JEE Advanced 2024 Question Paper 2 Question Paper with Solutions | JEE Advanced Mock Test | JeeMate

Q 1PhysicsSection B • Numericalelectrostatics

+4 / -0

An infinitely long thin wire, having a uniform charge density per unit length of $5 \mathrm{nC} / \mathrm{m}$ , is passing through a spherical shell of radius $1 \mathrm{~m}$ , as shown in the figure. A $10 \mathrm{nC}$ charge is distributed uniformly over the spherical shell. If the configuration of the charges remains static, the magnitude of the potential difference between points $\mathrm{P}$ and $\mathrm{R}$ , in Volt, is ________.

[Given: In SI units $\frac{1}{4 \pi \epsilon_0}=9 \times 10^9, \ln 2=0.7$ . Ignore the area pierced by the wire.]

qImage--electrostatics-electricity-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 2PhysicsSection B • Numericalelectrostatics

+4 / -0

A charge is kept at the central point $\mathrm{P}$ of a cylindrical region. The two edges subtend a half-angle $\theta$ at $\mathrm{P}$ , as shown in the figure. When $\theta=30^{\circ}$ , then the electric flux through the curved surface of the cylinder is $\Phi$ . If $\theta=60^{\circ}$ , then the electric flux through the curved surface becomes $\Phi / \sqrt{n}$ , where the value of $n$ is ___.

Answer Format:Enter numerical value rounded to the nearest integer

Q 3PhysicsSection A • MCQdual nature of radiation

+3 / -1

A metal target with atomic number

Z=46

is bombarded with a high energy electron beam. The emission of X-rays from the target is analyzed. The ratio

r

of the wavelengths of the

K_\alpha

-line and the cut-off is found to be

r=2

. If the same electron beam bombards another metal target with

Z=41

, the value of

r

will be

A

2.53

B

1.27

C

2.24

D

1.58

Official Correct Answer:Option (A)

Step-by-Step Solution:

To solve this problem, we need to understand the relationship between the atomic number $Z$ , the characteristic X-ray wavelengths, and the cut-off wavelength in X-ray spectra.

The cut-off wavelength, also known as the minimum wavelength ( $\lambda_{\text{min}}$ ), corresponds to the maximum energy of the X-rays produced and can be related to the accelerating voltage used in the X-ray tube. The characteristic X-rays, such as the $K_\alpha$ -line, correspond to specific electron transitions in the atom and are dependent on the atomic number $Z$ of the target material.

The wavelength of the $K_\alpha$ -line is given by Moseley's Law, which can be approximated as:

$\lambda_{K_\alpha} \propto \frac{1}{(Z - 1)^2}$

The cut-off wavelength is given by the relationship between the energy of the incident electrons and the X-ray produced:

$\lambda_{\text{min}} = \frac{hc}{eV}$

Now, given the ratio $r$ for the first metal target with $Z = 46$ :

$r = \frac{\lambda_{K_\alpha}}{\lambda_{\text{min}}} = 2$

For the second metal target with $Z = 41$ , let's denote the new ratio as $r_2$ . We assume the same cutoff wavelength since the same electron beam is used.

The ratio for $Z = 41$ can be calculated using the relationship between the $K_\alpha$ -line wavelenghts:

$\frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{Z_2 - 1}{Z_1 - 1} \right)^2$

Substituting $Z_1 = 46$ and $Z_2 = 41$ :

$\frac{\lambda_{K_\alpha (1)}}{\lambda_{K_\alpha (2)}} = \left( \frac{41 - 1}{46 - 1} \right)^2 = \left( \frac{40}{45} \right)^2 = \left( \frac{8}{9} \right)^2 = \frac{64}{81}$

Let $\lambda_{K_\alpha (1)} = 2 \lambda_{\text{min}}$ because $r = 2$ . Therefore:

$\lambda_{K_\alpha (2)} = \lambda_{K_\alpha (1)} \cdot \frac{81}{64} = 2 \lambda_{\text{min}} \cdot \frac{81}{64}$

Thus, the new ratio $r_2$ is:

$r_2 = \frac{\lambda_{K_\alpha (2)}}{\lambda_{\text{min}}} = 2 \cdot \frac{81}{64} = \frac{162}{64} = 2.53$

So, the value of $r$ for the second metal target with $Z = 41$ is:

Option A: 2.53

Q 4PhysicsSection B • Numericalwave optics

+3 / -0

Read the passage carefully:

In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of

0.8 \mathrm{~mm}

. The distance between the slits at time

t

is given by

d=(0.8+0.04 \sin \omega t) \mathrm{mm}

, where

\omega=0.08 \mathrm{rad} \mathrm{s}^{-1}

. The distance of the screen from the slits is

1 \mathrm{~m}

and the wavelength of the light used to illuminate the slits is

6000

Å . The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point

O

.

qImage-huygens-principle-and-interference-of-light-wave-optics-optics-Jeemate

The

8^{\text {th }}

bright fringe above the point

\mathrm{O}

oscillates with time between two extreme positions. The separation between these two extreme positions, in micrometer

(\mu \mathrm{m})

, is _________ .

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:601.50

Step-by-Step Solution:

The position of the $m^{\text{th}}$ bright fringe on the screen (measured from the central point $O$ ) in a Young’s double-slit experiment is given by the condition for constructive interference: $y_m = \dfrac{m \, \lambda \, D}{d}$ , where

$m$ = fringe order, here $m = 8$ ,
$\lambda$ = wavelength of light, $6000 \text{ Å} = 6000 \times 10^{-10}\,\text{m} = 6 \times 10^{-7}\,\text{m}$ ,
$D$ = slit-screen distance, $1\,\text{m}$ ,
$d$ = instantaneous slit separation.

The slit separation oscillates as $d = \bigl(0.8 + 0.04 \sin \omega t\bigr)\,\text{mm}$ with $\omega = 0.08\,\text{rad s}^{-1}$ . Hence

Maximum separation (when $\sin \omega t = +1$ ): $d_{\max} = (0.8 + 0.04)\,\text{mm} = 0.84\,\text{mm} = 0.84 \times 10^{-3}\,\text{m}$ .
Minimum separation (when $\sin \omega t = -1$ ): $d_{\min} = (0.8 - 0.04)\,\text{mm} = 0.76\,\text{mm} = 0.76 \times 10^{-3}\,\text{m}$ .

Because $y_m$ is inversely proportional to $d$ , the fringe lies farthest from $O$ when $d$ is minimum and closest when $d$ is maximum.

Extreme positions of the 8th bright fringe:

$y_{\max} = \dfrac{8 \times 6 \times 10^{-7} \times 1}{0.76 \times 10^{-3}} = \dfrac{4.8 \times 10^{-6}}{0.76 \times 10^{-3}} = \left( \dfrac{4.8}{0.76} \right) \times 10^{-3}\,\text{m} \approx 6.3158 \times 10^{-3}\,\text{m} = 6.3158\,\text{mm}$ .

$y_{\min} = \dfrac{8 \times 6 \times 10^{-7} \times 1}{0.84 \times 10^{-3}} = \dfrac{4.8 \times 10^{-6}}{0.84 \times 10^{-3}} = \left( \dfrac{4.8}{0.84} \right) \times 10^{-3}\,\text{m} \approx 5.7143 \times 10^{-3}\,\text{m} = 5.7143\,\text{mm}$ .

Separation between the two extreme positions:

$\Delta y = y_{\max} - y_{\min} = (6.3158 - 5.7143)\,\text{mm} = 0.6015\,\text{mm}$ .

Converting to micrometres: $0.6015\,\text{mm} = 0.6015 \times 10^{3}\,\mu\text{m} = 601.5\,\mu\text{m}$ .

Hence, the required separation is 601.50 µm.

Q 5PhysicsSection B • Numericalheat and thermodynamics

+4 / -0

A spherical soap bubble inside an air chamber at pressure

P_0=10^5 \mathrm{~Pa}

has a certain radius so that the excess pressure inside the bubble is

\Delta P=144 \mathrm{~Pa}

. Now, the chamber pressure is reduced to

8 P_0 / 27

so that the bubble radius and its excess pressure change. In this process, all the temperatures remain unchanged. Assume air to be an ideal gas and the excess pressure

\Delta P

in both the cases to be much smaller than the chamber pressure. The new excess pressure

\Delta P

in

\mathrm{Pa}

is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:96

Step-by-Step Solution:

Let the outside (chamber) pressure initially be $P_0 = 10^{5}\,\text{Pa}$ .
Inside the soap bubble the pressure is higher by the excess pressure $\Delta P$ produced by surface tension, so

$P_{\text{in,1}} = P_0 + \Delta P.$

For a soap bubble (two surfaces) the excess pressure is related to the surface tension $T$ and the bubble radius $R$ by

$\Delta P = \dfrac{4T}{R} \qquad -(1)$

The chamber pressure is now reduced to $P_2 = \dfrac{8P_0}{27}$ while temperature is kept constant. Because no air escapes from the bubble, the amount of gas inside remains the same, so the ideal-gas relation at constant temperature applies:

$P_{\text{in,1}}V_1 = P_{\text{in,2}}V_2,$

where $V = \dfrac{4}{3}\pi R^{3}$ is the bubble volume. Using $P_{\text{in,1}} = P_0 + \Delta P$ and $P_{\text{in,2}} = P_2 + \Delta P_2$ , we get

$(P_0 + \Delta P)R^{3} = (P_2 + \Delta P_2)R_2^{3}. \qquad -(2)$

At the new equilibrium the excess pressure is

$\Delta P_2 = \dfrac{4T}{R_2}. \qquad -(3)$

Divide $(1)$ by $(3)$ to relate the two radii:

$\dfrac{\Delta P}{\Delta P_2} = \dfrac{R_2}{R} \;\;\Longrightarrow\;\; R_2 = R\dfrac{\Delta P}{\Delta P_2}. \qquad -(4)$

The statement “ $\Delta P$ is much smaller than the chamber pressure” lets us neglect the terms $\Delta P$ and $\Delta P_2$ compared with $P_0$ and $P_2$ . Thus $P_{\text{in,1}}\approx P_0$ and $P_{\text{in,2}}\approx P_2$ , and equation $(2)$ simplifies to

$P_0 R^{3} = P_2 R_2^{3}. \qquad -(5)$

Substituting $R_2$ from $(4)$ into $(5)$ :

$P_0 R^{3} = P_2\!\left(R\dfrac{\Delta P}{\Delta P_2}\right)^{3} \;\;\Longrightarrow\;\; \dfrac{\Delta P}{\Delta P_2} = \left(\dfrac{P_2}{P_0}\right)^{1/3}. \qquad -(6)$

Rearranging $(6)$ gives

$\Delta P_2 = \Delta P\left(\dfrac{P_2}{P_0}\right)^{1/3}. \qquad -(7)$

Insert the numerical values: $\Delta P = 144\,\text{Pa}$ and $P_2/P_0 = \dfrac{8}{27}.$ Because $\left(\dfrac{8}{27}\right)^{1/3} = \dfrac{2}{3},$

$\Delta P_2 = 144 \times \dfrac{2}{3} = 96\,\text{Pa}.$

Therefore, the new excess pressure in the bubble is 96 Pa.

Q 6PhysicsSection B • Numericalray optics

+4 / -0

Two equilateral-triangular prisms $\mathrm{P}_1$ and $\mathrm{P}_2$ are kept with their sides parallel to each other, in vacuum, as shown in the figure. A light ray enters prism $\mathrm{P}_1$ at an angle of incidence $\theta$ such that the outgoing ray undergoes minimum deviation in prism $\mathrm{P}_2$ . If the respective refractive indices of $\mathrm{P}_1$ and $\mathrm{P}_2$ are $\sqrt{\frac{3}{2}}$ and $\sqrt{3}$ , then $\theta=\sin ^{-1}\left[\sqrt{\frac{3}{2}} \sin \left(\frac{\pi}{\beta}\right)\right]$ , where the value of $\beta$ is ____.

$qImage-refraction-tir-and-prism-geometrical-optics-optics-Jeemate$

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:12

Step-by-Step Solution:

For each equilateral-triangular prism the refracting (apex) angle is $A = 60^\circ = \dfrac{\pi}{3}$ .

Step 1: Condition for minimum deviation in prism $P_2$
For minimum deviation in any prism, the path of the ray is symmetrical, so $r = r' = \dfrac{A}{2} = 30^\circ$ inside the prism, and the angles of incidence and emergence are equal: $i = e$ .

Applying Snell’s law at the first face of $P_2$ (outside medium is vacuum, $n = 1$ ):
$\sin i = n_2 \sin r$

Given $n_2 = \sqrt{3}$ , therefore
$\sin i = \sqrt{3}\,\sin 30^\circ = \sqrt{3}\left(\dfrac{1}{2}\right)=\dfrac{\sqrt{3}}{2}$
so $i = 60^\circ = \dfrac{\pi}{3}$ .

Hence the light ray must strike the first face of $P_2$ with an incidence angle of $60^\circ$ .

Step 2: This same ray is the emergent ray from prism $P_1$
The two prisms are arranged with their corresponding faces parallel; therefore the angle the ray makes with the normal when it leaves $P_1$ is the same $60^\circ$ . Let that angle of emergence from $P_1$ be $e_1 = 60^\circ$ .

Step 3: Ray path inside prism $P_1$
Let the angles of refraction at the two faces of $P_1$ be $r_1$ and $r_2$ . For any prism
$r_1 + r_2 = A = 60^\circ \qquad -(1)$

Snell’s law at the second face (emergence) of $P_1$ :
$n_1 \sin r_2 = \sin e_1$
Given $e_1 = 60^\circ$ and $n_1 = \sqrt{\dfrac{3}{2}}$ , we get
$\sin r_2 = \dfrac{\sin 60^\circ}{n_1} = \dfrac{\dfrac{\sqrt{3}}{2}}{\sqrt{\dfrac{3}{2}}} = \dfrac{\sqrt{2}}{2} = \sin 45^\circ$
so $r_2 = 45^\circ$ .

Using $(1)$ :
$r_1 = 60^\circ - 45^\circ = 15^\circ.$

Step 4: Incidence angle $\theta$ on the first face of $P_1$
Apply Snell’s law at the first face of $P_1$ :
$\sin \theta = n_1 \sin r_1 = \sqrt{\dfrac{3}{2}}\;\sin 15^\circ$

Since $15^\circ = \dfrac{\pi}{12}$ ,
$\sin \theta = \sqrt{\dfrac{3}{2}}\;\sin\!\left(\dfrac{\pi}{12}\right).$

The question states this result as $\theta = \sin^{-1}\!\left[\sqrt{\dfrac{3}{2}}\;\sin\!\left(\dfrac{\pi}{\beta}\right)\right].$ Comparing, we must have $\dfrac{\pi}{\beta} = \dfrac{\pi}{12}\quad\Longrightarrow\quad \beta = 12.$

Hence the required value is 12.

