JEE Advanced 2026 Question Paper 1 Question Paper with Solutions | JEE Advanced Mock Test | JeeMate

Q 1PhysicsSection A • MCQrotational motion

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Consider a large disk of radius $R$ and two smaller disks, each of radius $r=R/50$ , lying on its circumference, as shown in the figure. The smaller disks are initially in contact with each other, with an angular separation $\Delta\theta$ between their centers. They are made to roll without slipping in opposite directions, with constant angular velocities $\omega$ and $2\omega$ while the large disk is held stationary. The time $\tau$ at which the smaller disks are again in contact is:

[Use $\sin(\Delta\theta)=\Delta\theta$ and ignore gravity.]

A

$\tau=51\times\left(2\pi-\dfrac{4}{51}\right)/\omega$

B

$\tau=51\times\left(2\pi-\dfrac{2}{51}\right)/3\omega$

C

$\tau=51\times\left(2\pi-\dfrac{4}{51}\right)/3\omega$

D

$\tau=51\times\left(2\pi-\dfrac{2}{51}\right)/\omega$

Q 2PhysicsSection A • MCQelectromagnetic induction

+3 / -1

Consider a circuit consisting of a capacitor of capacitance $C$ and a coil with $N$ turns per unit length, cross sectional area $S$ and length $d$ , where $d^2\gg S$ . There is another coil of length $d/2$ , cross sectional area $S/2$ and $2N$ turns per unit length completely inside the larger coil, as shown in the figure. The ends of this smaller coil are connected with each other by an insulated conducting wire. The self-inductance of the larger coil is $L$ . Neglecting edge effects and all the Ohmic resistances, the resonant frequency of the circuit is:

A

$\dfrac{4}{\sqrt{15\,LC}}$

B

$\dfrac{6}{\sqrt{5\,LC}}$

C

$\dfrac{2}{\sqrt{3\,LC}}$

D

$\sqrt{\dfrac{2}{3\,LC}}$

Q 3PhysicsSection A • MCQrotational motion

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A solid cylinder of radius $R$ rolls without slipping with a center of mass speed $v_0=\sqrt{\dfrac{gR}{3}}$ on a horizontal surface with a vertical edge, as shown in the figure. Here, $g$ is the acceleration due to the gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:

A

$0$

B

$\sqrt{\dfrac{5gR}{7}}$

C

$\sqrt{\dfrac{gR}{15}}$

D

$\sqrt{\dfrac{3gR}{7}}$

Official Correct Answer:Option (B)

Step-by-Step Solution:

Angular momentum about the corner is conserved during the collision transition, followed by conservation of mechanical energy during rotation about the corner until the normal force becomes zero.

Given: $v_0 = \sqrt{\frac{gR}{3}}$ , $I_{\text{cm}} = \frac{1}{2}MR^2$

From conservation of angular momentum about the corner $P$ :

$L_i = L_f$

$Mv_0R + I_{\text{cm}}\omega_0 = I_P\omega_1$

$Mv_0R + \left(\frac{1}{2}MR^2\right)\left(\frac{v_0}{R}\right) = \left(\frac{1}{2}MR^2 + MR^2\right)\omega_1$

$\frac{3}{2}Mv_0R = \frac{3}{2}MR^2\omega_1 \implies \omega_1 = \frac{v_0}{R}$

At the moment it loses contact at angle $\theta$ from the vertical:

$N = 0 \implies Mg\cos\theta = \frac{Mv^2}{R} \implies MgR\cos\theta = Mv^2$

From conservation of mechanical energy during rotation about the corner:

$K_i + U_i = K_f + U_f$

$\frac{1}{2}I_P\omega_1^2 + MgR = \frac{1}{2}I_P\left(\frac{v}{R}\right)^2 + MgR\cos\theta$

$\frac{1}{2}\left(\frac{3}{2}MR^2\right)\left(\frac{v_0}{R}\right)^2 + MgR = \frac{1}{2}\left(\frac{3}{2}MR^2\right)\left(\frac{v}{R}\right)^2 + Mv^2$

$\frac{3}{4}Mv_0^2 + MgR = \frac{7}{4}Mv^2$

Substituting $gR = 3v_0^2$ :

$\frac{3}{4}Mv_0^2 + M(3v_0^2) = \frac{7}{4}Mv^2$

$\frac{15}{4}v_0^2 = \frac{7}{4}v^2 \implies v^2 = \frac{15}{7}v_0^2 = \frac{15}{7}\left(\frac{gR}{3}\right) = \frac{5gR}{7}$

$v = \sqrt{\frac{5gR}{7}}$

Q 4PhysicsSection A • MCQray optics

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A double convex lens made of glass of refractive index $1.5$ and radii of curvature of the curved surfaces $20\,\mathrm{cm}$ each is immersed in a liquid of refractive index $n_L$ . The correct plot showing the variation of the power, in the units of diopter ( $D$ ), as a function of $n_L$ is:

A

qImage----Jeemate

B

qImage----Jeemate

C

qImage----Jeemate

D

qImage----Jeemate

Q 5PhysicsSection A • MCQatoms and nuclei

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Consider a hydrogen atom with $v_k$ , $r_k$ , and $K_k$ denoting the velocity, orbital radius and kinetic energy of the electron in the $k^{\text{th}}$ orbit, respectively. The electron undergoes a transition from the $n^{\text{th}}$ orbit, emitting radiation corresponding to the Lyman series. Considering $h$ to be the Planck's constant and $\epsilon_0$ the permittivity of the free space, the correct statement(s) is(are):

A

Magnitude of change in kinetic energy of electron can be expressed as $\dfrac{h}{4\pi}\left|\dfrac{n v_n}{r_n}-\dfrac{v_1}{r_1}\right|$ .

B

Magnitude of change in de Broglie wavelength of the electron can be expressed as $\dfrac{e^2}{4\epsilon_0}\left|\dfrac{1}{K_n}-\dfrac{1}{K_1}\right|$ .

C

Frequency of the radiation emitted can be expressed as $\dfrac{e^2}{8\pi\epsilon_0 h}\left(\dfrac{1}{r_1}-\dfrac{1}{r_n}\right)$ .

D

Magnitude of change in total energy of the electron can be expressed as $\dfrac{h}{2\pi}\left|\dfrac{v_1}{r_1}-\dfrac{n v_n}{r_n}\right|$ .

Official Correct Answer:Option (A, C)

Step-by-Step Solution:

For an electron of mass $m$ and charge $e$ orbiting a stationary proton (charge $+e$ ) in a circular path of radius $r_k$ with speed $\mathrm{v}_{\mathrm{k}}$ .

qImage--atoms-and-nuclei-modern-physics-Jeemate

The electrostatic force provides the necessary centripetal acceleration,

$\frac{\mathrm{m} \mathrm{v}_{\mathrm{k}}^2}{\mathrm{r}_{\mathrm{k}}}=\frac{\mathrm{e}^2}{4 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}^2} \Rightarrow \mathrm{~m} \mathrm{v}_{\mathrm{k}}^2=\frac{\mathrm{e}^2}{4 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}}$

The kinetic energy of an electron in its orbit is,

$\mathrm{K}_{\mathrm{k}}=\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{k}}^2=\frac{\mathrm{e}^2}{8 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}}$

The potential energy of an electron in its orbit is,

$\mathrm{U}_{\mathrm{k}}=-\frac{\mathrm{e}^2}{4 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}}$

So, the total energy of an electron in hydrogen atom in $\mathrm{k}^{\text {th }}$ orbit is,

$\mathrm{E}_{\mathrm{k}}=\mathrm{K}_{\mathrm{k}}+\mathrm{U}_{\mathrm{k}}=-\frac{\mathrm{e}^2}{8 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}}=-\mathrm{K}_{\mathrm{k}}$

According to Bohr's second postulate the orbital angular momentum is quantized as an integral multiple of $\frac{h}{2 \pi}$ :

$\mathrm{L}_{\mathrm{k}}=\mathrm{m} \mathrm{v}_{\mathrm{k}} \mathrm{r}_{\mathrm{k}}=\frac{\mathrm{kh}}{2 \pi} \Rightarrow \mathrm{~m}=\frac{\mathrm{kh}}{2 \pi \mathrm{v}_{\mathrm{k}} \mathrm{r}_{\mathrm{k}}}$

For the ground state ( $\mathrm{k}=1$ )

$\mathrm{m}=\frac{\mathrm{h}}{2 \pi \mathrm{v}_1 \mathrm{r}_1}$

For the $\mathrm{n}^{\text {th }}$ state ( $\mathrm{k}=\mathrm{n}$ )

$\mathrm{m}=\frac{\mathrm{nh}}{2 \pi \mathrm{v}_{\mathrm{n}} \mathrm{r}_{\mathrm{n}}}$

When the electron undergoes a transition from a higher state n to the ground state 1 the magnitude of change in kinetic energy is:

$|\Delta K|=\left|K_n-K_1\right|$

Since the mass of the electron is constant, $\mathrm{K}_{\mathrm{k}}=\frac{1}{2} \mathrm{~L}_{\mathrm{k}} \omega_{\mathrm{k}}$ , where angular velocity $\omega_{\mathrm{k}}=\frac{\mathrm{v}_{\mathrm{k}}}{\mathrm{r}_{\mathrm{k}}}$ :

$\mathrm{K}_{\mathrm{k}}=\frac{1}{2}\left(\frac{\mathrm{k} \mathrm{~h}}{2 \pi}\right)\left(\frac{\mathrm{v}_{\mathrm{k}}}{\mathrm{r}_{\mathrm{k}}}\right)=\frac{\mathrm{k} \mathrm{hv}_{\mathrm{k}}}{4 \pi \mathrm{r}_{\mathrm{k}}}$

For $\mathrm{k}=\mathrm{n}$

$\mathrm{K}_{\mathrm{n}}=\frac{\mathrm{nhv}_{\mathrm{n}}}{4 \pi \mathrm{r}_{\mathrm{n}}}$

And for $\mathrm{k}=1$ :

$\mathrm{K}_1=\frac{\mathrm{hv}_1}{4 \pi \mathrm{r}_1}$

Therefore, the magnitude of the change in kinetic energy is:

$|\Delta \mathrm{K}|=\left|\frac{\mathrm{nh}}{4 \pi} \frac{\mathrm{v}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}}-\frac{\mathrm{h}}{4 \pi} \frac{\mathrm{v}_1}{\mathrm{r}_1}\right|=\frac{\mathrm{h}}{4 \pi}\left|\frac{\mathrm{n}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}}-\frac{\mathrm{v}_1}{\mathrm{r}_1}\right|$

Thus, option (A) is correct

The de Broglie wavelength $\lambda_{\mathrm{k}}$ of the electron is given by:

$\lambda_{\mathrm{k}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{k}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{\mathrm{k}}}}$

From the kinetic energy equation

$\mathrm{K}_{\mathrm{k}}=\frac{\mathrm{e}^2}{8 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}} \Rightarrow \mathrm{r}_{\mathrm{k}}=\frac{\mathrm{e}^2}{8 \pi \epsilon_0 \mathrm{~K}_{\mathrm{k}}}$

Using Bohr's quantization condition $2 \pi \mathrm{r}_{\mathrm{k}}=\mathrm{k} \lambda_{\mathrm{k}}$ :

$\lambda_k=\frac{2 \pi r_k}{k}=\frac{2 \pi}{k}\left(\frac{e^2}{8 \pi \epsilon_0 K_k}\right)=\frac{e^2}{4 \epsilon_0 k_k}$

The change in wavelength between the two levels is :

$|\Delta \lambda|=\left|\lambda_n-\lambda_1\right|=\frac{\mathrm{e}^2}{4 \epsilon_0}\left|\frac{1}{\mathrm{nK}_{\mathrm{n}}}-\frac{1}{1 \cdot \mathrm{~K}_1}\right|$

$|\Delta \lambda|=\frac{\mathrm{e}^2}{4 \epsilon_0}\left|\frac{1}{\mathrm{nK}_{\mathrm{n}}}-\frac{1}{\mathrm{~K}_1}\right|$

Thus, option (B) is incorrect.

