2n & \text{if } n < 0. \end{cases}$

Define $(g \circ f)(n) = g(f(n))$ for all $n \in \mathbb{N}$ , and $(f \circ g)(n) = f(g(n))$ for all $n \in \mathbb{Z}$ .

Then which of the following statements is (are) TRUE?

A

$g \circ f$ is NOT one-one and $g \circ f$ is NOT onto

B

$f \circ g$ is NOT one-one but $f \circ g$ is onto

C

$g$ is one-one and $g$ is onto

D

$f$ is NOT one-one but $f$ is onto

Official Correct Answer:Option (A, D)

Step-by-Step Solution:

For convenience let $\mathbb{N}=\{1,2,3,\dots\}$ and $\mathbb{Z}=\{\dots,-2,-1,0,1,2,\dots\}$ .

1. Properties of $f:\mathbb{N}\to\mathbb{Z}$
• If $n$ is odd, write $n=2k-1\;(k\ge 1)$ . Then $f(n)=\frac{n+1}{2}=\frac{2k-1+1}{2}=k,$ which produces all positive integers $1,2,3,\dots$ .
• If $n$ is even, write $n=2k\;(k\ge 1)$ . Then $f(n)=\frac{4-n}{2}=\frac{4-2k}{2}=2-k,$ which produces the values $1,0,-1,-2,-3,\dots$ .
Combining the two cases, the image of $f$ is all of $\mathbb{Z}$ , so $f$ is onto.

However, $f(1)=\frac{1+1}{2}=1,\quad f(2)=\frac{4-2}{2}=1,$ and $1\ne 2$ . Hence distinct inputs can give the same output, so $f$ is not one-one.

Conclusion for $f$ : not one-one, but onto ⇒ Option D is true.

2. Properties of $g:\mathbb{Z}\to\mathbb{N}$
• $n\ge 0$ : $g(n)=3+2n=3,5,7,9,\dots$ (all odd numbers $\ge 3$ ).
• $n\lt 0$ : write $n=-k\;(k\ge 1)$ , then $g(n)=-2(-k)=2k=2,4,6,8,\dots$ (all even numbers $\ge 2$ ).
Thus $\operatorname{Im}(g)=\{2,3,4,5,6,\dots\}=\mathbb{N}\setminus\{1\}$ , so $g$ is not onto.

To test injectivity, suppose $g(n_1)=g(n_2)$ .
  • If both $n_1,n_2\ge 0$ , then $3+2n_1=3+2n_2\Rightarrow n_1=n_2$ (strictly increasing).
  • If both $n_1,n_2\lt 0$ , write them as $-k_1,-k_2$ . Then $2k_1=2k_2\Rightarrow k_1=k_2\Rightarrow n_1=n_2$ .
  • If one is $\ge 0$ and the other $\lt 0$ , their images have opposite parity (odd vs even), so equality is impossible.
Therefore $g$ is one-one.

Conclusion for $g$ : one-one, but not onto ⇒ Option C is false.

3. The composition $g\circ f:\mathbb{N}\to\mathbb{N}$
First compute the value explicitly.

Case 1:

n=2k-1

(odd,

k\ge 1

).

$f(n)=k\quad(\text{positive})$
$g(f(n))=g(k)=3+2k$ Hence $g\circ f(n)=3+2k$ , i.e. the numbers $5,7,9,\dots$ .

Case 2:

n=2k

(even,

k\ge 1

).

$f(n)=2-k$
If $k=1$ , $f(n)=1,\;g(1)=5$ (already obtained).
If $k=2$ , $f(n)=0,\;g(0)=3$ .
If $k\ge 3$ , $f(n)=2-k\le -1,\;g(2-k)=2k-4$ (even numbers $2,4,6,8,\dots$ ).

The total image of $g\circ f$ is therefore $\{2,3,4,5,6,7,8,9,\dots\}=\mathbb{N}\setminus\{1\},$ so $g\circ f$ is not onto.

Also $g\circ f(1)=5$ and $g\circ f(2)=5$ , yet $1\ne 2$ , so the composition is not one-one.

Conclusion for $g\circ f$ : neither one-one nor onto ⇒ Option A is true.

4. The composition $f\circ g:\mathbb{Z}\to\mathbb{Z}$

Case 1:

n\ge 0

.

$g(n)=3+2n\;(\text{odd})$
$f(g(n))=\frac{(3+2n)+1}{2}=n+2.$ So for $n=0,1,2,\dots$ we obtain $2,3,4,\dots$ .

Case 2:

n=-k

with

k\ge 1

.

$g(n)=2k\;(\text{even})$
$f(g(n))=\frac{4-2k}{2}=2-k.$ For $k=1,2,3,\dots$ this gives $1,0,-1,-2,\dots$ .

The union of the two cases yields every integer, so $f\circ g$ is onto. To check injectivity, observe: • $n\ge 0\; \Rightarrow f\circ g(n)=n+2$ (strictly increasing). • $n\lt 0\; \Rightarrow f\circ g(n)=2-n$ (strictly decreasing). No non-negative input can share an image with a negative input, so different arguments always give different values. Hence $f\circ g$ is one-one. Therefore Option B is false.

Final result: Only Option A and Option D are correct.

Q 39MathematicsSection A • MCQcomplex numbers

+4 / -0

Let $\mathbb{R}$ denote the set of all real numbers. Let $z_1 = 1 + 2i$ and $z_2 = 3i$ be two complex numbers, where $i = \sqrt{-1}$ . Let

$S = \{(x, y) \in \mathbb{R} \times \mathbb{R} : |x + iy - z_1| = 2|x + iy - z_2|\}.$

Then which of the following statements is (are) TRUE?