Q 7PhysicsSection A • MCQmagnetism

+3 / -1

A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass $m$ and radius $r$ and it is in a uniform vertical magnetic field $B_0$ , as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity $g$ , on two conducting supports at $\mathrm{P}$ and $\mathrm{Q}$ . When a current $I$ is passed through the loop, the loop turns about the line $\mathrm{PQ}$ by an angle $\theta$ given by

qImage-force-and-torque-on-current-carrying-conductor-magnetism-electricity-Jeemate

A

\tan \theta=\pi r I B_0 /(m g)

B

\tan \theta=2 \pi r I B_0 /(m g)

C

\tan \theta=\pi r I B_0 /(2 m g)

D

\tan \theta=m g /\left(\pi r I B_0\right)

Q 8PhysicsSection B • Numericalelectrostatics

+4 / -2

A small electric dipole $\vec{p}_0$ , having a moment of inertia $I$ about its center, is kept at a distance $r$ from the center of a spherical shell of radius $R$ . The surface charge density $\sigma$ is uniformly distributed on the spherical shell. The dipole is initially oriented at a small angle $\theta$ as shown in the figure. While staying at a distance $r$ , the dipole is free to rotate about its center.

If released from rest, then which of the following statement(s) is(are) correct?

[ $\varepsilon_0$ is the permittivity of free space.]

Q 9PhysicsSection B • Numericalproperties of matter

+4 / -2

A table tennis ball has radius $(3 / 2) \times 10^{-2} \mathrm{~m}$ and mass $(22 / 7) \times 10^{-3} \mathrm{~kg}$ . It is slowly pushed down into a swimming pool to a depth of $d=0.7 \mathrm{~m}$ below the water surface and then released from rest. It emerges from the water surface at speed $v$ , without getting wet, and rises up to a height $H$ . Which of the following option(s) is(are) correct?

[Given: $\pi=22 / 7, g=10 \mathrm{~m} \mathrm{~s}^{-2}$ , density of water $=1 \times 10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ , viscosity of water $=1 \times 10^{-3} \mathrm{~Pa}$ -s.]

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

Radius of the ball, $r = \frac{3}{2}\times 10^{-2}\;{\rm m}=0.015\;{\rm m}$
Mass of the ball, $m = \frac{22}{7}\times 10^{-3}\;{\rm kg}= \pi\times 10^{-3}\;{\rm kg}=3.1429\times 10^{-3}\;{\rm kg}$

Volume of the ball
$V = \frac{4}{3}\pi r^{3}$
$r^{3}=0.015^{3}=3.375\times 10^{-6}\,{\rm m^{3}}$
$V=\frac{4}{3}\pi(3.375\times 10^{-6})=4.18879\times3.375\times10^{-6}=1.414\times10^{-5}\;{\rm m^{3}}$

Buoyant force in water
$F_{b} = \rho_{\text w}gV = (1000)(10)(1.414\times10^{-5}) = 0.1414\;{\rm N}$

Weight of the ball
$W = mg = (3.1429\times10^{-3})(10)=0.03143\;{\rm N}$

Net upward force (viscosity neglected)
$F_{\text{net}} = F_{b}-W = 0.1414-0.03143 = 0.1100\;{\rm N}\approx 0.11\;{\rm N}$

Option A - Work done in pushing the ball down by $d=0.7\;{\rm m}$ (quasi-static)
To move slowly we must apply a downward force equal to the constant upward imbalance $F_{b}-W$ . $W_{\text{push}} = (F_{b}-W)\,d = (0.11)(0.7)=0.077\;{\rm J}$
Hence Option A is correct.

Option B - Speed $v$ at the water surface (viscosity neglected)
Acceleration inside water: $a=\dfrac{F_{\text{net}}}{m}=\dfrac{0.11}{3.1429\times10^{-3}}\approx35\;{\rm m\,s^{-2}}$
Using $v^{2}=2as$ with $s=d=0.7\;{\rm m}$ :
$v=\sqrt{2(35)(0.7)}=\sqrt{49}=7\;{\rm m\,s^{-1}}$
Thus Option B is correct.

Option C - Height $H$ above the surface
After emerging, only gravity acts. $H=\dfrac{v^{2}}{2g}=\dfrac{49}{20}=2.45\;{\rm m}$ , not $1.4\;{\rm m}$ . Therefore Option C is incorrect.

Option D - Ratio of (net force) to (maximum viscous force)
Maximum possible viscous (Stokes) drag inside water occurs at speed $v=7\;{\rm m\,s^{-1}}$ :
$F_{\eta,\max}=6\pi\eta rv = 6\left(\frac{22}{7}\right)(1\times10^{-3})(0.015)(7)$
$(\frac{22}{7})\times7 = 22$ , so $F_{\eta,\max}=6(22)(0.015)\times10^{-3}=1.98\times10^{-3}\;{\rm N}$

Hence $\dfrac{F_{\text{net}}}{F_{\eta,\max}}=\dfrac{0.11}{1.98\times10^{-3}}\approx55.56=\frac{500}{9}$
Option D is therefore correct.

Final result: Option A, Option B and Option D are correct.

Q 10PhysicsSection B • Numericalsimple harmonic motion

+3 / -0

Read the passage carefully:

Two particles, 1 and 2, each of mass $m$ , are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at $x_0$ , are oscillating with amplitude $a$ and angular frequency $\omega$ . Thus, their positions at time $t$ are given by $x_1(t)=\left(x_0+d\right)+a \sin \omega t$ and $x_2(t)=\left(x_0-d\right)-a \sin \omega t$ , respectively, where $d>2 a$ . Particle 3 of mass $m$ moves towards this system with speed $u_0=a \omega / 2$ , and undergoes instantaneous elastic collision with particle 2 , at time $t_0$ . Finally, particles 1 and 2 acquire a center of mass speed $v_{\mathrm{cm}}$ and oscillate with amplitude $b$ and the same angular frequency $\omega$ .

qImage--simple-harmonic-motion-mechanics-Jeemate

If the collision occurs at time

t_0=0

, the value of

v_{\mathrm{cm}} /(a \omega)

will be ______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 11PhysicsSection A • MCQelectromagnetic induction

+3 / -1

A region in the form of an equilateral triangle (in $x-y$ plane) of height $L$ has a uniform magnetic field $\vec{B}$ pointing in the $+z$ -direction. A conducting loop $\mathrm{PQR}$ , in the form of an equilateral triangle of the same height $L$ , is placed in the $x-y$ plane with its vertex $\mathrm{P}$ at $x=0$ in the orientation shown in the figure. At $t=0$ , the loop starts entering the region of the magnetic field with a uniform velocity $\vec{v}$ along the $+x$ -direction. The plane of the loop and its orientation remain unchanged throughout its motion.

qImage-motional-emf-and-eddy-current-electromagnetic-induction-electricity-Jeemate

Which of the following graph best depicts the variation of the induced emf $(E)$ in the loop as a function of the distance $(x)$ starting from $x=0$ ?

A

B

C

D

Q 12PhysicsSection B • Numericalmagnetism

+4 / -2

A positive, singly ionized atom of mass number $A_{\mathrm{M}}$ is accelerated from rest by the voltage $192 \mathrm{~V}$ . Thereafter, it enters a rectangular region of width $w$ with magnetic field $\vec{B}_0=0.1 \hat{k}$ Tesla, as shown in the figure. The ion finally hits a detector at the distance $x$ below its starting trajectory.

[Given: Mass of neutron/proton $=(5 / 3) \times 10^{-27} \mathrm{~kg}$ , charge of the electron $=1.6 \times 10^{-19} \mathrm{C}$ .]

qImage-motion-of-charged-particle-inside-magnetic-field-magnetism-electricity-Jeemate

Which of the following option(s) is(are) correct?

Q 13PhysicsSection B • Numericalmotion

+4 / -0

A ball is thrown from the location

\left(x_0, y_0\right)=(0,0)

of a horizontal playground with an initial speed

v_0

at an angle

\theta_0

from the

+x

-direction. The ball is to be hit by a stone, which is thrown at the same time from the location

\left(x_1, y_1\right)=(L, 0)

. The stone is thrown at an angle

\left(180-\theta_1\right)

from the

+x

-direction with a suitable initial speed. For a fixed

v_0

, when

\left(\theta_0, \theta_1\right)=\left(45^{\circ}, 45^{\circ}\right)

, the stone hits the ball after time

T_1

, and when

\left(\theta_0, \theta_1\right)=\left(60^{\circ}, 30^{\circ}\right)

, it hits the ball after time

T_2

. In such a case,

\left(T_1 / T_2\right)^2

is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:2

Step-by-Step Solution:

Let the ball start from the origin with speed $v_0$ and projection angle $\theta_0$ (measured from the +x-axis).
Its velocity components are $v_{bx}=v_0\cos\theta_0$ and $v_{by}=v_0\sin\theta_0$ .

The stone is thrown simultaneously from $(L,0)$ at an angle $(180^\circ-\theta_1)$ with the +x-axis. Hence its horizontal component is toward the negative x-direction:
$v_{sx}=-v_s\cos\theta_1$ , $v_{sy}=v_s\sin\theta_1$ , where $v_s$ is the stone’s launch speed.

Suppose they meet after time $t$ (same $t$ for both because they start together).

Horizontal co-ordinate equality
$x_{\text{ball}} = x_{\text{stone}}$ gives $v_0\cos\theta_0\,t = L + (-v_s\cos\theta_1)\,t$ $\Rightarrow t\,(v_0\cos\theta_0+v_s\cos\theta_1)=L \quad -(1)$

Vertical co-ordinate equality
Both experience the same downward acceleration $g$ , so the $-\dfrac12 g t^2$ term cancels: $v_0\sin\theta_0\,t = v_s\sin\theta_1\,t$ Since $t\gt 0$ , $v_0\sin\theta_0 = v_s\sin\theta_1 \quad -(2)$

From $(2)$ , express the stone’s speed:
$v_s = \dfrac{v_0\sin\theta_0}{\sin\theta_1} \quad -(3)$

Insert $(3)$ into $(1)$ to obtain the common time of flight:

$t = \dfrac{L}{v_0\cos\theta_0 + \left(\dfrac{v_0\sin\theta_0}{\sin\theta_1}\right)\cos\theta_1 } = \dfrac{L}{v_0\left(\cos\theta_0+\sin\theta_0\cot\theta_1\right)} \quad -(4)$

Thus, for any pair $(\theta_0,\theta_1)$ , $T(\theta_0,\theta_1)=\dfrac{L}{v_0\left(\cos\theta_0+\sin\theta_0\cot\theta_1\right)}$

Case 1: $(\theta_0,\theta_1)=(45^\circ,45^\circ)$ $\cos45^\circ=\sin45^\circ=\dfrac{\sqrt2}{2}, \;\cot45^\circ=1$
Denominator in $(4)$ : $\cos45^\circ+\sin45^\circ\cot45^\circ = \dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}= \sqrt2$ Hence $T_1=\dfrac{L}{v_0\sqrt2} \quad -(5)$

Case 2: $(\theta_0,\theta_1)=(60^\circ,30^\circ)$ $\cos60^\circ=\dfrac12,\; \sin60^\circ=\dfrac{\sqrt3}{2},\; \cot30^\circ=\sqrt3$
Denominator in $(4)$ : $\dfrac12+\dfrac{\sqrt3}{2}\,\sqrt3 = \dfrac12+\dfrac{3}{2}=2$ Thus $T_2=\dfrac{L}{2v_0} \quad -(6)$

Required ratio
$\left(\dfrac{T_1}{T_2}\right)^2 =\left(\dfrac{\,L/(v_0\sqrt2)\,}{\,L/(2v_0)\,}\right)^2 =\left(\dfrac{2}{\sqrt2}\right)^2 =\left(\sqrt2\right)^2 =2$

Therefore, $(T_1/T_2)^2 = 2$ .

Q 14PhysicsSection A • MCQgravitation

+3 / -1

A particle of mass $m$ is under the influence of the gravitational field of a body of mass $M(\gg m)$ . The particle is moving in a circular orbit of radius $r_0$ with time period $T_0$ around the mass $M$ . Then, the particle is subjected to an additional central force, corresponding to the potential energy $V_{\mathrm{c}}(r)=m \alpha / r^3$ , where $\alpha$ is a positive constant of suitable dimensions and $r$ is the distance from the center of the orbit. If the particle moves in the same circular orbit of radius $r_0$ in the combined gravitational potential due to $M$ and $V_{\mathrm{c}}(r)$ , but with a new time period $T_1$ , then $\left(T_1^2-T_0^2\right) / T_1^2$ is given by

[G is the gravitational constant.]

A

\frac{3 \alpha}{G M r_0^2}

B

\frac{\alpha}{2 G M r_0^2}

C

\frac{\alpha}{G M r_0^2}

D

\frac{2 \alpha}{G M r_0^2}

Official Correct Answer:Option (A)

Step-by-Step Solution:

The gravitational potential energy due to the central mass $M$ is
$V_g(r)= -\dfrac{G M m}{r}\,.$

For a circular orbit of radius $r_0$ under gravity alone, the inward (centripetal) force balance is
$\dfrac{m v_0^{\,2}}{r_0}= \dfrac{G M m}{r_0^{2}} \;\;\Longrightarrow\;\; v_0^{\,2}= \dfrac{G M}{r_0}\,.$

The corresponding time period is therefore
$T_0 =\dfrac{2\pi r_0}{v_0}=2\pi\sqrt{\dfrac{r_0^{3}}{G M}} \;\;\Longrightarrow\;\; T_0^{2}= \dfrac{4\pi^{2} r_0^{3}}{G M}\,.$

Now an additional central potential is applied:
$V_c(r)=\dfrac{m\alpha}{r^{3}}\qquad (\alpha\gt 0).$

The radial force associated with this potential is obtained from $F_r=-\dfrac{dV_c}{dr}$ :
$F_c(r)=-\frac{d}{dr}\left(\dfrac{m\alpha}{r^{3}}\right)=+\,\dfrac{3m\alpha}{r^{4}}.$
Since this force is positive in the outward radial direction, it is repulsive. Hence it opposes gravitation.

