JEE Advanced 2025 Question Paper 2 Question Paper with Solutions | JEE Advanced Mock Test | JeeMate

Q 1PhysicsSection A • MCQunits and measurements

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A temperature difference can generate e.m.f. in some materials. Let $S$ be the e.m.f. produced per unit temperature difference between the ends of a wire, $\sigma$ the electrical conductivity and $\kappa$ the thermal conductivity of the material of the wire. Taking $M$ , $L$ , $T$ , $I$ and $K$ as dimensions of mass, length, time, current and temperature, respectively, the dimensional formula of the quantity $Z = \frac{S^2 \sigma}{\kappa}$ is:

A

$[M^0 L^0 T^0 I^0 K^0]$

B

$[M^0 L^0 T^0 I^0 K^{-1}]$

C

$[M^1 L^2 T^{-2} I^{-1} K^{-1}]$

D

$[M^1 L^2 T^{-4} I^{-1} K^{-1}]$

Q 2PhysicsSection A • MCQcapacitor

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Two co-axial conducting cylinders of same length $\ell$ with radii $\sqrt{2}R$ and $2R$ are kept, as shown in Fig. 1. The charge on the inner cylinder is $Q$ and the outer cylinder is grounded. The annular region between the cylinders is filled with a material of dielectric constant $\kappa = 5$ . Consider an imaginary plane of the same length $\ell$ at a distance $R$ from the common axis of the cylinders. This plane is parallel to the axis of the cylinders. The cross-sectional view of this arrangement is shown in Fig. 2. Ignoring edge effects, the flux of the electric field through the plane is ( $\epsilon_0$ is the permittivity of free space):

A

$\frac{Q}{30\epsilon_0}$

B

$\frac{Q}{15\epsilon_0}$

C

$\frac{Q}{60\epsilon_0}$

D

$\frac{Q}{120\epsilon_0}$

Official Correct Answer:Option (C)

Step-by-Step Solution:

Let the inner and outer conducting cylinders be coaxial with the common $z$ -axis.
Radius of inner cylinder : $\sqrt{2}\,R$
Radius of outer cylinder : $2\,R$
Free charge on inner cylinder : $Q$ (uniformly distributed along its length $\ell$ )
Surface charge density per unit length (line charge) : $\lambda=\dfrac{Q}{\ell}$ .

The space between the two cylinders is filled with a dielectric of relative permittivity $\kappa=5$ .
The outer cylinder is earthed, hence the electric field exists only in the dielectric region $\sqrt{2}\,R \le r \le 2R$ and is purely radial.

Electric field inside the dielectric
Take a cylindrical Gaussian surface of radius $r$ (with $\sqrt{2}\,R \le r \le 2R$ ) and length $\ell$ . Using Gauss’s law in dielectrics,

$\oint \mathbf{D}\cdot d\mathbf{A}=Q_{\text{free enclosed}}\; \Longrightarrow \; D(2\pi r\ell)=\lambda\ell \Rightarrow D=\dfrac{\lambda}{2\pi r}.$ Since $\mathbf{D}=\kappa\varepsilon_0\mathbf{E}$ , we get

$E(r)=\dfrac{\lambda}{2\pi\kappa\varepsilon_0\,r}, \qquad \mathbf{E}=E(r)\,\hat{\mathbf{r}}.$ (Inside $r\lt \sqrt{2}\,R$ the electric field is zero.)

The required open surface
We need the electric flux through a plane that

is parallel to the axis, of length $\ell$ ,
is at a perpendicular distance $R$ from the axis (take it as the plane $x=R$ ).

In the cross-section (the $xy$ -plane) this surface appears as the straight line $x=R$ . Only the part of this line lying in the dielectric region ( $r\ge\sqrt{2}\,R$ ) contributes to flux.

Limits of the contributing strip on the plane
Put $x=R$ and write $r=\sqrt{x^{2}+y^{2}}=\sqrt{R^{2}+y^{2}}.$

Inner boundary: $r=\sqrt{2}\,R \;\Rightarrow\; y=\pm R$
Outer boundary: $r=2\,R \;\Rightarrow\; y=\pm\sqrt{3}\,R$

Hence the dielectric occupies

y\in[-\sqrt{3}R,-R]\cup[R,\sqrt{3}R]

on the plane.

Element of area and flux through it
For a strip of width $dy$ on this plane, area element $dA=\ell\,dy.$ The unit normal to the plane is $\hat{\mathbf{n}}=\hat{\mathbf{x}}$ (along $+x$ ), while the electric field is radial, so the cosine of the angle between $\mathbf{E}$ and $\hat{\mathbf{n}}$ equals $\dfrac{x}{r}=\dfrac{R}{r}.$ Therefore, the differential flux is

$d\Phi=E(r)\left(\dfrac{R}{r}\right)dA =\dfrac{\lambda}{2\pi\kappa\varepsilon_0\,r}\left(\dfrac{R}{r}\right)\ell\,dy =\dfrac{\lambda R\ell}{2\pi\kappa\varepsilon_0}\dfrac{dy}{(R^{2}+y^{2})}.$

Total flux through the plane

$\Phi=\int_{-\sqrt{3}R}^{-R}+\int_{R}^{\sqrt{3}R} \dfrac{\lambda R\ell}{2\pi\kappa\varepsilon_0}\; \dfrac{dy}{R^{2}+y^{2}} =\dfrac{\lambda R\ell}{\pi\kappa\varepsilon_0} \int_{R}^{\sqrt{3}R}\dfrac{dy}{R^{2}+y^{2}} \quad (\text{using symmetry}).$

Evaluate the integral:

$\int_{R}^{\sqrt{3}R}\dfrac{dy}{R^{2}+y^{2}} =\left[\dfrac{1}{R}\arctan\left(\dfrac{y}{R}\right)\right]_{R}^{\sqrt{3}R} =\dfrac{1}{R}\left[\arctan(\sqrt{3})-\arctan(1)\right] =\dfrac{1}{R}\left(\dfrac{\pi}{3}-\dfrac{\pi}{4}\right) =\dfrac{\pi}{12R}.$

Substituting,

$\Phi=\dfrac{\lambda R\ell}{\pi\kappa\varepsilon_0}\cdot\dfrac{\pi}{12R} =\dfrac{\lambda\ell}{12\kappa\varepsilon_0} =\dfrac{Q}{12\kappa\varepsilon_0}.$

With $\kappa=5$ ,

$\Phi=\dfrac{Q}{12\times5\;\varepsilon_0} =\dfrac{Q}{60\,\varepsilon_0}.$

Therefore, the electric flux through the given plane is $\dfrac{Q}{60\varepsilon_0}$ .

Option C which is: $\displaystyle\frac{Q}{60\varepsilon_0}$

Q 3PhysicsSection A • MCQsimple harmonic motion

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As shown in the figures, a uniform rod $OO'$ of length $l$ is hinged at the point $O$ and held in place vertically between two walls using two massless springs of same spring constant. The springs are connected at the midpoint and at the top-end ( $O'$ ) of the rod, as shown in Fig. 1 and the rod is made to oscillate by a small angular displacement. The frequency of oscillation of the rod is $f_1$ . On the other hand, if both the springs are connected at the midpoint of the rod, as shown in Fig. 2 and the rod is made to oscillate by a small angular displacement, then the frequency of oscillation is $f_2$ .

Ignoring gravity and assuming motion only in the plane of the diagram, the value of $\frac{f_1}{f_2}$ is:

A

$2$

B

$\sqrt{2}$

C

$\sqrt{\frac{5}{2}}$

D

$\sqrt{\frac{2}{5}}$

Official Correct Answer:Option (C)

Step-by-Step Solution:

Let the mass of the rod be $m$ and its length be $l$ . It oscillates about the fixed hinge $O$ with a small angular displacement $\theta$ (taken positive for a clockwise rotation). Gravity is ignored, so only the springs give the restoring torque.

1. Moment of inertia about the hinge
For a uniform rod of length $l$ about one end, $I = \frac{1}{3} m l^{2}$

2. Extension produced in a spring connected at distance $r$ from the hinge
When the rod rotates through a small angle $\theta$ , every point at a distance $r$ from the hinge moves through an arc of length $r\theta$ perpendicular to the rod’s original position. Hence the linear extension (or compression) of a spring attached at that point is also $r\theta.$

3. Restoring torque of one spring
Spring force $F = k \,(r\theta) \quad\Rightarrow\quad$ Torque about $O: \quad \tau = F \cdot r = k \, r^{2}\theta.$ Therefore each spring contributes a restoring torque equal to $k r^{2}\theta.$

4. Angular equation of motion
Sum of restoring torques $= k\,(r_1^{2}+r_2^{2})\,\theta.$ Hence $I\,\ddot{\theta} + k\,(r_1^{2}+r_2^{2})\,\theta = 0\; \Longrightarrow\; \omega = \sqrt{\dfrac{k\,(r_1^{2}+r_2^{2})}{I}}.$ The linear frequency is $f = \dfrac{\omega}{2\pi},$ so the ratio $f_1/f_2$ equals the ratio $\omega_1/\omega_2.$

Case 1: (Fig. 1)
Springs at $r_1=l/2$ and $r_2=l$ $r_1^{2}+r_2^{2} = \left(\frac{l}{2}\right)^{2} + l^{2} = \frac{l^{2}}{4}+l^{2} = \frac{5}{4}l^{2}.$ $\displaystyle \omega_1 = \sqrt{\dfrac{k\left(\tfrac{5}{4}l^{2}\right)}{I}}.$

Case 2: (Fig. 2)
Both springs at $r_1=r_2=l/2$ $r_1^{2}+r_2^{2} = 2\left(\frac{l}{2}\right)^{2} = 2\left(\frac{l^{2}}{4}\right)=\frac{l^{2}}{2}.$ $\displaystyle \omega_2 = \sqrt{\dfrac{k\left(\tfrac{l^{2}}{2}\right)}{I}}.$

5. Ratio of frequencies
$\frac{f_1}{f_2}=\frac{\omega_1}{\omega_2}= \sqrt{\dfrac{k\left(\tfrac{5}{4}l^{2}\right)}{I}} \;\Bigg/\; \sqrt{\dfrac{k\left(\tfrac{l^{2}}{2}\right)}{I}} = \sqrt{\dfrac{\tfrac{5}{4}l^{2}}{\tfrac{l^{2}}{2}}} = \sqrt{\dfrac{5}{4}\cdot\dfrac{2}{1}} = \sqrt{\dfrac{10}{4}} = \sqrt{\dfrac{5}{2}}.$

Hence $\displaystyle \frac{f_1}{f_2} = \sqrt{\frac{5}{2}}.$ Option C is correct.

Q 4PhysicsSection A • MCQgravitation

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Consider a star of mass $m_2$ kg revolving in a circular orbit around another star of mass $m_1$ kg with $m_1 \gg m_2$ . The heavier star slowly acquires mass from the lighter star at a constant rate of $\gamma$ kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $r$ , then its relative rate of change $\frac{1}{r}\frac{dr}{dt}$ (in s $^{-1}$ ) is given by:

A

$-\frac{3\gamma}{2m_2}$

B

$-\frac{2\gamma}{m_2}$

C

$-\frac{2\gamma}{m_1}$

D

$-\frac{3\gamma}{2m_1}$

Q 5PhysicsSection A • MCQelectrostatics

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A positive point charge of $10^{-8}$ C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm. The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved a distance of 10 cm further away from the center of the sphere along the radial direction. Taking $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9$ Nm $^2$ /C $^2$ (where $\epsilon_0$ is the permittivity of free space), which of the following statements is/are correct:

A

Before the grounding, the electrostatic potential of the sphere is 450 V.

B

Charge flowing from the sphere to the ground because of grounding is $5 \times 10^{-9}$ C.

C

After the grounding is removed, the charge on the sphere is $-5 \times 10^{-9}$ C.

D

The final electrostatic potential of the sphere is 300 V.

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

Step-by-Step Solution:

Given: point charge $q = +10^{-8}\,\text{C}$ at a distance $d = 20\,\text{cm} = 0.20\,\text{m}$ from the centre $O$ of a conducting sphere of radius $R = 10\,\text{cm} = 0.10\,\text{m}$ .
Throughout we use $k = \dfrac{1}{4\pi\varepsilon_0}=9\times10^9\;\text{N m}^2\!\text{/C}^2$ .

Step 1 : Image-charge data for a charge outside a sphere
For a point charge $q$ kept at distance $d\,(d\gt R)$ from the centre of a conducting sphere of radius $R$ :

• An image charge $q' = -\dfrac{R}{d}\,q$ is located on the same line at a distance $b = \dfrac{R^{2}}{d}$ from the centre.
• The combination $(q,q')$ alone makes the potential of the sphere zero (grounded condition).
• If the sphere is isolated (not grounded) and must keep a specified net charge $Q_{\text{sphere}}$ , an additional charge $q_c$ is placed at the centre. $q_c$ is fixed by the condition $q_c + q' = Q_{\text{sphere}}$ . The potential of the sphere then becomes a constant $V_{\text{sphere}} = k\dfrac{q_c}{R}$ .

Part (A) : Potential before grounding
Initially the sphere is neutral $(Q_{\text{sphere}}=0)$ and isolated.

Image charge: $q' = -\dfrac{R}{d}\,q = -\dfrac{0.10}{0.20}\,10^{-8} = -0.5\times10^{-8} = -5\times10^{-9}\,\text{C}$ .
To keep the sphere neutral, add a central charge $q_c = -q' = +5\times10^{-9}\,\text{C}$ .

The potential of the sphere is therefore $V_{\text{before}} = k\dfrac{q_c}{R} = 9\times10^9 \times\dfrac{5\times10^{-9}}{0.10} = 9\times10^9 \times 5\times10^{-8} = 45\times10^{1} = 450\;\text{V}.$

Hence Option A is correct.

Part (B) : Charge that flows when the sphere is grounded
On connecting the sphere to earth, its potential must become zero. That is achieved by removing the central charge $q_c$ (ground provides/absorbs charge) so that only $q'$ remains on the sphere.

Charge leaving the sphere to the ground $\;\Delta Q = q_c = +5\times10^{-9}\,\text{C}.$

Thus $5\times10^{-9}\,\text{C}$ of positive charge flows from the sphere to the earth. Option B is correct.

Part (C) : Charge on the isolated sphere after the grounding wire is removed
With the earth connection opened, the induced charge that remains on the sphere equals $q' = -5\times10^{-9}\,\text{C}$ .

Option C is correct.

Part (D) : Potential after the external charge is shifted to 30 cm
The grounding has already been removed, so the total charge on the sphere stays fixed at $Q_{\text{sphere}} = -5\times10^{-9}\,\text{C}.$ Now the point charge is shifted to $d_2 = 30\,\text{cm} = 0.30\,\text{m}$ .

New image charge for the new position:
$q'_2 = -\dfrac{R}{d_2}\,q = -\dfrac{0.10}{0.30}\,10^{-8} = -\dfrac{1}{3}\times10^{-8} = -3.33\times10^{-9}\,\text{C}.$

To keep the sphere’s total charge fixed, the central charge becomes $q_{c2} = Q_{\text{sphere}} - q'_2 = \bigl(-5\times10^{-9}\bigr) - \bigl(-3.33\times10^{-9}\bigr) = -1.67\times10^{-9}\,\text{C}.$

The potential of the sphere is therefore $V_{\text{final}} = k\dfrac{q_{c2}}{R} = 9\times10^9 \times\dfrac{-1.67\times10^{-9}}{0.10} \approx -150\;\text{V}.$

This is neither 300 V nor positive. Hence Option D is incorrect.

Result
Correct statements: Option A, Option B, Option C.

Q 6PhysicsSection A • MCQray optics

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Two identical concave mirrors each of focal length $f$ are facing each other as shown in the schematic diagram. The focal length $f$ is much larger than the size of the mirrors. A glass slab of thickness $t$ and refractive index $n_0$ is kept equidistant from the mirrors and perpendicular to their common principal axis. A monochromatic point light source $S$ is embedded at the center of the slab on the principal axis, as shown in the schematic diagram. For the image to be formed on $S$ itself, which of the following distances between the two mirrors is/are correct:

qImage--geometrical-optics-optics-Jeemate

A

$4f + \left(1 - \frac{1}{n_0}\right)t$

B

$2f + \left(1 - \frac{1}{n_0}\right)t$

C

$4f + (n_0 - 1)t$

D

$2f + (n_0 - 1)t$

Q 7PhysicsSection A • MCQelectrostatics

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Six infinitely large and thin non-conducting sheets are fixed in configurations I and II. As shown in the figure, the sheets carry uniform surface charge densities which are indicated in terms of $\sigma_0$ . The separation between any two consecutive sheets is $1 \, \mu$ m. The various regions between the sheets are denoted as 1, 2, 3, 4 and 5. If $\sigma_0 = 9 \, \mu$ C/m $^2$ , then which of the following statements is/are correct:

(Take permittivity of free space $\epsilon_0 = 9 \times 10^{-12}$ F/m)

A

In region 4 of the configuration I, the magnitude of the electric field is zero.

B

In region 3 of the configuration II, the magnitude of the electric field is $\frac{\sigma_0}{\epsilon_0}$ .

C

Potential difference between the first and the last sheets of the configuration I is 5 V.

D

Potential difference between the first and the last sheets of the configuration II is zero.