A

$S$ is a circle with centre $\left(-\frac{1}{3}, \frac{10}{3}\right)$

B

$S$ is a circle with centre $\left(\frac{1}{3}, \frac{8}{3}\right)$

C

$S$ is a circle with radius $\frac{\sqrt{2}}{3}$

D

$S$ is a circle with radius $\frac{2\sqrt{2}}{3}$

Official Correct Answer:Option (A, D)

Step-by-Step Solution:

Let a general point in the plane be $P(x,y)$ , so that the complex number corresponding to $P$ is $x+iy$ .

Given $z_1 = 1+2i$ and $z_2 = 3i = 0+3i$ , the set $S$ is defined by
$|x + iy - z_1| = 2\,|x + iy - z_2|.$

Convert each modulus to distance form:
$|x + iy - z_1| = |(x-1) + i(y-2)| = \sqrt{(x-1)^2 + (y-2)^2},$
$|x + iy - z_2| = |x + i(y-3)| = \sqrt{x^2 + (y-3)^2}.$

Hence the locus condition is
$\sqrt{(x-1)^2 + (y-2)^2} = 2\sqrt{x^2 + (y-3)^2}.$

Square both sides to remove the square roots:
$(x-1)^2 + (y-2)^2 = 4\bigl(x^2 + (y-3)^2\bigr).$

Expand each side:
Left side: $x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 + y^2 - 2x - 4y + 5.$
Right side: $4\bigl(x^2 + y^2 - 6y + 9\bigr) = 4x^2 + 4y^2 - 24y + 36.$

Move everything to the right side and combine like terms:
$0 = 4x^2 + 4y^2 - 24y + 36 - (x^2 + y^2 - 2x - 4y + 5).$
$0 = 3x^2 + 3y^2 + 2x - 20y + 31.$

Divide by $3$ to simplify:
$x^2 + y^2 + \frac{2}{3}x - \frac{20}{3}y + \frac{31}{3} = 0.$

Complete the square in both $x$ and $y$ .

For $x$ :
$x^2 + \frac{2}{3}x = \left(x + \frac{1}{3}\right)^2 - \left(\frac{1}{3}\right)^2.$

For $y$ :
$y^2 - \frac{20}{3}y = \left(y - \frac{10}{3}\right)^2 - \left(\frac{10}{3}\right)^2.$

Substitute these back:
$\left(x + \frac{1}{3}\right)^2 - \frac{1}{9} + \left(y - \frac{10}{3}\right)^2 - \frac{100}{9} + \frac{31}{3} = 0.$

Convert all constants to a common denominator $9$ :
$-\frac{1}{9} - \frac{100}{9} + \frac{31}{3} = -\frac{101}{9} + \frac{93}{9} = -\frac{8}{9}.$

Thus the equation of the locus becomes
$\left(x + \frac{1}{3}\right)^2 + \left(y - \frac{10}{3}\right)^2 = \frac{8}{9}.$

This is the standard form of a circle with

Centre : $\left(-\frac{1}{3},\,\frac{10}{3}\right),$
Radius : $\sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}.$

Comparing with the given options:

Option A (centre $(-\frac{1}{3},\frac{10}{3})$ ) is TRUE.
Option D (radius $\frac{2\sqrt{2}}{3}$ ) is TRUE.
Options B and C are false.

Hence, the correct statements are:
Option A and Option D.

Q 40MathematicsSection B • Numericalsets and relations

+4 / -0

Let the set of all relations $R$ on the set $\{a, b, c, d, e, f\}$ , such that $R$ is reflexive and symmetric, and $R$ contains exactly 10 elements, be denoted by $\mathcal{S}$ .

Then the number of elements in $\mathcal{S}$ is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 41MathematicsSection B • Numericalvector algebra

+4 / -0

For any two points $M$ and $N$ in the $XY$ -plane, let $\overrightarrow{MN}$ denote the vector from $M$ to $N$ , and $\vec{0}$ denote the zero vector. Let $P, Q$ and $R$ be three distinct points in the $XY$ -plane. Let $S$ be a point inside the triangle $\triangle PQR$ such that

$\overrightarrow{SP} + 5\,\overrightarrow{SQ} + 6\,\overrightarrow{SR} = \vec{0}.$

Let $E$ and $F$ be the mid-points of the sides $PR$ and $QR$ , respectively. Then the value of

$\frac{\text{length of the line segment } EF}{\text{length of the line segment } ES}$

is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

Let the position vectors of the vertices be $\mathbf{p},\;\mathbf{q},\;\mathbf{r}$ for the points $P,\;Q,\;R$ respectively, and let $\mathbf{s}$ be the position vector of the point $S$ .

The given vector condition is
$\overrightarrow{SP}+5\,\overrightarrow{SQ}+6\,\overrightarrow{SR}=\vec{0}.$

Using $\overrightarrow{SP}=\mathbf{p}-\mathbf{s},\;\overrightarrow{SQ}=\mathbf{q}-\mathbf{s},\;\overrightarrow{SR}=\mathbf{r}-\mathbf{s},$ we get

$\bigl(\mathbf{p}-\mathbf{s}\bigr)+5\bigl(\mathbf{q}-\mathbf{s}\bigr)+6\bigl(\mathbf{r}-\mathbf{s}\bigr)=\vec{0}.$

Simplifying,
$\mathbf{p}+5\mathbf{q}+6\mathbf{r}-12\mathbf{s}= \vec{0}\;\Longrightarrow\; \mathbf{s}= \dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}}{12}.$

Next, let $E$ and $F$ be the mid-points of $PR$ and $QR$ .