According to Bohr's frequency condition, the energy of the photon emitted during a transition from level n to level 1 is equal to the change in total energy :

$\mathrm{h} \nu=\mathrm{E}_{\mathrm{n}}-\mathrm{E}_1=\left(-\mathrm{K}_{\mathrm{n}}\right)-\left(-\mathrm{K}_1\right)=\mathrm{K}_1-\mathrm{K}_{\mathrm{n}}$

The kinetic energy in $k^{ ext{th}}$ energy level is $K_k = \frac{e^2}{8\pi \epsilon_0 r_k}$ ,

$h\nu = \frac{e^2}{8\pi \epsilon_0 r_1} - \frac{e^2}{8\pi \epsilon_0 r_n} = \frac{e^2}{8\pi \epsilon_0} \left( \frac{1}{r_1} - \frac{1}{r_n} \right)$

$\nu = \frac{e^2}{8\pi \epsilon_0 h} \left( \frac{1}{r_1} - \frac{1}{r_n} \right)$

Therefore, the frequency of radiation emitted can be expressed as $\frac{e^2}{8\pi \epsilon_0 h} \left( \frac{1}{r_1} - \frac{1}{r_n} \right)$ . Thus, option (C) is correct.

The magnitude of the change in total energy is exactly equal to the magnitude of the change in kinetic energy because $|E_k| = K_k$ .

$|\Delta E| = |\Delta K| = \frac{h}{4\pi} \left| \frac{n\,v_n}{r_n} - \frac{v_1}{r_1} \right|$

Hence, option (D) is incorrect.

Therefore, the correct options are (A) and (C).

Q 6PhysicsSection A • MCQmotion in a plane

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A particle is thrown with a speed $v$ from a point $O$ at an angle $\theta$ with the horizontal plane such that it passes through the point $P$ at a height of $1\,\mathrm{m}$ and horizontal distance of $5\,\mathrm{m}$ from $O$ , as shown in the figure. If acceleration due to gravity is $g\,\mathrm{ms^{-2}}$ , then the correct statement(s) is(are):

A

If $\theta=45^\circ$ , then $v=\dfrac{5\sqrt{g}}{2}\,\mathrm{ms^{-1}}$ .

B

If $\theta=45^\circ$ , the particle reaches its maximum height before it reaches $P$ .

C

If $\theta=30^\circ$ , the particle reaches its maximum height after reaching $P$ .

D

If $\theta=\tan^{-1}\!\left(\dfrac{1}{5}\right)$ , then $v=125\sqrt{g}\,\mathrm{ms^{-1}}$ .

Official Correct Answer:Option (A, B)

Step-by-Step Solution:

The trajectory equation of a projectile is given by $y = x \tan\theta - \frac{gx^2}{2v^2\cos^2\theta}$ , and the position of maximum height occurs at $x_H = \frac{R}{2} = \frac{v^2 \sin 2\theta}{2g}$ .

Given coordinates of point $P$ : $(x, y) = (5, 1)$

Evaluating Option (A) and (B) with $\theta = 45^\circ$ :

$1 = 5 \tan 45^\circ - \frac{g(5)^2}{2v^2 \cos^2 45^\circ}$

$1 = 5 - \frac{25g}{2v^2 \left(\frac{1}{2}\right)} = 5 - \frac{25g}{v^2}$

$\frac{25g}{v^2} = 4 \implies v^2 = \frac{25g}{4} \implies v = \frac{5\sqrt{g}}{2}\text{ ms}^{-1}$

$x_H = \frac{\left(\frac{25g}{4}\right) \sin 90^\circ}{2g} = \frac{25}{8} = 3.125\text{ m}$

$x_H = 3.125\text{ m} < 5\text{ m} \implies \text{maximum height is reached before } P$

Evaluating Option (C) with $\theta = 30^\circ$ :

$1 = 5 \tan 30^\circ - \frac{g(5)^2}{2v^2 \cos^2 30^\circ} = \frac{5}{\sqrt{3}} - \frac{25g}{2v^2 \left(\frac{3}{4}\right)}$

$\frac{50g}{3v^2} = \frac{5}{\sqrt{3}} - 1 \implies v^2 = \frac{50g}{3\left(\frac{5}{\sqrt{3}} - 1\right)} = \frac{50g}{5\sqrt{3} - 3}$

$x_H = \frac{v^2 \sin 60^\circ}{2g} = \frac{50g}{2g(5\sqrt{3} - 3)} \cdot \frac{\sqrt{3}}{2} = \frac{25\sqrt{3}}{2(5\sqrt{3} - 3)} = \frac{25}{10 - 2\sqrt{3}} \approx \frac{25}{6.536} \approx 3.82\text{ m}$

$x_H = 3.82\text{ m} < 5\text{ m} \implies \text{maximum height is reached before } P$

Evaluating Option (D) with $\tan\theta = \frac{1}{5}$ :

$1 = 5\left(\frac{1}{5}\right) - \frac{g(5)^2}{2v^2\cos^2\theta} \implies 1 = 1 - \frac{25g}{2v^2\cos^2\theta} \implies \frac{25g}{2v^2\cos^2\theta} = 0 \implies v \to \infty$

Q 7PhysicsSection A • MCQheat and thermodynamics

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A quasi-static cycle of a monoatomic ideal gas contains an isothermal process ( $\boldsymbol{ab}$ ), followed by an isochoric process ( $\boldsymbol{bc}$ ) and an adiabatic process ( $\boldsymbol{ca}$ ) as shown in the figure. The volumes of the gas are $V_1$ and $V_2$ at $\boldsymbol{a}$ and $\boldsymbol{b}$ , respectively. If the cycle has heat input $Q_{\text{in}}$ and output $Q_{\text{out}}$ , then the efficiency of the cycle is defined as $\eta=\dfrac{Q_{\text{in}}-Q_{\text{out}}}{Q_{\text{in}}}$ . The correct statement(s) is(are):

[Given: $\ln 2\approx 0.7$ ]

A

If $V_2/V_1=8$ , the heat released in the process $\boldsymbol{bc}$ is smaller than the heat absorbed in the process $\boldsymbol{ab}$ .

B

For a given value of $V_2/V_1$ , $\eta$ does not depend on the temperature of the isothermal process.

C

If $V_2/V_1=8$ , then the temperature of the gas at $\boldsymbol{a}$ is 4 times the temperature of the gas at $\boldsymbol{c}$ .

D

If $V_2/V_1=8$ , then the pressure of the gas at $\boldsymbol{a}$ is 4 times the pressure of the gas at $\boldsymbol{b}$ .

Q 8PhysicsSection A • MCQelectromagnetic waves

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The electric field associated with an electromagnetic wave travelling in vacuum is given by $E_0\sin(3y+4z+\omega t)\hat{i}$ , where $\omega$ is the angular frequency. All quantities are in SI units. The correct statement(s) about this wave is(are):

[Given: speed of light in vacuum $c=3\times 10^8\,\mathrm{ms^{-1}}$ .]

A

The wave is travelling in $-\dfrac{1}{5}\left(3\hat{j}+4\hat{k}\right)$ direction.

B

The magnitude of the wave vector is $0.5\,\mathrm{m^{-1}}$ .

C

The value of $\omega$ is $1.5\times 10^{9}\,\mathrm{rad\,s^{-1}}$ .

D

The magnetic field associated with this wave is given by $\dfrac{E_0}{c}\sin(3y+4z+\omega t)(4\hat{j}-3\hat{k})$ .

Official Correct Answer:Option (A, C)

Step-by-Step Solution:

The standard for a plane electromagnetic wave traveling in a general direction is

$\vec{E} = \vec{E}_0 \sin(\vec{k} \cdot \vec{r} + \omega t + \phi)$

Here, $\vec{k} = k_x \hat{i} + k_y \hat{j} + k_z \hat{k}$ is the wave vector, and $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ is the position vector.

$\vec{k} \cdot \vec{r} = k_x x + k_y y + k_z z$

The electric field phase is given as $(3y + 4z + \omega t)$ .

On comparing with $(\vec{k} \cdot \vec{r} + \omega t)$ , $k_x = 0$ , $k_y = 3$ , $k_z = 4$

Therefore, the wave vector is:

$\vec{k} = 3\hat{j} + 4\hat{k}$

The unit vector $\hat{n}$ in the direction of wave propagation:

$\hat{n} = -\frac{\vec{k}}{|\vec{k}|} = \frac{-3\hat{j} + 4\hat{k}}{\sqrt{3^2 + 4^2}} = \frac{-3\hat{j} + 4\hat{k}}{\sqrt{9 + 16}} = \frac{-3\hat{j} + 4\hat{k}}{5} = -\frac{1}{5}(3\hat{j} + 4\hat{k})$

Thus, option (A) is correct.

The magnitude of the wave vector $|\vec{k}|$ is,

${k} = |\vec{\text{k}}| = \sqrt{k_x^2 + k_y^2 + k_z^2} = \sqrt{0^2 + 3^2 + 4^2} = \sqrt{25} = 5\ \text{m}^{-1}$

So, the magnitude of the wave vector is $5\ \text{m}^{-1}$ . Thus, option (B) is incorrect.

In a vacuum, the angular frequency $\omega$ , the magnitude of the wave vector $\text{k}$ , and the speed of light $c$ are related by the wave speed equation:

$c = \frac{\omega}{\text{k}} \rightarrow \omega = c \cdot \text{k}$

Given the speed of light $c = 3 \times 10^8\ \text{ms}^{-1}$ , and our calculated wave vector magnitude $\text{k} = 5\ \text{m}^{-1}$ :

$\omega = (3 \times 10^8) \times (5) = 15 \times 10^8\ \text{rad}\ \text{s}^{-1} = 1.5 \times 10^9\ \text{rad}\ \text{s}^{-1}$

Thus, option (C) is correct.