To keep the particle in the same circular orbit of radius $r_0$ , the inward gravitational pull must now be reduced by this repulsion:
$\dfrac{m v_1^{\,2}}{r_0}= \dfrac{G M m}{r_0^{2}}-\dfrac{3m\alpha}{r_0^{4}} \;\;\Longrightarrow\;\; v_1^{\,2}= \dfrac{G M}{r_0}-\dfrac{3\alpha}{r_0^{3}}.$

The new time period becomes
$T_1 =\dfrac{2\pi r_0}{v_1}\,,\qquad T_1^{2}= \dfrac{4\pi^{2} r_0^{2}}{v_1^{\,2}} =\dfrac{4\pi^{2} r_0^{3}} {\,G M-\dfrac{3\alpha}{r_0^{2}}}\;.$ Define $A=4\pi^{2}r_0^{3}$ for brevity, so that
$T_0^{2}= \dfrac{A}{G M},\qquad T_1^{2}= \dfrac{A}{\,G M-\dfrac{3\alpha}{r_0^{2}}}\;.$

We now evaluate the required ratio:
$\dfrac{T_1^{2}-T_0^{2}}{T_1^{2}} =1-\dfrac{T_0^{2}}{T_1^{2}} =1-\dfrac{\dfrac{A}{G M}} {\dfrac{A}{\,G M-\dfrac{3\alpha}{r_0^{2}}}} =1-\dfrac{G M-\dfrac{3\alpha}{r_0^{2}}}{G M} =\dfrac{3\alpha}{G M r_0^{2}}\;.$

Thus
$\boxed{\dfrac{T_1^{2}-T_0^{2}}{T_1^{2}}=\dfrac{3\alpha}{G M r_0^{2}}}\,.$
Hence the correct option is:
Option A which is: $\dfrac{3\alpha}{G M r_0^{2}}.$

Q 15PhysicsSection B • Numericalsimple harmonic motion

+3 / -0

Read the passage carefully:

Two particles, 1 and 2, each of mass $m$ , are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at $x_0$ , are oscillating with amplitude $a$ and angular frequency $\omega$ . Thus, their positions at time $t$ are given by $x_1(t)=\left(x_0+d\right)+a \sin \omega t$ and $x_2(t)=\left(x_0-d\right)-a \sin \omega t$ , respectively, where $d>2 a$ . Particle 3 of mass $m$ moves towards this system with speed $u_0=a \omega / 2$ , and undergoes instantaneous elastic collision with particle 2 , at time $t_0$ . Finally, particles 1 and 2 acquire a center of mass speed $v_{\mathrm{cm}}$ and oscillate with amplitude $b$ and the same angular frequency $\omega$ .

If the collision occurs at time

t_0=\pi /(2 \omega)

, then the value of

4 b^2 / a^2

will be ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:4.25

Step-by-Step Solution:

At any instant the coordinates of the two identical masses are given as
$x_1(t)=\bigl(x_0+d\bigr)+a\sin\omega t,\qquad x_2(t)=\bigl(x_0-d\bigr)-a\sin\omega t$

Hence their centre-of-mass (C-O-M) position is
$x_{cm}=\frac{x_1+x_2}{2}=x_0$ which is fixed (initially the C-O-M is at rest).

For the system “two masses $m$ joined by a mass-less spring of force constant $k$ ” the oscillation (stretch-compress) mode has angular frequency
$\omega=\sqrt{\frac{2k}{m}}\;.\qquad -(1)$

At the instant of collision, $t_0=\dfrac{\pi}{2\omega}\;,$ we have
$\sin\omega t_0 =\sin\frac{\pi}{2}=1,\qquad \cos\omega t_0 =0.$

Therefore, just before collision (denote by the superscript “−”)
$v_1^-=\dot x_1= a\omega\cos\omega t_0 =0,$ $v_2^-=\dot x_2=-a\omega\cos\omega t_0 =0.$ Both particles of the spring system are instantaneously at rest.

A third particle, also of mass $m$ , approaches particle 2 with speed
$u_0=\frac{a\omega}{2}.$

Because the collision between 3 and 2 is perfectly elastic and the two masses are equal, the standard 1-D result gives
$v_3^+=0,\qquad v_2^+=u_0$ immediately after collision (superscript “+”). Particle 1 is untouched, so $v_1^+=0$ .

Hence the C-O-M velocity acquired by the spring system (masses 1 & 2) is
$v_{cm}=\frac{v_1^+ + v_2^+}{2} =\frac{u_0}{2} =\frac{a\omega}{4}.$

The internal (relative) motion is analysed in the C-O-M frame. Let the extension of the spring from its natural length be $x = (x_1-x_2)-2d.$ Initially, the oscillation had amplitude $a$ for each mass, so the relative amplitude was $2a$ . At $t=t_0$ the spring is at its extreme extension:
$x_0 = 2a,\qquad \dot x^- = 0.$

Immediately after collision the relative velocity becomes $\dot x^+ = v_1^+ - v_2^+ = 0 - u_0 = -u_0 = -\frac{a\omega}{2}.$ (The negative sign shows the spring will start to compress.)

Total internal energy in the C-O-M frame just after collision is the sum of
(i) spring potential energy and (ii) kinetic energy of relative motion.
For the two-body system the reduced mass is $\mu=\dfrac{m}{2}$ , and with $k$ related to $\omega$ by (1): $k=\dfrac{m\omega^2}{2}$ .

Thus
$E'=\frac12 kx_0^{\,2}+\frac12\mu\dot x^{+2} =\frac12\left(\frac{m\omega^2}{2}\right)(2a)^2 +\frac12\left(\frac{m}{2}\right)\left(\frac{a\omega}{2}\right)^2 =2k a^2+\frac{mu_0^{\,2}}{4}\;.$

After the collision the two masses again execute simple harmonic motion (same $\omega$ ) but with a larger amplitude $b$ . At an extreme, kinetic energy is zero, so
$E'=\frac12 k(2b)^2 = 2k b^2.$

Equating the two expressions for $E'$ :
$2k b^2 = 2k a^2 + \frac{m u_0^{\,2}}{4}.$

Divide by $2k$ and substitute $u_0=\dfrac{a\omega}{2}$ and $k=\dfrac{m\omega^2}{2}$ :
$b^2 = a^2 + \frac{m(a\omega/2)^2}{4}\,\frac{1}{2k} = a^2 + \frac{a^2\omega^2/4}{4\omega^2} = a^2 + \frac{a^2}{16} = a^2\left(1 + \frac1{16}\right) = a^2\frac{17}{16}.$

Therefore
$\frac{4b^2}{a^2} = 4\left(\frac{17}{16}\right) = \frac{17}{4}=4.25.$

Hence, the required value is 4.25.

Q 16PhysicsSection B • Numericalwave optics

+3 / -0

Read the passage carefully:

In a Young's double slit experiment, each of the two slits A and B, as shown in the figure, are oscillating about their fixed center and with a mean separation of

0.8 \mathrm{~mm}

. The distance between the slits at time

t

is given by

d=(0.8+0.04 \sin \omega t) \mathrm{mm}

, where

\omega=0.08 \mathrm{rad} \mathrm{s}^{-1}

. The distance of the screen from the slits is

1 \mathrm{~m}

and the wavelength of the light used to illuminate the slits is

6000

Å . The interference pattern on the screen changes with time, while the central bright fringe (zeroth fringe) remains fixed at point

O

.

The maximum speed in

\mu \mathrm{m} / \mathrm{s}

at which the

8^{\text {th }}

bright fringe will move is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:24

Step-by-Step Solution:

For the 8th bright fringe in Young’s double-slit experiment, the instantaneous position on the screen is

$y_m(t)=\frac{m\lambda D}{d(t)}$

with $m=8,\;D=1\text{ m},\;\lambda =6000\text{ \AA}=6\times10^{-7}\text{ m}$ and

$d(t)=\bigl(0.8+0.04\sin\omega t\bigr)\text{ mm} =(8\times10^{-4}+4\times10^{-5}\sin\omega t)\text{ m},\qquad \omega =0.08\text{ rad s}^{-1}$

The speed of this fringe is obtained by differentiating:

$v(t)=\frac{dy_m}{dt} =-\frac{m\lambda D}{\bigl[d(t)\bigr]^2}\,\frac{dd}{dt}$

First derivative of the slit separation:

$\frac{dd}{dt}=4\times10^{-5}\;\omega\cos\omega t =4\times10^{-5}\times0.08\cos\omega t =3.2\times10^{-6}\cos\omega t\;(\text{m s}^{-1})$

Hence

$v(t)=\frac{8\,(6\times10^{-7})(3.2\times10^{-6})\, |\cos\omega t|}{\bigl(8\times10^{-4}+4\times10^{-5}\sin\omega t\bigr)^2}$

To find the maximum value, put $\theta=\omega t$ and maximise

$f(\theta)=\frac{|\cos\theta|} {\bigl(a+b\sin\theta\bigr)^2},\qquad a=8\times10^{-4},\;b=4\times10^{-5}$

Within $0\le\theta\lt2\pi$ , choose the region $\cos\theta\gt0$ (same magnitude repeats elsewhere).
Set $\displaystyle\frac{d}{d\theta}\ln f(\theta)=0$ :

$-\tan\theta-\frac{2b\cos\theta}{a+b\sin\theta}=0$

After simplification, using $x=\sin\theta$ :

$b x^{2}-a x-2b=0$

Substituting $a=8\times10^{-4},\;b=4\times10^{-5}$ gives

$x^{2}-20x-2=0\; \Longrightarrow\; x=\sin\theta\approx-0.0995$

With this $\cos\theta=\sqrt{1-x^{2}}\approx0.995$ and

$d_{\text{max}}=a+b\sin\theta =8\times10^{-4}+4\times10^{-5}(-0.0995) \approx7.960\times10^{-4}\text{ m}$

Finally, the maximum fringe speed is

$v_{\max} =\frac{8(6\times10^{-7})(3.2\times10^{-6})(0.995)} {(7.960\times10^{-4})^{2}} \approx2.4\times10^{-5}\text{ m s}^{-1}$

Expressing in micrometres per second:

$v_{\max}\approx2.4\times10^{-5}\text{ m s}^{-1} =24\text{ }\mu\text{m s}^{-1}$

Therefore, the maximum speed of the 8th bright fringe is 24 μm/s.

Q 17PhysicsSection B • Numericalunits and measurements

+4 / -0

The dimensions of a cone are measured using a scale with a least count of

2 \mathrm{~mm}

. The diameter of the base and the height are both measured to be

20.0 \mathrm{~cm}

. The maximum percentage error in the determination of the volume is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:3

Step-by-Step Solution:

To determine the maximum percentage error in the volume of the cone, we first need to understand the dependence of the cone's volume on its dimensions: the diameter of the base and its height.

The volume $V$ of a cone is given by the formula:

$V = \frac{1}{3} \pi r^2 h$

where $r$ is the radius of the base and $h$ is the height.

Given that both the diameter ( $D$ ) of the base and the height ( $h$ ) are measured to be $20.0 \mathrm{~cm}$ , we can find the radius $r$ as follows:

$r = \frac{D}{2} = \frac{20.0 \mathrm{~cm}}{2} = 10.0 \mathrm{~cm}$

Now, let's denote the errors in measuring the diameter and height as $\Delta D$ and $\Delta h$ respectively.

Since the measurements are taken with a scale having a least count of $2 \mathrm{~mm}$ , we have:

$\Delta D = 2 \mathrm{~mm} = 0.2 \mathrm{~cm}$

and

$\Delta h = 2 \mathrm{~mm} = 0.2 \mathrm{~cm}$

To find the maximum percentage error in the volume, we need to use the formula for the propagation of relative errors. Considering the volume formula $V = \frac{1}{3} \pi r^2 h$ , the relative errors in $r$ and $h$ will propagate into the volume as follows:

The relative error in the radius $\Delta r / r$ is:

$\left(\frac{\Delta r}{r}\right) = \left(\frac{\Delta D / 2}{D / 2}\right) = \left(\frac{\Delta D}{D}\right) = \frac{0.2 \mathrm{~cm}}{20.0 \mathrm{~cm}} = 0.01$

The relative error in the height $\Delta h / h$ is:

$\left(\frac{\Delta h}{h}\right) = \frac{0.2 \mathrm{~cm}}{20.0 \mathrm{~cm}} = 0.01$

Since $V$ is proportional to $r^2$ and $h$ , the total relative error in the volume is given by:

$\left(\frac{\Delta V}{V}\right) = 2 \left(\frac{\Delta r}{r}\right) + \left(\frac{\Delta h}{h}\right)$

Substituting the relative errors, we get:

$\left(\frac{\Delta V}{V}\right) = 2 (0.01) + 0.01 = 0.02 + 0.01 = 0.03$

Hence, the maximum percentage error in the volume is:

$0.03 \times 100\% = 3\%$

Therefore, the maximum percentage error in the determination of the volume is 3%.