Mid-point position vectors:
$\mathbf{e}= \dfrac{\mathbf{p}+\mathbf{r}}{2}, \qquad \mathbf{f}= \dfrac{\mathbf{q}+\mathbf{r}}{2}.$

Length of $EF$
$\overrightarrow{EF}= \mathbf{f}-\mathbf{e}= \dfrac{\mathbf{q}+\mathbf{r}}{2}-\dfrac{\mathbf{p}+\mathbf{r}}{2} =\dfrac{\mathbf{q}-\mathbf{p}}{2}.$
Hence $|EF|=\dfrac{1}{2}\,|\mathbf{q}-\mathbf{p}|.$

Length of $ES$
$\overrightarrow{ES}= \mathbf{s}-\mathbf{e} =\dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}}{12}-\dfrac{\mathbf{p}+\mathbf{r}}{2} =\dfrac{\mathbf{p}+5\mathbf{q}+6\mathbf{r}-6\mathbf{p}-6\mathbf{r}}{12} =\dfrac{-5\mathbf{p}+5\mathbf{q}}{12} =\dfrac{5}{12}\,(\mathbf{q}-\mathbf{p}).$
Therefore $|ES|=\dfrac{5}{12}\,|\mathbf{q}-\mathbf{p}|.$

Required ratio
$\dfrac{|EF|}{|ES|}= \dfrac{\dfrac{1}{2}\,|\mathbf{q}-\mathbf{p}|}{\dfrac{5}{12}\,|\mathbf{q}-\mathbf{p}|} =\dfrac{1}{2}\times\dfrac{12}{5} =\dfrac{6}{5}=1.2.$

Hence, $\dfrac{\text{length of }EF}{\text{length of }ES}=1.2,$ which lies in the range 1.15-1.25.

Q 42MathematicsSection B • Numericalpermutations and combinations

+4 / -0

Let $S$ be the set of all seven-digit numbers that can be formed using the digits 0, 1 and 2. For example, 2210222 is in $S$ , but 0210222 is NOT in $S$ .

Then the number of elements $x$ in $S$ such that at least one of the digits 0 and 1 appears exactly twice in $x$ , is equal to ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

Let a seven-digit number $x$ contain $a_0$ zeros, $a_1$ ones and $a_2$ twos, so that

$a_0 + a_1 + a_2 = 7$ and $a_0,\,a_1,\,a_2 \ge 0$ .

Because the first (left-most) digit cannot be $0$ , we count the admissible arrangements for each triple $(a_0,a_1,a_2)$ as follows.

Total arrangements with these counts = $\dfrac{7!}{a_0!\,a_1!\,a_2!}$ .
Arrangements whose first digit is $0$ : fix a zero at the first place and permute the remaining $6$ digits, giving $\dfrac{6!}{(a_0-1)!\,a_1!\,a_2!}$ (this term appears only when $a_0\ge 1$ ).

Hence the required number for the triple $(a_0,a_1,a_2)$ is

$N(a_0,a_1,a_2)= \begin{cases} \dfrac{7!}{a_0!\,a_1!\,a_2!}-\dfrac{6!}{(a_0-1)!\,a_1!\,a_2!}, & a_0\ge 1\\[6pt] \dfrac{7!}{a_0!\,a_1!\,a_2!}, & a_0=0 \end{cases}$

For $a_0\ge 1$ , simplifying gives

$N(a_0,a_1,a_2)=\dfrac{6!}{a_0!\,a_1!\,a_2!}\,(7-a_0)\qquad -(1)$

The problem asks for the count of all numbers in which at least one of the digits $0$ or $1$ appears exactly twice. We split this into three disjoint cases.

Case 1:

a_0=2,\;a_1\neq 2

Let the possible values of $a_1$ be $0,1,3,4,5$ (they must stay within $0\le a_1\le 5$ ). For each value we compute $a_2=7-2-a_1$ and use $(1)$ .

$\begin{aligned} a_1=0,&\;a_2=5:&\;N=\dfrac{6!\,(7-2)}{2!\,0!\,5!}=15\\ a_1=1,&\;a_2=4:&\;N=\dfrac{6!\,(7-2)}{2!\,1!\,4!}=75\\ a_1=3,&\;a_2=2:&\;N=\dfrac{6!\,(7-2)}{2!\,3!\,2!}=150\\ a_1=4,&\;a_2=1:&\;N=\dfrac{6!\,(7-2)}{2!\,4!\,1!}=75\\ a_1=5,&\;a_2=0:&\;N=\dfrac{6!\,(7-2)}{2!\,5!\,0!}=15 \end{aligned}$

Total for Case 1 = $15+75+150+75+15 = 330$ .

Case 2:

a_1=2,\;a_0\neq 2

Now $a_0$ can be $0,1,3,4,5$ . Compute $a_2=7-a_0-2$ and apply the appropriate formula.

$\begin{aligned} a_0=0,&\;a_2=5:&\;N=\dfrac{7!}{0!\,2!\,5!}=21\\ a_0=1,&\;a_2=4:&\;N=\dfrac{6!\,(7-1)}{1!\,2!\,4!}=90\\ a_0=3,&\;a_2=2:&\;N=\dfrac{6!\,(7-3)}{3!\,2!\,2!}=120\\ a_0=4,&\;a_2=1:&\;N=\dfrac{6!\,(7-4)}{4!\,2!\,1!}=45\\ a_0=5,&\;a_2=0:&\;N=\dfrac{6!\,(7-5)}{5!\,2!\,0!}=6 \end{aligned}$

Total for Case 2 = $21+90+120+45+6 = 282$ .