For a transverse electromagnetic wave, the electric field vector $\vec{\text{E}}$ , the magnetic field vector $\vec{\text{B}}$ , and the propagation unit vector $\hat{\text{n}}$ are mutually orthogonal and follow a definitive right-hand rule cross product vector relation:

$\vec{\text{B}} = \frac{1}{c}\,(\hat{\text{n}} \times \vec{\text{E}})$

$\vec{\text{B}} = \frac{1}{\omega}\,(\vec{\text{k}} \times \vec{\text{E}})$

Here, $\vec{\text{k}} = 3\hat{\text{j}} + 4\hat{\text{k}}$ and $\vec{\text{E}} = E_0 \sin(3y + 4z + \omega\ t)\,\hat{\text{i}}$

$\vec{\text{k}} \times \vec{\text{E}} = (3\hat{\text{j}} + 4\hat{\text{k}}) \times [E_0 \sin(3y + 4z + \omega\ t)\,\hat{\text{i}}]$

$\vec{\text{k}} \times \vec{\text{E}} = E_0 \sin(3y + 4z + \omega\ t)\,[3(\hat{\text{j}} \times \hat{\text{i}}) + 4(\hat{\text{k}} \times \hat{\text{i}})]$

$\vec{\text{k}} \times \vec{\text{E}} = E_0 \sin(3y + 4z + \omega\ t)\,[3(-\hat{\text{k}}) + 4(\hat{\text{j}})] = E_0 \sin(3y + 4z + \omega\ t)\,(4\hat{\text{j}} - 3\hat{\text{k}})$

Putting in magnetic field expression,

$\vec{\text{B}} = \frac{1}{\omega}\left[E_0 \sin(3y + 4z + \omega\ t)\,(4\hat{\text{j}} - 3\hat{\text{k}})\right]$

Since $\omega = c \cdot \text{k}$ and $\text{k} = 5$ :

$\vec{\text{B}} = \frac{E_0}{5c}\sin(3y + 4z + \omega\ t)\,(4\hat{\text{j}} - 3\hat{\text{k}})$

Thus, option (D) is incorrect.

Therefore, the correct options are (A) and (C).

Q 9PhysicsSection B • Numericalfluid mechanics

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A tank contains two immiscible liquids of densities $6\rho$ and $2\rho$ . The higher density liquid is filled up to a height $L/2$ from the bottom. A thin rod of density $\rho$ and length $L$ is fully immersed and hinged at the bottom so that it can oscillate freely, as shown in the figure. If the rod is slightly disturbed from its equilibrium, the time period of small oscillations is $\dfrac{2\pi}{n}\sqrt{\dfrac{L}{g}}$ , where $g$ is the acceleration due to gravity. The value of $n$ is:

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The time period of small oscillations for a hinged rod under buoyant and gravitational forces is determined by the net restoring torque equation $\tau_{\text{net}} = -I\alpha = -\left(\tau_B - \tau_g\right)$ , leading to $T = 2\pi\sqrt{\frac{I}{\kappa_{\text{torsional}}}}$ .

Given: $\text{Rod length} = L$ , $\text{Rod density} = \rho$ , $\text{Lower liquid layer } \left(0 \le y \le \frac{L}{2}\right) = 6\rho$ , $\text{Upper liquid layer } \left(\frac{L}{2} \le y \le L\right) = 2\rho$

Let $A$ be the cross-sectional area of the rod.

Moment of inertia of the rod about the bottom hinge:

$I = \frac{1}{3}ML^2 = \frac{1}{3}(\rho A L)L^2 = \frac{1}{3}\rho A L^3$

For a small angular displacement $\theta$ , finding the restoring torque contributions about the hinge:

$\tau_g = Mg \cdot \frac{L}{2}\theta = (\rho A L)g \frac{L}{2}\theta = \frac{1}{2}\rho A g L^2\theta$

$\tau_{B1} \text{ (due to lower liquid)} = (6\rho A \frac{L}{2})g \cdot \frac{L}{4}\theta = \frac{3}{4}\rho A g L^2\theta$

$\tau_{B2} \text{ (due to upper liquid)} = (2\rho A \frac{L}{2})g \cdot \frac{3L}{4}\theta = \frac{3}{4}\rho A g L^2\theta$

Total buoyant torque:

$\tau_B = \tau_{B1} + \tau_{B2} = \frac{3}{4}\rho A g L^2\theta + \frac{3}{4}\rho A g L^2\theta = \frac{3}{2}\rho A g L^2\theta$

Net restoring torque:

$\tau_{\text{net}} = \tau_B - \tau_g = \left(\frac{3}{2} - \frac{1}{2}\right)\rho A g L^2\theta = \rho A g L^2\theta$

$\omega_0 = \sqrt{\frac{\kappa}{I}} = \sqrt{\frac{\rho A g L^2}{\frac{1}{3}\rho A L^3}} = \sqrt{\frac{3g}{L}}$

$T = \frac{2\pi}{\omega_0} = \frac{2\pi}{\sqrt{3}}\sqrt{\frac{L}{g}}$

$n = \sqrt{3} \approx 1.73$

Q 10PhysicsSection B • Numericalheat and thermodynamics

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As shown in the figure, five Carnot engines, each with efficiency $\eta$ and same number of cycles per unit time, are operating between six heat reservoirs. The amount of heat released per cycle by one engine is completely absorbed by the next engine. Consider $Q_0$ to be the amount of heat absorbed per cycle by the first engine and $W$ as the amount of total work done by all the engines per cycle, then the net efficiency of the system is found to be $\eta_{\text{net}}=\dfrac{W}{Q_0}=\dfrac{211}{243}$ . The value of $\eta$ is:

Answer Format:Enter numerical value rounded to the nearest integer

Q 11PhysicsSection B • Numericalheat and thermodynamics

+4 / -0

As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition ( $P_1$ ) and a freely movable but thermally insulated piston ( $P_2$ ). The partition $P_1$ with thermal conductivity $K$ , cross sectional area $A$ and width $x$ divides the container into two sections, $S_1$ and $S_2$ , each containing one mole of a monoatomic gas. The piston $P_2$ moves freely such that the gas in $S_2$ is always at the atmospheric pressure. Initially, the difference between the temperatures of $S_1$ and $S_2$ is $\Delta T_0$ . The time it takes for the temperature difference to become $\dfrac{\Delta T_0}{2}$ is $nxR/KA$ , where $R$ is the universal gas constant. The value of $n$ is:

[ Given: $\ln 2\approx 0.7$ ]

Answer Format:Enter numerical value rounded to the nearest integer

Q 12PhysicsSection B • Numericalmoving charges and magnetism

+4 / -0

A hollow, right circular cone of base radius $R$ and height $h$ , with its tip at the origin is rotating about the $Z$ -axis with an angular velocity $\omega$ , as shown in the figure. The cone carries a total charge $Q$ uniformly distributed on its curved surface. The magnitude of magnetic field at a point $(0,0,z)$ , where $z\gg R$ and $z\gg h$ , is $\dfrac{n\mu_0}{4\pi}\dfrac{QR^2\omega}{z^3}$ . The value of $n$ is:

Answer Format:Enter numerical value rounded to the nearest integer

Q 13PhysicsSection A • MCQwaves

+3 / -1

List-I shows four configurations made of straight and semi-circular narrow tubes containing air. A sound wave of wavelength $\lambda=0.29\,\mathrm{m}$ enters these structures at the point $S$ and a sound detector is placed at $D$ . Between the points $S$ and $D$ , the sound travels only through the tubes. List-II contains the possible smallest values of $l$ (refer to the figures) for which the detector $D$ records maximum amplitude. Ignore effects of sharp corners. [Given $\cos(15^\circ)=0.97$ ]

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

A

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

B

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

C

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

D

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

Q 14PhysicsSection A • MCQwave optics

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In the List-I, four optical effects are mentioned. The physical phenomena of light which are essential to describe these optical effects are given in List-II. Choose the option which describes the correct match between the entries in List-I to those in List-II.

A

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

B

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

C

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

D

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

Q 15PhysicsSection A • MCQelectromagnetic induction

+3 / -1

List-I contains four conducting loops lying in the $XY$ plane, as shown in the figures. The loops are rotating about $Z$ axis passing through the point $O$ with time period $T$ in clockwise direction. The region $x>0$ contains a uniform magnetic field $B$ in the $+z$ direction. List-II contains the qualitative variation of the induced current $i(t)$ for each of these loops. Choose the option which describes the correct match between the entries in List-I to those in List-II.

A

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

B

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

C

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

D

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

Q 16PhysicsSection A • MCQrotational motion

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List-I shows four planar structures made of uniform solid rods each of mass $m$ and length $l$ . In the List-II the possible moment of inertia of these structures about an axis $OCO'$ , which lies in the plane of the structures, are given. Choose the option that describes the correct match between the entries in List-I to those in List-II.

A

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

B

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

C

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

D

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

Official Correct Answer:Option (A)

Step-by-Step Solution:

For structure (P):

$I_{CA} = \frac{1}{3}ml^2\sin^2 45^\circ = \frac{1}{6}ml^2$

$I_{CB} = \frac{1}{3}ml^2\sin^2 45^\circ = \frac{1}{6}ml^2$

$I_P = \frac{1}{6}ml^2 + \frac{1}{6}ml^2 = \frac{1}{3}ml^2 \implies (\text{P}) \rightarrow (5)$

For structure (Q):

$I_{CA} = \frac{1}{3}ml^2\sin^2 60^\circ = \frac{1}{4}ml^2$

$I_{CB} = \frac{1}{3}ml^2\sin^2 60^\circ = \frac{1}{4}ml^2$

$I_{AB} = I_{\text{cm}} + md^2 = \frac{1}{12}ml^2 + m\left(l\sin 60^\circ\right)^2 = \frac{1}{12}ml^2 + \frac{3}{4}ml^2 = \frac{5}{6}ml^2$

$I_Q = \frac{1}{4}ml^2 + \frac{1}{4}ml^2 + \frac{5}{6}ml^2 = \frac{4}{3}ml^2 - \frac{1}{12}ml^2 = \frac{5}{4}ml^2 \implies (\text{Q}) \rightarrow (1)$

For structure (R):

$I_{CD} = I_{CB} = I_{AD} = I_{AB} = \frac{1}{3}ml^2\sin^2 45^\circ = \frac{1}{6}ml^2$

$I_R = 4 \times \frac{1}{6}ml^2 = \frac{2}{3}ml^2 \implies (\text{R}) \rightarrow (4)$

For structure (S):

$I_{CA} = \frac{1}{3}ml^2\sin^2 30^\circ = \frac{1}{12}ml^2$

$I_{CB} = \frac{1}{3}ml^2\sin^2 30^\circ = \frac{1}{12}ml^2$

$I_S = \frac{1}{12}ml^2 + \frac{1}{12}ml^2 = \frac{1}{6}ml^2 \implies (\text{S}) \rightarrow (2)$

Q 17ChemistrySection A • MCQthermodynamics

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An ideal gas $(0.5\ \mathrm{mol})$ , initially at $2$ bar pressure, is compressed at a constant temperature of $600\ \mathrm{K}$ in two steps: first, against a constant external pressure of $P$ bar $(2 \lt P \lt 8)$ , and then against constant external pressure of $8$ bar. At each step, the compression is stopped only when the pressure of the gas becomes equal to the external pressure. The total work done on the gas in these steps is $W$ . Considering all possible values of $P$ $(2 \lt P \lt 8)$ and taking the gas constant as $R$ (in $\mathrm{J\ K^{-1}\ mol^{-1}}$ ), the minimum value of $|W|$ (in $\mathrm{J}$ ) is

A

$207R$

B

$600R$

C

$630R$

D

$900R$

Official Correct Answer:Option (B)

Step-by-Step Solution:

Given data:

$n = 0.5\ \text{mol},\ P_0 = 2\ \text{bar},\ T = 600\ \text{K}$

The gas is compressed in two steps: first against constant pressure $P_1$ and then against constant pressure of 8 bar. Here $P_1$ can take any value between 2 and 8 bar.