Q 18ChemistrySection B • Numericalcoordination compounds

+4 / -0

Among

\left[\mathrm{Co}(\mathrm{CN})_4\right]^{4-},\left[\mathrm{Co}(\mathrm{CO})_3(\mathrm{NO})\right], \mathrm{XeF}_4,\left[\mathrm{PCl}_4\right]^{+},\left[\mathrm{PdCl}_4\right]^{2-},\left[\mathrm{ICl}_4\right]^{-},\left[\mathrm{Cu}(\mathrm{CN})_4\right]^{3-}

and

\mathrm{P}_4

the total number of species with tetrahedral geometry is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:4OR5

Step-by-Step Solution:

$[\mathrm{Co}(\mathrm{CN})_4]^{4-} \Rightarrow \mathrm{Co}^0 \Rightarrow 3 d^7 4 s^2$

Due to SFL, $\mathrm{CN}^{-}$ pairing and transference of electron takes place and hybridisation is $d s p^2$

Geometry $\Rightarrow$ Square planer

qImage--coordination-compounds-inorganic-chemistry-Jeemate

OR

$\begin{gathered}\Rightarrow \text { hybridisation }\left[\mathrm{Co}(\mathrm{CN})_4\right]^{4-} \end{gathered}$ $\to$ $\mathrm{sp}^3$

$\begin{aligned} & {\left[\mathrm{Co}(\mathrm{CO})_3(\mathrm{NO})\right] \rightarrow \mathrm{Co}^{-1} \rightarrow 3\mathrm{d}^8 4 \mathrm{s}^2 \rightarrow 3 \mathrm{d}^{10} \text { ( In presence of S.F.L) }} \\ & \quad \text { here NO present as }+1 \text { state } \\ & \Rightarrow \text { Hybridisation } \rightarrow\left[\mathrm{Co}(\mathrm{CO})_3(\mathrm{NO})\right] \rightarrow \mathrm{sp}^3\end{aligned}$

Geometry $=$ Tetrahedral

$\mathrm{XeF}_4 \Rightarrow 4 \mathrm{bp}+2 lp \Rightarrow s p^3 d^2$

Geometry = Square planer

$\mathrm{PCl}_4^{+} \Rightarrow 4 \mathrm{lb}+0 \mathrm{lp}$

$s p^3 \Rightarrow$ tetrahedral

$\left[\mathrm{PdCl}_4\right]^{2-} \Rightarrow \mathrm{Pd}^{2+}, \mathrm{Cl}^{-}$ behaves as $\mathrm{SFL}$

$\mathrm{Pd}^{2+} \Rightarrow 4 d^8 \Rightarrow d s p^2 \Rightarrow$ square planer

$\begin{array}{r} \mathrm{ICl}_4^{\ominus}\Rightarrow 4 \mathrm{bp}+2 \mathrm{lp} \\ Hybridisation = s p^3 d^2 \end{array}$

Geometry = square planer

$\begin{aligned} {\left.[\mathrm{Cu}(\mathrm{CN}))_4\right]^{3-} } & \Rightarrow \mathrm{Cu}^{+1} \Rightarrow 3 d^{10} \Rightarrow s p^3\end{aligned}$

Geometry = Tetrahedral

$P_4$ is tetrahedral

Q 19ChemistrySection B • Numericalcompounds containing nitrogen

+4 / -2

The option(s) with correct sequence of reagents for the conversion of $\mathbf{P}$ to $\mathbf{Q}$ is(are)

qImage-diazonium-salts-and-other-nitrogen-containing-compounds-compounds-containing-nitrogen-organic-chemistry-Jeemate

Q 20ChemistrySection B • Numericalaldehydes ketones and carboxylic acids

+3 / -0

Read the passage carefully:

An organic compound $\mathbf{P}$ with molecular formula $\mathrm{C}_9 \mathrm{H}_{18} \mathrm{O}_2$ decolorizes bromine water and also shows positive iodoform test. $\mathbf{P}$ on ozonolysis followed by treatment with $\mathrm{H}_2 \mathrm{O}_2$ gives $\mathbf{Q}$ and $\mathbf{R}$ . While compound $\mathbf{Q}$ shows positive iodoform test, compound $\mathbf{R}$ does not give positive iodoform test. $\mathbf{Q}$ and $\mathbf{R}$ on oxidation with pyridinium chlorochromate (PCC) followed by heating give $\mathbf{S}$ and $\mathbf{T}$ , respectively. Both $\mathbf{S}$ and $\mathbf{T}$ show positive iodoform test.

Complete copolymerization of 500 moles of $\mathbf{Q}$ and 500 moles of $\mathbf{R}$ gives one mole of a single acyclic copolymer $\mathbf{U}$ .

[Given, atomic mass: $\mathrm{H}=1, \mathrm{C}=12, \mathrm{O}=16$ ]

Sum of number of oxygen atoms in

\mathbf{S}

and

\mathbf{T}

is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 21ChemistrySection B • Numericalbiomolecules

+4 / -0

For a double strand DNA, one strand is given below:

qImage-vitamins-and-nucleic-acids-biomolecules-organic-chemistry-Jeemate

The amount of energy required to split the double strand DNA into two single strands is _______ kcal $\operatorname{mol}^{-1}$ .

[Given: Average energy per H-bond for A-T base pair $=1.0 ~\mathrm{kcal}~ \mathrm{mol}^{-1}$ , G-C base pair $=1.5 ~\mathrm{kcal}$ $\mathrm{mol}^{-1}$ , and A-U base pair $=1.25 ~\mathrm{kcal} ~\mathrm{mol}^{-1}$ . Ignore electrostatic repulsion between the phosphate groups.]

Answer Format:Enter numerical value rounded to the nearest integer

Q 22ChemistrySection B • Numericalp block elements

+4 / -2

The compound(s) having peroxide linkage is(are)

Q 23ChemistrySection B • Numericalsurface chemistry

+4 / -0

To form a complete monolayer of acetic acid on $1 \mathrm{~g}$ of charcoal, $100 \mathrm{~mL}$ of $0.5 \mathrm{M}$ acetic acid was used. Some of the acetic acid remained unadsorbed. To neutralize the unadsorbed acetic acid, 40 $\mathrm{mL}$ of $1 \mathrm{M} \mathrm{NaOH}$ solution was required. If each molecule of acetic acid occupies $\mathbf{P} \times 10^{-23} \mathrm{~m}^2$ surface area on charcoal, the value of $\mathbf{P}$ is _____.

[Use given data: Surface area of charcoal $=1.5 \times 10^2 \mathrm{~m}^2 \mathrm{~g}^{-1}$ ; Avogadro's number $\left(\mathrm{N}_{\mathrm{A}}\right)=6.0 \times 10^{23}$ $\left.\mathrm{mol}^{-1}\right]$

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:2500

Step-by-Step Solution:

Let's first understand and break down the given information. We are provided with the following details to solve for the value of $\mathbf{P}$ :

1. Volume of acetic acid solution used: $100 \mathrm{~mL}$ of $0.5 \mathrm{M}$ acetic acid.

2. Volume of $\mathrm{NaOH}$ solution used for neutralization: $40 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{NaOH}$ .

3. Surface area of the charcoal: $1.5 \times 10^2 \mathrm{~m}^2 \mathrm{~g}^{-1}$ .

4. Avogadro's number: $\mathrm{N}_{\mathrm{A}}=6.0 \times 10^{23} \mathrm{~mol}^{-1}$ .

Let's start with the concentration and volume of acetic acid used:

\text{Moles of acetic acid initially} = 0.5 \text{ M} \times \frac{100 \text{ mL}}{1000} = 0.05 \text{ moles}

Next, let's determine the moles of $\mathrm{NaOH}$ used for neutralization:

\text{Moles of } \mathrm{NaOH} = 1 \text{ M} \times \frac{40 \text{ mL}}{1000} = 0.04 \text{ moles}

Since $\mathrm{NaOH}$ completely neutralizes the unadsorbed acetic acid:

\text{Moles of unadsorbed acetic acid} = 0.04 \text{ moles}

Therefore, the moles of acetic acid adsorbed on the charcoal is given by:

\text{Moles of adsorbed acetic acid} = 0.05 - 0.04 = 0.01 \text{ moles}

Now, we need to calculate the total number of molecules of acetic acid adsorbed:

\text{Number of molecules} = 0.01 \text{ moles} \times 6.0 \times 10^{23} \text{ molecules/mole} = 6.0 \times 10^{21} \text{ molecules}

Given the surface area of charcoal, we can find the total surface area:

\text{Total surface area} = 1.5 \times 10^2 \text{ m}^2 / \text{g} \times 1 \text{ g} = 1.5 \times 10^2 \text{ m}^2

Finally, we need to determine the area occupied by each molecule of acetic acid:

\text{Area per molecule of acetic acid} = \frac{\text{Total surface area}}{\text{Number of molecules}} = \frac{1.5 \times 10^2 \text{ m}^2}{6.0 \times 10^{21} \text{ molecules}}

Let's perform the calculation:

\text{Area per molecule of acetic acid} = \frac{1.5 \times 10^2}{6.0 \times 10^{21}} = 0.25 \times 10^{-19} \text{ m}^2 = 2.5 \times 10^{-20} \text{ m}^2

Since the problem states that the area per molecule is $\mathbf{P} \times 10^{-23} \mathrm{~m}^2$ , we equate:

2.5 \times 10^{-20} = \mathbf{P} \times 10^{-23}

Solving for $\mathbf{P}$ :

\mathbf{P} = \frac{2.5 \times 10^{-20}}{10^{-23}} = 2.5 \times 10^3 = 2500

Therefore, the value of $\mathbf{P}$ is 2500.

Q 24ChemistrySection B • Numericalaldehydes ketones and carboxylic acids

+3 / -0

Read the passage carefully:

An organic compound $\mathbf{P}$ with molecular formula $\mathrm{C}_9 \mathrm{H}_{18} \mathrm{O}_2$ decolorizes bromine water and also shows positive iodoform test. $\mathbf{P}$ on ozonolysis followed by treatment with $\mathrm{H}_2 \mathrm{O}_2$ gives $\mathbf{Q}$ and $\mathbf{R}$ . While compound $\mathbf{Q}$ shows positive iodoform test, compound $\mathbf{R}$ does not give positive iodoform test. $\mathbf{Q}$ and $\mathbf{R}$ on oxidation with pyridinium chlorochromate (PCC) followed by heating give $\mathbf{S}$ and $\mathbf{T}$ , respectively. Both $\mathbf{S}$ and $\mathbf{T}$ show positive iodoform test.

Complete copolymerization of 500 moles of $\mathbf{Q}$ and 500 moles of $\mathbf{R}$ gives one mole of a single acyclic copolymer $\mathbf{U}$ .

[Given, atomic mass: $\mathrm{H}=1, \mathrm{C}=12, \mathrm{O}=16$ ]

The molecular weight of

\mathbf{U}

is ______ .

Answer Format:Enter numerical value rounded to the nearest integer

Q 25ChemistrySection B • Numericalchemical kinetics and nuclear chemistry

+4 / -0

A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into $\mathrm{Pb}-206$ while the rest of it remains undisintegrated.

When the age of the sample is $\mathbf{P} \times 10^8$ years, the ratio of mass of $\mathrm{Pb}-206$ to that of $\mathrm{U}-238$ in the sample is found to be 7. The value of $\mathbf{P}$ is _______.

[Given: Half-life of $\mathrm{U}-238$ is $4.5 \times 10^9$ years; $\log _e 2=0.693$ ]

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:143

Step-by-Step Solution:

Given Data:

Ratio of mass of $\mathrm{Pb}^{206}$ to $\mathrm{U}^{238}$ = $\frac{7}{1}$

Mass of $\mathrm{Pb}^{206}$ = 7 g

Mass of $\mathrm{U}^{238}$ = 1 g

Half-life of $\mathrm{U}^{238}$ = $4.5 \times 10^9$ years

$\log_e 2 = 0.693$

Calculate Moles:

Moles of $\mathrm{Pb}^{206}$ :

$\text{Moles of } \mathrm{Pb}^{206} = \frac{7}{206} = 34 \times 10^{-3} \text{ moles}$

Moles of $\mathrm{U}^{238}$ :

$\text{Moles of } \mathrm{U}^{238} = \frac{1}{238} = 4.2 \times 10^{-3} \text{ moles}$

Initial Moles of $\mathrm{U}^{238}$ ( $N_0$ ):

$N_0 = \text{Moles of } \mathrm{U}^{238} \text{ initially} = \text{Moles of } \mathrm{Pb}^{206} + \text{Moles of remaining } \mathrm{U}^{238}$

$N_0 = 34 \times 10^{-3} + 4.2 \times 10^{-3} = 38.2 \times 10^{-3} \text{ moles}$

Calculate the Time:

Using the decay formula:

$\lambda t = \ln \frac{N_0}{N_t}$

Where $\lambda$ (decay constant):

$\lambda = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{4.5 \times 10^9 \text{ years}}$

Calculate $\ln \frac{N_0}{N_t}$ :

$\ln \frac{N_0}{N_t} = \ln \frac{38.2}{4.2} = \ln 9.09$

Simplify $\ln 9.09$ :

$\ln 9.09 \approx 2.21$

Plug in the Values:

$t = \frac{4.5 \times 10^9 \text{ years}}{0.693} \times \ln 9.09$

$t = 4.5 \times 10^9 \text{ years} \times \frac{2.21}{0.693}$

$t \approx 4.5 \times 10^9 \times 3.184 \approx 14.33 \times 10^9 \text{ years}$

Convert to $P \times 10^8$ :

$t = P \times 10^8 \text{ years}$

$P \times 10^8 = 14.33 \times 10^9$

$P = 143.28$

Thus, the value of $P$ is $\boxed{143.28}$ .

Q 26ChemistrySection A • MCQp block elements

+3 / -1

The species formed on fluorination of phosphorus pentachloride in a polar organic solvent are

A

\left[\mathrm{PF}_4\right]^{+}\left[\mathrm{PF}_6\right]^{-}

and

\left[\mathrm{PCl}_4\right]^{+}\left[\mathrm{PF}_6\right]^{-}

B

\left[\mathrm{PCl}_4\right]^{+}\left[\mathrm{PCl}_4 \mathrm{~F}_2\right]^{-}

and

\left[\mathrm{PCl}_4\right]^{+}\left[\mathrm{PF}_6\right]^{-}

C

\mathrm{PF}_3

and

\mathrm{PCl}_3

D

\mathrm{PF}_5

and

\mathrm{PCl}_3

Q 27ChemistrySection A • MCQaldehydes ketones and carboxylic acids

+3 / -1

In the following reaction sequence, the major product $\mathbf{Q}$ is

qImage-chemical-reactions-for-aldehydes-and-ketones-aldehydes-ketones-and-carboxylic-acids-organic-chemistry-Jeemate

A

B

C

D

Q 28ChemistrySection B • Numericalsolutions

+4 / -0

Vessel-1 contains

\mathbf{w}_2 \mathrm{~g}

of a non-volatile solute

\mathbf{X}

dissolved in

\mathbf{w}_1 \mathrm{~g}

of water. Vessel- 2 contains

\mathbf{w}_2 \mathrm{~g}

of another non-volatile solute

\mathbf{Y}

dissolved in

\mathbf{w}_1 \mathrm{~g}

of water. Both the vessels are at the same temperature and pressure. The molar mass of

\mathbf{X}

is

80 \%

of that of

\mathbf{Y}

. The van't Hoff factor for

\mathbf{X}

is 1.2 times of that of

\mathbf{Y}

for their respective concentrations.

The elevation of boiling point for solution in Vessel-1 is ________

\%

of the solution in Vessel-2.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:150

Step-by-Step Solution:

To determine the elevation of the boiling point for the solutions in both vessels, we need to use the concept of boiling point elevation. The boiling point elevation is given by the formula:

$\Delta T_b = i \cdot K_b \cdot m$

Where:

$\Delta T_b$ is the boiling point elevation
$i$ is the van't Hoff factor
$K_b$ is the ebullioscopic constant (which is the same for both solutions since they are dissolved in water)
$m$ is the molality of the solution

We need to compare the boiling point elevations for both solutions in Vessel-1 and Vessel-2.

For Vessel-1, let the boiling point elevation be $\Delta T_{b1}$ , and for Vessel-2, let it be $\Delta T_{b2}$ .