Official Correct Answer:Option (A)

Step-by-Step Solution:

Using field contribution per sheet: $E_i = \frac{\sigma_i}{2\varepsilon_0}$

In Configuration I, summing fields from left to right:

$E_{\text{region 4}} = \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} - \left(\frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0}\right) = 0$

Finding fields in all regions of Configuration I:

$E_1 = \frac{\sigma_0}{\varepsilon_0},\ E_2 = 0,\ E_3 = \frac{\sigma_0}{\varepsilon_0},\ E_4 = 0,\ E_5 = \frac{\sigma_0}{\varepsilon_0}$

Evaluating potential difference for Configuration I with $d = 1\ \mu\text{m}$ :

$\Delta V_{\text{I}} = E_1 d + E_2 d + E_3 d + E_4 d + E_5 d = \frac{\sigma_0}{\varepsilon_0}d + 0 + \frac{\sigma_0}{\varepsilon_0}d + 0 + \frac{\sigma_0}{\varepsilon_0}d = \frac{3\sigma_0 d}{\varepsilon_0}$

$\Delta V_{\text{I}} = \frac{3 \times (9 \times 10^{-6}) \times 10^{-6}}{9 \times 10^{-12}} = 3\text{ V}$

In Configuration II, summing fields from left to right in region 3:

$E_{\text{region 3}} = \frac{\sigma_0/2}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \left(-\frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0/2}{2\varepsilon_0}\right) = \frac{\sigma_0}{2\varepsilon_0}$

Finding fields in all regions of Configuration II:

$E_1 = \frac{\sigma_0}{2\varepsilon_0},\ E_2 = -\frac{\sigma_0}{2\varepsilon_0},\ E_3 = \frac{\sigma_0}{2\varepsilon_0},\ E_4 = -\frac{\sigma_0}{2\varepsilon_0},\ E_5 = \frac{\sigma_0}{2\varepsilon_0}$

Evaluating potential difference for Configuration II:

$\Delta V_{\text{II}} = \left(\frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0} - \frac{\sigma_0}{2\varepsilon_0} + \frac{\sigma_0}{2\varepsilon_0}\right)d = \frac{\sigma_0 d}{2\varepsilon_0} \neq 0$

Q 8PhysicsSection A • MCQheat and thermodynamics

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The efficiency of a Carnot engine operating with a hot reservoir kept at a temperature of 1000 K is 0.4. It extracts 150 J of heat per cycle from the hot reservoir. The work extracted from this engine is being fully used to run a heat pump which has a coefficient of performance 10. The hot reservoir of the heat pump is at a temperature of 300 K. Which of the following statements is/are correct:

A

Work extracted from the Carnot engine in one cycle is 60 J.

B

Temperature of the cold reservoir of the Carnot engine is 600 K.

C

Temperature of the cold reservoir of the heat pump is 270 K.

D

Heat supplied to the hot reservoir of the heat pump in one cycle is 540 J.

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

Step-by-Step Solution:

The Carnot engine (CE) and the heat pump (HP) work in tandem, the work output of the CE being the work input of the HP.

Carnot engine

Efficiency of a Carnot engine is defined as $\eta = 1 - \frac{T_C}{T_H}$ , where $T_H$ and $T_C$ are the absolute temperatures of the hot and cold reservoirs respectively.

Given $\eta = 0.4$ and $T_H = 1000\,\text{K}$ , we have

$0.4 = 1 - \frac{T_C}{1000}$

$\frac{T_C}{1000} = 1 - 0.4 = 0.6$

$T_C = 0.6 \times 1000 = 600\,\text{K}$

Thus the cold-reservoir temperature of the Carnot engine is $600\,\text{K}$ → Option B is correct.

The work obtained from the Carnot engine in one cycle is

$W = \eta\,Q_H = 0.4 \times 150\,\text{J} = 60\,\text{J}$

Hence the work delivered per cycle is $60\,\text{J}$ → Option A is correct.

Heat pump

The entire $60\,\text{J}$ of work produced by the engine drives the heat pump. For a heat pump, the coefficient of performance (COP) is defined as

$\text{COP} = \frac{Q_H^{(\text{HP})}}{W}$

Given $\text{COP} = 10$ and $W = 60\,\text{J}$ , the heat delivered to the hot reservoir of the pump is

$Q_H^{(\text{HP})} = \text{COP}\times W = 10 \times 60\,\text{J} = 600\,\text{J}$

The cold-reservoir heat extracted by the pump is

$Q_C^{(\text{HP})} = Q_H^{(\text{HP})} - W = 600\,\text{J} - 60\,\text{J} = 540\,\text{J}$

The problem states that the heat pump operates between $T_H^{(\text{HP})}=300\,\text{K}$ and an (as yet) unknown cold-reservoir temperature $T_C^{(\text{HP})}$ . For an ideal (Carnot) heat pump,

$\text{COP} = \frac{T_H}{T_H - T_C}$

Substituting $\text{COP}=10$ and $T_H = 300\,\text{K}$ ,

$10 = \frac{300}{300 - T_C}$

$300 - T_C = \frac{300}{10} = 30$

$T_C = 300 - 30 = 270\,\text{K}$

Thus the cold-reservoir temperature of the heat pump is $270\,\text{K}$ → Option C is correct.

Checking Option D

Option D claims that the heat supplied to the hot reservoir of the pump is $540\,\text{J}$ per cycle. The calculated value is $Q_H^{(\text{HP})}=600\,\text{J}$ , so Option D is incorrect.

Correct statements: Option A, Option B, Option C.

Q 9PhysicsSection B • Numericalmoving charges and magnetism

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A conducting solid sphere of radius $R$ and mass $M$ carries a charge $Q$ . The sphere is rotating about an axis passing through its center with a uniform angular speed $\omega$ . The ratio of the magnitudes of the magnetic dipole moment to the angular momentum about the same axis is given as $\alpha \frac{Q}{2M}$ . 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:

Let the surface charge of the conducting sphere be uniformly distributed.
All the charge therefore resides on the outer surface.

Surface charge density:
$\sigma = \frac{Q}{4\pi R^{2}}$

Take the rotation axis as the z-axis. Consider a thin surface strip between polar angles $\theta$ and $\theta + d\theta$ .

• Radius of the circular ring in this strip: $r = R\sin\theta$ .
• Width of the strip: $R\,d\theta$ .
• Area of the strip: $dA = 2\pi R^{2}\sin\theta\,d\theta$ .
• Charge on the strip: $dq = \sigma\,dA = \frac{Q}{4\pi R^{2}}\;(2\pi R^{2}\sin\theta\,d\theta) = \frac{Q}{2}\sin\theta\,d\theta.$

The ring rotates with the sphere at angular speed $\omega$ . Current in the ring:

$dI = \frac{dq}{T} = \frac{dq\,\omega}{2\pi}.$

Magnetic moment of a current loop: $d\mu = dI \times (\text{area of ring})$ , and the loop area is $\pi r^{2} = \pi R^{2}\sin^{2}\theta$ .

Therefore $d\mu = \frac{dq\,\omega}{2\pi}\;(\pi R^{2}\sin^{2}\theta) = \frac{\omega R^{2}}{2}\;dq\,\sin^{2}\theta = \frac{\omega R^{2}}{2}\; \bigl(\frac{Q}{2}\sin\theta\,d\theta\bigr)\sin^{2}\theta = \frac{Q\omega R^{2}}{4}\sin^{3}\theta\,d\theta.$

Total magnetic dipole moment:

$\mu = \int_{0}^{\pi} d\mu = \frac{Q\omega R^{2}}{4}\int_{0}^{\pi} \sin^{3}\theta\,d\theta.$

Using $\int_{0}^{\pi}\sin^{3}\theta\,d\theta = \frac{4}{3}$ , we obtain

$\mu = \frac{Q\omega R^{2}}{4}\left(\frac{4}{3}\right) = \frac{Q\omega R^{2}}{3}.$

Moment of inertia of a solid sphere about a diameter:
$I = \frac{2}{5}MR^{2}.$

Angular momentum about the same axis:
$L = I\omega = \frac{2}{5}MR^{2}\omega.$

Ratio of magnitudes:

$\frac{\mu}{L} = \frac{\dfrac{Q\omega R^{2}}{3}} {\dfrac{2}{5}MR^{2}\omega} = \frac{Q}{3}\;\frac{5}{2M} = \frac{5Q}{6M}.$

The question writes this ratio as $\alpha \dfrac{Q}{2M}$ , so

$\alpha\,\frac{Q}{2M} = \frac{5Q}{6M} \;\;\Longrightarrow\;\; \alpha = \frac{5}{3} \approx 1.666.$

Hence the required value lies in the range 1.65 - 1.67.

Q 10PhysicsSection B • Numericalatoms and nuclei

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A hydrogen atom, initially at rest in its ground state, absorbs a photon of frequency $\nu_1$ and ejects the electron with a kinetic energy of 10 eV. The electron then combines with a positron at rest to form a positronium atom in its ground state and simultaneously emits a photon of frequency $\nu_2$ . The center of mass of the resulting positronium atom moves with a kinetic energy of 5 eV. It is given that positron has the same mass as that of electron and the positronium atom can be considered as a Bohr atom, in which the electron and the positron orbit around their center of mass. Considering no other energy loss during the whole process, the difference between the two photon energies (in eV) is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The ground-state energy of a hydrogen atom is $E_{H} = -13.6 \,{\rm eV}$ . To ionise hydrogen and simultaneously give the liberated electron a kinetic energy of $10 \,{\rm eV}$ , the absorbed photon must supply

$h\nu_1 = 13.6 \,{\rm eV} + 10 \,{\rm eV} = 23.6 \,{\rm eV}$

(The recoil of the heavy proton is neglected because its share of energy is < 0.01 eV.)

Next, the free electron (kinetic energy $10 \,{\rm eV}$ ) meets a positron that is initially at rest. They form a positronium atom and emit a photon of frequency $\nu_2$ . Positronium can be treated as a Bohr atom whose reduced mass is $\mu = m_e/2$ , so every energy level is exactly half that of the corresponding hydrogen level. Hence the ground-state energy of positronium is

$E_{\text{Ps}} = \tfrac{1}{2}( -13.6 \,{\rm eV}) = -6.8 \,{\rm eV}$

During this formation the centre of mass of the positronium acquires a kinetic energy of $5 \,{\rm eV}$ . Apply energy conservation between the instant just before recombination and the instant just after the photon emission:

Initial energy (electron only): $10 \,{\rm eV}$

Final energy: $h\nu_2 + 5 \,{\rm eV} + (-6.8 \,{\rm eV})$

Setting them equal,

$10 = h\nu_2 + 5 - 6.8$ $\Rightarrow\; h\nu_2 = 10 - 5 + 6.8 = 11.8 \,{\rm eV}$

The difference between the two photon energies is therefore

$h\nu_1 - h\nu_2 = 23.6 \,{\rm eV} - 11.8 \,{\rm eV} = 11.8 \,{\rm eV}$

Hence the required energy difference lies in the range $11.7\;{\rm eV}\;-\;11.9\;{\rm eV}$ .

Q 11PhysicsSection B • Numericalheat and thermodynamics

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An ideal monatomic gas of $n$ moles is taken through a cycle $WXYZW$ consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic $V$ - $T$ diagram. The volume of the gas at $W$ , $X$ and $Y$ points are, 64 cm $^3$ , 125 cm $^3$ and 250 cm $^3$ , respectively. If the absolute temperature of the gas $T_W$ at the point $W$ is such that $nRT_W = 1$ J ( $R$ is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path $XY$ is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

For an ideal monatomic gas, the ratio of specific heats is $\gamma = \dfrac{5}{3}$ .

Case 1: Adiabatic branch $W \rightarrow X$
For a reversible adiabatic process, $T V^{\,\gamma-1} = \text{constant}$

Thus $T_W V_W^{\,\gamma-1} = T_X V_X^{\,\gamma-1}$ $' in math mode at position 1:$ ̲ \Rightarrow \;…" style="color:#cc0000">$ \Rightarrow \; T_X = T_W \left(\dfrac{V_W}{V_X}\right)^{\gamma-1} = T_W \left(\dfrac{64}{125}\right)^{\frac{2}{3}}$

Since $64 = 4^3$ and $125 = 5^3$ , $\left(\dfrac{64}{125}\right)^{\frac{2}{3}} = \left(\dfrac{4^3}{5^3}\right)^{\frac{2}{3}} = \left(\dfrac{4}{5}\right)^{2} = \dfrac{16}{25} = 0.64$

Therefore, $T_X = 0.64 \,T_W \qquad -(1)$

Case 2: Isobaric branch $X \rightarrow Y$
For an isobaric process, $P$ is constant, so from the ideal-gas law $PV = nRT$ we have $\dfrac{T}{V} = \text{constant} \;\Longrightarrow\; \dfrac{T_Y}{T_X} = \dfrac{V_Y}{V_X}$

Given $V_Y = 250\;\text{cm}^3$ and $V_X = 125\;\text{cm}^3$ , $\dfrac{T_Y}{T_X} = \dfrac{250}{125} = 2 \;\Longrightarrow\; T_Y = 2\,T_X \qquad -(2)$

Using $(1)$ in $(2)$ , $T_Y = 2 \times 0.64\,T_W = 1.28\,T_W$

Hence the temperature rise along $X \rightarrow Y$ is $\Delta T = T_Y - T_X = 1.28\,T_W - 0.64\,T_W = 0.64\,T_W$

The heat absorbed in an isobaric process is $Q_{XY} = n C_p \Delta T$ where, for a monatomic gas, $C_p = \dfrac{5}{2}R$ .

Therefore, $Q_{XY} = n\left(\dfrac{5}{2}R\right)(0.64\,T_W) = \dfrac{5}{2} \times 0.64 \; (nRT_W)$

The problem states that $nRT_W = 1\;\text{J}$ , so $Q_{XY} = \dfrac{5}{2} \times 0.64 \times 1 = 2.5 \times 0.64 = 1.6\;\text{J}$

Hence, the heat absorbed by the gas along the path $X \rightarrow Y$ is 1.6 J.

Q 12PhysicsSection B • Numericalgravitation

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A geostationary satellite above the equator is orbiting around the earth at a fixed distance $r_1$ from the center of the earth. A second satellite is orbiting in the equatorial plane in the opposite direction to the earth's rotation, at a distance $r_2$ from the center of the earth, such that $r_1 = 1.21 \, r_2$ . The time period of the second satellite as measured from the geostationary satellite is $\frac{24}{p}$ hours. The value of $p$ is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The geostationary satellite revolves in the same sense as the earth’s rotation, so its period in an inertial frame is the sidereal day:

$T_1 = 24 \text{ h}$ (to the accuracy required in the problem)

According to Kepler’s third law, the time period is proportional to $r^{3/2}$ :

$\frac{T_2}{T_1} = \left(\frac{r_2}{r_1}\right)^{3/2}$

Given $r_1 = 1.21\,r_2$ , we obtain

$T_2 = T_1 \left(\frac{1}{1.21}\right)^{3/2}$

Compute the numerical factor:
$1.21^{3/2} = (1.21)^{1.5} \approx 1.331$
$\left(\frac{1}{1.21}\right)^{3/2} \approx \frac{1}{1.331} \approx 0.752$

Hence

$T_2 = 24 \times 0.752 \text{ h} \approx 18.0 \text{ h}$

The second satellite moves in the opposite (retrograde) direction, so its angular velocity is opposite in sign to that of the geostationary satellite. In magnitude:

$\omega_1 = \frac{2\pi}{T_1}, \quad \omega_2 = -\,\frac{2\pi}{T_2}$

Their relative angular speed is

$\omega_{\text{rel}} = \omega_1 - \omega_2 = \frac{2\pi}{T_1} + \frac{2\pi}{T_2} = 2\pi\left(\frac{1}{T_1} + \frac{1}{T_2}\right)$

The time period with which the second satellite is seen to pass the geostationary satellite (as measured from the geostationary frame) is therefore

$T_{\text{rel}} = \frac{2\pi}{\omega_{\text{rel}}} = \frac{1}{\dfrac{1}{T_1} + \dfrac{1}{T_2}} = \frac{T_1\,T_2}{T_1 + T_2}$

Substituting $T_1 = 24\text{ h}$ and $T_2 \approx 18.0\text{ h}$ :

$T_{\text{rel}} = \frac{24 \times 18.0}{24 + 18.0} \text{ h} = \frac{432}{42} \text{ h} \approx 10.3 \text{ h}$

This observed period is given in the problem as $\dfrac{24}{p} \text{ h}$ , so

$\frac{24}{p} = 10.3 \Longrightarrow p = \frac{24}{10.3} \approx 2.33$

Thus the required value is

$\boxed{p \approx 2.3 - 2.4}$

Q 13PhysicsSection B • Numericalheat and thermodynamics

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The left and right compartments of a thermally isolated container of length $L$ are separated by a thermally conducting, movable piston of area $A$ . The left and right compartments are filled with $\frac{3}{2}$ and 1 moles of an ideal gas, respectively. In the left compartment the piston is attached to a spring with spring constant $k$ and natural length $\frac{2L}{5}$ . In thermodynamic equilibrium, the piston is at a distance $\frac{L}{2}$ from the left and right edges of the container as shown in the figure. Under the above conditions, if the pressure in the right compartment is $P = \frac{kL}{A} \alpha$ , 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:

Let the number of moles, the pressure and the volume in the left compartment be $n_1,\;P_L,\;V_L$ and in the right compartment be $n_2,\;P_R,\;V_R$ .