We can find the volumes at each stage using the ideal gas equation $PV = nRT$ :

$V_0 = \frac{nRT}{2}, \quad V_1 = \frac{nRT}{P_1}, \quad V_2 = \frac{nRT}{8}$

Step 1: Work done on the gas during compression against external pressure $P_1$ is given by:

$W_1 = P_1(V_0 - V_1)$

Substituting the values of $V_0$ and $V_1$ :

$W_1 = P_1\left(\frac{nRT}{2} - \frac{nRT}{P_1}\right) = nRT\left(\frac{P_1}{2} - 1\right) \qquad \ldots(1)$

Step 2: Work done on the gas when compressed against 8 bar pressure:

$W_2 = P_2(V_1 - V_2) = 8\left(\frac{nRT}{P_1} - \frac{nRT}{8}\right) = nRT\left(\frac{8}{P_1} - 1\right) \qquad \ldots(2)$

Total work done on the gas:

$|W| = W_1 + W_2 = nRT\left(\frac{P_1}{2} - 1\right) + nRT\left(\frac{8}{P_1} - 1\right)$

$|W| = nRT\left(\frac{P_1}{2} + \frac{8}{P_1} - 2\right)$

Finding the condition for minimum work:

We need to minimize $\left(\frac{P_1}{2} + \frac{8}{P_1}\right)$ . Using the inequality AM ≥ GM :

$\frac{\left(\frac{P_1}{2} + \frac{8}{P_1}\right)}{2} \ge \sqrt{\frac{P_1}{2} \times \frac{8}{P_1}}$

Simplifying,

$\frac{P_1^2 + 16}{4P_1} \ge 2$

$P_1^2 - 8P_1 + 16 \ge 0$

$(P_1 - 4)^2 \ge 0$

This means the minimum value of $|W|$ occurs when $P_1 = 4$ bar.

Substitute $P_1 = 4$ :

$|W|_{\min} = nRT\left(\frac{4}{2} + \frac{8}{4} - 2\right)$

$|W|_{\min} = 0.5 \times R \times 600 \times (2 + 2 - 2)$

$|W|_{\min} = 600R$

Q 18ChemistrySection A • MCQchemical kinetics

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For a reversible reaction $\mathbf{R}\rightleftharpoons\mathbf{P}$ , at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constants $k_f$ and $k_b$ , respectively. At time zero, the concentration of $\mathbf{R}$ is $[\mathbf{R}]_0$ and the concentration of $\mathbf{P}$ is zero. At any given time, $[\mathbf{R}]$ and $[\mathbf{P}]$ are the concentrations of $\mathbf{R}$ and $\mathbf{P}$ , respectively. If $k_b=4k_f$ , the correct graphical representation of the reaction is

A

qImage----Jeemate

B

qImage----Jeemate

C

qImage----Jeemate

D

qImage----Jeemate

Official Correct Answer:Option (C)

Step-by-Step Solution:

Let the initial concentration of the reactant be $[\mathbf R]_0$ . At any time $t$ , let $x$ be the amount of $\mathbf R$ that has been converted to $\mathbf P$ .

Then $[\mathbf R]=[\mathbf R]_0-x,$ $[\mathbf P]=x.$

Because both the forward and backward steps are elementary first-order, the net rate of formation of $\mathbf P$ is

$\frac{dx}{dt}=k_f([\mathbf R]_0-x)-k_bx$ $\Longrightarrow\frac{dx}{dt}+(k_f+k_b)x=k_f[\mathbf R]_0\; -(1)$

This is a linear first-order differential equation whose solution is obtained using the integrating-factor method:

$x=\frac{k_f[\mathbf R]_0}{k_f+k_b}\Bigl(1-e^{-(k_f+k_b)t}\Bigr)\; -(2)$

Hence the time-dependent concentrations are

$[\mathbf R]=[\mathbf R]_0-x =[\mathbf R]_0\left[\frac{k_b}{k_f+k_b}+\frac{k_f}{k_f+k_b}\,e^{-(k_f+k_b)t}\right]\; -(3)$

$[\mathbf P]=x =[\mathbf R]_0\frac{k_f}{k_f+k_b}\Bigl(1-e^{-(k_f+k_b)t}\Bigr)\; -(4)$

At equilibrium $t\to\infty$ , the exponential terms vanish, giving

$[\mathbf R]_{eq}= [\mathbf R]_0\frac{k_b}{k_f+k_b},\qquad [\mathbf P]_{eq}= [\mathbf R]_0\frac{k_f}{k_f+k_b}.$

For the data in the problem, $k_b=4k_f$ , so

$[\mathbf R]_{eq}= [\mathbf R]_0\frac{4k_f}{5k_f}=0.8[\mathbf R]_0,$ $[\mathbf P]_{eq}= [\mathbf R]_0\frac{k_f}{5k_f}=0.2[\mathbf R]_0.$

Therefore the concentration-time curves must exhibit the following features:

$[\mathbf R]$ falls exponentially from $[\mathbf R]_0$ to $0.8[\mathbf R]_0$ ; the drop is only 20 % of the initial value.
$[\mathbf P]$ rises from 0 to $0.2[\mathbf R]_0$ with the same time constant.
The two curves never cross, because $[\mathbf R]$ always exceeds $[\mathbf P]$ (ratio 4 : 1 at equilibrium).
Both curves approach horizontal asymptotes as $t\to\infty$ (equilibrium).

Among the given diagrams, the only one that shows • $[\mathbf R]$ starting at $[\mathbf R]_0$ and leveling off near 0.8 $[\mathbf R]_0$ , • $[\mathbf P]$ starting at 0 and leveling off near 0.2 $[\mathbf R]_0$ , • no intersection of the two curves, is Option C.

Option C which is: the correct graph.

Q 19ChemistrySection A • MCQchemical bonding and molecular structure

+3 / -1

The correct order of dipole moments for the given species is

A

$BF_3=NH_4^{+}\lt NF_3\lt NH_3$

B

$BF_3\lt NH_4^{+}\lt NF_3\lt NH_3$

C

$NH_4^{+}\lt BF_3\lt NH_3\lt NF_3$

D

$BF_3\lt NH_4^{+}\lt NH_3\lt NF_3$

Q 20ChemistrySection A • MCQalcohols phenols and ethers

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Considering $LiBH_4$ reduces an ester group to the corresponding alcohol and does not reduce a carboxylic acid group, the correct statement about the major products $\mathbf{P}$ , $\mathbf{Q}$ , $\mathbf{R}$ and $\mathbf{S}$ is

A

$\mathbf{P}$ & $\mathbf{Q}$ are identical, and $\mathbf{R}$ & $\mathbf{S}$ are diastereomers.

B

$\mathbf{P}$ & $\mathbf{Q}$ are diastereomers, and $\mathbf{R}$ & $\mathbf{S}$ are identical.

C

$\mathbf{P}$ & $\mathbf{Q}$ are diastereomers, and $\mathbf{R}$ & $\mathbf{S}$ are diastereomers.

D

$\mathbf{P}$ & $\mathbf{Q}$ are identical, and $\mathbf{R}$ & $\mathbf{S}$ are identical.

Q 21ChemistrySection A • MCQatomic structure

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The $2s$ and the $2p$ orbital energies of hydrogen atom are $E_{2s}(\mathrm{H})$ and $E_{2p}(\mathrm{H})$ , respectively. The $2s$ and the $2p$ orbital energies of lithium atom are $E_{2s}(\mathrm{Li})$ and $E_{2p}(\mathrm{Li})$ , respectively. The correct option(s) about the orbital energies is(are)

A

$E_{2s}(\mathrm{Li})\lt E_{2p}(\mathrm{Li})$

B

$E_{2s}(\mathrm{H})=E_{2p}(\mathrm{H})$

C

$E_{2p}(\mathrm{H})\lt E_{2s}(\mathrm{Li})$

D

$E_{2s}(\mathrm{H})\gt E_{2s}(\mathrm{Li})$

Official Correct Answer:Option (A, B, D)

Step-by-Step Solution:

To compare orbital energies, we analyze hydrogen (single-electron system) and lithium (multi-electron system) separately.

(i) Hydrogen atom (single-electron system):
In hydrogen-like species, the energy of an orbital depends only on the principal quantum number $n$ , not on $l$ . Therefore, all orbitals having the same $n$ possess equal energy.

Hence, for hydrogen:

$E_{2s}(H) = E_{2p}(H)$

So, Option B is correct.

(ii) Lithium atom (multi-electron system):
In multi-electron atoms, orbital energies depend on the $(n + l)$ rule. The orbital with lower $(n + l)$ value has lower energy. If two orbitals have the same $n + l$ , the one with lower $n$ is lower in energy.

For lithium:

For $2s$ : $n = 2, \, l = 0 \Rightarrow n + l = 2$
For $2p$ : $n = 2, \, l = 1 \Rightarrow n + l = 3$

Since $(n + l)_{2s} < (n + l)_{2p}$ , we get:

$E_{2s}(Li) < E_{2p}(Li)$

This is further supported by the fact that the $2s$ orbital penetrates closer to the nucleus and experiences higher effective nuclear charge, making it more stable (lower energy).

Thus, Option A is correct.

(iii) Comparison between hydrogen and lithium:
Energy of an orbital is given by:

$E \propto -\frac{Z_{\text{eff}}^2}{n^2}$

In lithium, even after shielding by two inner $1s$ electrons, the effective nuclear charge experienced by the $2s$ electron is greater than that in hydrogen. As a result, the $2s$ electron in lithium is more strongly attracted and has lower (more negative) energy.

Therefore:

$E_{2s}(Li) < E_{2s}(H)$

So, Option D is also correct.

(iv) Checking Option C:
Since $E_{2p}(H) = E_{2s}(H)$ and $E_{2s}(Li)$ is lower than $E_{2s}(H)$ , we have:

$E_{2p}(H) > E_{2s}(Li)$

Thus, Option C is incorrect.

Final Answer: A, B and D

Q 22ChemistrySection A • MCQp block elements

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Correct statement(s) about the compounds $\mathbf{X}$ , $\mathbf{Y}$ and $\mathbf{Z}$ is(are)

qImage--p-block-elements-inorganic-chemistry-Jeemate

A

$\mathbf{X}$ is used for sterilizing drinking water.

B

$\mathbf{Y}$ has a planar structure.

C

$\mathbf{Z}$ is used in the enrichment of $^{235}\mathrm{U}$ .

D

$\mathbf{Y}$ is a stronger Lewis base than ammonia.

Q 23ChemistrySection A • MCQredox reactions

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Reaction of $PtF_6$ with oxygen $(O_2)$ gas results in the formation of an ionic compound, $\mathbf{X^{+}Y^{-}}$ . Correct statement(s) is(are)

A

The bond order of $\mathbf{X^{+}}$ is $1.5$ .

B

Valence $d$ -orbitals of the metal ion in $\mathbf{X^{+}Y^{-}}$ has $5$ electrons.

C

$PtF_6$ acts as an oxidant in this reaction.

D

$PtF_6$ acts as a fluorinating agent in this reaction.

Q 24ChemistrySection A • MCQhydrocarbons

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In the following reaction sequence, $\mathbf{Q}$ , $\mathbf{R}$ , $\mathbf{S}$ and $\mathbf{T}$ are the major products.

The correct statement(s) about $\mathbf{Q}$ , $\mathbf{R}$ , $\mathbf{S}$ and $\mathbf{T}$ is(are)

A

$\mathbf{S}$ on warming with ammoniacal $AgNO_3$ results in the formation of silver mirror.

