JEE Advanced 2022 Question Paper 2 Question Paper with Solutions | JEE Advanced Mock Test | JeeMate

Q 1PhysicsSection A • MCQmagnetism

+3 / -1

Which one of the following options represents the magnetic field $\vec{B}$ at $\mathrm{O}$ due to the current flowing in the given wire segments lying on the $x y$ plane?

qImage-biot-savarts-law-and-magnetic-field-due-to-current-carrying-wire-magnetism-electricity-Jeemate

A

\vec{B}=\frac{-\mu_{o} I}{L}\left(\frac{3}{2}+\frac{1}{4 \sqrt{2} \pi}\right) \hat{k}

B

\vec{B}=-\frac{\mu_{o} I}{L}\left(\frac{3}{2}+\frac{1}{2 \sqrt{2} \pi}\right) \hat{k}

C

\vec{B}=\frac{-\mu_{o} I}{L}\left(1+\frac{1}{4 \sqrt{2} \pi}\right) \hat{k}

D

\vec{B}=\frac{-\mu_{o} I}{L}\left(1+\frac{1}{4 \pi}\right) \hat{k}

Q 2PhysicsSection B • Numericalunits and measurements

+3 / -1

In a particular system of units, a physical quantity can be expressed in terms of the electric charge $e$ , electron mass $m_{e}$ , Planck's constant $h$ , and Coulomb's constant $k=\frac{1}{4 \pi \epsilon_{0}}$ , where $\epsilon_{0}$ is the permittivity of vacuum. In terms of these physical constants, the dimension of the magnetic field is $[B]=[e]^{\alpha}\left[m_{e}\right]^{\beta}[h]^{\gamma}[k]^{\delta}$ . The value of $\alpha+\beta+\gamma+\delta$ is _______.

Answer Format:Enter numerical value rounded to the nearest integer

Q 3PhysicsSection B • Numericalcurrent electricity

+3 / -1

Two resistances $R_{1}=X \Omega$ and $R_{2}=1 \Omega$ are connected to a wire $A B$ of uniform resistivity, as shown in the figure. The radius of the wire varies linearly along its axis from $0.2 \mathrm{~mm}$ at $A$ to $1 \mathrm{~mm}$ at $B$ . A galvanometer $(\mathrm{G})$ connected to the center of the wire, $50 \mathrm{~cm}$ from each end along its axis, shows zero deflection when $A$ and $B$ are connected to a battery. The value of $X$ is ____________.

qImage-potentiometer-current-electricity-electricity-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 4PhysicsSection B • Numericalsimple harmonic motion

+3 / -1

On a frictionless horizontal plane, a bob of mass $m=0.1 \mathrm{~kg}$ is attached to a spring with natural length $l_{0}=0.1 \mathrm{~m}$ . The spring constant is $k_{1}=0.009 \,\mathrm{Nm}^{-1}$ when the length of the spring $l>l_{0}$ and is $k_{2}=0.016 \,\mathrm{Nm}^{-1}$ when $l < l_{0}$ . Initially the bob is released from $l=$ $0.15 \mathrm{~m}$ . Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is $T=(n \pi) s$ , then the integer closest to $n$ is __________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 5PhysicsSection B • Numericalelectrostatics

+4 / -2

A disk of radius

\mathrm{R}

with uniform positive charge density

\sigma

is placed on the

x y

plane with its center at the origin. The Coulomb potential along the

z

-axis is

V(z)=\frac{\sigma}{2 \epsilon_{0}}\left(\sqrt{R^{2}+z^{2}}-z\right) .

A particle of positive charge

q

is placed initially at rest at a point on the

z

axis with

z=z_{0}

and

z_{0}>0

. In addition to the Coulomb force, the particle experiences a vertical force

\vec{F}=-c \hat{k}

with

c>0

. Let

\beta=\frac{2 c \epsilon_{0}}{q \sigma}

.