Let's denote the molar mass of solute $Y$ as $M_Y$ g/mol. According to the problem, the molar mass of solute $X$ is 80% of that of $Y$ , i.e.,

$M_X = 0.8 \cdot M_Y$

The van't Hoff factor for $X$ is 1.2 times that of $Y$ , i.e.,

$i_X = 1.2 \cdot i_Y$

The molality $m$ of each solution can be calculated using the formula:

$m = \dfrac{w_2}{M \cdot w_1}$

Thus, the molality of $X$ and $Y$ solutions are:

For solute $X$ :

$m_X = \dfrac{w_2}{M_X \cdot w_1} = \dfrac{w_2}{0.8 \cdot M_Y \cdot w_1}$

For solute $Y$ :

$m_Y = \dfrac{w_2}{M_Y \cdot w_1}$

Now, substituting the values into the boiling point elevation formula, we get for Vessel-1:

$\Delta T_{b1} = i_X \cdot K_b \cdot m_X = (1.2 \cdot i_Y) \cdot K_b \cdot \left( \dfrac{w_2}{0.8 \cdot M_Y \cdot w_1} \right)$

For Vessel-2:

$\Delta T_{b2} = i_Y \cdot K_b \cdot m_Y = i_Y \cdot K_b \cdot \left( \dfrac{w_2}{M_Y \cdot w_1} \right)$

To find the ratio of $\Delta T_{b1}$ to $\Delta T_{b2}$ :

$\dfrac{\Delta T_{b1}}{\Delta T_{b2}} = \dfrac{(1.2 \cdot i_Y) \cdot K_b \cdot \left( \dfrac{w_2}{0.8 \cdot M_Y \cdot w_1} \right)}{i_Y \cdot K_b \cdot \left( \dfrac{w_2}{M_Y \cdot w_1} \right)}$

Simplifying the ratio:

$\dfrac{\Delta T_{b1}}{\Delta T_{b2}} = \dfrac{1.2}{0.8} = 1.5$

This means the elevation of the boiling point for the solution in Vessel-1 is 150% of that in Vessel-2. Thus, the elevation of boiling point for solution in Vessel-1 is 150% of the elevation of the boiling point for the solution in Vessel-2.

Q 29ChemistrySection B • Numericalelectrochemistry

+4 / -2

An aqueous solution of hydrazine $\left(\mathrm{N}_2 \mathrm{H}_4\right)$ is electrochemically oxidized by $\mathrm{O}_2$ , thereby releasing chemical energy in the form of electrical energy. One of the products generated from the electrochemical reaction is $\mathrm{N}_2(\mathrm{~g})$ .

Choose the correct statement(s) about the above process

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

To determine the correct statement(s) regarding the electrochemical oxidation of hydrazine ( $\mathrm{N}_2 \mathrm{H}_4$ ) by $\mathrm{O}_2$ , let's analyze each option in detail.

Option A: $\mathrm{OH}^{-}$ ions react with $\mathrm{N}_2 \mathrm{H}_4$ at the anode to form $\mathrm{N}_2$ $(\mathrm{~g})$ and water, releasing 4 electrons to the anode. This statement is accurate because in an electrochemical cell, the anode is where oxidation occurs. Hydrazine can be oxidized in the presence of $\mathrm{OH}^{-}$ ions to yield nitrogen gas and water, releasing electrons as a part of this redox reaction:

\mathrm{N}_2 \mathrm{H}_4 + 4 \mathrm{OH}^{-} \rightarrow \mathrm{N}_2 + 4 \mathrm{H}_2 \mathrm{O} + 4e^{-}

Option B: At the cathode, $\mathrm{N}_2 \mathrm{H}_4$ breaks to $\mathrm{N}_2$ $(\mathrm{~g})$ and nascent hydrogen released at the electrode reacts with oxygen to form water. This statement is incorrect because, usually, the cathode is where reduction occurs, not a breaking down of hydrazine. Instead, reduction reactions involve the gain of electrons. Additionally, at the cathode, it's more common for oxygen to be reduced rather than hydrazine itself breaking down.

Option C: At the cathode, molecular oxygen gets converted to $\mathrm{OH}^{-}$ . This statement is correct. At the cathode in an alkaline medium, oxygen undergoes reduction to form hydroxide ions $\mathrm{OH}^{-}$ , as depicted by the half-reaction:

\mathrm{O}_2 + 2H_2O + 4e^{-} \rightarrow 4 \mathrm{OH}^{-}

Option D: Oxides of nitrogen are major by-products of the electrochemical process. This statement is incorrect because the primary product from the electrochemical reaction of hydrazine is nitrogen gas ( $\mathrm{N}_2$ ), and not oxides of nitrogen. The formation of oxides of nitrogen would be notable only under different specific conditions not mentioned in this context.

Therefore, the correct statements regarding the electrochemical oxidation of hydrazine by $\mathrm{O}_2$ are:

Option A and Option C.

Q 30ChemistrySection B • Numericalsalt analysis

+3 / -0

Read the passage carefully:

When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex

\mathbf{P}

is formed. In a strong acidic medium, the equilibrium shifts completely towards

\mathbf{P}

. Addition of zinc chloride to

\mathbf{P}

in a slightly acidic medium results in a sparingly soluble complex Q.

The number of moles of potassium iodide required to produce two moles of

\mathbf{P}

is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 31ChemistrySection A • MCQsolid state

+3 / -1

In a metal deficient oxide sample,

\mathbf{M}_{\mathbf{x}} \mathbf{Y}_2 \mathbf{O}_4

(

\mathbf{M}

and

\mathbf{Y}

are metals),

\mathbf{M}

is present in both +2 and +3 oxidation states and

\mathbf{Y}

is in +3 oxidation state. If the fraction of

\mathbf{M}^{2+}

ions present in

\mathbf{M}

is

\frac{1}{3}

, the value of

\mathbf{X}

is _______ .

A

0.25

B

0.33

C

0.67

D

0.75

Q 32ChemistrySection B • Numericalaldehydes ketones and carboxylic acids

+4 / -0

An organic compound $\mathbf{P}$ having molecular formula $\mathrm{C}_6 \mathrm{H}_6 \mathrm{O}_3$ gives ferric chloride test and does not have intramolecular hydrogen bond. The compound $\mathbf{P}$ reacts with 3 equivalents of $\mathrm{NH}_2 \mathrm{OH}$ to produce oxime $\mathbf{Q}$ . Treatment of $\mathbf{P}$ with excess methyl iodide in the presence of $\mathrm{KOH}$ produces compound $\mathbf{R}$ as the major product. Reaction of $\mathbf{R}$ with excess iso-butylmagnesium bromide followed by treatment with $\mathrm{H}_3 \mathrm{O}^{+}$ gives compound $\mathbf{S}$ as the major product.

The total number of methyl $\left(-\mathrm{CH}_3\right)$ group(s) in compound $\mathbf{S}$ is ___.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:6OR12

Step-by-Step Solution:

Step 1: Identification of Compound P

Molecular Formula: $\text{C}_6\text{H}_6\text{O}_3$
Ferric Chloride ( $\text{FeCl}_3$ ) Test: Indicates the presence of phenolic or enolic hydroxyl ( $-\text{OH}$ ) groups.
No Intramolecular Hydrogen Bonding: This rules out adjacent arrangements like pyrogallol ( $1,2,3$ -trihydroxybenzene).

Consequently, compound $P$ is phloroglucinol ( $1,3,5$ -trihydroxybenzene). Phloroglucinol readily exhibits keto-enol tautomerism to exist in its keto form, cyclohexane- $1,3,5$ -trione. Because it has $3$ carbonyl groups in this form, it reacts perfectly with $3$ equivalents of hydroxylamine ( $\text{NH}_2\text{OH}$ ) to yield the trioxime Q.

Step 2: Formation of Compound R

When phloroglucinol ( $P$ ) is treated with excess methyl iodide ( $\text{CH}_3\text{I}$ ) in the presence of a strong base like $\text{KOH}$ , it undergoes exhaustive C-alkylation at its active methylene positions rather than O-alkylation.

Each of the three $-\text{CH}_2-$ positions in the cyclohexane- $1,3,5$ -trione tautomer is methylated twice, leading to $2,2,4,4,6,6$ -hexamethylcyclohexane- $1,3,5$ -trione as compound $R$ .

Methyl groups in $R$ : $6$ methyl groups attached directly to the ring carbons.

Step 3: Formation of Compound S

Compound $R$ contains $3$ ketone carbonyl groups. When treated with excess iso-butylmagnesium bromide ( $\text{i-BuMgBr}$ , a Grignard reagent) followed by acid hydrolysis ( $\text{H}_3\text{O}^+$ ), each carbonyl carbon undergoes a nucleophilic addition reaction to form a tertiary alcohol.

Each ketone group adds one iso-butyl group ( $-\text{CH}_2-\text{CH}(\text{CH}_3)_2$ ).
Since there are $3$ ketone groups, a total of $3$ iso-butyl groups are added.

Step 4: Total Methyl Group Count in S

Let's add up all the methyl ( $-\text{CH}_3$ ) groups present in the final structure $S$ :

From the core structure of $R$ : $6 \text{ methyl groups}$
From the $3$ attached iso-butyl groups: $3 \times 2 = 6 \text{ methyl groups}$

$\text{Total number of methyl groups} = 6 + 6 = 12$

Q 33ChemistrySection A • MCQatomic structure

+3 / -1

According to Bohr's model, the highest kinetic energy is associated with the electron in the

A

first orbit of

\mathrm{H}

atom

B

first orbit of

\mathrm{He}^{+}

C

second orbit of

\mathrm{He}^{+}

D

second orbit of

\mathrm{Li}^{2+}

Q 34ChemistrySection B • Numericalsalt analysis

+3 / -0

Read the passage carefully:

When potassium iodide is added to an aqueous solution of potassium ferricyanide, a reversible reaction is observed in which a complex

\mathbf{P}

is formed. In a strong acidic medium, the equilibrium shifts completely towards

\mathbf{P}

. Addition of zinc chloride to

\mathbf{P}

in a slightly acidic medium results in a sparingly soluble complex Q.

The number of zinc ions present in the molecular formula of

\mathbf{Q}

is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 35MathematicsSection B • Numericallimits continuity and differentiability

+4 / -2

Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that

$\lim\limits_{x \rightarrow \infty} \frac{\sin \left(x^2\right)\left(\log _e x\right)^\alpha \sin \left(\frac{1}{x^2}\right)}{x^{\alpha \beta}\left(\log _e(1+x)\right)^\beta}=0 .$

Then which of the following is (are) correct?

Q 36MathematicsSection A • MCQinverse trigonometric functions

+3 / -1

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

$\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)$

is

A

\frac{7}{24}

B

\frac{-7}{24}

C

\frac{-5}{24}

D

\frac{5}{24}

Q 37MathematicsSection A • MCQlimits continuity and differentiability

+3 / -1

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by

$f(x)=\left\{\begin{array}{cc} x^2 \sin \left(\frac{\pi}{x^2}\right), & \text { if } x \neq 0, \\ 0, & \text { if } x=0 . \end{array}\right.$

Then which of the following statements is TRUE?

A

f(x)=0

has infinitely many solutions in the interval

\left[\frac{1}{10^{10}}, \infty\right)

.

B

f(x)=0

has no solutions in the interval

\left[\frac{1}{\pi}, \infty\right)

.

C

The set of solutions of

f(x)=0

in the interval

\left(0, \frac{1}{10^{10}}\right)

is finite.

D

f(x)=0

has more than 25 solutions in the interval

\left(\frac{1}{\pi^2}, \frac{1}{\pi}\right)

.

Q 38MathematicsSection B • Numericaldefinite integration

+3 / -0

Read the passage carefully:

Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$ .

The value of

2 \int\limits_0^{\frac{\pi}{2}} f(x) g(x) d x-\int\limits_0^{\frac{\pi}{2}} g(x) d x

is ____________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:0

Step-by-Step Solution:

We have to show that
$2\int_{0}^{\frac{\pi}{2}} f(x)\,g(x)\,dx-\int_{0}^{\frac{\pi}{2}} g(x)\,dx=0,$
where $f(x)=\sin^{2}x$ and $g(x)=\sqrt{\dfrac{\pi x}{2}-x^{2}}=\sqrt{x\!\left(\dfrac{\pi}{2}-x\right)}.$

Step 1: Evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}} g(x)\,dx$

Put $x=\dfrac{\pi}{2}\,t,\;0\le t\le1.$
Then $dx=\dfrac{\pi}{2}\,dt$ and
$g(x)=\sqrt{\dfrac{\pi x}{2}-x^{2}}=\sqrt{\dfrac{\pi^{2}}{4}\,t-\dfrac{\pi^{2}}{4}\,t^{2}}=\dfrac{\pi}{2}\sqrt{t(1-t)}.$

Thus
$\int_{0}^{\frac{\pi}{2}} g(x)\,dx =\int_{0}^{1}\dfrac{\pi}{2}\sqrt{t(1-t)}\;\dfrac{\pi}{2}\,dt =\dfrac{\pi^{2}}{4}\int_{0}^{1}t^{\frac12}(1-t)^{\frac12}\,dt.$

The integral is the Beta-function $B\!\left(\tfrac32,\tfrac32\right)=\dfrac{\Gamma\!\left(\tfrac32\right)^2}{\Gamma(3)}.$
Using $\Gamma\!\left(\tfrac32\right)=\dfrac{\sqrt{\pi}}{2}$ and $\Gamma(3)=2,$
$B\!\left(\tfrac32,\tfrac32\right)=\dfrac{\pi/4}{2}=\dfrac{\pi}{8}.$

Hence
$\int_{0}^{\frac{\pi}{2}} g(x)\,dx =\dfrac{\pi^{2}}{4}\cdot\dfrac{\pi}{8} =\dfrac{\pi^{3}}{32}.\quad -(1)$

Step 2: Evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}}\!f(x)\,g(x)\,dx$

With the same substitution $x=\dfrac{\pi}{2}t,$

$f(x)\,g(x)=\sin^{2}\!\left(\dfrac{\pi t}{2}\right)\! \left(\dfrac{\pi}{2}\sqrt{t(1-t)}\right),\qquad dx=\dfrac{\pi}{2}\,dt.$

Therefore
$\int_{0}^{\frac{\pi}{2}}\!f(x)g(x)\,dx =\dfrac{\pi^{2}}{4}\int_{0}^{1}\sin^{2}\!\left(\dfrac{\pi t}{2}\right) \sqrt{t(1-t)}\,dt.\quad -(2)$

Step 3: Simplify the inner integral

Write $\sin^{2}\theta=\dfrac{1-\cos2\theta}{2}.$
Thus

$\sin^{2}\!\left(\dfrac{\pi t}{2}\right) =\dfrac{1-\cos(\pi t)}{2}.$

Let
$I=\int_{0}^{1}\sin^{2}\!\left(\dfrac{\pi t}{2}\right)\sqrt{t(1-t)}\,dt =\dfrac12\int_{0}^{1}\sqrt{t(1-t)}\,dt -\dfrac12\underbrace{\int_{0}^{1}\cos(\pi t)\sqrt{t(1-t)}\,dt}_{K}.$

Step 4: Show $K=0$ by symmetry

Set $t=\dfrac12+u,\;dt=du,$ so $u\in\left[-\dfrac12,\dfrac12\right].$
Then $\sqrt{t(1-t)}=\sqrt{\dfrac14-u^{2}},$ an even function of $u$ , while

$\cos(\pi t)=\cos\!\left(\dfrac{\pi}{2}+\pi u\right)=-\sin(\pi u),$ which is an odd function of $u$ .