B

$\mathbf{Q}$ on treatment with $Cl_2$ (excess)/UV gives gammaxane.

C

$\mathbf{T}$ is a heterocyclic compound.

D

$\mathbf{R}$ on acid catalyzed intramolecular cyclization followed by treatment with $Zn\text{-}Hg/HCl$ gives $9,10$ -dihydroxyanthracene.

Official Correct Answer:Option (A, B, C)

Step-by-Step Solution:

The reagents used at each stage and the structures obtained are summarised first and then analysed with respect to the four statements.

Case 1: Formation of $\mathbf{Q}$ Benzene is treated with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ (Friedel-Crafts alkylation). The methyl group substitutes one hydrogen of benzene to give toluene, $C_6H_5CH_3$ . Hence $\mathbf{Q}$ = toluene.

Case 2: Formation of $\mathbf{R}$ Oxidation of any alkyl side-chain on an aromatic ring by alkaline, hot $KMnO_4$ converts the entire side-chain to a carboxyl group. Therefore toluene is oxidised to benzoic acid, $C_6H_5COOH$ . Hence $\mathbf{R}$ = benzoic acid.

Case 3: Formation of $\mathbf{S}$ Benzoic acid is converted to its acid chloride with $SOCl_2$ : $C_6H_5COOH \xrightarrow{SOCl_2} C_6H_5COCl$ The acid chloride is then subjected to Rosenmund reduction ( $H_2/Pd{-}BaSO_4$ ), which stops at the aldehyde stage: $C_6H_5COCl \xrightarrow[\text{Poisoned}\;Pd]{H_2} C_6H_5CHO$ Thus $\mathbf{S}$ = benzaldehyde.

Case 4: Formation of $\mathbf{T}$ Benzaldehyde undergoes benzoin condensation in the presence of catalytic $KCN$ /ethanol to give benzoin: $2\,C_6H_5CHO \xrightarrow{KCN/EtOH} C_6H_5CH(OH)COC_6H_5$ Benzoin contains two benzene rings joined by $-CH(OH)-CO-$ , but no hetero-atom in a ring. Hence $\mathbf{T}$ = benzoin (a non-heterocyclic acyloin).

Now each assertion is examined.

Option A “ $\mathbf{S}$ on warming with ammoniacal $AgNO_3$ results in the formation of a silver mirror.” Tollens’ reagent, $[Ag(NH_3)_2]^+$ , oxidises aldehydes to acids while reducing itself to metallic silver. Since $\mathbf{S}$ is benzaldehyde (an aldehyde), it gives a positive silver-mirror test. Hence Option A is TRUE.

Option B “ $\mathbf{Q}$ on treatment with $Cl_2$ (excess)/UV gives gammaxane.” Gammaxane (lindane, $C_6H_6Cl_6$ ) is obtained by photochemical addition of chlorine to benzene, not to toluene. $\mathbf{Q}$ is toluene, and side-chain/ring chlorination products are formed instead of BHC. Hence Option B is FALSE.

Option C “ $\mathbf{T}$ is a heterocyclic compound.” $\mathbf{T}$ (benzoin) is an open-chain acyloin; it contains no ring that has a hetero-atom as a member. Hence Option C is FALSE.

Option D “ $\mathbf{R}$ on acid-catalysed intramolecular cyclisation followed by $Zn{-}Hg/HCl$ gives $9,10$ -dihydroxy-anthracene.” $\mathbf{R}$ is benzoic acid, which possesses only one carbonyl group and therefore cannot furnish the two adjacent carbonyls required to produce anthraquinone (the precursor of 9,10-dihydroxy-anthracene). Consequently the sequence described cannot occur, so the statement is incorrect. Hence Option D is FALSE.

Thus, the only correct statement is:

Option A which is: $\mathbf{S}$ on warming with ammoniacal $AgNO_3$ results in the formation of silver mirror.

Q 25ChemistrySection B • Numericalgaseous state

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Two cylinders, both fitted with frictionless pistons, are filled with mixtures of $\mathrm{He}$ and $\mathrm{Ar}$ gases. In the first cylinder, the masses of $\mathrm{He}$ and $\mathrm{Ar}$ are $m_1$ and $m_2$ , respectively. In the second cylinder, the masses of $\mathrm{He}$ and $\mathrm{Ar}$ are $m_2$ and $m_1$ , respectively. The molar mass of $\mathrm{Ar}$ is $10$ times the molar mass of $\mathrm{He}$ . The external pressure applied by the piston on the first cylinder needs to be $5$ times that on the second cylinder so that the volume of the gas mixtures in both the cylinders are equal at the same temperature. Assuming $\mathrm{He}$ and $\mathrm{Ar}$ behave like ideal gases, the value of $(m_1/m_2)$ is ___.

Answer Format:Enter numerical value rounded to the nearest integer

Q 26ChemistrySection B • Numericalcoordination compounds

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The total number of all possible isomers for the square planar complex with formula $K[M(NCS)(NO_2)(gly)]$ is ___.

(M = metal ion and gly = $NH_2CH_2COO^{-}$ )

Answer Format:Enter numerical value rounded to the nearest integer

Q 27ChemistrySection B • Numericalaldehydes ketones and carboxylic acids

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The sum of total number of carbonyl groups $(>C=O)$ present in the major products $\mathbf{X}$ and $\mathbf{Y}$ in the following reactions is ___.

Answer Format:Enter numerical value rounded to the nearest integer

Q 28ChemistrySection B • Numericalhydrocarbons

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Treatment of buta- $1,3$ -diyne with $NaNH_2$ ( $2$ equivalents), followed by reaction with excess of trans- $CH_3\text{-}CH\!=\!CH\text{-}CH_2\text{-}Br$ gives $\mathbf{X}$ as the major product. The maximum number of carbon atoms that are collinear (in a straight line) in $\mathbf{X}$ is ___.

Answer Format:Enter numerical value rounded to the nearest integer

Q 29ChemistrySection A • MCQthermodynamics

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List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy $(\Delta H)$ and entropy $(\Delta S)$ . Match each entry in List-I to the appropriate entry in List-II, and choose the correct option.

A

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

B

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

C

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

D

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

Official Correct Answer:Option (C)

Step-by-Step Solution:

To decide the correct match we only have to figure out the signs of $\Delta H$ (enthalpy change) and $\Delta S$ (entropy change) for each process in List-I and then pick the List-II entry that has the same pair of signs.

Case P : condensation / freezing type process

During condensation or freezing the system releases heat to the surroundings, so $\Delta H \lt 0$ (exothermic).
Liquid or solid is more ordered than vapour, so entropy decreases, i.e. $\Delta S \lt 0$ .
Hence $\Delta H \lt 0, \; \Delta S \lt 0$ → List-II (2).

Case Q : isothermal free expansion of an ideal gas into vacuum

For an ideal gas expanding freely and isothermally, internal energy does not change; therefore $\Delta H = 0$ because $\Delta H = \Delta U + \Delta (pV)$ and at constant $T$ for an ideal gas $\Delta U = 0$ , $\Delta (pV)=0$ .
However the gas occupies a larger volume after expansion, hence disorder increases and $\Delta S \gt 0$ .
Therefore $\Delta H = 0, \; \Delta S \gt 0$ → List-II (5).

Case R : combustion reaction that produces gaseous products

Combustion reactions are strongly exothermic, so $\Delta H \lt 0$ .
If the number of moles of gaseous products exceeds that of the reactants (e.g. $2CO + O_2 \rightarrow 2CO_2$ performed at high temperature where $CO_2$ is gaseous), randomness increases and $\Delta S \gt 0$ .
Thus $\Delta H \lt 0, \; \Delta S \gt 0$ → List-II (1).

Case S : vaporisation / sublimation process

To convert a liquid or solid into vapour, heat must be supplied, so $\Delta H \gt 0$ (endothermic).
The gaseous state is far more disordered than the condensed state, so $\Delta S \gt 0$ .
Hence $\Delta H \gt 0, \; \Delta S \gt 0$ → List-II (4).

Collecting the matches:
(P)→(2), (Q)→(5), (R)→(1), (S)→(4).

The option that contains this combination is Option C.

Therefore the correct answer is:
Option C which is: (P)→(2), (Q)→(5), (R)→(1), (S)→(4).

Q 30ChemistrySection A • MCQchemical bonding and molecular structure

+3 / -1

Consider the following species:

$SOCl_2$ , $XeOF_4$ , $ClF_3$ , $ClF_5$ , $XeF_5^{+}$ , $SO_3^{2-}$ , $XeF_3^{+}$ , $SF_4$

List-I contains different molecular shapes and List-II contains total number of species with the same molecular shapes from the given species. Match each entry in List-I with the appropriate entry in List-II and choose the correct option.

A

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

B

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

C

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

D

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

Q 31ChemistrySection A • MCQaldehydes ketones and carboxylic acids

+3 / -1

The List-II contains products obtained from the reaction of compounds in List-I with $O_3$ / $Zn$ - $H_2O$ followed by cyclization (via more stable enolate) in the presence of aqueous $NaOH$ . Match each entry in List-I with 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) $\to$ (3), (Q) $\to$ (4), (R) $\to$ (5), (S) $\to$ (2)

C

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

D

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

Official Correct Answer:Option (C)

Step-by-Step Solution:

Ozonolysis in the presence of $O_3/Zn$ - $H_2O$ cleaves every $C=C$ bond and converts the two vinylic carbons into carbonyl groups (an aldehyde if the carbon was $\;-CH=$ and formaldehyde if it was $\;=CH_2$ ).
With the dienes given in List-I, this step therefore furnishes an ω-dialdehyde (or an ω-aldehyd-one when a double bond is internally substituted) plus two molecules of $HCHO$ that are of no further consequence.

When the dialdehyde is now treated with aqueous $NaOH$ the more stable enolate (the one that is more highly substituted) is generated first. That enolate performs an intramolecular aldol addition on the other carbonyl group, producing a β-hydroxy-carbonyl compound. The size of the ring obtained is decided purely by the number of carbon atoms that originally separate the two carbonyl groups:

Number of carbon atoms between the two $-CHO$ groups = $n$
Ring size produced in the aldol step = $(n+1)$ (because a new C-C bond is formed between the α-carbon of one carbonyl and the carbonyl carbon of the other).

Case P

The two double bonds of P are situated at C-1 and C-9 of a nonane chain (1,8-nonadiene). After ozonolysis

$CH_2=CH-(CH_2)_5-CH=CH_2\xrightarrow[Zn/H_2O]{O_3}OHC-(CH_2)_5-CHO+2HCHO$

Six methylenes are now counted from one carbonyl carbon to the other, so $n=5$ and the intramolecular aldol gives a six-membered β-hydroxy aldehyde. That product is entry (2) of List-II.

Case Q

Q is 1,7-octadiene. Ozonolysis furnishes

$CH_2=CH-(CH_2)_4-CH=CH_2\longrightarrow OHC-(CH_2)_4-CHO+2HCHO$

Now $n=4$ ; the aldol cyclisation therefore delivers a five-membered β-hydroxy aldehyde, which is shown as entry (1) in List-II.

Case R

R is a methyl-substituted analogue of Q (2-methyl-1,7-octadiene). Ozonolysis gives

$CH_2=CH-CH(CH_3)-(CH_2)_3-CH=CH_2\longrightarrow OHC-CH(CH_3)-(CH_2)_3-CHO+2HCHO$

Again $n=4$ , so a five-membered ring is formed. Because the more substituted enolate is generated at the α-carbon that bears the $CH_3$ group, the methyl ends up on the ring carbon bearing the newly formed hydroxyl group. This corresponds to entry (5) in List-II.