Case 3:

a_0=2,\;a_1=2,\;a_2=3

Both digits $0$ and $1$ appear exactly twice.

$N=\dfrac{6!\,(7-2)}{2!\,2!\,3!}=150$

Summing all three cases:

$330 + 282 + 150 = 762$

Therefore, the required number of seven-digit numbers is
762.

Q 43MathematicsSection B • Numericallimits continuity and differentiability

+4 / -0

Let $\alpha$ and $\beta$ be the real numbers such that

$\lim_{x \to 0} \frac{1}{x^3} \left( \frac{\alpha}{2} \int_0^x \frac{1}{1 - t^2}\, dt + \beta x \cos x \right) = 2.$

Then the value of $\alpha + \beta$ is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

Let $F(x)=\dfrac{1}{x^{3}}\left(\dfrac{\alpha}{2}\int_{0}^{x}\dfrac{1}{1-t^{2}}\;dt+\beta x\cos x\right)$ and we want $\displaystyle\lim_{x\to 0}F(x)=2$ .

Step 1 : Series expansion of the integral
Use the geometric series for small $|t|$ : $\dfrac{1}{1-t^{2}}=1+t^{2}+t^{4}+t^{6}+\dots$

Integrate term-by-term from $0$ to $x$ :
$\int_{0}^{x}\dfrac{1}{1-t^{2}}\;dt=\int_{0}^{x}\!\!(1+t^{2}+t^{4}+\dots)\;dt = x+\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}+O(x^{7}).$

Step 2 : Series expansion of $x\cos x$
The Maclaurin series of $\cos x$ is $1-\dfrac{x^{2}}{2}+\dfrac{x^{4}}{24}+O(x^{6})$ , hence
$x\cos x=x-\dfrac{x^{3}}{2}+\dfrac{x^{5}}{24}+O(x^{7}).$

Step 3 : Numerator as a power series
Write the numerator $N(x)$ up to order $x^{3}$ :
$N(x)=\dfrac{\alpha}{2}\!\left(x+\dfrac{x^{3}}{3}+O(x^{5})\right) +\beta\!\left(x-\dfrac{x^{3}}{2}+O(x^{5})\right).$

Collect coefficients:

• Coefficient of $x$ : $C_{1}= \dfrac{\alpha}{2}+\beta.$
• Coefficient of $x^{3}$ : $C_{3}= \dfrac{\alpha}{2}\cdot\dfrac{1}{3}+\beta\!\left(-\dfrac{1}{2}\right)=\dfrac{\alpha}{6}-\dfrac{\beta}{2}.$

Step 4 : Finite limit condition
Since we divide by $x^{3}$ , the term in $x$ must vanish for the limit to remain finite:
$C_{1}=0 \;\;\Longrightarrow\;\; \dfrac{\alpha}{2}+\beta=0\;\;\Longrightarrow\;\;\beta=-\dfrac{\alpha}{2}.$

With this relation, the numerator behaves as
$N(x)=\Bigl(\dfrac{\alpha}{6}-\dfrac{\beta}{2}\Bigr)x^{3}+O(x^{5}).$

Substitute $\beta=-\dfrac{\alpha}{2}$ :
$C_{3}= \dfrac{\alpha}{6}-\Bigl(-\dfrac{\alpha}{4}\Bigr)=\dfrac{\alpha}{6}+\dfrac{\alpha}{4}=\dfrac{5\alpha}{12}.$

Step 5 : Impose the required limit
Thus,
$F(x)=\dfrac{N(x)}{x^{3}}=\dfrac{5\alpha}{12}+O(x^{2}).$
Taking $x\to 0$ gives $\displaystyle\lim_{x\to 0}F(x)=\dfrac{5\alpha}{12}.$

We need this limit to equal $2$ , so
$\dfrac{5\alpha}{12}=2\;\;\Longrightarrow\;\;\alpha=\dfrac{24}{5}=4.8.$

Step 6 : Find $\beta$ and $\alpha+\beta$
$\beta=-\dfrac{\alpha}{2}=-\dfrac{4.8}{2}=-2.4.$
Therefore,
$\alpha+\beta=4.8-2.4=2.4.$

The required value is $\boxed{2.4}$ , which lies in the interval 2.35 - 2.45.

Q 44MathematicsSection B • Numericalfunctions

+4 / -0

Let $\mathbb{R}$ denote the set of all real numbers. Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that $f(x) > 0$ for all $x \in \mathbb{R}$ , and $f(x + y) = f(x)f(y)$ for all $x, y \in \mathbb{R}$ .

Let the real numbers $a_1, a_2, \ldots, a_{50}$ be in an arithmetic progression. If $f(a_{31}) = 64f(a_{25})$ , and

$\sum_{i=1}^{50} f(a_i) = 3(2^{25} + 1),$

then the value of

$\sum_{i=6}^{30} f(a_i)$

is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The functional equation $f(x+y)=f(x)f(y)$ with the extra condition $f(x)\gt 0$ for all real $x$ has the well-known positive solutions
$f(x)=b^{x}$ for some constant $b\gt 0$ .

Let the arithmetic progression be
$a_i=a_1+(i-1)d,\qquad i=1,2,\dots ,50.$

Then the corresponding function values form a geometric progression:

$f(a_i)=b^{a_i}=b^{\,a_1+(i-1)d}=b^{a_1}\bigl(b^{d}\bigr)^{\,i-1}.$

Set $r=b^{d}\;( \gt 0).$ So
$f(a_i)=f(a_1)\,r^{\,i-1}\qquad (1).$

Step 1: Find the common ratio $r$ .