The product $-\sin(\pi u)\sqrt{\dfrac14-u^{2}}$ is odd, hence its integral over the symmetric interval vanishes: $K=0.$

Step 5: Evaluate $I$ and then the required integral

Using $K=0$ and $\displaystyle\int_{0}^{1}\sqrt{t(1-t)}\,dt=\dfrac{\pi}{8},$

$I=\dfrac12\cdot\dfrac{\pi}{8}=\dfrac{\pi}{16}.$

Substituting into $(2):$

$\int_{0}^{\frac{\pi}{2}}\!f(x)g(x)\,dx =\dfrac{\pi^{2}}{4}\cdot\dfrac{\pi}{16} =\dfrac{\pi^{3}}{64}.\quad -(3)$

Step 6: Form the required expression

$2\int_{0}^{\frac{\pi}{2}} f(x)g(x)\,dx -\int_{0}^{\frac{\pi}{2}} g(x)\,dx =2\left(\dfrac{\pi^{3}}{64}\right)-\dfrac{\pi^{3}}{32} =\dfrac{\pi^{3}}{32}-\dfrac{\pi^{3}}{32}=0.$

Hence the required value is 0.

Q 39MathematicsSection B • Numericaldefinite integration

+3 / -0

Read the passage carefully:

Let $f:\left[0, \frac{\pi}{2}\right] \rightarrow[0,1]$ be the function defined by $f(x)=\sin ^2 x$ and let $g:\left[0, \frac{\pi}{2}\right] \rightarrow[0, \infty)$ be the function defined by $g(x)=\sqrt{\frac{\pi x}{2}-x^2}$ .

The value of

\frac{16}{\pi^3} \int\limits_0^{\frac{\pi}{2}} f(x) g(x) d x

is ______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:0.25

Step-by-Step Solution:

We have to show that
$2\int_{0}^{\frac{\pi}{2}} f(x)\,g(x)\,dx-\int_{0}^{\frac{\pi}{2}} g(x)\,dx=0,$
where $f(x)=\sin^{2}x$ and $g(x)=\sqrt{\dfrac{\pi x}{2}-x^{2}}=\sqrt{x\!\left(\dfrac{\pi}{2}-x\right)}.$

Step 1: Evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}} g(x)\,dx$

Put $x=\dfrac{\pi}{2}\,t,\;0\le t\le1.$
Then $dx=\dfrac{\pi}{2}\,dt$ and
$g(x)=\sqrt{\dfrac{\pi x}{2}-x^{2}}=\sqrt{\dfrac{\pi^{2}}{4}\,t-\dfrac{\pi^{2}}{4}\,t^{2}}=\dfrac{\pi}{2}\sqrt{t(1-t)}.$

Thus
$\int_{0}^{\frac{\pi}{2}} g(x)\,dx =\int_{0}^{1}\dfrac{\pi}{2}\sqrt{t(1-t)}\;\dfrac{\pi}{2}\,dt =\dfrac{\pi^{2}}{4}\int_{0}^{1}t^{\frac12}(1-t)^{\frac12}\,dt.$

The integral is the Beta-function $B\!\left(\tfrac32,\tfrac32\right)=\dfrac{\Gamma\!\left(\tfrac32\right)^2}{\Gamma(3)}.$
Using $\Gamma\!\left(\tfrac32\right)=\dfrac{\sqrt{\pi}}{2}$ and $\Gamma(3)=2,$
$B\!\left(\tfrac32,\tfrac32\right)=\dfrac{\pi/4}{2}=\dfrac{\pi}{8}.$

Hence
$\int_{0}^{\frac{\pi}{2}} g(x)\,dx =\dfrac{\pi^{2}}{4}\cdot\dfrac{\pi}{8} =\dfrac{\pi^{3}}{32}.\quad -(1)$

Step 2: Evaluate $\displaystyle\int_{0}^{\frac{\pi}{2}}\!f(x)\,g(x)\,dx$

With the same substitution $x=\dfrac{\pi}{2}t,$

$f(x)\,g(x)=\sin^{2}\!\left(\dfrac{\pi t}{2}\right)\! \left(\dfrac{\pi}{2}\sqrt{t(1-t)}\right),\qquad dx=\dfrac{\pi}{2}\,dt.$

Therefore
$\int_{0}^{\frac{\pi}{2}}\!f(x)g(x)\,dx =\dfrac{\pi^{2}}{4}\int_{0}^{1}\sin^{2}\!\left(\dfrac{\pi t}{2}\right) \sqrt{t(1-t)}\,dt.\quad -(2)$

Step 3: Simplify the inner integral

Write $\sin^{2}\theta=\dfrac{1-\cos2\theta}{2}.$
Thus

$\sin^{2}\!\left(\dfrac{\pi t}{2}\right) =\dfrac{1-\cos(\pi t)}{2}.$

Let
$I=\int_{0}^{1}\sin^{2}\!\left(\dfrac{\pi t}{2}\right)\sqrt{t(1-t)}\,dt =\dfrac12\int_{0}^{1}\sqrt{t(1-t)}\,dt -\dfrac12\underbrace{\int_{0}^{1}\cos(\pi t)\sqrt{t(1-t)}\,dt}_{K}.$

Step 4: Show $K=0$ by symmetry

Set $t=\dfrac12+u,\;dt=du,$ so $u\in\left[-\dfrac12,\dfrac12\right].$
Then $\sqrt{t(1-t)}=\sqrt{\dfrac14-u^{2}},$ an even function of $u$ , while

$\cos(\pi t)=\cos\!\left(\dfrac{\pi}{2}+\pi u\right)=-\sin(\pi u),$ which is an odd function of $u$ .

The product $-\sin(\pi u)\sqrt{\dfrac14-u^{2}}$ is odd, hence its integral over the symmetric interval vanishes: $K=0.$

Step 5: Evaluate $I$ and then the required integral

Using $K=0$ and $\displaystyle\int_{0}^{1}\sqrt{t(1-t)}\,dt=\dfrac{\pi}{8},$

$I=\dfrac12\cdot\dfrac{\pi}{8}=\dfrac{\pi}{16}.$

Substituting into $(2):$

$\int_{0}^{\frac{\pi}{2}}\!f(x)g(x)\,dx =\dfrac{\pi^{2}}{4}\cdot\dfrac{\pi}{16} =\dfrac{\pi^{3}}{64}.\quad -(3)$

Step 6: Form the required expression

$2\int_{0}^{\frac{\pi}{2}} f(x)g(x)\,dx -\int_{0}^{\frac{\pi}{2}} g(x)\,dx =2\left(\dfrac{\pi^{3}}{64}\right)-\dfrac{\pi^{3}}{32} =\dfrac{\pi^{3}}{32}-\dfrac{\pi^{3}}{32}=0.$

Hence the required value is 0.

Q 40MathematicsSection B • Numericalapplication of integration

+4 / -0

Let the function $f:[1, \infty) \rightarrow \mathbb{R}$ be defined by

$f(t)=\left\{\begin{array}{cc} (-1)^{n+1} 2, & \text { if } t=2 n-1, n \in \mathbb{N}, \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1 < t < 2 n+1, n \in \mathbb{N} . \end{array}\right.$

Define $g(x)=\int_1^x f(t) d t, x \in(1, \infty)$ . Let $\alpha$ denote the number of solutions of the equation $g(x)=0$ in the interval $(1,8]$ and $\beta=\lim \limits_{x \rightarrow l+} \frac{g(x)}{x-1}$ .

Then the value of $\alpha+\beta$ is equal to _______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:5

Step-by-Step Solution:

The values of $f(t)$ at the odd integers are obtained directly from the first line of the definition.

$f(2n-1)=(-1)^{\,n+1}\,2 \quad\Rightarrow\quad f(1)=2,\;f(3)=-2,\;f(5)=2,\;f(7)=-2,\dots$

Between two successive odd integers $2n-1$ and $2n+1$ the function is a straight-line joining these end-values, i.e. it is linear on every interval $[2n-1,\,2n+1]$ .

The integral required is $g(x)=\displaystyle\int_{1}^{x}f(t)\,dt,\;x\gt 1.$ Because $g'(x)=f(x)$ , the right-hand derivative of $g$ at $x=1$ equals $f(1)$ : $\beta=\lim_{x\to1^{+}}\frac{g(x)}{x-1}=g'(1^{+})=f(1)=2.\tag{1}$

Next we look for the zeros of $g(x)$ in $(1,8].$ Observe first that on each interval $[2n-1,\,2n+1]$ the end-values of $f$ are equal in magnitude and opposite in sign, so the average value of $f$ over the full length $2$ is zero. Hence

$\int_{2n-1}^{2n+1}f(t)\,dt=0 \;\Longrightarrow\; g(2n+1)=g(2n-1).$

Starting from $g(1)=0$ , this gives successively

$g(3)=0,\;g(5)=0,\;g(7)=0,\dots$

Thus $x=3,5,7$ are certainly zeros lying in $(1,8].$ To see whether any additional zeros occur inside the open sub-intervals, evaluate $g(x)$ on each of them.

Case 1:

x\in[1,3]

The straight-line expression for $f$ here is $f(t)=2-2(t-1).$ Integrating,

$g(x)=\int_{1}^{x}\!\bigl[2-2(t-1)\bigr]dt =(2x-(x-1)^2)-2 =-x^2+4x-3.$ This quadratic satisfies $g(1)=g(3)=0$ and $g(2)=1\gt0,$ so no other root occurs in $(1,3).$

Case 2:

x\in[3,5]

Here $f(t)=-2+2(t-3).$ Put $y=x-3\;(0\le y\le2).$ Then

$g(x)=\int_{3}^{x}\!\bigl[-2+2(t-3)\bigr]dt =-2y+y^{2}=y(y-2).$ Hence $g(3)=g(5)=0$ and $g(x)\lt0$ for $x\in(3,5).$ Again, no extra root.

Case 3:

x\in[5,7]

With $u=x-5\;(0\le u\le2)$ one obtains $g(x)=2u-u^{2}=u(2-u)\gt0$ in $(5,7),$ and only the end-points give zeros.

Case 4:

x\in[7,8]

Let $v=x-7$ . The integral gives $g(x)=-2v+v^{2}=v(v-2)\lt0$ for $v\in(0,1],$ so no zero arises in $(7,8]$ except the already counted one at $x=7.$

Collecting all solutions in $(1,8]$ :

$x=3,\;5,\;7\quad\Rightarrow\quad\alpha=3.\tag{2}$

Finally, from $\eqref{1}$ and $\eqref{2}$

$\alpha+\beta=3+2=5.$

Hence the required value equals 5.

Q 41MathematicsSection A • MCQapplication of integration

+3 / -1

Let

S=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x \geq 0, y \geq 0, y^2 \leq 4 x, y^2 \leq 12-2 x\right.

and

\left.3 y+\sqrt{8} x \leq 5 \sqrt{8}\right\}

. If the area of the region

S

is

\alpha \sqrt{2}

, then

\alpha

is equal to

A

\frac{17}{2}

B

\frac{17}{3}

C

\frac{17}{4}

D

\frac{17}{5}

Official Correct Answer:Option (B)

Step-by-Step Solution:

We have to find the area of the region

$S=\left\{(x,y)\in\mathbb{R}\times\mathbb{R}:\;x\ge 0,\;y\ge 0,\;y^{2}\le 4x,\;y^{2}\le 12-2x,\;3y+\sqrt 8\,x\le 5\sqrt 8\right\}$

(1) From $y^{2}\le 4x$ we get $x\ge\dfrac{y^{2}}{4}$ . This is the region to the right of the right-opening parabola $y^{2}=4x$ .

(2) From $y^{2}\le 12-2x$ we get $x\le 6-\dfrac{y^{2}}{2}$ . This is the region to the left of the left-opening parabola $y^{2}=12-2x$ whose vertex is at $(6,0)$ .

(3) From $3y+\sqrt 8\,x\le 5\sqrt 8$ write $\sqrt 8 = 2\sqrt 2$ :

$3y+2\sqrt 2\,x\le 10\sqrt 2 \;\Longrightarrow\; x\le \dfrac{10\sqrt 2-3y}{2\sqrt 2}=5-\dfrac{3y}{2\sqrt 2}$

This is the region to the left of the straight line $3y+2\sqrt 2\,x=10\sqrt 2$ .