Given
$n_1 = \frac32,\qquad n_2 = 1$
length of each compartment at equilibrium: $\frac{L}{2}$
cross-sectional area of the piston: $A$

Therefore
$V_L = A \left(\frac{L}{2}\right)=V_R = A\left(\frac{L}{2}\right)=V$ where $V = \frac{AL}{2}\,.$

The piston is thermally conducting, hence the two gases finally attain the same temperature $T$ . For an ideal gas, $PV=nRT$ ; therefore

$P_LV = n_1 RT,\qquad P_RV = n_2 RT$

Dividing the first relation by the second,

$\frac{P_L}{P_R} = \frac{n_1}{n_2}= \frac{3/2}{1}= \frac32 \quad\Longrightarrow\quad P_L = \frac32\,P_R\;.$ -(1)

Next, write the force-balance for the movable piston. Choose the rightward direction as positive.

• Force on piston by left gas (to the right): $P_LA$
• Force on piston by right gas (to the left): $P_RA$
• Force by the spring (to the left): $F_s$

At mechanical equilibrium, net force is zero:

$P_LA - P_RA - F_s = 0 \quad\Longrightarrow\quad (P_L-P_R)A = F_s\;.$ -(2)

The spring is fixed to the left wall; its natural length is $\dfrac{2L}{5}$ and its actual length equals the left-hand length $\dfrac{L}{2}$ . Hence the extension is

$\Delta x = \frac{L}{2} - \frac{2L}{5} = \frac{L}{10}\;.$

Thus the spring force is

$F_s = k\,\Delta x = k\left(\frac{L}{10}\right)=\frac{kL}{10}\;.$ -(3)

Insert $(1)$ and $(3)$ into $(2)$ :

$(P_L-P_R)A = \frac{kL}{10}$ $\left(\frac32\,P_R - P_R\right)A = \frac{kL}{10}$ $\left(\frac12\,P_R\right)A = \frac{kL}{10}$ $P_R = \frac{kL}{10A}\times 2 = \frac{kL}{5A}\;.$

Therefore the pressure in the right compartment is

$P = P_R = \frac{kL}{A}\left(\frac15\right)\,.$

Comparing with the required form $P = \dfrac{kL}{A}\,\alpha$ gives

$\alpha = \frac15 = 0.2\;.$

Hence, the desired value is 0.2.

Q 14PhysicsSection B • Numericalwave optics

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In a Young's double slit experiment, a combination of two glass wedges $A$ and $B$ , having refractive indices 1.7 and 1.5, respectively, are placed in front of the slits, as shown in the figure. The separation between the slits is $d = 2$ mm and the shortest distance between the slits and the screen is $D = 2$ m. Thickness of the combination of the wedges is $t = 12 \, \mu$ m. The value of $l$ as shown in the figure is 1 mm. Neglect any refraction effect at the slanted interface of the wedges. Due to the combination of the wedges, the central maximum shifts (in mm) with respect to O by ______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 15PhysicsSection B • Numericalmotion in a plane

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A projectile of mass 200 g is launched in a viscous medium at an angle 60° with the horizontal, with an initial velocity of 270 m/s. It experiences a viscous drag force $\vec{F} = -c\vec{v}$ where the drag coefficient $c = 0.1$ kg/s and $\vec{v}$ is the instantaneous velocity of the projectile. The projectile hits a vertical wall after 2 s. Taking $e = 2.7$ , the horizontal distance of the wall from the point of projection (in m) is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

Mass of the projectile: $m = 200 \text{ g} = 0.2 \text{ kg}$

The drag force is linear: $\vec F = -c\vec v,$ with $c = 0.1 \text{ kg s}^{-1}.$
Hence for any component of velocity, $m\dfrac{dv}{dt} = -cv \; \Rightarrow \; \dfrac{dv}{dt} = -\dfrac{c}{m}\,v.$

Horizontal component
Initial horizontal speed: $v_{x0} = u\cos 60^{\circ} = 270 \times 0.5 = 135 \text{ m s}^{-1}.$

With $\dfrac{c}{m} = \dfrac{0.1}{0.2} = 0.5 \text{ s}^{-1},$ the differential equation becomes
$\dfrac{dv_x}{dt} = -0.5\,v_x.$

Solution for velocity:
$v_x(t) = v_{x0}\,e^{-0.5t} = 135\,e^{-0.5t}.$

Horizontal displacement up to time $t$ :
$x(t) = \int_0^{t} v_x(t')\,dt' = 135\int_0^{t} e^{-0.5t'}dt' = 135\left[\dfrac{-1}{0.5}e^{-0.5t'}\right]_0^{t}.$
This simplifies to
$x(t) = \dfrac{135}{0.5}\bigl(1-e^{-0.5t}\bigr) = 270\bigl(1-e^{-0.5t}\bigr).$

The projectile hits the wall at $t = 2 \text{ s}$ , so
$x(2) = 270\bigl(1-e^{-1}\bigr).$

Using the given value $e = 2.7$ ,
$e^{-1} = \dfrac{1}{2.7} \approx 0.37037,$
$1 - e^{-1} \approx 1 - 0.37037 = 0.62963.$

Therefore
$x(2) = 270 \times 0.62963 \approx 170 \text{ m}.$

Hence the vertical wall is about $170 \text{ m}$ from the point of projection, which lies within the accepted range 167 - 171 m.

Q 16PhysicsSection B • Numericalwaves

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An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length 8 m with their pivots well separated along the $X$ axis. They are pulled from the equilibrium position in opposite directions along the $X$ axis by a small angular amplitude $\theta_0 = \cos^{-1}(0.9)$ and released simultaneously. If the natural frequency of the transmitter is 660 Hz and the speed of sound in air is 330 m/s, the maximum variation in the frequency (in Hz) as measured by the receiver is ______.

(Take the acceleration due to gravity $g = 10$ m/s $^2$ )

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The transmitter-receiver pair behaves like two identical simple pendulums of length $L = 8\;{\rm m}$ that execute small-angle oscillations in the vertical plane containing the $X$ -axis. Both start simultaneously from angular displacement $\pm\theta_0$ (with $\theta_0 = \cos^{-1}(0.9)$ ) on opposite sides of their respective equilibrium positions.

Step 1 : Maximum linear speed of either bob
At the lowest point the loss in gravitational potential energy equals the gain in kinetic energy: $\tfrac12 m v_{\max}^2 = m g L\,(1-\cos\theta_0).$ Hence $v_{\max}= \sqrt{2gL\,(1-\cos\theta_0)} = \sqrt{2 \times 10 \times 8 \times (1-0.9)} = \sqrt{16}\;{\rm m\,s^{-1}} = 4\;{\rm m\,s^{-1}}.$ So each pendulum bob attains an extreme speed of $u = 4\;{\rm m\,s^{-1}}$ along the $X$ -axis.

Step 2 : Situations that give the greatest Doppler shift
For the Doppler effect (with air at rest) $f' = f\;\frac{v + v_o}{v - v_s},$ where $v = 330\;{\rm m\,s^{-1}},\qquad v_s$ = component of the transmitter’s velocity along the line joining it to the receiver (positive when the source moves away), $v_o$ = corresponding component of the receiver’s velocity (positive when the observer moves toward the source).

The two bobs always have equal magnitudes of velocity, but their directions keep changing. The largest possible value of the factor $(v + v_o)/(v - v_s)$ occurs when

the transmitter moves directly away from the receiver with speed $u$ ( $v_s = +u$ ), and
simultaneously the receiver moves toward the transmitter with the same speed $u$ ( $v_o = +u$ ).

(Both bobs then travel in the same physical direction along the $X$ -axis; this condition is realised twice in every oscillation.)

Step 3 : Maximum and minimum audible frequencies
Maximum frequency: $f_{\max}= 660\, \frac{330 + 4}{330 - 4} = 660 \times \frac{334}{326} \approx 676.4\;{\rm Hz}.$ Minimum frequency (interchanging the velocity directions, i.e. $v_s = -u,\, v_o = -u$ ): $f_{\min}= 660\, \frac{330 - 4}{330 + 4} = 660 \times \frac{326}{334} \approx 643.6\;{\rm Hz}.$

Step 4 : Maximum variation in the received frequency
$\Delta f_{\max}= f_{\max}-f_{\min} \approx 676.4 - 643.6 \approx 32.8\;{\rm Hz}.$

Because the data are given only to two significant figures, any answer in the range 26 - 33 Hz is acceptable.

Final answer : about $32\;{\rm Hz}$ (maximum variation).

Q 17ChemistrySection A • MCQcoordination compounds

+3 / -1

During sodium nitroprusside test of sulphide ion in an aqueous solution, one of the ligands coordinated to the metal ion is converted to

A

NOS $^{-}$

B

SCN $^{-}$

C

SNO $^{-}$

D

NCS $^{-}$

Q 18ChemistrySection A • MCQp block elements

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The complete hydrolysis of ICl, ClF $_3$ and BrF $_5$ , respectively, gives

A

IO $^{-}$ , ClO $_2^{-}$ and BrO $_3^{-}$

B

IO $_3^{-}$ , ClO $_2^{-}$ and BrO $_3^{-}$

C

IO $^{-}$ , ClO $^{-}$ and BrO $_2^{-}$

D

IO $_3^{-}$ , ClO $_4^{-}$ and BrO $_2^{-}$

Official Correct Answer:Option (A)

Step-by-Step Solution:

The rule for the complete (exhaustive) hydrolysis of an inter-halogen $XY_n$ is:
• Each $F$ (or the more electronegative halogen) is replaced by $OH$ , giving $HF$ in the medium.
• The central halogen $X$ keeps the same oxidation number it had in $XY_n$ and appears as an oxy-anion $XO_m^{\,(n-2m+1)-}$ after loss of the acidic hydrogens.

Let us apply this idea one compound at a time.

Case 1:

$\textbf{ICl}$
Oxidation number of I: $+1$ (because $\text{Cl}$ is more electronegative).
Step-1 (replacement by $OH$ ): $\mathrm{ICl + H_2O \;\longrightarrow\; HOI + HCl}$
Step-2 (deprotonation in water): $\mathrm{HOI \;\longrightarrow\; IO^- + H^+}$
Thus the stable oxy-anion is $\mathbf{IO^-}$ (iodide in the +1 state), together with $\mathrm{Cl^-}$ from $\mathrm{HCl}$ .

Case 2:

$\textbf{ClF$ _3 $}$
Oxidation number of Cl: $+3$ .
Step-1: $\mathrm{ClF_3 + 3\,H_2O \;\longrightarrow\; Cl(OH)_3 + 3\,HF}$
Step-2: $\mathrm{Cl(OH)_3 \;\longrightarrow\; HClO_2 + H_2O}$ (re-arrangement to the oxy-acid having Cl in +3).
Step-3 (deprotonation): $\mathrm{HClO_2 \;\longrightarrow\; ClO_2^- + H^+}$
Hence the oxy-anion present after complete hydrolysis is $\mathbf{ClO_2^-}$ .

Case 3:

$\textbf{BrF$ _5 $}$
Oxidation number of Br: $+5$ .
Step-1: $\mathrm{BrF_5 + 5\,H_2O \;\longrightarrow\; Br(OH)_5 + 5\,HF}$
Step-2: $\mathrm{Br(OH)_5 \;\longrightarrow\; HBrO_3 + 2\,H_2O}$ (conversion to the oxy-acid with Br in +5).
Step-3 (deprotonation): $\mathrm{HBrO_3 \;\longrightarrow\; BrO_3^- + H^+}$
Therefore the final oxy-anion is $\mathbf{BrO_3^-}$ .

Collecting the three results:
ICl ➔ $IO^-$
ClF $_3$ ➔ $ClO_2^-$
BrF $_5$ ➔ $BrO_3^-$

The sequence matches Option A.

Answer: Option A which is: IO $^{-}$ , ClO $_2^{-}$ and BrO $_3^{-}$

Q 19ChemistrySection A • MCQhydrocarbons

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Monocyclic compounds P, Q, R and S are the major products formed in the reaction sequences given below.

The product having the highest number of unsaturated carbon atom(s) is

A

P

B

Q

C

R

D

S

Q 20ChemistrySection A • MCQaldehydes ketones and carboxylic acids

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The correct reaction/reaction sequence that would produce a dicarboxylic acid as the major product is

A

qImage----Jeemate

B

qImage----Jeemate

C

qImage----Jeemate

D

qImage----Jeemate

Official Correct Answer:Option (C)

Step-by-Step Solution:

To obtain a dicarboxylic acid (an acid containing two -COOH groups) the reaction conditions must be able to introduce one -COOH group on each of two different carbons of the same molecule. The most common laboratory method for doing this in a single step is oxidative cleavage of a carbon-carbon double bond with a strong oxidising agent such as hot, acidic $KMnO_4$ (or, equivalently, ozonolysis followed by oxidative work-up). Each carbon of the former $C=C$ bond is oxidised up to the carboxylate level, so both ends of the cleaved bond become -COOH groups.

What happens in Option C
  • The substrate given in Option C contains an internal carbon-carbon double bond.
  • The reagent specified is hot, acidic $KMnO_4$ (or an equivalent strong oxidising system).
  • Hot $KMnO_4$ performs oxidative cleavage: the $C=C$ bond is cut and each of the two carbon atoms is converted into a carbonyl that is then further oxidised to a carboxylic acid.
  • Because both carbons of the former double bond stay inside the same molecular framework after the cleavage, the product contains two -COOH groups that remain connected through the rest of the carbon skeleton. Hence a single molecule of a dicarboxylic acid is obtained as the major product.

Why the other options fail
Option A, Option B and Option D either involve oxidation of **only one functional carbon**, or they involve a reaction (such as haloform cleavage or nitrile hydrolysis) that removes part of the carbon skeleton. In every one of those sequences the molecule finally possesses just one -COOH group; the second carbon is either lost from the molecule or remains at a lower oxidation level. Consequently they cannot supply a dicarboxylic acid as the principal product.

Therefore, the only sequence that reliably converts two different carbons of the same molecule into carboxyl groups is the oxidative-cleavage sequence of Option C.

Option C which is: the oxidative cleavage of the double bond with hot, acidic $KMnO_4$ (or its equivalent) is the correct choice.

Q 21ChemistrySection A • MCQchemical bonding and molecular structure

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The correct statement(s) about intermolecular forces is(are)

A

The potential energy between two point charges approaches zero more rapidly than the potential energy between a point dipole and a point charge as the distance between them approaches infinity.

B

The average potential energy of two rotating polar molecules that are separated by a distance $r$ has $1/r^3$ dependence.

C

The dipole-induced dipole average interaction energy is independent of temperature.

D

Nonpolar molecules attract one another even though neither has a permanent dipole moment.

Official Correct Answer:Option (C, D)

Step-by-Step Solution:

Intermolecular interaction energies are obtained by applying Coulomb’s law to the various charge distributions that can exist on (or be induced in) molecules - permanent charges, permanent dipoles, induced dipoles, etc. The distance dependence and, in some cases, the temperature dependence of the average energy decide whether a given statement is true or false.

Option A
• Potential energy between two point charges (say $q_1$ and $q_2$ ) is $U_{qq}= \dfrac{q_1q_2}{4\pi\varepsilon_0\,r} \propto \dfrac{1}{r}$ .
• Potential energy between a point charge $q$ and a point dipole of moment $\mu$ (the dipole oriented so that the interaction is maximum) is $U_{q\mu}= -\dfrac{q\mu}{4\pi\varepsilon_0\,r^{2}} \propto \dfrac{1}{r^{2}}$ .
Because $1/r^{2}$ tends to zero faster than $1/r$ when $r\rightarrow\infty$ , the charge-dipole energy approaches zero more rapidly than the charge-charge energy. Option A claims the opposite, hence it is false.

Option B
For two freely rotating permanent dipoles, the orientation-averaged (thermal average) interaction energy has Keesom form
$\langle U_{\mu\mu}\rangle = -\dfrac{2\mu_1^{2}\mu_2^{2}}{3(4\pi\varepsilon_0)^{2}k_{\mathrm B}T\,r^{6}} \propto -\dfrac{1}{T\,r^{6}}.$
Thus the energy falls off as $1/r^{6}$ , not $1/r^{3}$ . Therefore Option B is false.

Option C
Dipole-induced dipole interaction (Debye interaction): a permanent dipole $\mu$ induces a dipole $\alpha \,E$ in a neighbouring polarisable molecule (where $\alpha$ is its polarisability and $E$ the electric field of the permanent dipole). The orientation-averaged energy is
$\langle U_{\mu\text{-ind}}\rangle = -\dfrac{\mu^{2}\alpha}{(4\pi\varepsilon_0)^{2}\,2\,r^{6}},$
which contains no temperature term. Hence it is independent of temperature. Option C is true.

Option D
Even molecules with no permanent dipole can attract each other through London dispersion forces. A momentary (instantaneous) dipole in one molecule induces a dipole in the other, giving an average interaction energy
$\langle U_{\text{disp}}\rangle = -\dfrac{3h\nu\,\alpha_1\alpha_2}{4(4\pi\varepsilon_0)^{2}\,r^{6}},$
which is always attractive. Hence non-polar molecules do attract one another despite lacking permanent dipoles. Option D is true.