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

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

\mathrm{V}(z) =\frac{\sigma}{2 \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right)

\mathrm{U}(z)_{\mathrm{net}} =\frac{\sigma q}{2 \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right)+\mathrm{cz}

=c\left[\frac{\sigma q}{2 c \varepsilon_0}\left(\sqrt{\mathrm{R}^2+z^2}-z\right)+z\right]

When,

z=0

and

\beta=1 / 4

\mathrm{U}_{(z) \text { net }}=c(4 \mathrm{R})=4 R c

When,

z=z_0=\frac{25}{7} \mathrm{R}

and;

\mathrm{B}=\frac{1}{4}

U_{(z) \text { net }}=C\left[4 \times \frac{26 R}{7}-3 \times \frac{25 R}{7}\right]=\frac{29 R c}{7}

When,

z=z_0=\frac{3 R}{7}

and

\beta=\frac{1}{4}

\mathrm{U}_{(z) \text { net }}=3 R c

When,

Z=\frac{R}{\sqrt{3}}

and

\beta=\frac{1}{4}

\mathrm{U}_{(z) n \mathrm{t}}=2.887 \mathrm{Rc}

As per option (A), particle reaches at origin with the

\mathrm{KE}

\frac{d(U)_z}{d z}=0 \text { at } z=\frac{3 R}{\sqrt{7}}

So, at

\beta = \frac{1}{4}

and

z=\frac{3 R}{\sqrt{7}}

\mathrm{U}_{z(\text { net })}=\sqrt{7} R c=2.645 R c

Similarly, for option (B),

\mathrm{U}_{(z) \text { net }}

at

z=\frac{\mathrm{R}}{\sqrt{3}}

the kinetic energy at origin

will become negative and mearly equal to

3 R c

Also, for option (C)

\mathrm{U}_{(z) \text { net }}

at

\mathrm{Z}=\frac{\mathrm{R}}{\sqrt{3}}>\mathrm{U}_{(z) \text { net }}

at

z=\frac{3 \mathrm{R}}{\sqrt{7}}

Particle will return to

z_0

As per option

D,(\beta>1

and

z_0>0)

U_{z(\text { net })}

increases continuously with

\mathrm{Z}

. So, particle reaches the origin.

Q 6PhysicsSection B • Numericalwave optics

+4 / -2

A double slit setup is shown in the figure. One of the slits is in medium 2 of refractive index $n_{2}$ . The other slit is at the interface of this medium with another medium 1 of refractive index $n_{1}\left(\neq n_{2}\right)$ . The line joining the slits is perpendicular to the interface and the distance between the slits is $d$ . The slit widths are much smaller than $d$ . A monochromatic parallel beam of light is incident on the slits from medium 1. A detector is placed in medium 2 at a large distance from the slits, and at an angle $\theta$ from the line joining them, so that $\theta$ equals the angle of refraction of the beam. Consider two approximately parallel rays from the slits received by the detector.

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

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

Q 7PhysicsSection B • Numericalelectrostatics

+4 / -2

In the figure, the inner (shaded) region $A$ represents a sphere of radius $r_{A}=1$ , within which the electrostatic charge density varies with the radial distance $r$ from the center as $\rho_{A}=k r$ , where $k$ is positive. In the spherical shell $B$ of outer radius $r_{B}$ , the electrostatic charge density varies as $\rho_{B}=$ $\frac{2 k}{r}$ . Assume that dimensions are taken care of. All physical quantities are in their SI units.

qImage-electric-flux-and-gauss-law-electrostatics-electricity-Jeemate

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

Q 8PhysicsSection B • Numericalproperties of matter

+4 / -2

A bubble has surface tension

S

. The ideal gas inside the bubble has ratio of specific heats

\gamma=

\frac{5}{3}

. The bubble is exposed to the atmosphere and it always retains its spherical shape. When the atmospheric pressure is

P_{a 1}

, the radius of the bubble is found to be

r_{1}

and the temperature of the enclosed gas is

T_{1}

. When the atmospheric pressure is

P_{a 2}

, the radius of the bubble and the temperature of the enclosed gas are

r_{2}

and

T_{2}

, respectively.

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

Q 9PhysicsSection A • MCQunits and measurements

+3 / -1

Area of the cross-section of a wire is measured using a screw gauge. The pitch of the main scale is $0.5 \mathrm{~mm}$ . The circular scale has 100 divisions and for one full rotation of the circular scale, the main scale shifts by two divisions. The measured readings are listed below.

Measurement condition	Main scale reading	Circular scale reading
Two arms of gauge touching each other without wire	0 division	4 divisions
Attempt-1: With wire	4 divisions	20 divisions
Attempt-2: With wire	4 divisions	16 divisions

What are the diameter and cross-sectional area of the wire measured using the screw gauge?