Case S

S is 2-methyl-1,8-nonadiene. Its ozonolysis product is

$CH_2=CH-CH(CH_3)-(CH_2)_4-CH=CH_2\longrightarrow OHC-CH(CH_3)-(CH_2)_4-CHO+2HCHO$

Now $n=5$ , hence the aldol condensation builds a six-membered ring. As in case R the enolate forms preferentially at the $CH(CH_3)$ end, so the methyl group becomes attached to the β-hydroxy carbon of the ring, giving entry (3) of List-II.

Collecting the results:

(P) → (2), (Q) → (1), (R) → (5), (S) → (3)

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

Q 32ChemistrySection A • MCQcompounds containing nitrogen

+3 / -1

Match the major products obtained in the reactions given in List-I with the corresponding structures in List-II and choose the correct option.

A

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

B

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

C

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

D

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

Official Correct Answer:Option (B)

Step-by-Step Solution:

Four classic name reactions of nitrogen-containing compounds are involved: Sandmeyer cyanation, Hoffmann bromamide degradation, Gabriel phthalimide synthesis and the carbylamine test. By recalling the characteristic product of each reaction, we can match every entry of List-I (P, Q, R, S) with one structure from List-II (1 - 5).

Case P : Sandmeyer cyanation

An aryl diazonium salt reacts with $CuCN$ to give the corresponding aryl nitrile.

$C_6H_5N_2^+Cl^- \;\xrightarrow[\,]{\,CuCN\,}\; C_6H_5CN + N_2\uparrow$

The product is benzonitrile, structure (1). Hence $P \to (1)$ .

Case Q : Hoffmann bromamide degradation

When an amide is treated with $Br_2$ and alcoholic $KOH$ , the carbonyl carbon is lost as $CO_3^{2-}$ and a primary amine containing one carbon fewer is produced.

$CH_3CONH_2 \;\xrightarrow[\,4\,KOH\,]{\,Br_2\,}\; CH_3NH_2 + K_2CO_3 + 2\,KBr + 2\,H_2O$

The product is methylamine, structure (2). Hence $Q \to (2)$ .

Case R : Gabriel phthalimide synthesis followed by hydrazinolysis

Step 1 - phthalimide is converted to its potassium salt, which undergoes $S_N2$ substitution with an alkyl halide (say $C_2H_5Br$ ).
Step 2 - hydrazinolysis (or acid/base hydrolysis) liberates the primary amine.

Overall: $C_6H_4(CO)_2NH \;\xrightarrow[\,KOH\,]{\,C_2H_5Br\,}\; C_6H_4(CO)_2N-C_2H_5 \;\xrightarrow[\,NH_2NH_2\,]{}\; C_2H_5NH_2$

The product is ethylamine, structure (4). Hence $R \to (4)$ .

Case S : Carbylamine (isocyanide) reaction

Primary amines, chloroform and alcoholic $KOH$ give the offensive-smelling isocyanide (carbylamine).

$C_6H_5NH_2 + CHCl_3 + 3\,KOH \rightarrow C_6H_5NC + 3\,KCl + 3\,H_2O$

The product is phenyl isocyanide, structure (5). Hence $S \to (5)$ .

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

Therefore the correct option is
Option B which is: (P) → (1), (Q) → (2), (R) → (4), (S) → (5).

Q 33MathematicsSection A • MCQapplication of derivatives

+3 / -1

Consider the function $f:(0,\infty)\to(-\infty,\infty)$ given by

$f(x)=\sqrt{x}\,\log_e(x)-x+1.$

Then which one of the following statements is TRUE?

A

The derivative of the function $f$ is decreasing in the interval $(0,1)$

B

The function $f$ has a local maximum at some point $a\in(0,\infty)$

C

The function $f$ has a local minimum at some point $b\in(0,\infty)$

D

The function $f$ has NEITHER a point of local maximum NOR a point of local minimum in the interval $(0,\infty)$

Q 34MathematicsSection A • MCQcircle

+3 / -1

Let $P$ be the point on the parabola $y=x^2$ such that the slope of the tangent to the parabola at the point $P$ is $4$ . Let $Q$ be the point in the first quadrant lying on the circle $x^2+y^2=2$ such that the slope of the tangent to the circle at the point $Q$ is $-1$ . Let $R$ be the point in the first quadrant lying on the ellipse $x^2+4y^2=8$ such that the slope of the tangent to the ellipse at the point $R$ is $-\tfrac{1}{2}$ . Then the radius of the circle passing through the points $P,Q$ and $R$ is

A

$\sqrt{10}$

B

$\sqrt{5}$

C

$\sqrt{\tfrac{5}{2}}$

D

$2\sqrt{5}$

Q 35MathematicsSection A • MCQmatrices and determinants

+3 / -1

Which one of the following matrices can be obtained by performing elementary row transformations on the $3\times 3$ identity matrix?

A

$\begin{bmatrix}1&1&1\\1&1&1\\1&1&1\end{bmatrix}$

B

$\begin{bmatrix}1&1&1\\2&3&4\\1&2&1\end{bmatrix}$

C

$\begin{bmatrix}1&1&1\\2&3&4\\2&5&8\end{bmatrix}$

D

$\begin{bmatrix}1&1&1\\-1&1&2\\0&2&3\end{bmatrix}$

Q 36MathematicsSection A • MCQinverse trigonometric functions

+3 / -1

Considering only the principal values of the inverse trigonometric functions, the value of

$\cot^{-1}(\cot(-11))+10\,\sin\!\left(2\cos^{-1}\!\left(\tfrac{1}{\sqrt{2}}\right)\right)+10\sin(2\tan^{-1}(2))$

is

A

$3\pi+7$

B

$7$

C

$4\pi+7$

D

$3\pi-5$

Q 37MathematicsSection A • MCQprobability

+4 / -0

Suppose that Box I contains 6 red balls and 9 green balls, and Box II contains 8 red balls and 12 green balls. All the balls of Box I and Box II are mixed together and a ball is chosen at random from them. Let $E_1$ be the event that the ball chosen belonged to Box I and let $E_2$ be the event that the ball chosen belonged to Box II. Let $F_1$ be the event that the ball chosen is red and let $F_2$ be the event that the ball chosen is green.

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

A

The events $E_1$ and $F_1$ are independent

B

The events $E_2$ and $F_2$ are dependent

C

The conditional probability $P(F_1\mid E_1)$ is less than the conditional probability $P(F_1\mid E_2)$

D

The conditional probability $P(F_1\mid E_1)$ is greater than the conditional probability $P(F_2\mid E_2)$

Official Correct Answer:Option (A, C)

Step-by-Step Solution:

Let us determine the total counts of the balls and find the probabilities for each event.

Box I contains:

$\text{Red balls} = 6$

$\text{Green balls} = 9$

$\text{Total balls in Box I} = 6 + 9 = 15$

Box II contains:

$\text{Red balls} = 8$

$\text{Green balls} = 12$

$\text{Total balls in Box II} = 8 + 12 = 20$

All balls are mixed together:

$\text{Total Red balls } (F_1) = 6 + 8 = 14$

$\text{Total Green balls } (F_2) = 9 + 12 = 21$

$\text{Overall Total balls} = 15 + 20 = 35$

Step 1: Calculate Individual Probabilities

The probability that a chosen ball belongs to Box I is:

$P(E_1) = \frac{15}{35} = \frac{3}{7}$

The probability that a chosen ball belongs to Box II is:

$P(E_2) = \frac{20}{35} = \frac{4}{7}$

The probability that a chosen ball is red is:

$P(F_1) = \frac{14}{35} = \frac{2}{5}$

The probability that a chosen ball is green is:

$P(F_2) = \frac{21}{35} = \frac{3}{5}$

Step 2: Evaluate Independence of $E_1$ and $F_1$

The intersection probability $P(E_1 \cap F_1)$ represents choosing a red ball that came from Box I:

$P(E_1 \cap F_1) = \frac{6}{35}$

Now, let us calculate the product of their individual probabilities:

$P(E_1) \cdot P(F_1) = \frac{3}{7} \cdot \frac{2}{5} = \frac{6}{35}$

Since $P(E_1 \cap F_1) = P(E_1) \cdot P(F_1)$ , the events $E_1$ and $F_1$ are independent. Therefore, Statement A is TRUE.

Step 3: Evaluate Independence of $E_2$ and $F_2$

The intersection probability $P(E_2 \cap F_2)$ represents choosing a green ball that came from Box II:

$P(E_2 \cap F_2) = \frac{12}{35}$

Now, let us calculate the product of their individual probabilities:

$P(E_2) \cdot P(F_2) = \frac{4}{7} \cdot \frac{3}{5} = \frac{12}{35}$

Since $P(E_2 \cap F_2) = P(E_2) \cdot P(F_2)$ , the events $E_2$ and $F_2$ are independent. This means any statement claiming they are dependent is FALSE.

Step 4: Evaluate Conditional Probabilities

Let us calculate $P(F_1 \mid E_1)$ :

$P(F_1 \mid E_1) = \frac{P(E_1 \cap F_1)}{P(E_1)} = \frac{\frac{6}{35}}{\frac{15}{35}} = \frac{6}{15} = \frac{2}{5}$

Let us calculate $P(F_1 \mid E_2)$ :

$P(F_1 \mid E_2) = \frac{P(E_2 \cap F_1)}{P(E_2)} = \frac{\frac{8}{35}}{\frac{20}{35}} = \frac{8}{20} = \frac{2}{5}$

Since $P(F_1 \mid E_1) = P(F_1 \mid E_2) = \frac{2}{5}$ , the conditional probabilities are equal. Therefore, Statement C is FALSE.

Let us check the last comparison involving $P(F_2 \mid E_2)$ :

$P(F_2 \mid E_2) = \frac{P(E_2 \cap F_2)}{P(E_2)} = \frac{\frac{12}{35}}{\frac{20}{35}} = \frac{12}{20} = \frac{3}{5}$

Comparing the values, $\frac{2}{5}$ is not greater than $\frac{3}{5}$ , making any statement asserting $P(F_1 \mid E_1) > P(F_2 \mid E_2)$ FALSE.

Conclusion

The correct statements are:

A. The events $E_1$ and $F_1$ are independent.

Q 38MathematicsSection A • MCQ3d geometry

+4 / -0

Let $P$ be the plane such that it contains the straight line $\dfrac{x-1}{2}=\dfrac{y-3}{3}=\dfrac{z+2}{1}$ and is perpendicular to the plane $x+2y+3z=4$ . Let $P_1$ be the plane which passes through the point $(4,2,2)$ and is parallel to $P$ .

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

A

The equation of the plane $P$ is $7x-5y+z=-10$

B

The distance between the planes $P$ and $P_1$ is $30$

C

The distance of the plane $P$ from the origin is $2\sqrt{3}$

D

The acute angle between the plane $P$ and the plane $2x+2y+z=3$ is $\cos^{-1}\!\left(\tfrac{1}{3\sqrt{3}}\right)$

Official Correct Answer:Option (A, D)

Step-by-Step Solution:

The plane $P$ contains the line

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

and is perpendicular to the plane

$x+2y+3z=4$

Let the direction ratios of the line be $\vec{d}=(2,3,1)$ and the normal to the given plane be $\vec{n}=(1,2,3)$

Since the line lies in plane $P$ and plane $P$ is perpendicular to the plane $x+2y+3z=4$

the normal to $P$ must be perpendicular to both $\vec{d}$ and $\vec{n}$

Hence,

$\vec{N}=\vec{d}\times\vec{n}$

$=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 3 \end{vmatrix}$

$=7\hat{i}-5\hat{j}+\hat{k}$

Therefore, the equation of plane $P$ is

$7(x-1)-5(y-3)+(z+2)=0$

$7x-5y+z=-10$

Hence Statement 1 is TRUE.