Combining (1) and (2):

$\dfrac{y^{2}}{4}\le x\le 6-\dfrac{y^{2}}{2}$ For such an $x$ to exist we need

$\dfrac{y^{2}}{4}\le 6-\dfrac{y^{2}}{2} \;\Longrightarrow\; 3y^{2}\le 24 \;\Longrightarrow\; y^{2}\le 8 \;\Longrightarrow\; 0\le y\le 2\sqrt 2$

Now decide which of the two upper bounds $x_{1}=6-\dfrac{y^{2}}{2},\qquad x_{2}=5-\dfrac{3y}{2\sqrt 2}$ is smaller (the region must satisfy both). Define

$D(y)=x_{2}-x_{1} =\Bigl(5-\dfrac{3y}{2\sqrt 2}\Bigr)-\Bigl(6-\dfrac{y^{2}}{2}\Bigr) =-1-\dfrac{3y}{2\sqrt 2}+\dfrac{y^{2}}{2}$

$D(y)=0 \;\Longrightarrow\; \dfrac{y^{2}}{2}-\dfrac{3y}{2\sqrt 2}-1=0 \;\Longrightarrow\; y^{2}-\dfrac{3}{\sqrt 2}y-2=0$ Solving gives $y=2\sqrt 2\quad\text{or}\quad y=-\dfrac{1}{\sqrt 2}$ . On $0\le y\lt 2\sqrt 2$ , we have $D(y)\lt 0$ , so $x_{2} Hence for every$ 0\le y\le 2\sqrt 2 $the$ x $-limits are $ x_{\text{min}}=\dfrac{y^{2}}{4},\qquad x_{\text{max}}=5-\dfrac{3y}{2\sqrt 2} $ The area$ A $of the region is therefore $ A=\int_{0}^{2\sqrt 2} \left[x_{\text{max}}-x_{\text{min}}\right]\,dy =\int_{0}^{2\sqrt 2} \left(5-\dfrac{3y}{2\sqrt 2}-\dfrac{y^{2}}{4}\right)dy $ Integrate term by term: $ \int 5\,dy = 5y\int -\dfrac{3y}{2\sqrt 2}\,dy = -\dfrac{3}{2\sqrt 2}\cdot\dfrac{y^{2}}{2} = -\dfrac{3y^{2}}{4\sqrt 2}\int -\dfrac{y^{2}}{4}\,dy = -\dfrac{1}{4}\cdot\dfrac{y^{3}}{3} = -\dfrac{y^{3}}{12} $ Thus $ A=\Bigl[\,5y-\dfrac{3y^{2}}{4\sqrt 2}-\dfrac{y^{3}}{12}\Bigr]_{0}^{2\sqrt 2} $ At$ y=2\sqrt 2 $: $ y^{2}=8,\quad y^{3}=16\sqrt 2 $ $ A=5(2\sqrt 2)-\dfrac{3\cdot8}{4\sqrt 2}-\dfrac{16\sqrt 2}{12} =10\sqrt 2-\dfrac{24}{4\sqrt 2}-\dfrac{4\sqrt 2}{3} $ $ \dfrac{24}{4\sqrt 2}=6/\sqrt 2=3\sqrt 2 $ $ \therefore\;A=10\sqrt 2-3\sqrt 2-\dfrac{4\sqrt 2}{3} =7\sqrt 2-\dfrac{4\sqrt 2}{3} =\dfrac{21\sqrt 2-4\sqrt 2}{3} =\dfrac{17\sqrt 2}{3} $ Thus$ A=\alpha\sqrt 2 $with$ \alpha=\dfrac{17}{3} $. Option B which is:$ \dfrac{17}{3}$$

Q 42MathematicsSection B • Numerical3d geometry

+4 / -2

A straight line drawn from the point

P(1,3,2)

, parallel to the line

\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}

, intersects the plane

L_1: x-y+3 z=6

at the point

Q

. Another straight line which passes through

Q

and is perpendicular to the plane

L_1

intersects the plane

L_2: 2 x-y+z=-4

at the point

R

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

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The given point is $P(1,3,2)$ and the line parallel to which we must draw a line through $P$ is
$\frac{x-2}{1}=\frac{y-4}{2}=\frac{z-6}{1}$

The direction ratios of this line are $1,2,1$ , so a line through $P$ with the same direction has parametric form
$x = 1 + t,\; y = 3 + 2t,\; z = 2 + t \qquad -(1)$

This line meets the plane $L_1 : x - y + 3z = 6$ at the point $Q$ . Substitute the coordinates from $(1)$ into the plane:

$\bigl(1+t\bigr) - \bigl(3+2t\bigr) + 3\bigl(2+t\bigr) = 6$
$1+t - 3 - 2t + 6 + 3t = 6$
$4 + 2t = 6 \;\;\Longrightarrow\;\; t = 1$

Putting $t=1$ back in $(1)$ gives
$Q(2,\,5,\,3)$

Vector $\overrightarrow{PQ} = (2-1,\,5-3,\,3-2) = (1,2,1)$ , so
$PQ = \sqrt{1^{2}+2^{2}+1^{2}} = \sqrt{6}$
Thus statement A is true.

A line through $Q$ perpendicular to plane $L_1$ must have direction equal to the normal of $L_1$ .
Since $L_1$ is $x - y + 3z = 6$ , its normal vector is $(1,-1,3)$ .

Parametric equations of this perpendicular line are
$x = 2 + s,\; y = 5 - s,\; z = 3 + 3s \qquad -(2)$

This line meets the plane $L_2 : 2x - y + z = -4$ at point $R$ . Insert the coordinates from $(2)$ into $L_2$ :

$2(2+s) - (5 - s) + (3 + 3s) = -4$
$4 + 2s - 5 + s + 3 + 3s = -4$
$2 + 6s = -4$
$6s = -6 \;\;\Longrightarrow\;\; s = -1$

Substituting $s=-1$ in $(2)$ gives
$R(1,\,6,\,0)$

Thus statement B, which claimed $R(1,6,3)$ , is false.

Centroid $G$ of $\triangle PQR$ is
$G\bigl(\,\tfrac{1+2+1}{3},\;\tfrac{3+5+6}{3},\;\tfrac{2+3+0}{3}\bigr) = \Bigl(\tfrac{4}{3},\;\tfrac{14}{3},\;\tfrac{5}{3}\Bigr)$

Hence statement C is true.

For statement D, compute the remaining two sides:

$QR = \sqrt{(1)^{2} + (1)^{2} + (-3)^{2}} = \sqrt{11}$
$RP = \sqrt{(0)^{2} + (-3)^{2} + 2^{2}} = \sqrt{13}$

Perimeter $= PQ + QR + RP = \sqrt{6} + \sqrt{11} + \sqrt{13}$ , not $\sqrt{2} + \sqrt{6} + \sqrt{11}$ , so statement D is false.

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

Q 43MathematicsSection B • Numericalparabola

+4 / -0

A normal with slope

\frac{1}{\sqrt{6}}

is drawn from the point

(0,-\alpha)

to the parabola

x^2=-4 a y

, where

a>0

. Let

L

be the line passing through

(0,-\alpha)

and parallel to the directrix of the parabola. Suppose that

L

intersects the parabola at two points

A

and

B

. Let

r

denote the length of the latus rectum and

s

denote the square of the length of the line segment

A B

. If

r: s=1: 16

, then the value of

24 a

is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:12

Step-by-Step Solution:

The parabola is $x^{2} = -4ay$ with $a \gt 0$ .
For this parabola:

• Vertex: $(0,0)$ • Focus: $(0,-a)$ • Directrix: $y = a$
• Length of the latus rectum $r = 4a$

1. Equation of the required normal
Let the point of contact on the parabola be $(x_1,y_1)$ .
Write the parabola as $y = -\dfrac{x^{2}}{4a}$ , so

$\frac{dy}{dx} = -\frac{x}{2a}$
Therefore, the slope of the tangent at $(x_1,y_1)$ is $m_t = -\frac{x_1}{2a}$ .
The slope of the normal is the negative reciprocal: $m_n = \frac{2a}{x_1}$ .

Given that the required normal has slope $\dfrac{1}{\sqrt6}$ , equate:

$\frac{2a}{x_1} = \frac{1}{\sqrt6} \;\Longrightarrow\; x_1 = 2a\sqrt6$

Find the corresponding $y_1$ on the parabola:

$y_1 = -\frac{x_1^{2}}{4a} = -\frac{(2a\sqrt6)^{2}}{4a} = -\frac{4a^{2}\,6}{4a} = -6a$

Thus the point of contact is $(2a\sqrt6,\,-6a)$ .

2. Locate the external point
The normal passes through the given point $(0,-\alpha)$ and has slope $\dfrac{1}{\sqrt6}$ .
Equation of the normal through $(0,-\alpha)$ :

$y + \alpha = \frac{1}{\sqrt6}\,x \;\Longrightarrow\; y = \frac{x}{\sqrt6} - \alpha$

Substitute $(x_1,y_1) = (2a\sqrt6,-6a)$ into this line:

$-6a = \frac{2a\sqrt6}{\sqrt6} - \alpha = 2a - \alpha$

Hence $\alpha = 2a + 6a = 8a$ .

3. Intersection of line $L$ with the parabola
Line $L$ passes through $(0,-\alpha)$ and is parallel to the directrix $y=a$ , so $L$ is horizontal:

$y = -\alpha = -8a$

Intersect with $x^{2} = -4ay$ :

$x^{2} = -4a(-8a) = 32a^{2}$

Thus the two intersection points are
$A(4a\sqrt2,\,-8a), \quad B(-4a\sqrt2,\,-8a).$

4. Lengths $r$ and $s$
Latus-rectum length: $r = 4a$ .
Length $AB = 4a\sqrt2 - (-4a\sqrt2) = 8a\sqrt2$ , so

$s = (AB)^{2} = (8a\sqrt2)^{2} = 128a^{2}$

Given $\dfrac{r}{s} = \dfrac{1}{16}$ :

$\frac{4a}{128a^{2}} = \frac{1}{16} \;\Longrightarrow\; \frac{1}{32a} = \frac{1}{16} \;\Longrightarrow\; 16 = 32a \;\Longrightarrow\; a = \frac12$

5. Required value
$24a = 24 \times \frac12 = 12$

Hence the answer is 12.

Q 44MathematicsSection B • Numericalprobability

+4 / -0

A bag contains

N

balls out of which 3 balls are white, 6 balls are green, and the remaining balls are blue. Assume that the balls are identical otherwise. Three balls are drawn randomly one after the other without replacement. For

i=1,2,3

, let

W_i, G_i

, and

B_i

denote the events that the ball drawn in the

i^{\text {th }}

draw is a white ball, green ball, and blue ball, respectively. If the probability

P\left(W_1 \cap G_2 \cap B_3\right)=\frac{2}{5 N}

and the conditional probability

P\left(B_3 \mid W_1 \cap G_2\right)=\frac{2}{9}

, then

N

equals ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:11

Step-by-Step Solution:

Let the total number of balls be $N$ .
Numbers of each colour:
    White   = $3$
    Green = $6$
    Blue   = $N-9$

Because the draws are without replacement, multiply the successive probabilities.

Step 1 : Compute $P(W_1 \cap G_2 \cap B_3)$
$P(W_1)=\frac{3}{N}$
After removing one white ball, total balls become $N-1$ and green balls remain $6$ , so
$P(G_2 \mid W_1)=\frac{6}{N-1}$
After removing a white and a green ball, total balls become $N-2$ and blue balls are still $N-9$ , so
$P(B_3 \mid W_1 \cap G_2)=\frac{N-9}{N-2}$
Therefore
$P(W_1 \cap G_2 \cap B_3)=\frac{3}{N}\cdot\frac{6}{N-1}\cdot\frac{N-9}{N-2}=\frac{18(N-9)}{N(N-1)(N-2)}$

This probability is given to be $\dfrac{2}{5N}$ , hence
$\frac{18(N-9)}{N(N-1)(N-2)}=\frac{2}{5N} \quad -(1)$

Step 2 : Compute the conditional probability $P(B_3 \mid W_1 \cap G_2)$
First find $P(W_1 \cap G_2)$ :
$P(W_1 \cap G_2)=\frac{3}{N}\cdot\frac{6}{N-1}=\frac{18}{N(N-1)}$
Thus
$P(B_3 \mid W_1 \cap G_2)=\frac{P(W_1 \cap G_2 \cap B_3)}{P(W_1 \cap G_2)} =\frac{\dfrac{18(N-9)}{N(N-1)(N-2)}}{\dfrac{18}{N(N-1)}} =\frac{N-9}{N-2}$
This conditional probability is given to be $\dfrac{2}{9}$ , so
$\frac{N-9}{N-2}=\frac{2}{9} \quad -(2)$

Step 3 : Solve equation $(2)$ for $N$
Cross-multiplying:
$9(N-9)=2(N-2)$
$9N-81=2N-4$
$7N=77$
$N=11$

Step 4 : Verification with equation $(1)$ (optional but reassuring)
Substitute $N=11$ in $(1)$ :
Left side $=\dfrac{18(11-9)}{11\cdot10\cdot9}=\dfrac{18\cdot2}{990}=\dfrac{36}{990}=\dfrac{2}{55}$
Right side $=\dfrac{2}{5\cdot11}=\dfrac{2}{55}$
Both sides match, confirming the value.

Hence the total number of balls is
11

Q 45MathematicsSection B • Numericalfunctions

+4 / -0

Let the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by

$f(x)=\frac{\sin x}{e^{\pi x}} \frac{\left(x^{2023}+2024 x+2025\right)}{\left(x^2-x+3\right)}+\frac{2}{e^{\pi x}} \frac{\left(x^{2023}+2024 x+2025\right)}{\left(x^2-x+3\right)} .$

Then the number of solutions of $f(x)=0$ in $\mathbb{R}$ is _________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:1

Step-by-Step Solution:

First, rewrite the given function in a factored form.

$f(x)=\frac{\sin x}{e^{\pi x}}\cdot\frac{x^{2023}+2024x+2025}{x^{2}-x+3}+\frac{2}{e^{\pi x}}\cdot\frac{x^{2023}+2024x+2025}{x^{2}-x+3}$

Factor out the common terms $\dfrac{x^{2023}+2024x+2025}{x^{2}-x+3}\cdot e^{-\pi x}$ :

$f(x)=\frac{x^{2023}+2024x+2025}{x^{2}-x+3}\,e^{-\pi x}\,(\sin x+2)$

To solve $f(x)=0$ we must examine when each factor can be zero.

1. The exponential factor $e^{-\pi x}$ is never zero for any real $x$ .
2. The quadratic denominator $x^{2}-x+3$ has discriminant $(-1)^{2}-4(1)(3)=-11\lt 0$ , so it is always positive and never zero.
3. The factor $\sin x+2$ cannot be zero because $\sin x\in[-1,1]$ , so $\sin x+2\in[1,3]$ for all $x\in\mathbb{R}$ .

Therefore the only possible zeros come from

$x^{2023}+2024x+2025=0$

Define $g(x)=x^{2023}+2024x+2025$ . To count its real roots, study monotonicity:

$g'(x)=2023x^{2022}+2024$

Since $x^{2022}\ge 0$ for every real $x$ , we have $g'(x)\ge 2024\gt 0$ . Thus $g(x)$ is strictly increasing on the entire real line.

A strictly increasing odd-degree polynomial with positive leading coefficient satisfies

$\lim_{x\to-\infty}g(x)=-\infty,\qquad \lim_{x\to+\infty}g(x)=+\infty$

By the Intermediate Value Theorem, it crosses the $x$ -axis exactly once. Hence $g(x)=0$ has exactly one real solution.

Because no other factor of $f(x)$ can become zero, this single root of $g(x)$ is the only root of $f(x)$ .