A

2.22 \pm 0.02 \mathrm{~mm}, \pi(1.23 \pm 0.02) \mathrm{mm}^{2}

B

2.22 \pm 0.01 \mathrm{~mm}, \pi(1.23 \pm 0.01) \mathrm{mm}^{2}

C

2.14 \pm 0.02 \mathrm{~mm}

,

\pi(1.14 \pm 0.02) \mathrm{mm}^{2}

D

2.14 \pm 0.01 \mathrm{~mm}, \pi(1.14 \pm 0.01) \mathrm{mm}^{2}

Q 10PhysicsSection B • Numericalatoms and nuclei

+3 / -1

In a radioactive decay chain reaction,

{ }_{90}^{230} \mathrm{Th}

nucleus decays into

{ }_{84}^{214} \mathrm{Po}

nucleus. The ratio of the number of

\alpha

to number of

\beta^{-}

particles emitted in this process is ________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:2

Step-by-Step Solution:

To solve this problem, we need to analyze the changes in the atomic number and mass number during the decay chain reaction from ${ }_{90}^{230}\mathrm{Th}$ (Thorium) to ${ }_{84}^{214}\mathrm{Po}$ (Polonium). Let's denote the number of alpha particles emitted by $n_{\alpha}$ and the number of beta-minus particles emitted by $n_{\beta^{-}}$ .

First, consider the changes in the mass number (A) and atomic number (Z). For alpha decay, which emits an $\alpha$ particle ( ${ }_{2}^{4}\mathrm{He}$ ), the mass number decreases by 4, and the atomic number decreases by 2. For beta-minus decay, which emits a $\beta^{-}$ particle (an electron, $e^{-}$ ), the atomic number increases by 1, but the mass number remains unchanged.

The initial and final nuclei are given as:

Initial: ${ }_{90}^{230}\mathrm{Th}$

Final: ${ }_{84}^{214}\mathrm{Po}$

The changes in the mass number (A) and atomic number (Z) are:

Change in mass number (A): $230 - 214 = 16$
Change in atomic number (Z): $90 - 84 = 6$

Since each alpha particle decreases the mass number by 4 and the atomic number by 2, if $n_{\alpha}$ is the number of alpha particles emitted, we have:

$\Delta A = 4n_{\alpha}$

$\Delta Z_{\alpha} = 2n_{\alpha}$

From the given changes, we can write:

$4n_{\alpha} = 16 \Rightarrow n_{\alpha} = \frac{16}{4} = 4$

The contribution to the change in atomic number due to the alpha particles is:

$\Delta Z_{\alpha} = 2 \times 4 = 8$

However, since the change in atomic number is 6, the number of beta-minus particles emitted must account for the difference:

$\Delta Z - \Delta Z_{\alpha} = n_{\beta^{-}} \Rightarrow 6 - 8 = n_{\beta^{-}} \Rightarrow n_{\beta^{-}} = 2 \text{ (increase)}$

Note that an increase of 2 units due to $\beta^-$ (beta-minus decay) particles means:

$\Delta Z_{\beta^{-}} = n_{\beta^{-}} = 2$

Thus the number of $\beta^{-}$ particles emitted is 2.

Therefore, the ratio of the number of $\alpha$ particles to the number of $\beta^{-}$ particles emitted is:

$\frac{n_{\alpha}}{n_{\beta^{-}}} = \frac{4}{2} = 2$

So, the ratio is 2.

Q 11PhysicsSection B • Numericalsimple harmonic motion

+3 / -1

A particle of mass

1 \mathrm{~kg}

is subjected to a force which depends on the position as

\vec{F}=

-k(x \hat{\imath}+y \hat{\jmath}) \mathrm{kg}\, \mathrm{m} \mathrm{s}^{-2}

with

k=1 \mathrm{~kg} \mathrm{~s}^{-2}

. At time

t=0

, the particle's position

\vec{r}=

\left(\frac{1}{\sqrt{2}} \hat{\imath}+\sqrt{2} \hat{\jmath}\right) m

and its velocity

\vec{v}=\left(-\sqrt{2} \hat{\imath}+\sqrt{2} \hat{\jmath}+\frac{2}{\pi} \hat{k}\right) m s^{-1}

. Let

v_{x}

and

v_{y}

denote the

x

and the

y

components of the particle's velocity, respectively. Ignore gravity. When

z=0.5 \mathrm{~m}

, the value of

\left(x v_{y}-y v_{x}\right)

is __________

m^{2} s^{-1}

.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:3

Step-by-Step Solution:

F_x=-x=m a_x

So

a_x=\frac{d^2 x}{d t^2}=-x

\Rightarrow x=A_x \sin \left(\omega t+\phi_x\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s})

and

v_x=A_x \omega \cos \left(\omega t+\phi_x\right)

at

t=0, x=\frac{1}{\sqrt{2}} \mathrm{~m}

and

v_x=-\sqrt{2} \mathrm{~m} / \mathrm{s}

So

\frac{1}{\sqrt{2}}=A_x \sin \phi_x

and

-\sqrt{2}=A_x \cos \phi_x

\begin{aligned} & \Rightarrow \tan \phi_x=-\frac{1}{2} ......(1) \\ & \text { and } A_x=\sqrt{\frac{5}{2}} \mathrm{~m} ......(2) \end{aligned}

Similarly

\begin{aligned} & F_y=-y=m a_y . \\ & \Rightarrow \frac{d^2 y}{d t^2}=-y \end{aligned}

So,

y=A_y \sin \left(\omega \mathrm{t}+\phi_{\mathrm{y}}\right) \quad(\omega=1 \mathrm{rad} / \mathrm{s})

and

v_y=A_y \omega \cos \left(\omega t+\phi_y\right)

at

t=0,y=\sqrt{2} \mathrm{~m}

and

v_y=\sqrt{2} \mathrm{~m} / \mathrm{s}

So

\sqrt{2}=A_y \sin \phi

and

\sqrt{2}=A_y \cos \phi

\Rightarrow \phi=\frac{\pi}{4} \text { and } A_y=2

So,

\left(x v_y-y v_x\right)

$ =\sqrt{\frac{5}{2}} \sin \left(\omega t+\phi_x\right) \times 2 \cos \left(\omega t+\phi_y\right)-2 \sin \left(\omega t+\phi_y\right) \times \sqrt{\frac{5}{2}} \cos \left(\omega t+\phi_x\right) $

=\sqrt{\frac{5}{2}} \times 2\left(\sin \left(\omega t+\phi_x\right) \cos \left(\omega t+\phi_y\right)-\sin \left(\omega t+\phi_y\right) \times \cos \left(\omega t+\phi_x\right)\right.

=\sqrt{10} \sin \left(\phi_x-\phi_y\right)

=\sqrt{10}\left(\sin \phi_x \cos \phi_y-\cos \phi_x \sin \phi_y\right)

=\sqrt{10}\left(\frac{1}{\sqrt{5}} \times \frac{1}{\sqrt{2}}-\left(-\frac{2}{\sqrt{5}}\right) \times \frac{1}{\sqrt{2}}\right)

=3

Q 12PhysicsSection B • Numericalray optics

+3 / -1

Consider a configuration of $n$ identical units, each consisting of three layers. The first layer is a column of air of height $h=\frac{1}{3} \mathrm{~cm}$ , and the second and third layers are of equal thickness $d=$ $\frac{\sqrt{3}-1}{2} \mathrm{~cm}$ , and refractive indices $\mu_{1}=\sqrt{\frac{3}{2}}$ and $\mu_{2}=\sqrt{3}$ , respectively. A light source 0 is placed on the top of the first unit, as shown in the figure. A ray of light from 0 is incident on the second layer of the first unit at an angle of $\theta=60^{\circ}$ to the normal. For a specific value of $n$ , the ray of light emerges from the bottom of the configuration at a distance $l=\frac{8}{\sqrt{3}} \mathrm{~cm}$ , as shown in the figure. The value of $n$ is ________.

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

Answer Format:Enter numerical value rounded to the nearest integer

Q 13PhysicsSection B • Numericalheat and thermodynamics

+4 / -2

In the given $P-V$ diagram, a monoatomic gas $\left(\gamma=\frac{5}{3}\right)$ is first compressed adiabatically from state $A$ to state $B$ . Then it expands isothermally from state $B$ to state $C$ . [Given: $\left(\frac{1}{3}\right)^{0.6} \simeq 0.5, \ln 2 \simeq 0.7$ ].

qImage-thermodynamics-process-heat-and-thermodynamics-mechanics-Jeemate

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

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

For adiabatic process

(\mathrm{A} \rightarrow \mathrm{B})

\begin{aligned} & \mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^\gamma=\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^\gamma \\ & 10^5 \times(0.8)^{\frac{5}{3}}=3 \times 10^5\left(\mathrm{~V}_{\mathrm{B}}\right)^{\frac{5}{3}} \\ & \Rightarrow \mathrm{V}_{\mathrm{B}}=0.8 \times\left(\frac{1}{3}\right)^{0.6}=0.4 \end{aligned}