Now,

$P: 7x-5y+z+10=0$

and since $P_1$ passes through $(4,2,2)$ and is parallel to $P$ ,

$7(4)-5(2)+2+D=0$

$20+D=0$

$D=-20$

Therefore,

$P_1: 7x-5y+z-20=0$

Distance between the planes is

$\frac{|10-(-20)|}{\sqrt{7^2+(-5)^2+1^2}}$

$=\frac{30}{\sqrt{75}}$

$=\frac{30}{5\sqrt{3}}$

$=2\sqrt{3}$

Hence Statement 2 is FALSE.

Distance of plane $P$ from the origin is

$\frac{|10|}{\sqrt{75}}$

$=\frac{2}{\sqrt{3}}$

$=\frac{2\sqrt{3}}{3}$

Hence Statement 3 is FALSE.

For the plane

$2x+2y+z=3$

the normal vector is

$\vec{N}_1=(2,2,1)$

The acute angle between the planes equals the angle between their normals.

Therefore,

$\cos\theta=\frac{|7(2)+(-5)(2)+1(1)|}{\sqrt{75}\sqrt{9}}$

$=\frac{5}{15\sqrt{3}}$

$=\frac{1}{3\sqrt{3}}$

Hence,

$\theta=\cos^{-1}\left(\frac{1}{3\sqrt{3}}\right)$

Therefore Statement 4 is TRUE.

Hence, the TRUE statements are

$\boxed{\text{Statements 1 and 4}}$

Q 39MathematicsSection A • MCQlimits continuity and differentiability

+4 / -0

Let $\mathbb{R}$ denote the set of all real numbers. Let $f:\mathbb{R}\to\mathbb{R}$ be an arbitrary function and let $g:\mathbb{R}\to\mathbb{R}$ be the function defined by

$g(x)=x\,f(x),\quad\text{for all }x\in\mathbb{R}.$

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

A

The function $g$ is always continuous at $x=0$

B

If $f$ is continuous at $x=0$ , then $g$ is differentiable at $x=0$

C

If $g$ is differentiable at $x=0$ , then $f$ is continuous at $x=0$

D

If $g$ is differentiable at $x=0$ , then $\lim_{x\to 0}f(x)$ exists

Q 40MathematicsSection A • MCQmatrices and determinants

+4 / -0

Consider the matrix $M=\begin{bmatrix}2&-1\\1&0\end{bmatrix}.$

Let $p,q,r,s,a,b,c$ and $d$ be integers such that $M^{26}=\begin{bmatrix}p&q\\r&s\end{bmatrix}$ and $\displaystyle\sum_{k=1}^{26}M^k=\begin{bmatrix}a&b\\c&d\end{bmatrix}.$

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

A

There exists a $2\times 2$ invertible matrix $N$ with real entries such that $MN=N\begin{bmatrix}1&1\\0&1\end{bmatrix}$

B

The value of $a$ is $378$

C

For any two given integers $m$ and $n$ , there exist unique integers $x$ and $y$ such that $px+qy=m$ and $rx+sy=n$

D

For each positive real number $t$ , the system of linear equations $(a+t)x+by=1,\;cx+(d+t)y=-1$ has a unique solution

Official Correct Answer:Option (A, C, D)

Step-by-Step Solution:

We can write the matrix as

$𝑀 = [\begin{matrix} 2 & - 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}] + [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

So, let $𝐵 = [\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}]$ . Then $𝑀 = 𝐵 + 𝐼$ .

Find $𝐵^{2}$

$𝐵^{2} = [\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}] [\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}] = [\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}]$

This means $𝐵$ is a nilpotent matrix, i.e., its square is a zero matrix.

Expand $𝑀^{26}$ using the Binomial Theorem

$𝑀^{26} = (𝐵 + 𝐼)^{26} =^{26} 𝐶_{0} 𝐼 +^{26} 𝐶_{1} 𝐵 +^{26} 𝐶_{2} 𝐵^{2} + \dots +^{26} 𝐶_{26} 𝐵^{26}$

As $𝐵^{2} = 0$ , all higher powers of $𝐵$ are zero. So,

$𝑀^{26} = 𝐼 + 26 𝐵$

Substitute the values

$𝑀^{26} = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + 26 [\begin{matrix} 1 & - 1 \\ 1 & - 1 \end{matrix}] = [\begin{matrix} 27 & - 26 \\ 26 & - 25 \end{matrix}]$

So, $𝑝 = 27, 𝑞 = - 26, 𝑟 = 26, 𝑠 = - 25.$

Find the sum $𝐷 = \sum_{26}^{𝑘 = 1} 𝑀^{𝑘}$

$𝐷 = 𝑀 + 𝑀^{2} + 𝑀^{3} + \dots + 𝑀^{26}$

Substitute $𝑀 = 𝐵 + 𝐼$ :

$𝐷 = (𝐵 + 𝐼) + (𝐵 + 𝐼)^{2} + (𝐵 + 𝐼)^{3} + \dots + (𝐵 + 𝐼)^{26}$

Since $(𝐵 + 𝐼)^{𝑘} = 𝐼 + 𝑘 𝐵$ , we have

$𝐷 = (𝐼 + 𝐵) + (𝐼 + 2 𝐵) + (𝐼 + 3 𝐵) + \dots + (𝐼 + 26 𝐵)$

Combine like terms:

$𝐷 = 26 𝐼 + 𝐵 (1 + 2 + 3 + \dots + 26)$

The sum $1 + 2 + 3 + \dots + 26 = 351$ , so

$𝐷 = 26 𝐼 + 351 𝐵$

Substitute the matrices:

$\sum_{26}^{𝑘 = 1} 𝑀^{𝑘} = [\begin{matrix} 377 & - 351 \\ 351 & - 325 \end{matrix}] = [\begin{matrix} 𝑎 & 𝑏 \\ 𝑐 & 𝑑 \end{matrix}]$

Hence, $𝑎 = 377, 𝑏 = - 351, 𝑐 = 351, 𝑑 = - 325.$

So, option (B) is incorrect.

Check option (C)

Given system of equations:

$27 𝑥 - 26 𝑦 = 𝑚$

$26 𝑥 - 25 𝑦 = 𝑛$

Determinant of coefficients:

$∣ \begin{matrix} 27 & - 26 \\ 26 & - 25 \end{matrix} ∣ \neq 0$

Since the determinant is non-zero, a unique solution exists.

On solving, we get:

$𝑥 = - 24 𝑛 - 25 𝑚, 𝑦 = - 23 𝑛 - 26 𝑚$

Both $𝑥$ and $𝑦$ are integers. Hence, option (C) is correct.

Check option (D)

We have:

$∣ \begin{matrix} 𝑎 + 𝑡 & 𝑏 \\ 𝑐 & 𝑑 + 𝑡 \end{matrix} ∣ = 𝑎 𝑑 + 𝑎 𝑡 + 𝑑 𝑡 + 𝑡^{2} - 𝑏 𝑐$

Substitute $𝑎 = 377, 𝑏 = - 351, 𝑐 = 351, 𝑑 = - 325$ :

$= (377) (- 325) + 𝑡 (52) + 𝑡^{2} + (351) (351) = 𝑡^{2} + 52 𝑡 + 676 = (𝑡 + 26)^{2}$

Since $(𝑡 + 26)^{2} \neq 0$ for all $𝑡 > 0$ , the determinant is non-zero, so a unique solution exists. Option (D) is correct.

Check option (A)

Let $𝑁 = [\begin{matrix} 𝑗 & 𝑘 \\ 𝑙 & 𝑚 \end{matrix}]$ .

Compute $𝑀 𝑁$ :

$𝑀 𝑁 = [\begin{matrix} 2 & - 1 \\ 1 & 0 \end{matrix}] [\begin{matrix} 𝑗 & 𝑘 \\ 𝑙 & 𝑚 \end{matrix}] = [\begin{matrix} 2 𝑗 - 𝑙 & 2 𝑘 - 𝑚 \\ 𝑗 & 𝑘 \end{matrix}]$

Compute $𝑁 [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}]$ :

$𝑁 [\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}] = [\begin{matrix} 𝑗 & 𝑗 + 𝑘 \\ 𝑙 & 𝑙 + 𝑚 \end{matrix}]$

Comparing both, we get $𝑗 = 𝑙$ and $2 𝑘 - 𝑚 = 𝑗 + 𝑘$ ⟹ $𝑚 - 𝑘 = 𝑗$ .

Determinant of $𝑁$ :

$| 𝑁 | = 𝑗 𝑚 - 𝑙 𝑘 = 𝑗 (𝑚 - 𝑘)$

This is not always zero if $𝑗 \neq 0$ and $𝑚 \neq 𝑘$ , which means $𝑁$ can be invertible. Hence, option (A) is correct.

Q 41MathematicsSection B • Numericalpermutations and combinations

+4 / -0

Let $S=\{1,2,3,\dots,10\}$ . Consider the set

$X=\{R:R\text{ is an equivalence relation on the set }S\text{ such that }R\text{ has exactly 42 elements}\}.$

Then the number of elements in $X$ is ___.

Answer Format:Enter numerical value rounded to the nearest integer

Q 42MathematicsSection B • Numericallimits continuity and differentiability

+4 / -0

Consider the function $f:\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)\to(-\infty,\infty)$ defined by

$f(x)=(|x|+|x-1|)\sin x+[x\sin x],$

where $[x\sin x]$ is the greatest integer less than or equal to $x\sin x$ .

Let $\alpha$ be the total number of points in the interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ at which $f$ is NOT continuous, and let $\beta$ be the total number of points in the interval $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ at which $f$ is NOT differentiable. Then the value of $\alpha+\beta$ is ___.

Answer Format:Enter numerical value rounded to the nearest integer

Q 43MathematicsSection B • Numericalpermutations and combinations

+4 / -0

The number of ways to distribute 10 identical red pens and 14 identical blue pens among four persons such that each person gets 6 pens, is ___.

Answer Format:Enter numerical value rounded to the nearest integer

Q 44MathematicsSection B • Numericalcomplex numbers

+4 / -0

Let $\alpha=\left(1-2\cos\tfrac{\pi}{11}\right)\left(1-2\cos\tfrac{3\pi}{11}\right)\left(1-2\cos\tfrac{9\pi}{11}\right)\left(1-2\cos\tfrac{27\pi}{11}\right)\left(1-2\cos\tfrac{81\pi}{11}\right).$

Then the value of $5-\alpha^2$ is ___.