Given $f(a_{31})=64\,f(a_{25}).$ Using (1):

$\frac{f(a_{31})}{f(a_{25})}=r^{31-25}=r^{6}=64=2^{6}\; \Longrightarrow\; r=2.$ (2)

Step 2: Determine $f(a_1).$

The first 50 terms sum to $3\bigl(2^{25}+1\bigr):$

Using (1) with $r=2$ ,

$\sum_{i=1}^{50} f(a_i)=f(a_1)\sum_{i=0}^{49} 2^{\,i}=f(a_1)\,\frac{2^{50}-1}{2-1} =f(a_1)\,\bigl(2^{50}-1\bigr).$

Equate this to the given value:

$f(a_1)\,\bigl(2^{50}-1\bigr)=3\bigl(2^{25}+1\bigr).$

Write $A=2^{25}.$ Then $2^{50}=A^{2}$ , and

$f(a_1)\,(A^{2}-1)=3(A+1).$

Because $A^{2}-1=(A-1)(A+1),$ we get

$f(a_1)=\frac{3(A+1)}{(A-1)(A+1)}=\frac{3}{A-1}=\frac{3}{2^{25}-1}.\qquad(3)$

Step 3: Sum from $i=6$ to $i=30$ .

This sum contains $30-6+1=25$ terms. The first term in this part is $f(a_6)=f(a_1)\,2^{\,5}$ . Hence

$\sum_{i=6}^{30} f(a_i)=f(a_1)\,2^{5}\sum_{k=0}^{24} 2^{\,k} =f(a_1)\,32\;\frac{2^{25}-1}{2-1} =32\,f(a_1)\,\bigl(2^{25}-1\bigr).$

Using (3), $f(a_1)\,\bigl(2^{25}-1\bigr)=3,$ so

$\sum_{i=6}^{30} f(a_i)=32\times 3=96.$

Therefore, the required value is 96.

Q 45MathematicsSection B • Numericaldifferential equations

+4 / -0

For all $x > 0$ , let $y_1(x)$ , $y_2(x)$ , and $y_3(x)$ be the functions satisfying

$\frac{dy_1}{dx} - (\sin x)^2\, y_1 = 0, \quad y_1(1) = 5,$

$\frac{dy_2}{dx} - (\cos x)^2\, y_2 = 0, \quad y_2(1) = \frac{1}{3},$

$\frac{dy_3}{dx} - \left(\frac{2 - x^3}{x^3}\right) y_3 = 0, \quad y_3(1) = \frac{3}{5e},$

respectively. Then

$\lim_{x \to 0^+} \frac{y_1(x)\, y_2(x)\, y_3(x) + 2x}{e^{3x} \sin x}$

is equal to ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The three linear first-order differential equations have the common form
$\frac{dy}{dx}-P(x)\,y=0 \; \Longrightarrow \; \frac{dy}{y}=P(x)\,dx \; \Longrightarrow \; y=C\,e^{\int P(x)\,dx}$

For $y_1(x):$
$P_1(x)=\sin^2 x=\frac{1-\cos 2x}{2}$
$\int P_1(x)\,dx=\int\frac{1-\cos 2x}{2}\,dx =\frac{x}{2}-\frac{\sin 2x}{4}$
Hence $y_1(x)=C_1\,e^{\,\frac{x}{2}-\frac{\sin 2x}{4}}$

Using $y_1(1)=5$ ,
$C_1=5\,e^{-\left(\frac{1}{2}-\frac{\sin 2}{4}\right)}$
Therefore
$y_1(x)=5\,\exp\!\Bigl(\!-\!\!\int_{x}^{1}\!\sin^2 t\,dt\Bigr) =5\,e^{\frac{x}{2}-\frac{\sin 2x}{4}}\,e^{-\left(\frac{1}{2}-\frac{\sin 2}{4}\right)}$

For $y_2(x):$
$P_2(x)=\cos^2 x=\frac{1+\cos 2x}{2}$
$\int P_2(x)\,dx=\int\frac{1+\cos 2x}{2}\,dx =\frac{x}{2}+\frac{\sin 2x}{4}$
So $y_2(x)=C_2\,e^{\,\frac{x}{2}+\frac{\sin 2x}{4}}$

Using $y_2(1)=\dfrac13$ ,
$C_2=\frac13\,e^{-\left(\frac{1}{2}+\frac{\sin 2}{4}\right)}$
Thus
$y_2(x)=\frac13\,\exp\!\Bigl(\!-\!\!\int_{x}^{1}\!\cos^2 t\,dt\Bigr) =\frac13\,e^{\frac{x}{2}+\frac{\sin 2x}{4}}\,e^{-\left(\frac{1}{2}+\frac{\sin 2}{4}\right)}$

For $y_3(x):$
$P_3(x)=\frac{2-x^3}{x^3}=\frac{2}{x^3}-1$
$\int P_3(x)\,dx=\int\!\left(\frac{2}{x^3}-1\right)dx =-x^{-2}-x$
Hence $y_3(x)=C_3\,e^{-x^{-2}-x}$

With $y_3(1)=\dfrac{3}{5e}$ ,
$C_3\,e^{-2}=\frac{3}{5e}\;\Longrightarrow\; C_3=\frac{3e}{5}$
Therefore
$y_3(x)=\frac{3e}{5}\,e^{-x^{-2}-x}$

Behaviour as $x\to 0^+$
The exponents in $y_1(x)$ and $y_2(x)$ approach fixed finite values, so
$\lim_{x\to 0^+}y_1(x)=A,\qquad \lim_{x\to 0^+}y_2(x)=B$
where $A,B$ are positive constants.