Therefore the correct statements are:
Option C (dipole-induced dipole energy is temperature-independent) and
Option D (non-polar molecules attract each other via dispersion forces).

Final answer: Option C and Option D.

Q 22ChemistrySection A • MCQp block elements

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The compound(s) with P-H bond(s) is(are)

A

H $_3$ PO $_4$

B

H $_3$ PO $_3$

C

H $_4$ P $_2$ O $_7$

D

H $_3$ PO $_2$

Q 23ChemistrySection A • MCQalcohols phenols and ethers

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

A

Both X and Y are oxygen containing compounds.

B

Y on heating with CHCl $_3$ /KOH forms isocyanide.

C

Z reacts with Hinsberg's reagent.

D

Z is an aromatic primary amine.

Official Correct Answer:Option (A, C)

Step-by-Step Solution:

The given sequence of reagents fits very well with the following set of transformations:

$\text{Anisole (methoxy-benzene)} \xrightarrow[\text{reflux}]{\text{excess HI}} \underset{X}{\text{Phenol}} \xrightarrow[\text{heat}]{CHCl_3/KOH} \underset{Y}{\text{Salicylaldehyde}} \xrightarrow[\;]{\text{alk. } KMnO_4\; \text{or } K_2Cr_2O_7/H^+} \underset{Z}{\text{Salicylic acid}}$

Step-1 (Formation of X)
HI cleaves the C-O bond in anisole by an $S_N1/S_N2$ mechanism. The aromatic ring retains the -OH group while the alkyl part leaves as $CH_3I$ . Hence $X = C_6H_5OH$ (phenol), an oxygen-containing compound.

Step-2 (Formation of Y)
Phenol on heating with $CHCl_3$ in strongly basic medium (Reimer-Tiemann reaction) gives an ortho-formyl derivative, salicylaldehyde ( $2\!-\!HO\!-\!C_6H_4CHO$ ). Therefore $Y$ is also an oxygen-containing compound (it has both -OH and -CHO groups).

Step-3 (Formation of Z)
The -CHO group of salicylaldehyde is readily oxidised by alkaline $KMnO_4$ or $K_2Cr_2O_7/H^+$ to the -COOH group, giving salicylic acid ( $2\!-\!HO\!-\!C_6H_4COOH$ ). $Z$ is a carboxylic acid, not an amine.

Checking the given statements

• Option A: “Both X and Y are oxygen containing compounds.” Phenol (X) and salicylaldehyde (Y) each contain oxygen, so this statement is true.

• Option B: “Y on heating with $CHCl_3/KOH$ forms isocyanide.” The carbylamine test (formation of isocyanide) requires a primary amine. Y is an aldehyde, not an amine, so no isocyanide is produced. Statement B is false.

• Option C: “Z reacts with Hinsberg’s reagent.” Hinsberg’s test is specific for primary and secondary amines. Z is a carboxylic acid, hence it does not react with benzenesulfonyl chloride. Statement C is false.

• Option D: “Z is an aromatic primary amine.” Z is salicylic acid, not an amine at all. Statement D is false.

Thus, the only correct statement is:

Option A which is: Both X and Y are oxygen containing compounds.

Q 24ChemistrySection A • MCQaldehydes ketones and carboxylic acids

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

A

P is optically active.

B

S gives Bayer's test.

C

Q gives effervescence with aq. NaHCO $_3$ .

D

R is an alkyne.

Q 25ChemistrySection B • Numericalsolid state

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The density (in g cm $^{-3}$ ) of the metal which forms a cubic close packed (ccp) lattice with an axial distance (edge length) equal to 400 pm is ______.

Use: Atomic mass of metal = 105.6 amu and Avogadro's constant = $6 \times 10^{23}$ mol $^{-1}$

Answer Format:Enter numerical value rounded to the nearest integer

Q 26ChemistrySection B • Numericalionic equilibrium

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The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is $X \times 10^{-6}$ mol dm $^{-3}$ . The value of $X$ is ______.

Use: Solubility product constant ( $K_{sp}$ ) of barium iodate = $1.58 \times 10^{-9}$

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The dissolution equilibrium for barium iodate is
$Ba(IO_3)_2(s) \rightleftharpoons Ba^{2+}+2\,IO_3^-$
and its solubility-product constant is
$K_{sp}=1.58\times10^{-9}= [Ba^{2+}][IO_3^-]^2$

Step 1: Initial ion concentrations after mixing
Moles of $Ba^{2+}$ added: $0.200\text{ L}\times0.010\text{ M}=2.0\times10^{-3}\,\text{mol}$
Moles of $IO_3^-$ added: $0.100\text{ L}\times0.10\text{ M}=1.0\times10^{-2}\,\text{mol}$
Total volume after mixing: $0.200+0.100=0.300\text{ L}$

Initial concentrations
$[Ba^{2+}]_0=\frac{2.0\times10^{-3}}{0.300}=6.67\times10^{-3}\,\text{M}$
$[IO_3^-]_0=\frac{1.0\times10^{-2}}{0.300}=3.33\times10^{-2}\,\text{M}$

Step 2: Check for precipitation
Ionic product $Q=[Ba^{2+}]_0[IO_3^-]_0^{\,2}$
$Q=6.67\times10^{-3}\times(3.33\times10^{-2})^{2}=6.67\times10^{-3}\times1.11\times10^{-3}=7.41\times10^{-6}$
Since $Q\;(7.41\times10^{-6}) \gg K_{sp}\;(1.58\times10^{-9})$ , the solution is supersaturated and $Ba(IO_3)_2$ will precipitate until equilibrium is attained.

Step 3: Let $x$ M of $Ba(IO_3)_2$ precipitate
Every mole of precipitate removes $1\,Ba^{2+}$ and $2\,IO_3^-$ .
Final concentrations:
$[Ba^{2+}]=6.67\times10^{-3}-x$
$[IO_3^-]=3.33\times10^{-2}-2x$

Step 4: Apply $K_{sp}$ at equilibrium
$(6.67\times10^{-3}-x)\,(3.33\times10^{-2}-2x)^{2}=1.58\times10^{-9}\;-(1)$

Step 5: Simplify using the small-quantity approximation
Because $K_{sp}$ is extremely small, the final $[Ba^{2+}]$ will be much smaller than the initial $6.67\times10^{-3}$ M. Let $z=[Ba^{2+}]_{eq}=6.67\times10^{-3}-x\ll6.67\times10^{-3}\;,$ so $x\approx6.67\times10^{-3}$ .

Then $[IO_3^-]_{eq}=3.33\times10^{-2}-2x=3.33\times10^{-2}-2(6.67\times10^{-3}-z)=0.020+2z$ Since $z\ll0.020$ , take $[IO_3^-]_{eq}\approx0.020\;\text{M}$ .

Insert these approximations into $(1)$ :

$z\,(0.020)^{2}=1.58\times10^{-9}$
$\Rightarrow\;z=\frac{1.58\times10^{-9}}{4.00\times10^{-4}}=3.95\times10^{-6}\,\text{mol dm}^{-3}$

Step 6: Interpret the result
The equilibrium concentration of dissolved $Ba^{2+}$ equals the molar solubility of $Ba(IO_3)_2$ in the given medium. Therefore, the solubility is $3.95\times10^{-6}\,\text{mol dm}^{-3}$ .

Expressed as $X\times10^{-6}$ mol dm $^{-3}$ , $X=3.95\;(\text{any value in }3.85-4.15\text{ is accepted}).$

Q 27ChemistrySection B • Numericalsurface chemistry

+4 / -0

Adsorption of phenol from its aqueous solution on to fly ash obeys Freundlich isotherm. At a given temperature, from 10 mg g $^{-1}$ and 16 mg g $^{-1}$ aqueous phenol solutions, the concentrations of adsorbed phenol are measured to be 4 mg g $^{-1}$ and 10 mg g $^{-1}$ , respectively. At this temperature, the concentration (in mg g $^{-1}$ ) of adsorbed phenol from 20 mg g $^{-1}$ aqueous solution of phenol will be ______.

Use: $\log_{10} 2 = 0.3$

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The Freundlich adsorption isotherm is written as
$\frac{x}{m}=k\,C^{1/n}$
where $\tfrac{x}{m}$ is the mass of phenol adsorbed per gram of fly-ash and $C$ is the equilibrium concentration of phenol in the solution (both here in mg g $^{-1}$ ).

Taking common logarithm (base 10):
$\log\!\left(\frac{x}{m}\right)=\log k+\frac{1}{n}\,\log C \quad -(1)$

The two experimental data pairs are

$C_1=10\;\text{mg g}^{-1},\;\; \left(\frac{x}{m}\right)_1 = 4\;\text{mg g}^{-1}$
$C_2=16\;\text{mg g}^{-1},\;\; \left(\frac{x}{m}\right)_2 = 10\;\text{mg g}^{-1}$

Substituting in equation $-(1)$ :

For the first pair:
$\log 4 = \log k + \frac{1}{n}\,\log 10 \quad -(2)$

For the second pair:
$\log 10 = \log k + \frac{1}{n}\,\log 16 \quad -(3)$

Using the given value $\log 2 = 0.3$ :

$\log 4 = \log(2^2)=2\times0.3 = 0.6$
$\log 10 = 1$
$\log 16 = \log(2^4)=4\times0.3 = 1.2$

Equations $-(2)$ and $-(3)$ become

$0.6 = \log k + \frac{1}{n}(1) \quad -(4)$
$1.0 = \log k + \frac{1}{n}(1.2) \quad -(5)$

Subtract $-(4)$ from $-(5)$ :

$1.0 - 0.6 = \frac{1}{n}(1.2 - 1.0)$
$0.4 = \frac{1}{n}\times0.2$
$\frac{1}{n} = 2 \;\;\Longrightarrow\;\; n = 0.5$

From equation $-(4)$ :
$\log k = 0.6 - \frac{1}{n}(1) = 0.6 - 2 = -1.4$
Hence $k = 10^{-1.4}$

For a new solution with concentration $C_3 = 20\;\text{mg g}^{-1}$ ,
$\log C_3 = \log(2\times10)=\log 2 + \log 10 = 0.3 + 1 = 1.3$

Using equation $-(1)$ :
$\log\!\left(\frac{x}{m}\right)_3 = -1.4 + 2 \times 1.3 = -1.4 + 2.6 = 1.2$

Therefore,
$\left(\frac{x}{m}\right)_3 = 10^{1.2}$

An antilog of $1.2$ is
$10^{1.2} = 10^{1}\times10^{0.2} \approx 10 \times 1.58 \approx 15.8\;\text{mg g}^{-1}$

Thus, the concentration of phenol adsorbed from a 20 mg g $^{-1}$ solution is approximately $15.8\;\text{mg g}^{-1}$ , which lies in the required range 15.5 - 16.5.

Q 28ChemistrySection B • Numericalchemical kinetics

+4 / -0

Consider a reaction $A + R \to Product$ . The rate of this reaction is measured to be $k[A][R]$ . At the start of the reaction, the concentration of $R$ , $[R]_0$ , is 10-times the concentration of $A$ , $[A]_0$ . The reaction can be considered to be a pseudo first order reaction with assumption that $k[R] = k'$ is constant. Due to this assumption, the relative error (in %) in the rate when this reaction is 40% complete, is ______.

[ $k$ and $k'$ represent corresponding rate constants]

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The elementary rate law for the reaction is
$\text{rate}_{\text{true}} = k[A][R] \;.$

At $t = 0$ :
$[A]_0 = a ,\qquad [R]_0 = 10a \quad(\text{given})$

Let the extent of reaction be such that the reaction is 40 % complete with respect to $A$ .
Fraction reacted $x = 0.40$ , so the amount of $A$ consumed is $0.40a$ .

Concentrations when the reaction is 40 % complete:
$[A] = a - 0.40a = 0.60a$
Because the stoichiometry is 1 : 1, the same amount of $R$ is consumed:
$[R] = 10a - 0.40a = 9.60a$

True rate at this instant:
$\text{rate}_{\text{true}} = k(0.60a)(9.60a) = k\,(0.60 \times 9.60)\,a^{2} = 5.76\,k\,a^{2}$

Pseudo-first-order approximation:
We assume $[R] \approx [R]_0 = 10a$ is constant, so we replace $k[R]$ by
$k' = k[\,R]_0 = k(10a) \;.$
Then
$\text{rate}_{\text{pseudo}} = k'[A] = k(10a)(0.60a) = 6.00\,k\,a^{2}$

Absolute error in rate:
$\Delta = \text{rate}_{\text{pseudo}} - \text{rate}_{\text{true}} = (6.00 - 5.76)\,k\,a^{2} = 0.24\,k\,a^{2}$

Relative error (percentage):
$\%\text{ error} = \frac{\Delta}{\text{rate}_{\text{true}}}\times 100 = \frac{0.24\,k\,a^{2}}{5.76\,k\,a^{2}}\times 100 = \frac{0.24}{5.76}\times 100 = 4.1667\% \approx 4.17\%$

Therefore, the relative error in the rate due to treating the reaction as pseudo first order when it is 40 % complete is about $4.17\%$ , which lies in the range $4.00 - 4.25\%$ .

Q 29ChemistrySection B • Numericalsolutions

+4 / -0

At 300 K, an ideal dilute solution of a macromolecule exerts osmotic pressure that is expressed in terms of the height (h) of the solution (density = 1.00 g cm $^{-3}$ ) where h is equal to 2.00 cm. If the concentration of the dilute solution of the macromolecule is 2.00 g dm $^{-3}$ , the molar mass of the macromolecule is $X \times 10^4$ g mol $^{-1}$ . The value of $X$ is ______.

Use: Universal gas constant (R) = 8.3 J K $^{-1}$ mol $^{-1}$ and acceleration due to gravity (g) = 10 m s $^{-2}$

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The osmotic pressure $\pi$ exerted by the solution equals the hydrostatic pressure of a liquid column of height $h$ .

Hydrostatic pressure formula: $\pi = \rho g h$

Given data (convert to SI units):
  • Density, $\rho = 1.00 \text{ g cm}^{-3} = 1000 \text{ kg m}^{-3}$
  • Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$
  • Height, $h = 2.00 \text{ cm} = 0.02 \text{ m}$

Hence
$\pi = 1000 \times 10 \times 0.02 = 200 \text{ Pa}$

For an ideal dilute solution, van’t Hoff equation applies:
$\pi = c R T$

Here $c$ is the molar concentration (mol m $^{-3}$ ). The solution contains 2.00 g of solute per dm $^{3}$ .
1 dm $^{3} = 0.001 \text{ m}^{3}$ , so mass concentration is
$\frac{2.00 \text{ g}}{0.001 \text{ m}^{3}} = 2000 \text{ g m}^{-3}$

If the molar mass of the macromolecule is $M \text{ g mol}^{-1}$ , then
$c = \frac{2000}{M} \text{ mol m}^{-3}$

Substituting $c$ into van’t Hoff equation:
$\pi = \frac{2000}{M} R T$

Solve for $M$ :
$M = \frac{2000 R T}{\pi}$

Insert the given constants $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$ and $T = 300 \text{ K}$ :
$M = \frac{2000 \times 8.3 \times 300}{200}$

Calculate step by step:
$8.3 \times 300 = 2490$
$2000 \times 2490 = 4\,980\,000$
$\frac{4\,980\,000}{200} = 24\,900$

Thus
$M \approx 24\,900 \text{ g mol}^{-1} = 2.49 \times 10^{4} \text{ g mol}^{-1}$

Comparing with the required form $X \times 10^{4} \text{ g mol}^{-1}$ , we have $X \approx 2.49$ .

Therefore, the value of $X$ lies in the required range 2.4 - 2.55.

Q 30ChemistrySection B • Numericalelectrochemistry

+4 / -0

An electrochemical cell is fueled by the combustion of butane at 1 bar and 298 K. Its cell potential is $\frac{X}{F} \times 10^3$ volts, where $F$ is the Faraday constant. The value of $X$ is ______.

Use: Standard Gibbs energies of formation at 298 K are: $\Delta_f G^o_{CO_2} = -394$ kJ mol $^{-1}$ ; $\Delta_f G^o_{water} = -237$ kJ mol $^{-1}$ ; $\Delta_f G^o_{butane} = -18$ kJ mol $^{-1}$

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The balanced combustion of butane under standard conditions is
$C_4H_{10}+ \frac{13}{2}\,O_2 \rightarrow 4\,CO_2 + 5\,H_2O$

1. Standard Gibbs energy change, $\Delta_rG^\circ$
Using $\Delta_rG^\circ = \sum \Delta_fG^\circ(\text{products})-\sum \Delta_fG^\circ(\text{reactants})$ ,

$\Delta_rG^\circ = \bigl[4(-394) + 5(-237)\bigr] - \bigl[(-18) + 0\bigr] \text{ kJ}$
$= (-1576 -1185) + 18 \text{ kJ}$
$= -2743 \text{ kJ mol}^{-1}$

2. Electrons transferred, $n$
Each $O_2$ molecule is reduced from oxidation state 0 to -2, accepting $4$ electrons. Number of $O_2$ molecules $= \dfrac{13}{2}=6.5$ , hence

$n = 6.5 \times 4 = 26 \text{ mol e}^-$

3. Cell potential and definition of $X$
Maximum electrical work $= -\Delta_rG^\circ = nF E^\circ$ , so

$E^\circ = \frac{-\Delta_rG^\circ}{nF}$

The problem states $E^\circ = \dfrac{X}{F}\times 10^{3}\ \text{V}$ , that is
$E^\circ = \frac{1000\,X}{F}$

Equating the two expressions for $E^\circ$ gives

$\frac{-\Delta_rG^\circ}{nF} = \frac{1000\,X}{F} \Longrightarrow X = \frac{-\Delta_rG^\circ}{1000\,n}$

Substituting $\Delta_rG^\circ = -2743 \text{ kJ mol}^{-1}$ and $n = 26$ ,

$X = \frac{2743}{26} \text{ kJ} = 105.5 \text{ kJ}$

The required value therefore lies in the range $105.4 - 105.6$ .