Work done in process

\mathrm{A} \rightarrow \mathrm{B}

\begin{aligned} & \mathrm{W}_{\mathrm{AB}}=\frac{\mathrm{P}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}-\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}}{\gamma-1} \\ & \Rightarrow \mathrm{W}_{\mathrm{AB}}=\frac{10^5 \times 0.8-3 \times 10^5 \times 0.4}{\frac{5}{3}-1} \\ & \Rightarrow \mathrm{W}_{\mathrm{AB}}=-60 \mathrm{~kJ}=\Rightarrow\left|\mathrm{W}_{\mathrm{AB}}\right|=60 \mathrm{~kJ} \end{aligned}

Work done in process

\mathrm{B} \rightarrow \mathrm{C}

(Isothermal process)

\begin{aligned} & \mathrm{W}_{\mathrm{BC}}=\mathrm{nRT} \ell \mathrm{n} \frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}}=\mathrm{P}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}} \ell \mathrm{n} \frac{\mathrm{V}_{\mathrm{C}}}{\mathrm{V}_{\mathrm{B}}} \\ & \Rightarrow \mathrm{W}_{\mathrm{BC}}=3 \times 10^5 \times 0.4 \ell \mathrm{n} \frac{0.8}{0.4} \\ & \Rightarrow \mathrm{W}_{\mathrm{BC}}=84 \mathrm{~kJ} \end{aligned}

Work done in process

\mathrm{C} \rightarrow \mathrm{A}

\mathrm{W}_{\mathrm{CA}}=\mathrm{P} \Delta \mathrm{V}=0 \quad(\because \Delta \mathrm{V}=0)

So total work done in the process

\mathrm{A} \rightarrow \mathrm{B} \rightarrow \mathrm{C}

\begin{aligned} & \mathrm{W}_{\mathrm{ABC}}=\mathrm{W}_{\mathrm{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{\mathrm{CA}}=-60+84+0 \\ & \mathrm{~W}_{\mathrm{ABC}}=24 \mathrm{~kJ} \end{aligned}

So correct options are (B,C,D)

Q 14PhysicsSection B • Numericalray optics

+3 / -1

An object and a concave mirror of focal length $f=10 \mathrm{~cm}$ both move along the principal axis of the mirror with constant speeds. The object moves with speed $V_{0}=15 \mathrm{~cm} \mathrm{~s}^{-1}$ towards the mirror with respect to a laboratory frame. The distance between the object and the mirror at a given moment is denoted by $u$ . When $u=30 \mathrm{~cm}$ , the speed of the mirror $V_{m}$ is such that the image is instantaneously at rest with respect to the laboratory frame, and the object forms a real image. The magnitude of $V_{m}$ is _________ $\mathrm{cm} \,\mathrm{s}^{-1}$ .

qImage-reflection-of-light-geometrical-optics-optics-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 15PhysicsSection B • Numericalelectrostatics

+3 / -1

A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$ , and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 \epsilon_{0}}$ (in SI units). The value of $n$ is _______.

qImage-electric-flux-and-gauss-law-electrostatics-electricity-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 16PhysicsSection A • MCQrotational motion

+3 / -1

A flat surface of a thin uniform disk $A$ of radius $R$ is glued to a horizontal table. Another thin uniform disk $B$ of mass $M$ and with the same radius $R$ rolls without slipping on the circumference of $A$ , as shown in the figure. A flat surface of $B$ also lies on the plane of the table. The center of mass of $B$ has fixed angular speed $\omega$ about the vertical axis passing through the center of $A$ . The angular momentum of $B$ is $n M \omega R^{2}$ with respect to the center of $A$ . Which of the following is the value of $n$ ?

qImage-combined-translational-and-rotational-motion-rotational-motion-mechanics-Jeemate

A

2

B

5

C

\frac{7}{2}

D

\frac{9}{2}

Q 17PhysicsSection B • Numericalcurrent electricity

+4 / -2

In Circuit-1 and Circuit- 2 shown in the figures,

R_{1}=1 \,\Omega, R_{2}=2 \,\Omega

and

R_{3}=3 \,\Omega

.

P_{1}

and

P_{2}

are the power dissipations in Circuit-1 and Circuit-2 when the switches

\mathrm{S}_{1}

and

\mathrm{S}_{2}

are in open conditions, respectively.

Q_{1}

and

Q_{2}

are the power dissipations in Circuit-1 and Circuit-2 when the switches

\mathrm{S}_{1}

and

\mathrm{S}_{2}

are in closed conditions, respectively.

qImage-electric-power-and-heating-effect-of-current-current-electricity-electricity-Jeemate

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

Q 18PhysicsSection A • MCQdual nature of radiation

+3 / -1

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0 \mathrm{~V}$ . This potential drops to $0.6 \mathrm{~V}$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? [Take $\frac{h c}{e}=1.24 \times$ $10^{-6} \mathrm{JmC}^{-1}$ .]