For $y_3(x)$ we have the dominant term $e^{-x^{-2}}$ ; thus
$0\lt y_3(x)\lt K\,e^{-1/x^2}\;\;(\text{for some }K)$
and consequently
$y_1(x)\,y_2(x)\,y_3(x)=O\!\bigl(e^{-1/x^2}\bigr)\qquad \bigl(x\to 0^+\bigr)$
which goes to zero faster than any positive power of $x$ .

Numerator near $x=0$
$y_1(x)\,y_2(x)\,y_3(x)+2x=2x+o(x)$

Denominator near $x=0$
$e^{3x}=1+3x+O(x^2),\qquad \sin x=x-\frac{x^{3}}{6}+O(x^{5})$
Thus
$e^{3x}\sin x =\bigl(1+3x+O(x^{2})\bigr) \bigl(x-\tfrac{x^{3}}{6}+O(x^{5})\bigr) =x+3x^{2}+O(x^{3})$

Limit
$\lim_{x\to 0^+}\frac{y_1(x)\,y_2(x)\,y_3(x)+2x}{e^{3x}\sin x} =\lim_{x\to 0^+}\frac{2x+o(x)}{x+3x^{2}+O(x^{3})} =\lim_{x\to 0^+}\frac{2+o(1)}{1+3x+O(x^{2})}=2$

Hence the required limit equals 2.

Q 46MathematicsSection A • MCQstatistics

+3 / -1

Consider the following frequency distribution:

Value	4	5	8	9	6	12	11
Frequency	5	$f_1$	$f_2$	2	1	1	3

Suppose that the sum of the frequencies is 19 and the median of this frequency distribution is 6.

For the given frequency distribution, let $\alpha$ denote the mean deviation about the mean, $\beta$ denote the mean deviation about the median, and $\sigma^2$ denote the variance.

Match each entry in List-I to the correct entry in List-II and choose the correct option.

List-I List-II

(P) $7f_1 + 9f_2$ is equal to (1) 146

(Q) $19\alpha$ is equal to (2) 47

(R) $19\beta$ is equal to (3) 48

(S) $19\sigma^2$ is equal to (4) 145

(5) 55

A

(P) $\to$ (5), (Q) $\to$ (3), (R) $\to$ (2), (S) $\to$ (4)

B

(P) $\to$ (5), (Q) $\to$ (2), (R) $\to$ (3), (S) $\to$ (1)

C

(P) $\to$ (5), (Q) $\to$ (3), (R) $\to$ (2), (S) $\to$ (1)

D

(P) $\to$ (3), (Q) $\to$ (2), (R) $\to$ (5), (S) $\to$ (4)

Official Correct Answer:Option (C)

Step-by-Step Solution:

The given information can be arranged in a two-row table.

Values (x): 4 5 8 9 6 12 11
Frequencies (f): 5 $f_1$ $f_2$ 2 1 1 3

Adding all the frequencies must give 19:

$5 + f_1 + f_2 + 2 + 1 + 1 + 3 = 19 \;\Longrightarrow\; f_1 + f_2 = 7 \qquad -(1)$

The median is stated to be 6. Since the total number of observations is 19 (an odd number), the median is the 10-th item in the ordered data.

Arranging the values in ascending order:
4 (5), 5 ( $f_1$ ), 6 (1), 8 ( $f_2$ ), 9 (2), 11 (3), 12 (1)

Cumulative frequency just before 6 is $5 + f_1$ and including the single 6 it becomes $6 + f_1$ . To make the 10-th observation land on the value 6 we require

$5 + f_1 \lt 10 \le 6 + f_1 \;\Longrightarrow\; f_1 = 4$

Using $f_1 + f_2 = 7$ from $(1)$ gives $f_2 = 3$ .

The completed frequency table is

4(5), 5(4), 6(1), 8(3), 9(2), 11(3), 12(1) with $\sum f = 19$ .

1. Mean $\mu$
$\sum fx = 4\cdot5 + 5\cdot4 + 6\cdot1 + 8\cdot3 + 9\cdot2 + 11\cdot3 + 12\cdot1 = 133$
$\mu = \frac{133}{19} = 7$

2. Mean deviation about the mean $\alpha$

$\sum f|x-\mu| = 15 + 8 + 1 + 3 + 4 + 12 + 5 = 48$
$\alpha = \frac{48}{19}\;\Longrightarrow\; 19\alpha = 48$

3. Mean deviation about the median $\beta$ (median = 6)

$\sum f|x-6| = 10 + 4 + 0 + 6 + 6 + 15 + 6 = 47$
$\beta = \frac{47}{19}\;\Longrightarrow\; 19\beta = 47$

4. Variance $\sigma^{2}$

$\sum f(x-\mu)^2 = 45 + 16 + 1 + 3 + 8 + 48 + 25 = 146$
$\sigma^{2} = \frac{146}{19}\;\Longrightarrow\; 19\sigma^{2} = 146$

5. Expression $7f_1 + 9f_2$
$7f_1 + 9f_2 = 7\cdot4 + 9\cdot3 = 28 + 27 = 55$

Matching with List-II
(P) $7f_1 + 9f_2 = 55$ → (5)
(Q) $19\alpha = 48$ → (3)
(R) $19\beta = 47$ → (2)
(S) $19\sigma^{2} = 146$ → (1)

Thus the correct matching is
(P)→(5), (Q)→(3), (R)→(2), (S)→(1)

Option C which is: (P)→(5), (Q)→(3), (R)→(2), (S)→(1)

Q 47MathematicsSection A • MCQlimits continuity and differentiability

+3 / -1

Let $\mathbb{R}$ denote the set of all real numbers. For a real number $x$ , let $[x]$ denote the greatest integer less than or equal to $x$ . Let $n$ denote a natural number.