Answer Format:Enter numerical value rounded to the nearest integer

Q 45MathematicsSection A • MCQcomplex numbers

+3 / -1

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

A

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

B

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

C

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

D

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

Official Correct Answer:Option (C)

Step-by-Step Solution:

Given $x^2 + x + 1 = 0 \implies \alpha = \omega, \ \beta = \omega^2$

$\text{Since } 1 + \omega + \omega^2 = 0 \implies \alpha + 1 = -\omega^2, \ \beta + 1 = -\omega$

Evaluating entry (P):

$\text{Roots: } \frac{1}{(\alpha+1)^{2026}} = \frac{1}{(-\omega^2)^{2026}} = \frac{1}{\omega^{4052}} = \frac{1}{\omega^2} = \omega$

$\frac{1}{(\beta+1)^{2026}} = \frac{1}{(-\omega)^{2026}} = \frac{1}{\omega^{2026}} = \frac{1}{\omega} = \omega^2$

$\text{Equation with roots } \omega, \omega^2 \text{ is } x^2 + x + 1 = 0 \implies \text{(P)} \rightarrow \text{(1)}$

Evaluating entry (Q):

$\text{Roots: } \frac{1}{(\alpha+1)^{2027}} = \frac{1}{(-\omega^2)^{2027}} = \frac{1}{-\omega^{4054}} = \frac{1}{-\omega} = -\omega^2$

$\frac{1}{(\beta+1)^{2027}} = \frac{1}{(-\omega)^{2027}} = \frac{1}{-\omega^{2027}} = \frac{1}{-\omega^2} = -\omega$

$\text{Equation with roots } -\omega, -\omega^2 \text{ is } x^2 - (-\omega - \omega^2)x + (-\omega)(-\omega^2) = 0 \implies x^2 - x + 1 = 0 \implies \text{(Q)} \rightarrow \text{(2)}$

Evaluating entry (R):

$\text{For } x^2 - x + 1 = 0 \implies \gamma = -\omega, \ \delta = -\omega^2$

$\gamma - 1 = -\omega - 1 = \omega^2, \quad \delta - 1 = -\omega^2 - 1 = \omega$

$\frac{1}{(\gamma-1)^{2026}} + \frac{1}{(\delta-1)^{2026}} = \frac{1}{(\omega^2)^{2026}} + \frac{1}{(\omega)^{2026}} = \frac{1}{\omega^2} + \frac{1}{\omega} = \omega + \omega^2 = -1 \implies \text{(R)} \rightarrow \text{(4)}$

Evaluating entry (S):

$\text{For } x^2 + x - 1 = 0 \implies p^2 + p = 1 \implies p + 1 = \frac{1}{p}, \quad r + 1 = \frac{1}{r}$

$\frac{1}{(p+1)^3} + \frac{1}{(r+1)^3} = p^3 + r^3 = (p+r)(p^2 - pr + r^2) = (p+r)((p+r)^2 - 3pr)$

$\text{Since } p+r = -1, \ pr = -1:$

$= (-1)((1) - 3(-1)) = (-1)(4) = -4 \implies \text{(S)} \rightarrow \text{(5)}$

Q 46MathematicsSection A • MCQtrigonometric equations

+3 / -1

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

A

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

B

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

C

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

D

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

Official Correct Answer:Option (B)

Step-by-Step Solution:

Evaluating entry (P):

$\sin^6 x + \cos^4 x = 1, \quad x \in [-\pi, \pi]$

$\sin^6 x = 1 - \cos^4 x = (1 - \cos^2 x)(1 + \cos^2 x) = \sin^2 x(1 + \cos^2 x)$

$\sin^2 x (\sin^4 x - 1 - \cos^2 x) = 0 \implies \sin^2 x (\sin^4 x - 1 - (1 - \sin^2 x)) = 0$

$\sin^2 x (\sin^4 x + \sin^2 x - 2) = 0 \implies \sin^2 x (\sin^2 x + 2)(\sin^2 x - 1) = 0$

$\sin x = 0 \quad \text{or} \quad \sin^2 x = 1 \implies \cos x = 0$

$x = 0, \pm\pi, \pm\frac{\pi}{2} \implies n(P) = 5 \implies \text{(P)} \rightarrow \text{(5)}$

Evaluating entry (Q):

$\sin^2 x + \cos^6 x = 1, \quad x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

$\cos^6 x = 1 - \sin^2 x = \cos^2 x \implies \cos^2 x (\cos^4 x - 1) = 0$

$\cos x = 0 \quad \text{or} \quad \cos^2 x = 1 \implies \sin x = 0$

$x = 0, \pm\frac{\pi}{2} \implies n(Q) = 3 \implies \text{(Q)} \rightarrow \text{(3)}$

Evaluating entry (R):

$\cos^2\left(\frac{x}{2}\right) - \sin^2 x = \frac{1}{2}, \quad x \in [-\pi, \pi]$

$\frac{1 + \cos x}{2} - (1 - \cos^2 x) = \frac{1}{2} \implies 1 + \cos x - 2 + 2\cos^2 x = 1$

$2\cos^2 x + \cos x - 2 = 0 \implies \cos x = \frac{-1 \pm \sqrt{17}}{4}$

$\text{Since } \frac{-1 - \sqrt{17}}{4} < -1, \quad \cos x = \frac{\sqrt{17} - 1}{4} \approx 0.78$

$x = \pm\cos^{-1}\left(\frac{\sqrt{17}-1}{4}\right) \implies n(R) = 2 \implies \text{(R)} \rightarrow \text{(2)}$

Evaluating entry (S):

$6\sin^2\left(\frac{x}{2}\right) - \cos 3x = 3, \quad x \in [-2\pi, 2\pi]$

$3(1 - \cos x) - \cos 3x = 3 \implies 3 - 3\cos x - (4\cos^3 x - 3\cos x) = 3$

$-4\cos^3 x = 0 \implies \cos x = 0$

$x = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2} \implies n(S) = 4 \implies \text{(S)} \rightarrow \text{(4)}$

Q 47MathematicsSection A • MCQmatrices and determinants

+3 / -1

For real numbers $\alpha,\beta,\gamma,\delta$ and $\mu$ , consider the matrix $M=\begin{bmatrix}\alpha&\tfrac{1}{\sqrt{2}}&-\tfrac{1}{\sqrt{2}}\\[2pt]\tfrac{1}{\sqrt{3}}&\beta&\tfrac{1}{\sqrt{3}}\\[2pt]\gamma&\delta&\mu\end{bmatrix}.$

Suppose that $MM^{T}=I$ , where $M^{T}$ is the transpose of $M$ and $I$ is the $3\times 3$ identity matrix. Let $\vec{u}=\alpha\hat{i}+\tfrac{1}{\sqrt{3}}\hat{j}+\gamma\hat{k},\quad \vec{v}=\tfrac{1}{\sqrt{2}}\hat{i}+\beta\hat{j}+\delta\hat{k},\quad \vec{w}=-\tfrac{1}{\sqrt{2}}\hat{i}+\tfrac{1}{\sqrt{3}}\hat{j}+\mu\hat{k}.$

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

A

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

B

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

C

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

D

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

Official Correct Answer:Option (A)

Step-by-Step Solution:

$M = \begin{bmatrix} \alpha & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \beta & \frac{1}{\sqrt{3}} \\ \gamma & \delta & \mu \end{bmatrix}$

$\vec{u} = \alpha\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \gamma\hat{k}$

$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \beta\hat{j} + \delta\hat{k}$

$\vec{w} = -\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \mu\hat{k}$

$\alpha^2 + \frac{1}{3} + \gamma^2 = 1 \quad \therefore \alpha^2 + \gamma^2 = \frac{2}{3} \quad \dots(1)$

$\frac{1}{2} + \beta^2 + \delta^2 = 1 \quad \beta^2 + \delta^2 = \frac{1}{2} \quad \dots(2)$

$\frac{1}{2} + \frac{1}{3} + \mu^2 = 1 \quad \mu^2 = \frac{1}{6} \quad \dots(3)$

$\gamma^2 + \delta^2 + \mu^2 = 1$

$\gamma^2 + \delta^2 = 1 - \frac{1}{6} = \frac{5}{6}$

For Q:

$x\vec{u} + y\vec{v} + z\vec{w} = \hat{j}$

$x\vec{u} \cdot \vec{u} + y\vec{v} \cdot \vec{u} + z\vec{w} \cdot \vec{u} = \vec{u} \cdot \hat{j}$

$x = \frac{1}{\sqrt{3}}$

For R:

$\vec{u}, \vec{v}, \vec{w}$ are mutually perpendicular unit vectors.

$\vert{}\vec{u} \cdot (\vec{v} \times \vec{w})\vert{}$

$\begin{bmatrix} \vec{u} & \vec{v} & \vec{w} \end{bmatrix} = 1$

For S:

$\vec{u} \times (\vec{v} \times \vec{w})$

$(\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$

$= 0 - 0 = 0$

Q 48MathematicsSection A • MCQcircle

+3 / -1

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

A

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

B

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

C

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

D

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

Official Correct Answer:Option (B)

Step-by-Step Solution:

Evaluating entry (P):

Radius $R$ equals the perpendicular distance from center $(1,2)$ to $3x + 4y - 1 = 0$ :

$R = \frac{\vert{}3(1) + 4(2) - 1\vert{}}{\sqrt{3^2 + 4^2}} = \frac{10}{5} = 2$

$\text{Equation of circle: } (x-1)^2 + (y-2)^2 = 4$

Testing point (3,2) from List-II: $(3-1)^2 + (2-2)^2 = 4 \implies 4 = 4 \implies \text{(P)} \rightarrow \text{(3)}$

Evaluating entry (Q):

Equation of tangent to $y^2 = 8x$ with slope $m$ : $y = mx + \frac{2}{m}$

Since it touches $x^2 + y^2 = 2$ : $\frac{\vert{}2/m\vert{}}{\sqrt{1 + m^2}} = \sqrt{2} \implies \frac{4}{m^2(1+m^2)} = 2 \implies m^4 + m^2 - 2 = 0 \implies m = 1 \quad (m > 0)$

$\text{Tangent line: } y = x + 2$

Testing point (7,9) from List-II: $9 = 7 + 2 \implies \text{(Q)} \rightarrow \text{(2)}$

Evaluating entry (R):

$\frac{x^2}{16} + \frac{y^2}{12} = 1 \implies a=4, \ b=\sqrt{12}$

$e = \sqrt{1 - \frac{12}{16}} = \frac{1}{2} \implies ae = 2$

$M = \left(ae, \frac{b^2}{a}\right) = \left(2, \frac{12}{4}\right) = (2, 3)$

Equation of normal at $M(2,3)$ : $\frac{a^2x}{x_1} - \frac{b^2y}{y_1} = a^2 - b^2 \implies \frac{16x}{2} - \frac{12y}{3} = 16 - 12 \implies 2x - y = 1$

Testing point (1,1) from List-II: $2(1) - 1 = 1 \implies \text{(R)} \rightarrow \text{(1)}$

Evaluating entry (S):

$ae = 5, \quad \frac{a}{e} = \frac{16}{5} \implies a^2 = 16 \implies b^2 = a^2(e^2 - 1) = (ae)^2 - a^2 = 25 - 16 = 9$

$\text{Equation of hyperbola: } \frac{x^2}{16} - \frac{y^2}{9} = 1$

Testing point $(8, 3\sqrt{3})$ from List-II: $\frac{64}{16} - \frac{27}{9} = 4 - 3 = 1 \implies \text{(S)} \rightarrow \text{(5)}$

JEE Advanced 2026 Paper 1 Question Paper with Solutions

Case 1: Partition {5, 4, 1}

Case 2: Partition {6, 2, 1, 1}