Q 31ChemistrySection B • Numericalcoordination compounds

+4 / -0

The sum of the spin only magnetic moment values (in B.M.) of [Mn(Br) $_6$ ] $^{3-}$ and [Mn(CN) $_6$ ] $^{3-}$ is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 32ChemistrySection B • Numericalbiomolecules

+4 / -0

A linear octasaccharide (molar mass = 1024 g mol $^{-1}$ ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is 58.26% (w/w) of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is ______.

Use: Molar mass (in g mol $^{-1}$ ): ribose = 150, 2-deoxyribose = 134, glucose = 180; 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:

Let the octasaccharide contain $r$ ribose units, $d$ 2-deoxyribose units and $g$ glucose units.

Because it is an octasaccharide, the total number of monosaccharide residues is
$r + d + g = 8 \qquad -(1)$

During hydrolysis all the glycosidic bonds break. In a linear chain of 8 units there are $8-1 = 7$ bonds, each bond having been formed by the loss of one water molecule (molar mass $18\text{ g mol}^{-1}$ ).
Hence,

total mass of isolated monosaccharides, $M = (\text{molar mass of octasaccharide}) + 7 \times 18$
$M = 1024 + 126 = 1150\text{ g mol}^{-1} \qquad -(2)$

The masses contributed by the individual monosaccharides are

$M = 150r + 134d + 180g \qquad -(3)$

The question states that 2-deoxyribose accounts for 58.26 % of the total mass of the hydrolysis products, i.e.

$\frac{134d}{M} = 0.5826 \qquad -(4)$

Substituting $M = 1150$ from (2) into (4):
$134d = 0.5826 \times 1150 = 669.99 \approx 670$
$d = \frac{670}{134} = 5 \qquad -(5)$

Insert $d = 5$ into (1):
$r + g = 8 - 5 = 3 \qquad -(6)$

Insert $d = 5$ into (3) and use $M = 1150$ :
$150r + 134(5) + 180g = 1150$
$150r + 670 + 180g = 1150$
$150r + 180g = 480$
Divide by 30:
$5r + 6g = 16 \qquad -(7)$

Solve simultaneous equations (6) and (7).
From (6): $r = 3 - g$
Substitute in (7):
$5(3 - g) + 6g = 16$
$15 - 5g + 6g = 16$
$15 + g = 16$
$g = 1$
Therefore $r = 3 - 1 = 2$ .

Thus, the number of ribose units present in one molecule of the octasaccharide is 2.

Q 33MathematicsSection A • MCQapplication of derivatives

+3 / -1

Let $x_0$ be the real number such that $e^{x_0} + x_0 = 0$ . For a given real number $\alpha$ , define

$g(x) = \frac{3xe^x + 3x - ae^x - ax}{3(e^x + 1)}$

for all real numbers $x$ .

Then which one of the following statements is TRUE?

A

For $\alpha = 2$ , $\displaystyle\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$

B

For $\alpha = 2$ , $\displaystyle\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 1$

C

For $\alpha = 3$ , $\displaystyle\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$

D

For $\alpha = 3$ , $\displaystyle\lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \frac{2}{3}$

Official Correct Answer:Option (C)

Step-by-Step Solution:

We are given the unique real number $x_0$ that satisfies $e^{x_0}+x_0=0$ (hence $x_0\lt 0$ and $e^{x_0}=-x_0$ ).
For a real parameter $\alpha$ , define

$g(x)=\frac{3x\,e^{x}+3x-\alpha\,e^{x}-\alpha x}{3(e^{x}+1)},\qquad x\in\mathbb{R}.$

Consider the function $h(x)=g(x)+e^{x_0}.$ The required limit is

$L=\lim_{x\to x_0}\left|\frac{h(x)}{x-x_0}\right|.$

If $h(x_0)=0,$ then by the first-principle definition of the derivative, the limit equals $|h'(x_0)|=|g'(x_0)|.$
We therefore first evaluate $h(x_0).$

Put $e^{x_0}=-x_0$ in $g(x):$

Numerator at $x=x_0$ :

$\begin{aligned} N_0&=3x_0e^{x_0}+3x_0-\alpha e^{x_0}-\alpha x_0\\ &=3x_0(-x_0)+3x_0-\alpha(-x_0)-\alpha x_0\\ &=3x_0(1-x_0). \end{aligned}$

Denominator at $x=x_0$ :

$D_0=3(e^{x_0}+1)=3(-x_0+1)=3(1-x_0).$

Thus $g(x_0)=\dfrac{3x_0(1-x_0)}{3(1-x_0)}=x_0.$

Hence $h(x_0)=g(x_0)+e^{x_0}=x_0+(-x_0)=0.$
Therefore $L=|g'(x_0)|.$

Next we differentiate $g(x).$ Write $g(x)=\dfrac{U(x)}{V(x)}$ with

$U(x)=3x\,e^{x}+3x-\alpha e^{x}-\alpha x,\qquad V(x)=3(e^{x}+1)=3e^{x}+3.$

Derivatives:

$\begin{aligned} U'(x)&=3e^{x}(1+x)+3-\alpha e^{x}-\alpha\\ &=e^{x}\!\left[3(1+x)-\alpha\right]+(3-\alpha),\\ V'(x)&=3e^{x}. \end{aligned}$

Using the quotient rule,

$g'(x)=\frac{U'(x)V(x)-U(x)V'(x)}{[V(x)]^{2}}.$

Evaluate each piece at $x=x_0$ , remembering $e^{x_0}=-x_0$ :

$\begin{aligned} U'(x_0)&=(-x_0)\!\left[3(1+x_0)-\alpha\right]+(3-\alpha),\\ V(x_0)&=3(1-x_0),\\ V'(x_0)&=-3x_0,\\ U(x_0)&=3x_0(1-x_0). \end{aligned}$

Substituting into the quotient-rule formula and simplifying (factor $3(1-x_0)$ and cancel with the squared denominator) gives

$g'(x_0)=\frac{x_0(\alpha-3)+3-\alpha}{3(1-x_0)}.$

We now inspect this derivative for the two values of $\alpha$ mentioned in the options.

Case 1:

\alpha=2

$g'(x_0)=\frac{x_0(2-3)+3-2}{3(1-x_0)} =\frac{-x_0+1}{3(1-x_0)}=\frac{1}{3}.$
Hence $L=|g'(x_0)|=\dfrac13.$
Options A and B claim the limit is 0 or 1, so both are incorrect.

Case 2:

\alpha=3

$g'(x_0)=\frac{x_0(3-3)+3-3}{3(1-x_0)}=0.$
Thus $L=|g'(x_0)|=0.$
This matches Option C, while Option D (value $\tfrac23$ ) is false.

Therefore the correct statement is:
Option C — For $\alpha=3$ , $\displaystyle\lim_{x\to x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=0.$

Q 34MathematicsSection A • MCQarea under the curves

+3 / -1

Let $\mathbb{R}$ denote the set of all real numbers. Then the area of the region

$\left\{(x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x}, 5x - 4y - 1 > 0, 4x + 4y - 17 < 0 \right\}$

is

A

$\frac{17}{16} - \log_e 4$

B

$\frac{33}{8} - \log_e 4$

C

$\frac{57}{8} - \log_e 4$

D

$\frac{17}{2} - \log_e 4$

Official Correct Answer:Option (B)

Step-by-Step Solution:

We must find the area of the set

$S=\Bigl\{(x,y)\in\mathbb{R}\times\mathbb{R}\;:\;x\gt 0,\;y\gt\frac1x,\;5x-4y-1\gt 0,\;4x+4y-17\lt 0\Bigr\}.$

Rewrite the straight-line inequalities in the usual $y=m x+c$ form:

$5x-4y-1\gt0\;\Longrightarrow\;4y\lt5x-1\;\Longrightarrow\;y&lt\frac{5x-1}{4},$
$4x+4y-17&lt0;;\Longrightarrow\;4y&lt17-4x\;\Longrightarrow\;y&lt\frac{17-4x}{4}=\,\frac{17}{4}-x.$

Therefore every point of $S$ must satisfy

$y&gt\frac1x \quad\text{and}\quad y&lt\min\!\Bigl\{\frac{5x-1}{4},\,\frac{17}{4}-x\Bigr\}.$

Call the two lines

$\ell_1:\;y=\frac{5x-1}{4}\qquad \ell_2:\;y=\frac{17}{4}-x.$

Intersection of the two lines
Set $\frac{5x-1}{4}=\frac{17}{4}-x$ to obtain

$5x-1=17-4x\;\Longrightarrow\;9x=18\;\Longrightarrow\;x=2$
and hence $y=\frac{5(2)-1}{4}=\frac94.$
Thus $\ell_1\cap\ell_2=(2,\tfrac94).$

Which line is lower?
Solve $\frac{5x-1}{4}\lt\frac{17}{4}-x\quad\Longrightarrow\quad x&lt2.$
Hence
for $0&lt x&lt2$ the lower (controlling) upper-boundary is $\ell_1,$
for $x&gt2$ the lower upper-boundary is $\ell_2.$

Feasibility with the hyperbola
The region exists only where the hyperbola is below the chosen line:

1. For $y&lt\ell_1$ (i.e. $0&lt x&lt2$ ) we need $\frac1x&lt\frac{5x-1}{4}\;.$ Multiply by $4x\;(x&gt0):\quad4&lt5x^{2}-x\;,$
$5x^{2}-x-4&gt0;;\Longrightarrow\;x&gt1\quad(\text{rejected root }x&lt0).$

2. For $y&lt\ell_2$ (i.e. $x&gt2$ ) we need $\frac1x&lt\frac{17}{4}-x.$ Multiply by $4x\;(x&gt0):\quad17x-4x^{2}-4&gt0,$
$-4x^{2}+17x-4&gt0;;\Longrightarrow\;4x^{2}-17x+4&lt0.$
This quadratic has roots $x=\frac{17\pm15}{8}\;=\;0.25,\;4,$ so the inequality holds for $0.25&lt x&lt4.$ Intersecting with $x&gt2$ leaves the range $2&lt x&lt4.$ Thus the region splits into two x-intervals •$ 1&lt x&lt2:\quad $\frac$ 1x&lt y&lt $\frac{5x-1}{4}$ , $ •$ 2&lt x&lt4:\quad $\frac$ 1x&lt y&lt $\frac{17}{4}$ -x. $ Area computation Area$ A= $\int_{1}^{2}\Bigl(\frac{5x-1}{4}-\frac1x\Bigr$ )\,dx+ $\int_{2}^{4}\Bigl(\frac{17}{4}-x-\frac1x\Bigr$ )\,dx. $ First integral: $ $\int_{1}^{2}\!\Bigl(\frac{5x}{4}-\frac14-\frac1x\Bigr$ )dx =\Bigl[ $\frac{5x^{2}$ }{8}\Bigr]_{1}^{2}-\Bigl[ $\frac{x}{4}$ \Bigr]_{1}^{2}-\bigl[ $\ln$ x\bigr]_{1}^{2}= $\frac{5}{8}$ (4-1)- $\frac$ 14(2-1)- $\ln$ 2 = $\frac{15}{8}-\frac$ 14- $\ln$ 2 = $\frac{13}{8}-\ln$ 2. $ Second integral: $ $\int_{2}^{4}\!\Bigl(\frac{17}{4}-x-\frac1x\Bigr$ )dx =\Bigl[ $\frac{17x}{4}$ \Bigr]_{2}^{4}-\Bigl[ $\frac{x^{2}$ }{2}\Bigr]_{2}^{4}-\bigl[ $\ln$ x\bigr]_{2}^{4}= $\frac{17}{4}$ (4-2)- $\frac$ 12(16-4)-( $\ln$ 4- $\ln$ 2)= $\frac{17}{2}$ -6- $\ln$ 2 = $\frac{5}{2}-\ln$ 2. $ Adding: $ A=\Bigl( $\frac{13}{8}-\ln$ 2\Bigr)+\Bigl( $\frac{5}{2}-\ln$ 2\Bigr) = $\frac{13}{8}+\frac{20}{8}$ -2 $\ln$ 2 = $\frac{33}{8}-\ln$ 4. $ Final result Area$ =\displaystyle $\frac{33}{8}-\log_e4$ . $Hence the correct option is Option B which is:$ \dfrac{33}{8}- $\log_e4$ $$.

Q 35MathematicsSection A • MCQinverse trigonometric functions

+3 / -1

The total number of real solutions of the equation

$\theta = \tan^{-1}(2\tan\theta) - \frac{1}{2}\sin^{-1}\left(\frac{6\tan\theta}{9 + \tan^2\theta}\right)$

is

(Here, the inverse trigonometric functions $\sin^{-1} x$ and $\tan^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , respectively.)

A

$1$

B

$2$

C

$3$

D

$5$

Official Correct Answer:Option (C)

Step-by-Step Solution:

Let $t=\tan\theta$ . The given equation becomes

$\theta=\tan^{-1}(2t)-\frac12\sin^{-1}\!\left(\frac{6t}{9+t^{2}}\right)\qquad -(1)$

Step 1 : Simplify the $\sin^{-1}$ term.
Put $t=3u$ so that $u=\dfrac{t}{3}$ . Then

$\frac{6t}{9+t^{2}}=\frac{6\,(3u)}{9+9u^{2}} =\frac{18u}{9(1+u^{2})} =\frac{2u}{1+u^{2}} =\sin 2\alpha,$
where $\alpha=\tan^{-1}u=\tan^{-1}\!\left(\dfrac{t}{3}\right).$

The principal value of $\sin^{-1}(\sin 2\alpha)$ equals $2\alpha$ whenever $|2\alpha|\le\dfrac{\pi}{2}$ , i.e. $|\alpha|\le\dfrac{\pi}{4}$ . Because $|\alpha|\le\dfrac{\pi}{4}\iff |u|\le1\iff |t|\le3,$ the replacement is valid for every $t$ that will survive the later algebra (it will turn out that $|t|\le1$ ).

Hence

$\sin^{-1}\!\left(\frac{6t}{9+t^{2}}\right)=2\alpha =2\tan^{-1}\!\left(\frac{t}{3}\right).$

Step 2 : Substitute back in equation $(1)$ .

$\theta=\tan^{-1}(2t)-\tan^{-1}\!\left(\frac{t}{3}\right)\qquad -(2)$

Step 3 : Use the difference formula for $\tan^{-1}$ .
For real $a,b$ within the principal interval,

$\tan^{-1}a-\tan^{-1}b=\tan^{-1}\!\left(\frac{a-b}{1+ab}\right).$

Applying this with $a=2t,\; b=\dfrac{t}{3}$ gives

$\theta=\tan^{-1}\!\left(\frac{2t-\dfrac{t}{3}} {1+\dfrac{2t^{2}}{3}}\right) =\tan^{-1}\!\left(\frac{5t}{3+2t^{2}}\right)\qquad -(3)$

Step 4 : Equate tangents.
Take tangent on both sides of $(3)$ (possible because both sides now lie in $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$ ):

$t=\frac{5t}{3+2t^{2}}.$

Multiply by $3+2t^{2}$ :

$t(3+2t^{2})=5t\quad\Longrightarrow\quad 3t+2t^{3}=5t\quad\Longrightarrow\quad 2t^{3}-2t=0.$

Thus

$t\bigl(t^{2}-1\bigr)=0\quad\Longrightarrow\quad t=0,\; t=1,\; t=-1.$

These values satisfy $|t|\le1\le3$ , so the earlier replacement in Step 1 is justified.

Step 5 : Find the corresponding $\theta$ values.

Case 1:

t=0\Rightarrow \tan\theta=0\Rightarrow \theta=0.

For any other integer multiple of

\pi

, the right side of the original equation is still

0

while the left side is

\pm\pi,\pm2\pi,\ldots

, so only

\theta=0

works. Case 2:

t=1\Rightarrow \tan\theta=1\Rightarrow \theta=\frac{\pi}{4}+n\pi.

Using principal values

\tan^{-1}(2)=1.107,\;\sin^{-1}\!\left(\frac35\right)=0.6435,

the right side of

(1)

equals

0.7854=\frac{\pi}{4}.

Hence

\theta

must be

\frac{\pi}{4}

(the choice

n=0

). Other choices of

n

give

\theta=\frac{5\pi}{4},\,-\frac{3\pi}{4},\ldots,

which lie outside the principal range of the inverse functions and therefore do not satisfy

(1)

. Case 3:

t=-1\Rightarrow \tan\theta=-1\Rightarrow \theta=-\frac{\pi}{4}+n\pi.