Match each entry in List-I to the correct entry in List-II and choose the correct option.

A

(P) $\to$ (1), (Q) $\to$ (3), (R) $\to$ (2), (S) $\to$ (5)

B

(P) $\to$ (2), (Q) $\to$ (1), (R) $\to$ (4), (S) $\to$ (3)

C

(P) $\to$ (5), (Q) $\to$ (1), (R) $\to$ (4), (S) $\to$ (3)

D

(P) $\to$ (2), (Q) $\to$ (3), (R) $\to$ (1), (S) $\to$ (5)

Official Correct Answer:Option (B)

Step-by-Step Solution:

Case P: We need the smallest natural number $n$ such that the function $f(x)=\left[\dfrac{10x^{3}-45x^{2}+60x+35}{n}\right]$ is continuous on $[1,2]$ .

The inner polynomial is $p(x)=10x^{3}-45x^{2}+60x+35$ .
Evaluate at the end-points:

$p(1)=10-45+60+35=60,\qquad p(2)=80-180+120+35=55.$

Since $p'(x)=30(x-1)(x-2)$ is negative for $1\lt x\lt 2,$ $p(x)$ decreases strictly from $60$ to $55$ , so on $[1,2]$ we have $55\le p(x)\le 60.$ (The full range is actually $[55,60]$ .)

Hence $\dfrac{p(x)}{n}$ lies in $\left[\dfrac{55}{n},\dfrac{60}{n}\right]$ . For $f(x)=[p(x)/n]$ to be continuous, the entire interval $[55/n,\,60/n]$ must be contained in one open interval $(k,k+1)$ for some integer $k$ (otherwise the greatest-integer function jumps inside $[1,2]$ ).

The length of this interval is $\dfrac{60-55}{n}=\dfrac{5}{n},$ so we first need $\dfrac{5}{n}\lt 1\;\Longrightarrow\;n\gt 5.$ Now test successive natural numbers $n\gt 5$ .

• $n=6:\;[55/6,60/6]=[9.166\ldots,10].$ The right end equals an integer; $f(x)$ jumps at $x=1$ , so discontinuous.
• $n=7:\;[7.857\ldots,8.571\ldots]$ crosses the integer $8$ ⇒ discontinuous.
• $n=8:\;[6.875,7.5]$ crosses the integer $7$ ⇒ discontinuous.
• $n=9:\;[6.111\ldots,6.666\ldots]\subset(6,7).$ No integer is touched; hence $f(x)$ is continuous.

Thus the minimum $n$ is $9$ . In List-II the number $9$ is entry (2). So (P) → (2).

Case Q: For $g(x)=(2n^{2}-13n-15)(x^{3}+3x)$ to be increasing on all $\mathbb{R}$ we need $g'(x)\ge 0$ everywhere.

$g'(x)=(2n^{2}-13n-15)\bigl(3x^{2}+3\bigr)=3(2n^{2}-13n-15)(x^{2}+1).$ Since $x^{2}+1\gt 0$ for all $x,$ the sign of $g'(x)$ is governed by $A(n)=2n^{2}-13n-15.$ Set $A(n)\gt 0$ :

$2n^{2}-13n-15\gt 0.$

The roots of $2n^{2}-13n-15=0$ are $n=\dfrac{13\pm\sqrt{169+120}}{4}=\dfrac{13\pm17}{4}\; \Rightarrow\; n=-1,\;7.5.$ The quadratic is positive for $n\gt 7.5$ (or $n\lt -1$ , which is irrelevant here). Hence the least natural $n$ making $g$ increasing is $n=8$ .

Entry $8$ is (1) in List-II. So (Q) → (1).

Case R: We need the smallest natural $n\gt 5$ such that $x=3$ is a point of local minima of $h(x)=(x^{2}-9)^{n}(x^{2}+2x+3).$

Notice $(x^{2}-9)^{n}=(x-3)^{n}(x+3)^{n},$ and near $x=3,$ the factor $(x+3)^{n}(x^{2}+2x+3)$ is positive (its value at $x=3$ is $6^{n}\cdot18\gt 0$ ). Thus near $x=3$ , $h(x)$ behaves like a positive constant times $(x-3)^{n}$ .

• If $n$ is odd, $(x-3)^{n}$ changes sign about $x=3$ , so $x=3$ is neither a local minimum nor a maximum.
• If $n$ is even, $(x-3)^{n}\ge 0$ with a zero at $x=3$ and positive values on both sides, giving a local minimum.

Therefore $n$ must be the smallest even natural number $\gt 5,$ i.e. $n=6$ .

Entry $6$ is (4) in List-II. So (R) → (4).

Case S: Define $l(x)=\sum_{k=0}^{4}\Bigl(\sin|x-k|+\cos\bigl|x-k+\tfrac12\bigr|\Bigr).$ A finite sum is nondifferentiable exactly where at least one summand is nondifferentiable.

1. $\sin|t|$ is nondifferentiable at $t=0$ because $\dfrac{d}{dt}\sin|t|=\cos|t|\operatorname{sgn}(t),$ whose one-sided limits at $t=0$ are $+1$ and $-1$ .

2. $\cos|t|$ is differentiable everywhere since $\dfrac{d}{dt}\cos|t|=-\sin|t|\operatorname{sgn}(t),$ and both one-sided limits at $t=0$ are $0$ .

Hence nondifferentiability of $l(x)$ occurs only where $|x-k|=0$ , i.e. at $x=k,\quad k=0,1,2,3,4.$ (The cos-terms never introduce new singularities.)