Therefore, the number of solutions of $f(x)=0$ in $\mathbb{R}$ is

1

Q 46MathematicsSection B • Numericalfunctions

+4 / -0

Let

f: \mathbb{R} \rightarrow \mathbb{R}

be a function such that

f(x+y)=f(x)+f(y)

for all

x, y \in \mathbb{R}

, and

g: \mathbb{R} \rightarrow(0, \infty)

be a function such that

g(x+y)=g(x) g(y)

for all

x, y \in \mathbb{R}

. If

f\left(\frac{-3}{5}\right)=12

and

g\left(\frac{-1}{3}\right)=2

, then the value of

\left(f\left(\frac{1}{4}\right)+g(-2)-8\right) g(0)

is _________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:51

Step-by-Step Solution:

The function $f : \mathbb{R} \to \mathbb{R}$ satisfies the additive rule
$f(x+y)=f(x)+f(y)\quad\forall\,x,y\in\mathbb{R}.$ For such a function:

• $f(0)=f(0+0)=f(0)+f(0)\;\Rightarrow\;f(0)=0$ .
• For any integer $n$ , $f(nx)=nf(x)$ (apply the rule $n$ times).
• For any rational $r=\dfrac{p}{q}$ , $f(rx)=rf(x)$ (use the integer result on $q x$ and divide).

Given $f\!\left(-\dfrac35\right)=12$ , we first find $f\!\left(\dfrac35\right)$ :
$f\!\left(\dfrac35\right)=-f\!\left(-\dfrac35\right)=-12.$ Now choose $x=\dfrac35$ and a rational multiplier $r=\dfrac{1/4}{3/5}=\dfrac{5}{12}$ so that $r\,x=\dfrac14$ . Using the rational-multiple rule,

$f\!\left(\dfrac14\right)=r\,f(x)=\dfrac{5}{12}\,(-12)=-5.$

The function $g : \mathbb{R} \to (0,\infty)$ satisfies the multiplicative rule
$g(x+y)=g(x)\,g(y)\quad\forall\,x,y\in\mathbb{R}.$ For such a function:

• $g(0)=g(0+0)=g(0)^2\;,\;g(0)\gt 0\;\Rightarrow\;g(0)=1.$
• For any integer $n$ , $g(nx)=g(x)^n$ (apply the rule $n$ times).
• Consequently, for rational $r=\dfrac{p}{q}$ , $g(rx)=g(x)^r$ (use the integer result and take roots).

Given $g\!\left(-\dfrac13\right)=2$ , we obtain $g\!\left(\dfrac13\right)=\dfrac1{g(-1/3)}=\dfrac12.$

Write $-2=6\!\left(-\dfrac13\right)$ . Using the integer-multiple rule,

$g(-2)=g\!\left(-\dfrac13\right)^{\,6}=2^{6}=64.$

Now evaluate the required expression:

$ $\bigl(f(1/4)+g(-2)-8\bigr)\,g(0)=\bigl((-5)+64-8\bigr)\,(1)=51.$

Hence the value is 51.

Q 47MathematicsSection B • Numericalvector algebra

+4 / -0

Let $\vec{p}=2 \hat{i}+\hat{j}+3 \hat{k}$ and $\vec{q}=\hat{i}-\hat{j}+\hat{k}$ . If for some real numbers $\alpha, \beta$ , and $\gamma$ , we have

$15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(2 \vec{p}+\vec{q})+\beta(\vec{p}-2 \vec{q})+\gamma(\vec{p} \times \vec{q}),$

then the value of $\gamma$ is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:2

Step-by-Step Solution:

Write the given vectors in component form:
$\vec{p}=2\hat i+\hat j+3\hat k \; \rightarrow \; (2,1,3), \qquad \vec{q}=\hat i-\hat j+\hat k \; \rightarrow \; (1,-1,1).$

First compute the three auxiliary vectors that appear on the right-hand side.

1. Vector $2\vec p+\vec q$
$2\vec p=(4,2,6) \quad\Longrightarrow\quad 2\vec p+\vec q=(4+1,\,2+(-1),\,6+1)=(5,1,7).$

2. Vector $\vec p-2\vec q$
$2\vec q=(2,-2,2) \quad\Longrightarrow\quad \vec p-2\vec q=(2-2,\,1-(-2),\,3-2)=(0,3,1).$

3. Vector $\vec p\times\vec q$
Using the determinant rule:
$\vec p\times\vec q= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2 & 1 & 3\\ 1 & -1 & 1 \end{vmatrix} = \hat i(1\cdot1-3\cdot(-1))-\hat j(2\cdot1-3\cdot1)+\hat k(2\cdot(-1)-1\cdot1) =4\hat i+\hat j-3\hat k.$
Thus $\vec p\times\vec q=(4,1,-3).$

Denote
$\vec a_1=(5,1,7),\qquad \vec a_2=(0,3,1),\qquad \vec a_3=(4,1,-3).$
The equation in component form becomes

$\alpha\vec a_1+\beta\vec a_2+\gamma\vec a_3=(15,10,6).$

Equate the $\hat i, \hat j, \hat k$ components:

$5\alpha+4\gamma=15 \quad-(1)$
$\alpha+3\beta+\gamma=10 \quad-(2)$
$7\alpha+\beta-3\gamma=6 \quad-(3)$

Solve sequentially. From $(1):$
$\alpha=\frac{15-4\gamma}{5}=3-0.8\gamma.$

Substitute this $\alpha$ into $(2):$
$(3-0.8\gamma)+3\beta+\gamma=10 \;\Longrightarrow\; 3\beta=7-0.2\gamma,$
so $\beta=\dfrac{7-0.2\gamma}{3}.$

Now use $(3):$
$7\alpha+\beta-3\gamma=6.$ Insert $\alpha=3-0.8\gamma$ and the above $\beta$ :

$7(3-0.8\gamma)+\beta-3\gamma=6$ $\Longrightarrow 21-5.6\gamma+\beta-3\gamma=6$ $\Longrightarrow \beta+21-8.6\gamma=6$ $\Longrightarrow \beta=-15+8.6\gamma.$

Equate the two expressions for $\beta$ :

$\frac{7-0.2\gamma}{3}=-15+8.6\gamma.$ Multiply by 3: $7-0.2\gamma=-45+25.8\gamma$ $\Longrightarrow 52=26\gamma$ $\Longrightarrow \gamma=2.$

Hence, the required value is $\boxed{2}$ .

Q 48MathematicsSection B • Numericalpermutations and combinations

+3 / -0

Read the passage carefully:

Let $S=\{1,2,3,4,5,6\}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties:

i. $R$ has exactly 6 elements.

ii. For each $(a, b) \in R$ , we have $|a-b| \geq 2$ .

Let $Y=\{R \in X$ : The range of $R$ has exactly one element $\}$ and $Z=\{R \in X: R$ is a function from $S$ to $S\}$ .

Let $n(A)$ denote the number of elements in a set $A$ .

If the value of

n(Y)+n(Z)

is

k^2

, then

|k|

is _________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 49MathematicsSection A • MCQlimits continuity and differentiability

+3 / -1

Let

k \in \mathbb{R}

. If

\lim \limits_{x \rightarrow 0+}(\sin (\sin k x)+\cos x+x)^{\frac{2}{x}}=e^6

, then the value of

k

is

A

1

B

2

C

3

D

4

Official Correct Answer:Option (B)

Step-by-Step Solution:

Let

$F(x)=\left(\sin(\sin kx)+\cos x+x\right)^{\frac{2}{x}}$

and put

$L=\lim_{x\to 0^{+}}F(x).$

We know the standard limit property: if $u(x)\to 0$ and $v(x)\to\infty$ such that $u(x)\,v(x)\to \ell,$ then

$\lim_{x\to 0}\bigl(1+u(x)\bigr)^{v(x)}=e^{\ell}.$

Hence we must rewrite the base inside the bracket in the form $1+u(x).$

Expand each elementary function near $x=0$ (retain terms up to $x^{2}$ because the final coefficient of $x$ decides the limit):

1. $\cos x = 1-\dfrac{x^{2}}{2}+O(x^{4}).$

2. $\sin kx = kx-\dfrac{k^{3}x^{3}}{6}+O(x^{5}).$

3. For small $y$ , $\sin y = y-\dfrac{y^{3}}{6}+O(y^{5}).$ Put $y=\sin kx$ in this expansion:

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

Now add the three series along with the term $x$ :

$\sin(\sin kx)+\cos x+x =\left(kx-\dfrac{k^{3}x^{3}}{3}\right)+\left(1-\dfrac{x^{2}}{2}\right)+x+O(x^{3}).$

Combine like powers of $x$ :

$=1+\bigl(k+1\bigr)x-\dfrac{x^{2}}{2}+O(x^{3}).$

Thus the bracket equals $1+u(x)$ where

$u(x)=\bigl(k+1\bigr)x-\dfrac{x^{2}}{2}+O(x^{3})$

and indeed $u(x)\to 0$ as $x\to 0^{+}.$

Take natural logarithm to prepare the exponent:

$\ln F(x)=\frac{2}{x}\,\ln\!\bigl(1+u(x)\bigr).$

For small $u(x)$ , $\ln(1+u)=u-\dfrac{u^{2}}{2}+O(u^{3}).$ Using only the linear part (higher powers vanish after multiplying by $\dfrac{2}{x}$ and then letting $x\to 0$ ):

$\ln\!\bigl(1+u(x)\bigr)=u(x)+O\!\bigl(u(x)^{2}\bigr) =\bigl(k+1\bigr)x+O(x^{2}).$

Therefore

$\ln F(x)=\frac{2}{x}\,\Bigl[\bigl(k+1\bigr)x+O(x^{2})\Bigr] =2\bigl(k+1\bigr)+O(x).$

Sending $x\to 0^{+}$ yields

$\lim_{x\to 0^{+}}\ln F(x)=2\bigl(k+1\bigr).$

Consequently

$L=\exp\!\Bigl[2\bigl(k+1\bigr)\Bigr] =e^{2(k+1)}.$

The problem states that $L=e^{6},$ so

$2(k+1)=6\quad\Longrightarrow\quad k+1=3\quad\Longrightarrow\quad k=2.$

Hence the required value of $k$ is $2$ .

Option B which is: $2$

Q 50MathematicsSection B • Numericalparabola

+4 / -2

Let

A_1, B_1, C_1

be three points in the

x y

-plane. Suppose that the lines

A_1 C_1

and

B_1 C_1

are tangents to the curve

y^2=8 x

at

A_1

and

B_1

, respectively. If

O=(0,0)

and

C_1=(-4,0)

, then which of the following statements is (are) TRUE?

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

The given parabola is $y^{2}=8x$ .
Rewrite it in standard form $y^{2}=4ax$ to identify the parameter $a$ .

Here $4a=8 \;\Rightarrow\; a=2$ .
For the parabola $y^{2}=4ax$ , a point on the curve can be written in parametric form as $P(t)\;:\;\bigl(at^{2},\,2at\bigr)$ and the tangent at this point is $ty = x + at^{2}$ .

Thus for our curve $a=2$ , so
Point $P(t)\;=\;\bigl(2t^{2},\,4t\bigr)$ , Tangent $t\,y = x + 2t^{2}\;\,\;-(1)$ .

The tangents at points $A_{1}$ and $B_{1}$ pass through the fixed point $C_{1}=(-4,0)$ . Substituting $x=-4,\;y=0$ in equation $(1)$ :

$t\,(0) = (-4) + 2t^{2}\;\Longrightarrow\;2t^{2}=4\;\Longrightarrow\;t^{2}=2.$ Hence $t=\sqrt{2}$ or $t=-\sqrt{2}$ .

Two distinct parameters give the two points of contact:
If $t_{1}=+\sqrt{2}$ → $A_{1}\bigl(2(\sqrt{2})^{2},\,4\sqrt{2}\bigr)=(4,\,4\sqrt{2})$ .
If $t_{2}=-\sqrt{2}$ → $B_{1}\bigl(2(\sqrt{2})^{2},\,4(-\sqrt{2})\bigr)=(4,\,-4\sqrt{2})$ .

Thus $A_{1}=(4,\,4\sqrt{2}),\qquad B_{1}=(4,\,-4\sqrt{2}),\qquad C_{1}=(-4,0).$

Checking Option A
Length $OA_{1} = \sqrt{(4-0)^{2} + (4\sqrt{2}-0)^{2}} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}.$ Option A is TRUE.

Checking Option B
Since $A_{1}$ and $B_{1}$ have the same $x$ -coordinate, $A_{1}B_{1}$ is vertical: $A_{1}B_{1}= |\,4\sqrt{2}-(-4\sqrt{2})\,| = 8\sqrt{2}\neq16.$ Option B is FALSE.

Checking the orthocentre for Options C and D
Side $A_{1}B_{1}$ is the vertical line $x=4$ . Therefore, the altitude from $C_{1}$ is the horizontal line $y=0$ (the $x$ -axis).

Find the altitude from $A_{1}$ :
Slope of side $B_{1}C_{1}$ $m_{BC} = \frac{-4\sqrt{2}-0}{4-(-4)}=\frac{-4\sqrt{2}}{8}=-\frac{\sqrt{2}}{2}.$ Hence slope of the altitude from $A_{1}$ (perpendicular to $B_{1}C_{1}$ ) is $m_{A} = \frac{2}{\sqrt{2}} = \sqrt{2}.$

Equation of this altitude through $A_{1}(4,4\sqrt{2})$ : $y-4\sqrt{2} = \sqrt{2}\,(x-4).$

Intersect this with $y=0$ (altitude from $C_{1}$ ):
$0-4\sqrt{2} = \sqrt{2}\,(x-4) \;\Longrightarrow\; -4 = x-4 \;\Longrightarrow\; x=0.$

The two altitudes meet at $H=(0,0)$ . Thus the orthocentre is the origin.
Option C (orthocentre $(0,0)$ ) is TRUE, while Option D (orthocentre $(1,0)$ ) is FALSE.

Final correct statements:
Option A and Option C.

Q 51MathematicsSection B • Numericalpermutations and combinations

+3 / -0

Read the passage carefully:

Let $S=\{1,2,3,4,5,6\}$ and $X$ be the set of all relations $R$ from $S$ to $S$ that satisfy both the following properties:

i. $R$ has exactly 6 elements.

ii. For each $(a, b) \in R$ , we have $|a-b| \geq 2$ .

Let $Y=\{R \in X$ : The range of $R$ has exactly one element $\}$ and $Z=\{R \in X: R$ is a function from $S$ to $S\}$ .

Let $n(A)$ denote the number of elements in a set $A$ .

If

n(X)={ }^m C_6

, then the value of

m

is _____

Answer Format:Enter numerical value rounded to the nearest integer

JEE Advanced 2024 Paper 2 Question Paper with Solutions