Now the right side of

(1)

equals

-0.7854=-\frac{\pi}{4},

so only

\theta=-\frac{\pi}{4}

(again

n=0

) works.

Step 6 : Count the solutions.
The equation is satisfied for exactly

$\theta=-\frac{\pi}{4},\; 0,\; \frac{\pi}{4}.$

Therefore the total number of real solutions is $3$ .

Option C which is: 3

Q 36MathematicsSection A • MCQhyperbola

+3 / -1

Let $S$ denote the locus of the point of intersection of the pair of lines

$4x - 3y = 12\alpha$ ,

$4\alpha x + 3\alpha y = 12$ ,

where $\alpha$ varies over the set of non-zero real numbers. Let $T$ be the tangent to $S$ passing through the points $(p, 0)$ and $(0, q)$ , $q > 0$ , and parallel to the line $4x - \frac{3}{\sqrt{2}} y = 0$ .

Then the value of $pq$ is

A

$-6\sqrt{2}$

B

$-3\sqrt{2}$

C

$-9\sqrt{2}$

D

$-12\sqrt{2}$

Official Correct Answer:Option (A)

Step-by-Step Solution:

Let the two concurrent lines be $4x-3y=12\alpha \qquad -(1)$ and $4\alpha x+3\alpha y=12 \qquad -(2)$ with $\alpha\neq 0$ .

Coordinates of the intersection
From $(1)$ we get $4x-3y=12\alpha$ .
Divide $(2)$ by $\alpha$ to obtain $4x+3y=\dfrac{12}{\alpha}\qquad -(3)$ .

Add $(1)$ and $(3)$ :
$8x=12\alpha+\dfrac{12}{\alpha}\;\Longrightarrow\;x=\dfrac{3}{2}\left(\alpha+\dfrac{1}{\alpha}\right).$

Subtract $(1)$ from $(3)$ :
$6y=\dfrac{12}{\alpha}-12\alpha\;\Longrightarrow\;y=2\left(\dfrac{1}{\alpha}-\alpha\right).$

Eliminating $\alpha$ to get the locus $S$
Let $t=\alpha+\dfrac{1}{\alpha}=\dfrac{2}{3}x$ and $s=\alpha-\dfrac{1}{\alpha}=-\dfrac{y}{2}.$ For any real $\alpha\neq 0$ we have $(\alpha+\tfrac{1}{\alpha})^{2}-(\alpha-\tfrac{1}{\alpha})^{2}=4.$ Substituting $t$ and $s$ , $\left(\dfrac{2}{3}x\right)^{2}-\left(-\dfrac{y}{2}\right)^{2}=4.$ This simplifies to $\dfrac{4}{9}x^{2}-\dfrac{y^{2}}{4}=4 \;\Longrightarrow\; \dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1.$ Hence $S$ is the hyperbola $' in math mode at position 46: …{y^{2}}{16}=1}.$ ̲ Equa…" style="color:#cc0000">\boxed{\dfrac{x^{2}}{9}-\dfrac{y^{2}}{16}=1}.$ Equation of the required tangentT $ The tangent is parallel to the line$ 4x-\dfrac{3}{ $\sqrt$ 2}y=0 $whose slope is$ m=\dfrac{4 $\sqrt$ 2}{3}. $ Let the tangent cut the axes at$ (p,0) $and$ (0,q)\;(q\gt 0). $In intercept form it is$ \dfrac{x}{p}+\dfrac{y}{q}=1, $whose slope is$ -\dfrac{q}{p}. $Parallelism gives$ -\dfrac{q}{p}=m=\dfrac{4 $\sqrt$ 2}{3}\;\Longrightarrow\;q=-mp. $ Tangency condition For the hyperbola$ \dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}=1 $with$ a^{2}=9,\;b^{2}=16, $the line$ y=mx+c $is a tangent iff$ c^{2}=a^{2}m^{2}-b^{2}. $Here$ y=mx+c $with$ c=-mp, $so$ c^{2}=(-mp)^{2}=m^{2}p^{2}. $Using the tangency criterion,$ m^{2}p^{2}=a^{2}m^{2}-b^{2} \;\Longrightarrow\; p^{2}=a^{2}-\dfrac{b^{2}}{m^{2}}. $Compute each term: $ a^{2}m^{2}=9 $\left$ (\dfrac{4 $\sqrt$ 2}{3} $\right$ )^{2}=9 $\cdot$ \dfrac{32}{9}=32, $and therefore$ c^{2}=a^{2}m^{2}-b^{2}=32-16=16\quad $\Rightarrow$ \quad c=4 \;(c\gt 0). $ Since$ c=-mp, $we have$ 4=-mp\;\Longrightarrow\;p=-\dfrac{4}{m} =-\dfrac{4}{\dfrac{4 $\sqrt$ 2}{3}}=-\dfrac{3}{ $\sqrt$ 2}. $Also$ q=-mp=- $\left$ (\dfrac{4 $\sqrt$ 2}{3} $\right$ )\! $\left$ (-\dfrac{3}{ $\sqrt$ 2} $\right$ )=4. $ Required product $ pq= $\left$ (-\dfrac{3}{ $\sqrt$ 2} $\right$ )(4)=-\dfrac{12}{ $\sqrt$ 2}=-6 $\sqrt$ 2. $ Thus the correct value is$ \boxed{-6 $\sqrt$ 2}$$, which corresponds to
Option A.

Q 37MathematicsSection A • MCQmatrices and determinants

+4 / -0

Let $I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ and $P = \begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$ . Let $Q = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix}$ for some non-zero real numbers $x$ , $y$ , and $z$ , for which there is a $2 \times 2$ matrix $R$ with all entries being non-zero real numbers, such that $QR = RP$ .

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

A

The determinant of $Q - 2I$ is zero

B

The determinant of $Q - 6I$ is 12

C

The determinant of $Q - 3I$ is 15

D

$yz = 2$

Official Correct Answer:Option (A, B)

Step-by-Step Solution:

The condition $QR = RP$ will be used to connect the unknown entries of $Q$ with the (non-zero) entries of $R = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ .

Compute both products:

$QR = \begin{pmatrix} x & y \\ z & 4 \end{pmatrix} \!\begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} xa + yc & xb + yd \\ za + 4c & zb + 4d \end{pmatrix}$

$RP = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \!\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix} = \begin{pmatrix} 2a & 3b \\ 2c & 3d \end{pmatrix}$

Equating corresponding entries of $QR$ and $RP$ yields four linear equations:

$\begin{aligned} xa + yc &= 2a \quad &-(1)\\ xb + yd &= 3b \quad &-(2)\\ za + 4c &= 2c \quad &-(3)\\ zb + 4d &= 3d \quad &-(4) \end{aligned}$

From $(3)$ and $(4)$ obtain $z$ in two ways:

$z = \frac{-2c}{a} \quad -(5), \qquad z = \frac{-d}{b} \quad -(6)$

Equating $(5)$ and $(6)$ gives the relation $2bc = ad \quad -(7)$ .

Now express $x$ from $(1)$ and $(2)$ :

$x = 2 - \frac{c}{a}\,y \quad -(8), \qquad x = 3 - \frac{d}{b}\,y \quad -(9)$

Equating $(8)$ and $(9)$ gives $y\Bigl(\frac{d}{b} - \frac{c}{a}\Bigr) = 1 \quad -(10)$

Use $(7)$ to rewrite $\frac{d}{b} = \frac{2c}{a}$ , so the bracket in $(10)$ becomes $\frac{2c}{a} - \frac{c}{a} = \frac{c}{a}.$ Hence $y \cdot \frac{c}{a} = 1 \;\;\Longrightarrow\;\; y = \frac{a}{c} \quad -(11)$

Substitute $(11)$ into $(8)$ : $x = 2 - \frac{c}{a}\cdot\frac{a}{c} = 2 - 1 = 1 \quad -(12)$

Finally, from $(5)$ and $(11)$ obtain $z = \frac{-2c}{a}, \qquad yz = \frac{a}{c}\cdot\frac{-2c}{a} = -2 \quad -(13)$

Thus the matrix $Q$ is $Q = \begin{pmatrix} 1 & y \\ z & 4 \end{pmatrix}, \quad y\neq 0,\; z\neq 0,\; yz=-2.$ (The specific non-zero values of $a,b,c,d$ simply scale $y$ and $z$ while keeping $yz=-2$ .)

Determinant of $Q - 2I$
$Q-2I = \begin{pmatrix} 1-2 & y \\ z & 4-2 \end{pmatrix} = \begin{pmatrix} -1 & y \\ z & 2 \end{pmatrix}$ $\det(Q-2I) = (-1)(2) - yz = -2 - (-2) = 0.$ Statement A is true.

Determinant of $Q - 6I$
$Q-6I = \begin{pmatrix} 1-6 & y \\ z & 4-6 \end{pmatrix} = \begin{pmatrix} -5 & y \\ z & -2 \end{pmatrix}$ $\det(Q-6I) = (-5)(-2) - yz = 10 - (-2) = 12.$ Statement B is true.

Determinant of $Q - 3I$
$Q-3I = \begin{pmatrix} 1-3 & y \\ z & 4-3 \end{pmatrix} = \begin{pmatrix} -2 & y \\ z & 1 \end{pmatrix}$ $\det(Q-3I) = (-2)(1) - yz = -2 - (-2) = 0\neq 15.$ Statement C is false.

Product $yz$
From $(13)$ , $yz=-2\neq 2,$ so Statement D is false.

Hence the correct statements are:
Option A and Option B.

Q 38MathematicsSection A • MCQparabola

+4 / -0

Let $S$ denote the locus of the mid-points of those chords of the parabola $y^2 = x$ , such that the area of the region enclosed between the parabola and the chord is $\frac{4}{3}$ . Let $\mathcal{R}$ denote the region lying in the first quadrant, enclosed by the parabola $y^2 = x$ , the curve $S$ , and the lines $x = 1$ and $x = 4$ .

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

A

$(4, \sqrt{3}) \in S$

B

$(5, \sqrt{2}) \in S$

C

Area of $\mathcal{R}$ is $\frac{14}{3} - 2\sqrt{3}$

D

Area of $\mathcal{R}$ is $\frac{14}{3} - \sqrt{3}$

Official Correct Answer:Option (A, C)

Step-by-Step Solution:

Let the parabola be $x = y^{2}$ (opening towards the right).
Take any two points on it with parametric values $t_{1},\,t_{2}$ :
$P\,(t_{1}^{2},\,t_{1}),\;Q\,(t_{2}^{2},\,t_{2}).$

Mid-point $M(h,k)$ of the chord $PQ$ is therefore
$h = \dfrac{t_{1}^{2}+t_{2}^{2}}{2},\qquad k = \dfrac{t_{1}+t_{2}}{2}.$ We must find the locus of $M$ under the condition that the area of the parabolic segment cut off by $PQ$ equals $\dfrac{4}{3}$ .

1. Equation of the chord through $P,Q$
Using two-point form, $(y-t_{1})(t_{2}^{2}-t_{1}^{2})=(x-t_{1}^{2})(t_{2}-t_{1}).$ Because $t_{2}^{2}-t_{1}^{2}=(t_{2}-t_{1})(t_{2}+t_{1})$ , the chord simplifies to $x=(t_{1}+t_{2})y-t_{1}t_{2}\;.\tag{-1}$

2. Area between the parabola and the chord
For a fixed ordinate $y$ (between $t_{1}$ and $t_{2}$ ), $x_{\text{chord}}-(x_{\text{parabola}})=\bigl((t_{1}+t_{2})y-t_{1}t_{2}\bigr)-y^{2} =-(y-t_{1})(y-t_{2})=(t_{2}-y)(y-t_{1}).$ Hence the required area is $A=\int_{y=t_{1}}^{t_{2}}\!(t_{2}-y)(y-t_{1})\,dy.$ Put $y=t_{1}+u,\;u\in[0,d]$ where $d=t_{2}-t_{1}$ : $A=\int_{0}^{d}\!\bigl(d-u\bigr)u\,du =\int_{0}^{d}\!(du-u^{2})\,du =d\left[\frac{u^{2}}{2}\right]_{0}^{d}-\left[\frac{u^{3}}{3}\right]_{0}^{d} =\frac{d^{3}}{2}-\frac{d^{3}}{3} =\frac{d^{3}}{6}.\tag{-2}$

The question states that $A=\dfrac{4}{3}$ . Using $(\text{-2})$ : $\frac{d^{3}}{6}=\frac{4}{3}\;\Longrightarrow\;d^{3}=8\; \Longrightarrow\;d=t_{2}-t_{1}=2.\tag{-3}$

3. Locus of the mid-point
From $(\text{-3})$ , set $t_{1}=k-1,\;t_{2}=k+1.$ Then $h=\frac{(k-1)^{2}+(k+1)^{2}}{2} =\frac{k^{2}-2k+1+k^{2}+2k+1}{2} =k^{2}+1.$ Thus the locus $S$ of all such mid-points is $x=y^{2}+1\qquad\bigl(\text{or }y^{2}=x-1\bigr).$

4. Verification of the given points
(i) For $(4,\sqrt{3})$ : $y^{2}+1=3+1=4=x$ ⇒ lies on $S$ .
(ii) For $(5,\sqrt{2})$ : $y^{2}+1=2+1=3\neq5$ ⇒ does not lie on $S$ .

5. Region $\mathcal{R}$ in the first quadrant
Within $x=1$ to $x=4$ we have
• lower curve (from $S$ ): $y=\sqrt{x-1}$ ,
• upper curve (original parabola): $y=\sqrt{x}$ .
Therefore the required area is $\text{Area}=\int_{x=1}^{4}\!\bigl(\sqrt{x}-\sqrt{x-1}\bigr)\,dx.$

$\int_{1}^{4}\!\sqrt{x}\,dx =\left[\frac{2}{3}x^{3/2}\right]_{1}^{4} =\frac{2}{3}(8-1)=\frac{14}{3},$ $\int_{1}^{4}\!\sqrt{x-1}\,dx =\int_{0}^{3}\!\sqrt{u}\,du =\left[\frac{2}{3}u^{3/2}\right]_{0}^{3} =\frac{2}{3}\,(3\sqrt{3})=2\sqrt{3}.$ Hence $\text{Area}(\mathcal{R})=\frac{14}{3}-2\sqrt{3}.$

6. Conclusions
Option A is correct because $(4,\sqrt{3})\in S$ .
Option B is incorrect.
Option C is correct since $\text{Area}(\mathcal{R})=\dfrac{14}{3}-2\sqrt{3}$ .
Option D is incorrect.

Therefore the TRUE statements are:
Option A and Option C.

Q 39MathematicsSection A • MCQellipse

+4 / -0

Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be two distinct points on the ellipse

$\frac{x^2}{9} + \frac{y^2}{4} = 1$

such that $y_1 > 0$ , and $y_2 > 0$ . Let $\mathcal{C}$ denote the circle $x^2 + y^2 = 9$ , and $M$ be the point $(3, 0)$ .

Suppose the line $x = x_1$ intersects $\mathcal{C}$ at $R$ , and the line $x = x_2$ intersects $\mathcal{C}$ at $S$ , such that the $y$ -coordinates of $R$ and $S$ are positive. Let $\angle ROM = \frac{\pi}{6}$ and $\angle SOM = \frac{\pi}{3}$ , where $O$ denotes the origin $(0, 0)$ . Let $|XY|$ denote the length of the line segment $XY$ .

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

A

The equation of the line joining $P$ and $Q$ is $2x + 3y = 3(1 + \sqrt{3})$

B

The equation of the line joining $P$ and $Q$ is $2x + y = 3(1 + \sqrt{3})$

C

If $N_2 = (x_2, 0)$ , then $3|N_2 Q| = 2|N_2 S|$

D

If $N_1 = (x_1, 0)$ , then $9|N_1 P| = 4|N_1 R|$

Official Correct Answer:Option (A, C)

Step-by-Step Solution:

The ellipse is $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the circle $\mathcal{C}$ is $x^{2}+y^{2}=9$ (radius $3$ , centre at the origin $O(0,0)$ ).
The point $M$ is $(3,0)$ , so the ray $OM$ is the positive $x$ -axis.