Thus there are $5$ real numbers $x_{0}$ at which $l(x)$ is not differentiable.

Entry $5$ in List-II corresponds to the number $5$ . So (S) → (3).

Collecting all the matches:
(P) → (2), (Q) → (1), (R) → (4), (S) → (3).

Therefore the correct option is:
Option B which is: (P)→(2), (Q)→(1), (R)→(4), (S)→(3).

Q 48MathematicsSection A • MCQvector algebra

+3 / -1

Let $\vec{w} = \hat{i} + \hat{j} - 2\hat{k}$ , and $\vec{u}$ and $\vec{v}$ be two vectors, such that $\vec{u} \times \vec{v} = \vec{w}$ and $\vec{v} \times \vec{w} = \vec{u}$ . Let $\alpha, \beta, \gamma$ , and $t$ be real numbers such that

$\vec{u} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}, \quad -t\alpha + \beta + \gamma = 0, \quad \alpha - t\beta + \gamma = 0, \quad \text{and} \quad \alpha + \beta - t\gamma = 0.$

Match each entry in List-I to the correct entry in List-II and choose the correct option.

A

(P) $\to$ (2), (Q) $\to$ (1), (R) $\to$ (4), (S) $\to$ (5)

B

(P) $\to$ (2), (Q) $\to$ (4), (R) $\to$ (3), (S) $\to$ (5)

C

(P) $\to$ (2), (Q) $\to$ (1), (R) $\to$ (4), (S) $\to$ (3)

D

(P) $\to$ (5), (Q) $\to$ (4), (R) $\to$ (1), (S) $\to$ (3)

Official Correct Answer:Option (A)

Step-by-Step Solution:

The three vectors are connected by
$\vec{u}\times\vec{v}= \vec{w},\qquad \vec{v}\times\vec{w}= \vec{u},\qquad \vec{w}= \hat{i}+ \hat{j}-2\hat{k}.$ The cross-product of two vectors is perpendicular to each of them, so

$\vec{u}\cdot\vec{v}=0, \qquad \vec{v}\cdot\vec{w}=0, \qquad \vec{w}\cdot\vec{u}=0.$

Hence the three vectors are pair-wise perpendicular.

Magnitude of $\vec{w}$ :
$|\vec{w}|^{2}=1^{2}+1^{2}+(-2)^{2}=6.$

Using the identity $|\vec{a}\times\vec{b}|^{2}=|\,\vec{a}\,|^{2}|\,\vec{b}\,|^{2}-(\vec{a}\cdot\vec{b})^{2},$
$|\vec{w}|^{2}=|\vec{u}\times\vec{v}|^{2}=|\vec{u}|^{2}|\vec{v}|^{2}\quad(\because\vec{u}\cdot\vec{v}=0)$ and
$|\vec{u}|^{2}=|\vec{v}\times\vec{w}|^{2}=|\vec{v}|^{2}|\vec{w}|^{2}\quad(\because\vec{v}\cdot\vec{w}=0).$

Let $|\vec{u}|^{2}=a,\; |\vec{v}|^{2}=b.$
Then $6=ab \quad\text{and}\quad a=6b.$ Substituting, $6=(6b)b\;\Longrightarrow\; b^{2}=1\;\Longrightarrow\; b=1.$ Therefore

Case P: $|\vec{v}|^{2}=1\; \Rightarrow\; (P)\to(2).$

The components of $\vec{u}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ obey the three equations

$-t\alpha+\beta+\gamma=0\; -(1)$
$\alpha-t\beta+\gamma=0\; -(2)$
$\alpha+\beta-t\gamma=0\; -(3).$

For a non-trivial solution, the determinant of the coefficient matrix must vanish:

$\begin{vmatrix}-t&1&1\\[2pt]1&-t&1\\[2pt]1&1&-t\end{vmatrix}=0 \;\Longrightarrow\; -t^{3}+3t+2=0 \;\Longrightarrow\; t^{3}-3t-2=0 =(t-2)(t+1)^{2}=0.$

Thus $t=2\quad\text{or}\quad t=-1.$ We must also satisfy $\vec{w}\cdot\vec{u}=0\;\Rightarrow\; \alpha+\beta-2\gamma=0\; -(4).$

Taking $t=-1$
Equations (1)-(3) all reduce to $\alpha+\beta+\gamma=0.$ With $\alpha=\sqrt{3}$ , $\beta+\gamma=-\sqrt{3}\; -(5).$ Using (4): $\sqrt{3}+\beta-2\gamma=0 \;\Longrightarrow\; -3\gamma=0 \;\Longrightarrow\; \gamma=0,\;\beta=-\sqrt{3}.$ Hence

Case Q: $\gamma^{2}=0\;\Rightarrow\; (Q)\to(1).$

Case R: $(\beta+\gamma)^{2}=(-\sqrt{3})^{2}=3\;\Rightarrow\; (R)\to(4).$

Taking $t=2$
From (1): $\beta=2\alpha-\gamma.$ From (2): $\alpha-2\beta+\gamma=0 \;\Longrightarrow\; \gamma=\alpha.$ Therefore $\beta=\alpha.$ With $\alpha=\sqrt{2}$ we get $t=2,$ hence

Case S: $t+3=2+3=5\;\Rightarrow\; (S)\to(5).$

Collecting all matches:
(P) $\to$ (2), (Q) $\to$ (1), (R) $\to$ (4), (S) $\to$ (5).

Thus the correct option is
Option A which is: (P) $\to$ (2), (Q) $\to$ (1), (R) $\to$ (4), (S) $\to$ (5).

JEE Advanced 2025 Paper 1 Question Paper with Solutions