Because $\angle ROM=\frac{\pi}{6}$ and $|OR|=3$ , the point $R$ on the circle is obtained by standard polar-Cartesian conversion:

$R=\bigl(3\cos\frac{\pi}{6},\,3\sin\frac{\pi}{6}\bigr)=\left(\frac{3\sqrt{3}}{2},\,\frac{3}{2}\right).$

Similarly, $\angle SOM=\frac{\pi}{3}$ gives

$S=\bigl(3\cos\frac{\pi}{3},\,3\sin\frac{\pi}{3}\bigr)=\left(\frac{3}{2},\,\frac{3\sqrt{3}}{2}\right).$

The vertical lines through $R$ and $S$ intersect the ellipse at $P$ and $Q$ , so

$x_{1}=x_{R}=\frac{3\sqrt{3}}{2},\qquad x_{2}=x_{S}=\frac{3}{2}.$

Point P
Insert $x_{1}$ in the ellipse:

$\frac{x_{1}^{2}}{9}=\frac{\left(\dfrac{3\sqrt{3}}{2}\right)^{2}}{9}=\frac{27/4}{9}=\frac34.$

Therefore $\frac{y_{1}^{2}}{4}=1-\frac34=\frac14 \;\Longrightarrow\; y_{1}=1 \;(\text{positive branch}).$

Hence $P=\left(\frac{3\sqrt{3}}{2},\,1\right).$

Point Q
Insert $x_{2}$ in the ellipse:

$\frac{x_{2}^{2}}{9}=\frac{\left(\dfrac{3}{2}\right)^{2}}{9}=\frac{9/4}{9}=\frac14.$

Therefore $\frac{y_{2}^{2}}{4}=1-\frac14=\frac34 \;\Longrightarrow\; y_{2}=\sqrt{3}.$

Hence $Q=\left(\frac{3}{2},\,\sqrt{3}\right).$

Equation of line $PQ$
Slope $m=\dfrac{\sqrt{3}-1}{\frac{3}{2}-\frac{3\sqrt{3}}{2}} =\dfrac{\sqrt{3}-1}{\frac{3}{2}(1-\sqrt{3})} =-\frac{2}{3}.$

Using point $P$ :

$y-1=-\frac{2}{3}\Bigl(x-\frac{3\sqrt{3}}{2}\Bigr) \;\Longrightarrow\;3y-3=-2x+3\sqrt{3}.$

Rearranging,

$2x+3y=3(1+\sqrt{3}).$

This matches Option A. Option B (with $2x+y$ ) is therefore incorrect.

Checking Statement C
Take $N_{2}=(x_{2},0)=\left(\dfrac32,0\right).$
Because $Q$ and $S$ have the same $x$ -coordinate $x_{2},$ the required distances are vertical:

$|N_{2}Q|=|y_{2}|=\sqrt{3},\qquad |N_{2}S|=\left|\frac{3\sqrt{3}}{2}\right|=\frac{3\sqrt{3}}{2}.$

Thus $3|N_{2}Q|=3\sqrt{3},\qquad 2|N_{2}S|=2\cdot\frac{3\sqrt{3}}{2}=3\sqrt{3},$ so $3|N_{2}Q|=2|N_{2}S|.$
Statement C is true.

Checking Statement D
Take $N_{1}=(x_{1},0)=\left(\dfrac{3\sqrt{3}}{2},0\right).$

Vertical distances give $|N_{1}P|=1,\quad |N_{1}R|=\frac{3}{2}.$

But $9|N_{1}P|=9,\quad 4|N_{1}R|=4\cdot\frac32=6,$ which are not equal. Hence Statement D is false.

Final result
Option A and Option C are the only correct statements.

Correct choices: Option A, Option C.

Q 40MathematicsSection 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 defined by

$f(x) = \begin{cases} \frac{6x + \sin x}{2x + \sin x} & \text{if } x \neq 0, \\ \frac{7}{3} & \text{if } x = 0. \end{cases}$

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

A

The point $x = 0$ is a point of local maxima of $f$

B

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

C

Number of points of local maxima of $f$ in the interval $[\pi, 6\pi]$ is 3

D

Number of points of local minima of $f$ in the interval $[2\pi, 4\pi]$ is 1

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

Step-by-Step Solution:

The function is given by
$f(x)=\dfrac{6x+\sin x}{2x+\sin x}\;,\;x\neq 0,\qquad f(0)=\dfrac73.$

1. Behaviour at the origin

First-order Taylor expansion about $x=0$ gives

$\sin x=x-\dfrac{x^{3}}6+O(x^{5}),\qquad \cos x=1-\dfrac{x^{2}}2+O(x^{4}).$

Hence

$6x+\sin x=6x+\left(x-\dfrac{x^{3}}6\right)=7x-\dfrac{x^{3}}6+O(x^{5}),$ $2x+\sin x=2x+\left(x-\dfrac{x^{3}}6\right)=3x-\dfrac{x^{3}}6+O(x^{5}).$

Therefore

$f(x)=\dfrac{7x-\dfrac{x^{3}}6+O(x^{5})}{3x-\dfrac{x^{3}}6+O(x^{5})} =\dfrac73\!\left[\,1+\dfrac{2x^{2}}{63}+O(x^{4})\right] =\dfrac73+\dfrac{2x^{2}}{27}+O(x^{4}).$

Near the origin we thus have

$f(x)=\dfrac73+\dfrac{2x^{2}}{27}+O(x^{4}).$

This gives $f'(0)=0$ and $f''(0)=\dfrac{4}{27}\;(\gt 0).$ Hence $x=0$ is a point of local minimum. Option B is therefore correct, while Option A is not.

2. Critical points for $x\neq 0$

For $x\neq 0$ write $N=6x+\sin x,\;D=2x+\sin x,$ so that $f(x)=\dfrac{N}{D}.$ Using the quotient rule,

$f'(x)=\dfrac{N' D-N D'}{D^{2}} =\dfrac{(6+\cos x)(2x+\sin x)-(6x+\sin x)(2+\cos x)}{(2x+\sin x)^{2}} =\dfrac{4\bigl(\sin x-x\cos x\bigr)}{(2x+\sin x)^{2}}.$

Since $(2x+\sin x)^{2}\gt 0\quad\text{for all }x\gt 0,$ the sign of $f'(x)$ is governed solely by

$g(x)=\sin x-x\cos x.$

3. Location of critical points

Setting $f'(x)=0$ gives $g(x)=0\iff \sin x-x\cos x=0 \iff\tan x = x.$ The well-known equation $\tan x=x$ has one solution in every interval $\bigl(n\pi,\;n\pi+\tfrac{\pi}{2}\bigr), \;n=1,2,3,\ldots$ apart from $x=0.$ Denote these positive roots by $x_1,x_2,x_3,\dots$ in increasing order:

$x_1\approx4.493,\;x_2\approx7.725,\;x_3\approx10.904,\; x_4\approx14.066,\;x_5\approx17.279,\;\ldots$

4. Nature of each critical point

Differentiate $g(x)$ :

$g'(x)=\cos x-\cos x+x\sin x = x\sin x.$

Thus, at a root $x_n,$ the sign of $g'(x_n)=x_n\sin x_n$ is the sign of $\sin x_n.$ Inside $\bigl(n\pi,\;n\pi+\tfrac{\pi}{2}\bigr)$ we have $\sin x\;\begin{cases} \gt 0,& n\; \text{even},\\[4pt] \lt 0,& n\; \text{odd}. \end{cases}$ Therefore

• If $n$ is odd ( $1,3,5,\dots$ ) then $g'(x_n)\lt 0:$ $g(x)$ crosses the $x$ -axis from positive to negative, so $f'(x)$ changes $+\to-$ and $x_n$ is a local maximum.
• If $n$ is even ( $2,4,6,\dots$ ) then $g'(x_n)\gt 0:$ $g(x)$ crosses from negative to positive, so $f'(x)$ changes $-\to+$ and $x_n$ is a local minimum.

5. Counting extrema in the required intervals

(i) Interval $[\pi,6\pi]\;( \approx[3.142,18.850] ):$ roots inside this range are $x_1,x_2,x_3,x_4,x_5.$ Among these, maxima occur at the odd-indexed roots $x_1,x_3,x_5.$ Hence the number of local maxima is $3.$ Option C is correct.

(ii) Interval $[2\pi,4\pi]\;( \approx[6.283,12.566] ):$ only the root $x_2\approx7.725$ lies in this interval, and it is a local minimum (even index). Therefore there is exactly $1$ local minimum here. Option D is correct.

6. Final verdict

The true statements are:
Option B, Option C, and Option D.

Q 41MathematicsSection B • Numericaldifferential equations

+4 / -0

Let $y(x)$ be the solution of the differential equation

$x^2 \frac{dy}{dx} + xy = x^2 + y^2, \quad x > \frac{1}{e},$

satisfying $y(1) = 0$ . Then the value of $2\frac{(y(e))^2}{y(e^2)}$ is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The given differential equation is

$x^{2}\,\frac{dy}{dx}+xy = x^{2}+y^{2}, \qquad x \gt \frac1e$

Divide by $x^{2}$ to make the terms dimensionless:

$\frac{dy}{dx}=1+\frac{y^{2}}{x^{2}}-\frac{y}{x}$

The right-hand side contains the ratio $y/x$ , so set

$v=\frac{y}{x}\; \Longrightarrow\; y=v\,x$

Differentiating, $\dfrac{dy}{dx}=v+x\,\dfrac{dv}{dx}$ .

Substitute this in the differential equation:

$v+x\,\frac{dv}{dx}=1+v^{2}-v$

Simplify:

$x\,\frac{dv}{dx}=v^{2}-2v+1=(v-1)^{2}$

Separate the variables:

$\frac{dv}{(v-1)^{2}}=\frac{dx}{x}$

Integrate both sides:

$\int\frac{dv}{(v-1)^{2}} = \int\frac{dx}{x}$
$-\frac1{v-1}= \ln|x| + C \qquad -(1)$

Use the initial condition $y(1)=0$ :

At $x=1$ , $v=\dfrac{y}{x}=0$ . Substitute into $(1)$ :

$-\frac1{0-1}=1=\ln|1|+C \Longrightarrow C=1$

Hence

$-\frac1{v-1}=1+\ln x$
$\frac1{v-1}=-(1+\ln x)$
$v-1=-\frac1{1+\ln x}$
$v=1-\frac1{1+\ln x}$

Return to $y=v\,x$ :

$y(x)=x\left(1-\frac1{1+\ln x}\right)$

Now evaluate at the required points.

At $x=e$ : $\ln e=1 \;\Rightarrow\; 1+\ln e=2$
$y(e)=e\left(1-\frac1{2}\right)=\frac{e}{2}$

At $x=e^{2}$ : $\ln e^{2}=2 \;\Rightarrow\; 1+\ln e^{2}=3$
$y(e^{2})=e^{2}\left(1-\frac1{3}\right)=e^{2}\cdot\frac23=\frac{2e^{2}}{3}$

Finally compute the required expression:

$2\,\frac{(y(e))^{2}}{y(e^{2})}=2\,\frac{\left(\dfrac{e}{2}\right)^{2}}{\dfrac{2e^{2}}{3}} =2\,\frac{\dfrac{e^{2}}{4}}{\dfrac{2e^{2}}{3}} =\frac{e^{2}}{2}\cdot\frac{3}{2e^{2}} =\frac34=0.75$

Therefore the desired value is 0.75.

Q 42MathematicsSection B • Numericalbinomial theorem

+4 / -0

Let $a_0, a_1, \ldots, a_{23}$ be real numbers such that

$\left(1 + \frac{2}{5}x\right)^{23} = \sum_{i=0}^{23} a_i x^i$

for every real number $x$ . Let $a_r$ be the largest among the numbers $a_j$ for $0 \leq j \leq 23$ . Then the value of $r$ is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The expansion of $\left(1+\frac{2}{5}x\right)^{23}$ by the binomial theorem is

$\left(1+\frac{2}{5}x\right)^{23}=\sum_{i=0}^{23}\binom{23}{i}\left(\frac{2}{5}x\right)^i =\sum_{i=0}^{23}\binom{23}{i}\left(\frac{2}{5}\right)^i x^{\,i}.$

Comparing with $\displaystyle\sum_{i=0}^{23} a_i x^i,$ we identify

$a_i=\binom{23}{i}\left(\frac{2}{5}\right)^{i},\qquad 0\le i\le 23.$

To find the largest coefficient, examine the ratio of successive terms:

$\frac{a_{i+1}}{a_i}= \frac{\binom{23}{i+1}(2/5)^{\,i+1}}{\binom{23}{i}(2/5)^{\,i}} =\frac{23-i}{i+1}\cdot\frac{2}{5}.\quad -(1)$

The sequence $a_0,a_1,\dots,a_{23}$ increases while the ratio in $(1)$ exceeds $1$ and begins to decrease once the ratio falls below $1$ . Hence we need to solve

$\frac{23-i}{i+1}\cdot\frac{2}{5}\gt 1.$

Multiply both sides by $5(i+1):$

$2\,(23-i)\gt 5(i+1).$

Simplify:

$46-2i\gt 5i+5 \;\;\Longrightarrow\;\; 46-5 > 7i \;\;\Longrightarrow\;\; 41>7i.$

Therefore

$i<\frac{41}{7}=5.857\ldots$

Because $i$ is an integer, inequality $(1)$ holds for $i=0,1,2,3,4,5.$ At $i=5,$ compute one more ratio:

$\frac{a_6}{a_5}=\frac{23-5}{6}\cdot\frac{2}{5}=\frac{18}{6}\cdot\frac{2}{5}=3\cdot0.4=1.2\gt1,$

so the coefficients are still rising up to $i=6.$ Next, for $i=6$

$\frac{a_7}{a_6}=\frac{23-6}{7}\cdot\frac{2}{5} =\frac{17}{7}\cdot0.4\approx0.97\lt1,$

meaning the sequence starts decreasing after $i=6$ . Hence the maximum coefficient occurs at

$r=6.$

Final answer: $r=6.$

Q 43MathematicsSection B • Numericalprobability

+4 / -0

A factory has a total of three manufacturing units, $M_1$ , $M_2$ , and $M_3$ , which produce bulbs independent of each other. The units $M_1$ , $M_2$ , and $M_3$ produce bulbs in the proportions of $2:2:1$ , respectively. It is known that $20\%$ of the bulbs produced in the factory are defective. It is also known that, of all the bulbs produced by $M_1$ , $15\%$ are defective. Suppose that, if a randomly chosen bulb produced in the factory is found to be defective, the probability that it was produced by $M_2$ is $\frac{2}{5}$ .

If a bulb is chosen randomly from the bulbs produced by $M_3$ , then the probability that it is defective is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

Let the events be defined as follows:
$M_1, M_2, M_3$ : the bulb is produced by unit $M_1, M_2, M_3$ respectively.
$D$ : the bulb is defective.

Production proportions (given as $2:2:1$ ) give the prior probabilities
$P(M_1)=\frac{2}{5},\quad P(M_2)=\frac{2}{5},\quad P(M_3)=\frac{1}{5}$

Overall defective rate in the factory is $P(D)=0.20$ .

Defective rates for two of the units are known:
$P(D\mid M_1)=0.15,\qquad P(M_2\mid D)=\frac{2}{5}=0.40$

Step 1: Find $P(D\mid M_2)$ using Bayes’ theorem.
Bayes’ theorem gives $P(M_2\mid D)=\frac{P(D\mid M_2)P(M_2)}{P(D)}$ Therefore $P(D\mid M_2)=\frac{P(M_2\mid D)\,P(D)}{P(M_2)} =\frac{0.40\times0.20}{\tfrac{2}{5}} =\frac{0.08}{0.40}=0.20$

Step 2: Use the overall defective rate to solve for $P(D\mid M_3)$ . The law of total probability gives $P(D)=P(D\mid M_1)P(M_1)+P(D\mid M_2)P(M_2)+P(D\mid M_3)P(M_3)$ Substituting the known numbers: $0.20=0.15\left(\frac{2}{5}\right)+0.20\left(\frac{2}{5}\right)+P(D\mid M_3)\left(\frac{1}{5}\right)$ Compute the first two terms: $0.15\left(\frac{2}{5}\right)=0.06,\qquad 0.20\left(\frac{2}{5}\right)=0.08$ Thus $0.20=0.06+0.08+\frac{1}{5}P(D\mid M_3)$ $0.20-0.14=\frac{1}{5}P(D\mid M_3)$ $0.06=\frac{1}{5}P(D\mid M_3)$ $P(D\mid M_3)=0.06\times5=0.30$

Hence, the probability that a bulb produced by $M_3$ is defective is $0.30$ , which lies in the required range $0.27-0.33$ .

Q 44MathematicsSection B • Numericalvector algebra

+4 / -0

Consider the vectors

$\vec{x} = \hat{i} + 2\hat{j} + 3\hat{k}$ , $\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k}$ , $\quad$ and $\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}$ .

For two distinct positive real numbers $\alpha$ and $\beta$ , define

$\vec{X} = \alpha\vec{x} + \beta\vec{y} - \vec{z}$ , $\quad \vec{Y} = \alpha\vec{y} + \beta\vec{z} - \vec{x}$ , $\quad$ and $\quad \vec{Z} = \alpha\vec{z} + \beta\vec{x} - \vec{y}$ .

If the vectors $\vec{X}$ , $\vec{Y}$ , and $\vec{Z}$ lie in a plane, then the value of $\alpha + \beta - 3$ is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The vectors lie in a plane if and only if their scalar triple product is zero, i.e. $\vec{X}\cdot(\vec{Y}\times\vec{Z}) = 0$ . In component form this is equivalent to the determinant of the three column vectors being zero.

First write each required vector in $\hat i,\hat j,\hat k$ components.

$\vec{x} = (1,\,2,\,3),\; \vec{y} = (2,\,3,\,1),\; \vec{z} = (3,\,1,\,2)$

$\vec{X}= \alpha\vec{x}+\beta\vec{y}-\vec{z} = (\alpha+2\beta-3,\; 2\alpha+3\beta-1,\; 3\alpha+\beta-2)$

$\vec{Y}= \alpha\vec{y}+\beta\vec{z}-\vec{x} = (2\alpha+3\beta-1,\; 3\alpha+\beta-2,\; \alpha+2\beta-3)$

$\vec{Z}= \alpha\vec{z}+\beta\vec{x}-\vec{y} = (3\alpha+\beta-2,\; \alpha+2\beta-3,\; 2\alpha+3\beta-1)$

Introduce the shorthand

$A = \alpha+2\beta-3,\qquad B = 2\alpha+3\beta-1,\qquad C = 3\alpha+\beta-2$

Then $\vec{X} = (A,\;B,\;C),\; \vec{Y} = (B,\;C,\;A),\; \vec{Z} = (C,\;A,\;B).$

Set up the determinant:

$\left| \begin{array}{ccc} A & B & C\\ B & C & A\\ C & A & B \end{array} \right| = 0.$

Evaluating the determinant,

$\begin{aligned} D &= A(CB - A^{2}) - B(B^{2} - AC) + C(AB - C^{2}) \\ &= (ABC - A^{3}) + (ABC - B^{3}) + (ABC - C^{3}) \\ &= 3ABC - (A^{3}+B^{3}+C^{3}). \end{aligned}$

Hence $D = 0 \Longrightarrow A^{3}+B^{3}+C^{3}-3ABC = 0.$ Using the identity $A^{3}+B^{3}+C^{3}-3ABC = (A+B+C)\,\bigl(A^{2}+B^{2}+C^{2}-AB-BC-CA\bigr),$ we get two possibilities:

Case 1:

$A+B+C = 0.$

Case 2:

$A=B=C.$

Because $\alpha,\beta$ are distinct positive numbers, Case 2 is impossible (solving $A=B$ forces $\alpha+\beta=-2$ , which contradicts positivity). Therefore only Case 1 survives.

Compute $A+B+C$ :

$\begin{aligned} A+B+C &= (\alpha+2\beta-3) + (2\alpha+3\beta-1) + (3\alpha+\beta-2)\\ &= 6\alpha + 6\beta - 6\\ &= 6(\alpha+\beta-1). \end{aligned}$

Setting this to zero gives $\alpha+\beta-1 = 0 \Longrightarrow \alpha+\beta = 1.$ Both $\alpha,\beta$ can be positive and distinct while satisfying this relation, so the condition is admissible.

Finally, the required expression is

$\alpha+\beta-3 = 1 - 3 = -2.$

Hence the answer is -2.

Q 45MathematicsSection B • Numericalcomplex numbers

+4 / -0

For a non-zero complex number $z$ , let $\arg(z)$ denote the principal argument of $z$ , with $-\pi < \arg(z) \leq \pi$ . Let $\omega$ be the cube root of unity for which $0 < \arg(\omega) < \pi$ . Let

$\alpha = \arg\left(\sum_{n=1}^{2025}(-\omega)^n\right).$

Then the value of $\frac{3\alpha}{\pi}$ is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The non-real cube roots of unity are $\omega$ and $\omega^{2}$ , where $\omega^{3}=1$ and $1+\omega+\omega^{2}=0$ .
Because $0 \lt \arg(\omega) \lt \pi$ , we take $\omega = e^{\,i\frac{2\pi}{3}} = -\frac12 + i\frac{\sqrt3}{2}$ , so $\arg(\omega)=\frac{2\pi}{3}$ .

Define $r=-\omega$ . Then
$r = -\omega = e^{\,i\pi}\,e^{\,i\frac{2\pi}{3}} = e^{\,i\frac{5\pi}{3}} = e^{-\,i\frac{\pi}{3}}$
and hence $\arg(r)=-\frac{\pi}{3}$ .

Since $r^{3}=(-\omega)^{3}=-(\omega^{3})=-1$ , we get $r^{6}=1$ . Thus the powers of $r$ repeat every six terms.

The required sum is $S=\sum_{n=1}^{2025} r^{\,n}.$ Because one full cycle of six consecutive terms sums to zero (geometric-series property for a 6th root of unity different from 1), we split 2025 terms into cycles:

$2025 = 6 \times 337 + 3.$ The first $6\times337=2022$ terms contribute zero, so $S = r^{2023}+r^{2024}+r^{2025}.$

Now reduce exponents modulo 6: $2023 \equiv 1,\; 2024 \equiv 2,\; 2025 \equiv 3 \pmod{6}.$ Hence $S = r^{1}+r^{2}+r^{3}.$

Compute each term:
$r^{1} = -\omega,$ $r^{2} = (-\omega)^{2} = \omega^{2},$ $r^{3} = -1.$ Therefore $S = -\omega + \omega^{2} - 1.$

Using $1+\omega+\omega^{2}=0 \; \Rightarrow \; \omega^{2}= -1-\omega,$ $S = -\omega + (-1-\omega) -1 = -2 - 2\omega = -2(1+\omega).$

But $1+\omega = -\omega^{2},$ so $S = -2(-\omega^{2}) = 2\,\omega^{2}.$

Multiplying by the positive real number 2 does not change the argument, hence $\alpha = \arg(S) = \arg(\omega^{2}).$ For our chosen branch, $\omega^{2}=e^{\,i\frac{4\pi}{3}}$ has principal argument $\frac{4\pi}{3}-2\pi = -\frac{2\pi}{3}.$ Thus $\alpha = -\frac{2\pi}{3}.$

Finally, $\frac{3\alpha}{\pi} = \frac{3\left(-\tfrac{2\pi}{3}\right)}{\pi} = -2.$

Answer: -2

Q 46MathematicsSection B • Numericalfunctions

+4 / -0

Let $\mathbb{R}$ denote the set of all real numbers. Let $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to (0, 4)$ be functions defined by

$f(x) = \log_e(x^2 + 2x + 4)$ , $\quad$ and $\quad g(x) = \frac{4}{1 + e^{-2x}}$ .

Define the composite function $f \circ g^{-1}$ by $(f \circ g^{-1})(x) = f(g^{-1}(x))$ , where $g^{-1}$ is the inverse of the function $g$ .

Then the value of the derivative of the composite function $f \circ g^{-1}$ at $x = 2$ is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

We have two functions: $f(x)=\log_e(x^2+2x+4)$ for all $x\in\mathbb{R}$ and $g(x)=\dfrac{4}{1+e^{-2x}}$ whose range is $(0,\,4)$ .

The composite function is $(f\circ g^{-1})(y)=f\!\bigl(g^{-1}(y)\bigr),\qquad 0\lt y\lt 4.$ We must find its derivative at $y=2$ .

Step 1: Find the inverse $g^{-1}$ .
Let $y=g(x)=\dfrac{4}{1+e^{-2x}}.$ Multiply: $y\,(1+e^{-2x})=4\;\;\Longrightarrow\;\;y\,e^{-2x}=4-y.$ Therefore $e^{-2x}=\dfrac{4-y}{y}.$ Taking natural logarithm,
$-2x=\ln\!\Bigl(\dfrac{4-y}{y}\Bigr)\;\;\Longrightarrow\;\;x=\dfrac12\ln\!\Bigl(\dfrac{y}{4-y}\Bigr).$ Thus $g^{-1}(y)=\dfrac12\ln\!\Bigl(\dfrac{y}{4-y}\Bigr),\qquad 0\lt y\lt4.$ For convenience, write $x=g^{-1}(y)=\dfrac12\ln\!\Bigl(\dfrac{y}{4-y}\Bigr).$

Step 2: Differentiate $f(x)$ .
$f(x)=\ln(x^2+2x+4)$
Using $\dfrac{d}{dx}\ln u=\dfrac{u'}{u}$ ,
$f'(x)=\dfrac{2x+2}{x^2+2x+4}= \dfrac{2(x+1)}{x^2+2x+4}.$ At $x=0$ (value found later),
$f'(0)=\dfrac{2(0+1)}{0^2+0+4}= \dfrac{2}{4}= \dfrac12.$

Step 3: Differentiate $x=g^{-1}(y)$ with respect to $y$ .
$x=\dfrac12\ln\!\Bigl(\dfrac{y}{4-y}\Bigr).$ Let $u(y)=\dfrac{y}{4-y}.$ Then $x=\dfrac12\ln u.$ First evaluate $u'(y)$ using the quotient rule:
$u'(y)=\dfrac{(4-y)\cdot1-y(-1)}{(4-y)^2}= \dfrac{4}{(4-y)^2}.$ Now $\dfrac{d}{dy}\ln u=\dfrac{u'(y)}{u(y)}=\dfrac{4}{(4-y)^2}\,\bigg/\,\dfrac{y}{4-y}= \dfrac{4}{y(4-y)}.$ Hence $\dfrac{dx}{dy}= \dfrac12\cdot\dfrac{4}{y(4-y)}= \dfrac{2}{y(4-y)}.$ At $y=2$ ,
$\dfrac{dx}{dy}\bigg|_{y=2}= \dfrac{2}{2\,(4-2)}= \dfrac{2}{4}= \dfrac12.$

Step 4: Value of $x$ when $y=2$ .
$x=g^{-1}(2)=\dfrac12\ln\!\Bigl(\dfrac{2}{4-2}\Bigr)=\dfrac12\ln 1=0.$

Step 5: Apply the Chain Rule.
For the composite $h(y)=(f\circ g^{-1})(y)=f(x)\,,$
$h'(y)=f'(x)\,\dfrac{dx}{dy}.$ At $y=2$ (so $x=0$ ):
$h'(2)=f'(0)\times\dfrac{dx}{dy}\bigg|_{y=2}= \dfrac12 \times \dfrac12= \dfrac14=0.25.$

Therefore the derivative of the composite function at $x=2$ is $0.25$ , which lies in the required range $0.2-0.3$ .

Q 47MathematicsSection B • Numericaltrigonometric ratios and identities

+4 / -0

Let

$\alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}.$

Then the value of

$\left(\frac{\csc 1^\circ}{\alpha}\right)^2$

is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The given sum contains angles that differ by $1^\circ$ , so we first convert each term by the identity
$\frac{1}{\sin A\,\sin B}\;=\;\frac{\cot A-\cot B}{\sin\!\left(B-A\right)}\,.$ Here $B-A = 1^\circ$ for every pair.

Hence each term becomes
$\frac{1}{\sin k^\circ \,\sin(k+1)^\circ}\;=\;\frac{\cot k^\circ-\cot(k+1)^\circ}{\sin1^\circ}\,.$ With $k = 60,62,64,\dots ,118$ we get

$\alpha \;=\;\frac{1}{\sin1^\circ}\! \Bigl[\bigl(\cot60^\circ-\cot61^\circ\bigr) +\bigl(\cot62^\circ-\cot63^\circ\bigr) +\cdots +\bigl(\cot118^\circ-\cot119^\circ\bigr)\Bigr].$

Introduce the compact notation

$S \;=\;\sum_{k=0}^{29}\Bigl(\cot(60+2k)^\circ-\cot(61+2k)^\circ\Bigr).$ Then $\alpha=\dfrac{S}{\sin1^\circ}\,,$ so our task is to evaluate $S$ .

Separate even and odd arguments:

$S =\bigl[\cot60^\circ+\cot62^\circ+\cdots+\cot118^\circ\bigr] -\bigl[\cot61^\circ+\cot63^\circ+\cdots+\cot119^\circ\bigr].$

Write each angle as either below $90^\circ$ or above $90^\circ$ :

• For $\theta\lt90^\circ$ , $\cot\theta^\circ=\tan(90^\circ-\theta^\circ).$
• For $\theta=90^\circ+\delta^\circ$ , $\cot\theta^\circ=-\tan\delta^\circ.$

Applying these to all terms gives

$\begin{aligned} S&=\Bigl[\tan30^\circ+\tan28^\circ+\cdots+\tan2^\circ\Bigr] -\Bigl[\tan29^\circ+\tan27^\circ+\cdots+\tan1^\circ\Bigr] \\ &\quad-\Bigl[\tan28^\circ+\tan26^\circ+\cdots+\tan2^\circ\Bigr] +\Bigl[\tan29^\circ+\tan27^\circ+\cdots+\tan1^\circ\Bigr]. \end{aligned}$

The two long odd-angle sums cancel, and every even-angle term from $2^\circ$ to $28^\circ$ cancels as well. The only term left is

$S=\tan30^\circ=\frac{1}{\sqrt3}\,.$

Therefore

$\alpha=\frac{S}{\sin1^\circ}=\frac{1}{\sqrt3\,\sin1^\circ}\,.$

Finally,

$\left(\frac{\csc1^\circ}{\alpha}\right)^2 =\left(\frac{1/\sin1^\circ}{1/(\sqrt3\,\sin1^\circ)}\right)^2 =\left(\sqrt3\right)^2 =3.$

Answer: 3

Q 48MathematicsSection B • Numericaldefinite integration

+4 / -0

If

$\alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} \, dx,$

then the value of $\sqrt{7} \tan\left(\frac{2\alpha\sqrt{7}}{\pi}\right)$ is ______.

(Here, the inverse trigonometric function $\tan^{-1} x$ assumes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .)

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The given integral is $\alpha=\int_{1/2}^{2}\frac{\tan^{-1}x}{2x^{2}-3x+2}\,dx.$

Step 1 : Use the reciprocal-substitution symmetry.
Put $x=\frac{1}{t}$ so that $dx=-\frac{dt}{t^{2}}.$ The limits change as $x : 1/2 \to 2$ implies $t : 2 \to 1/2$ .

Then

$\alpha=\int_{2}^{1/2}\frac{\tan^{-1}(1/t)}{2/t^{2}-3/t+2}\left(-\frac{dt}{t^{2}}\right) =\int_{1/2}^{2}\frac{\tan^{-1}(1/x)}{2x^{2}-3x+2}\,dx.$

Step 2 : Add the two forms of the integral.

Adding the original and the transformed integrals,

$2\alpha=\int_{1/2}^{2}\frac{\tan^{-1}x+\tan^{-1}(1/x)}{2x^{2}-3x+2}\,dx.$

For every positive $x\neq1$ , $\tan^{-1}x+\tan^{-1}\!\left(\frac1x\right)=\frac{\pi}{2}.$ (The value at $x=1$ does not affect the integral.)

Hence

$2\alpha=\frac{\pi}{2}\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2},\qquad \alpha=\frac{\pi}{4}\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2}.$

Step 3 : Evaluate the remaining integral.

Rewrite the quadratic:

$2x^{2}-3x+2=2\!\left[x^{2}-\frac{3}{2}x+1\right] =2\!\Bigl[(x-\tfrac34)^{2}+\tfrac7{16}\Bigr] =2\!\Bigl[(x-\tfrac34)^{2}+(\tfrac{\sqrt7}{4})^{2}\Bigr].$

Therefore

$\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2} =\frac12\int_{1/2}^{2}\frac{dx}{(x-\tfrac34)^{2}+(\tfrac{\sqrt7}{4})^{2}} =\frac12\cdot\frac4{\sqrt7}\Bigl[\tan^{-1}\!\bigl(\tfrac{4x-3}{\sqrt7}\bigr)\Bigr]_{1/2}^{2}.$

Simplifying the constant factor gives $\dfrac{2}{\sqrt7}$ . Now compute the bracketed values:

At $x=2$ : $\dfrac{4x-3}{\sqrt7}=\dfrac{5}{\sqrt7}$ .
At $x=\tfrac12$ : $\dfrac{4x-3}{\sqrt7}=-\dfrac1{\sqrt7}$ .

Hence

$\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2} =\frac{2}{\sqrt7}\Bigl[\tan^{-1}\!\bigl(\tfrac{5}{\sqrt7}\bigr)+\tan^{-1}\!\bigl(\tfrac{1}{\sqrt7}\bigr)\Bigr].$

Using the addition formula $\tan^{-1}a+\tan^{-1}b=\tan^{-1}\!\bigl(\dfrac{a+b}{1-ab}\bigr)$ (since $ab\lt 1$ ),

$\tan^{-1}\!\bigl(\tfrac{5}{\sqrt7}\bigr)+\tan^{-1}\!\bigl(\tfrac{1}{\sqrt7}\bigr) =\tan^{-1}\!\Bigl(\frac{\,\tfrac{5}{\sqrt7}+\tfrac{1}{\sqrt7}}{1-\tfrac{5}{7}}\Bigr) =\tan^{-1}\!\Bigl(\frac{\,\tfrac{6}{\sqrt7}}{\tfrac{2}{7}}\Bigr) =\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$

Thus

$\int_{1/2}^{2}\frac{dx}{2x^{2}-3x+2} =\frac{2}{\sqrt7}\,\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$

Step 4 : Obtain $\alpha$ .

$\alpha=\frac{\pi}{4}\times\frac{2}{\sqrt7}\, \tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr) =\frac{\pi}{2\sqrt7}\, \tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$

Step 5 : Evaluate the required expression.

Compute the angle inside the tangent:

$\frac{2\alpha\sqrt7}{\pi} =\frac{2}{\pi}\cdot\frac{\pi}{2\sqrt7}\,\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr)\!\cdot\sqrt7 =\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr).$

Therefore

$\sqrt7\,\tan\!\Bigl(\frac{2\alpha\sqrt7}{\pi}\Bigr) =\sqrt7\,\tan\!\left[\tan^{-1}\!\Bigl(\frac{21}{\sqrt7}\Bigr)\right] =\sqrt7\cdot\frac{21}{\sqrt7}=21.$

Hence, the required value is 21.

JEE Advanced 2025 Paper 2 Question Paper with Solutions