A

1.72 \times 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}

B

1.72 \times 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}

C

3.78 \times 10^{-7} \mathrm{~m}, 5.60 \mathrm{eV}

D

3.78 \times 10^{-7} \mathrm{~m}, 1.20 \mathrm{eV}

Q 19ChemistrySection B • Numericalthermodynamics

+4 / -2

The correct option(s) about entropy (S) is(are)

[\mathrm{R}=

gas constant,

\mathrm{F}=

Faraday constant,

\mathrm{T}=

Temperature

]

Q 20ChemistrySection B • Numericalp block elements

+4 / -2

The compound(s) which react(s) with

\mathrm{NH}_{3}

to give boron nitride (BN) is(are)

Q 21ChemistrySection A • MCQsolid state

+3 / -1

Atom

\mathrm{X}

occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in %) of the resultant solid is closest to

A

25

B

35

C

55

D

75

Q 22ChemistrySection A • MCQp block elements

+3 / -1

The reaction of

\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}

and

\mathrm{NaCl}

in water produces a precipitate that dissolves upon the addition of

\mathrm{HCl}

of appropriate concentration. The dissolution of the precipitate is due to the formation of

A

\mathrm{PbCl}_{2}

B

\mathrm{PbCl}_{4}

C

\left[\mathrm{PbCl}_{4}\right]^{2-}

D

\left[\mathrm{PbCl}_{6}\right]^{2-}

Q 23ChemistrySection B • Numericalpolymers

+4 / -2

Among the following, the correct statement(s) about polymers is(are)

Q 24ChemistrySection B • Numericalhydrocarbons

+3 / -1

The total number of chiral molecules formed from one molecule of

\mathbf{P}

on complete ozonolysis

\left(\mathrm{O}_{3}, \mathrm{Zn} / \mathrm{H}_{2} \mathrm{O}\right)

is _________.

qImage-reaction-of-alkenes-hydrocarbons-organic-chemistry-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 25ChemistrySection A • MCQp block elements

+3 / -1

The reaction of

\mathrm{HClO}_{3}

with

\mathrm{HCl}

gives a paramagnetic gas, which upon reaction with

\mathrm{O}_{3}

produces

A

\mathrm{Cl}_{2} \mathrm{O}

B

\mathrm{ClO}_{2}

C

\mathrm{Cl}_{2} \mathrm{O}_{6}

D

\mathrm{Cl}_{2} \mathrm{O}_{7}

Q 26ChemistrySection B • Numericalsome basic concepts of chemistry

+4 / -2

To check the principle of multiple proportions, a series of pure binary compounds

\left(\mathrm{P}_{\mathrm{m}} \mathrm{Q}_{\mathrm{n}}\right)

were analyzed and their composition is tabulated below. The correct option(s) is(are)

Compound	Weight % of $\mathrm{P}$	Weight % of $\mathrm{Q}$
1	50	50
2	44.4	55.6
3	40	60

Official Correct Answer:Declared in Solution

Step-by-Step Solution:

qImage-laws-of-chemical-combination-some-basic-concepts-of-chemistry-physical-chemistry-Jeemate

(A) If emperical formula of compound 3 is

P_3 Q_4

then its molar ratio will be

\frac{40}{3} : \frac{60}{4}

=\frac{40}{3} \times \frac{4}{60}=\frac{16}{18}=0.88

If empirical formula of compound 2 is

\mathrm{P}_3 \mathrm{Q}_5

, then its molar ratio

\begin{aligned} & =\frac{44.4}{2} : {\frac{55.6}{5}} \\ & =\frac{44.4}{2} \times \frac{5}{55.6}=2 \end{aligned}

Since molar ratio of both the compound is not equal

So, option (A) is not correct.

(B) If empirical formula of compound 3 is

\mathrm{P}_3 \mathrm{Q}_2

, i.e.,

\begin{aligned} & \frac{40}{M_P}: \frac{60}{M_Q}=\frac{3}{2} \\ & \Rightarrow \frac{40}{M_P} \times \frac{M_Q}{60}=\frac{3}{2} \end{aligned}

\begin{aligned} \Rightarrow \frac{4 \mathrm{M}_{\mathrm{Q}}}{6 \mathrm{M}_{\mathrm{P}}} & =\frac{3}{2} \\ \Rightarrow \frac{\mathrm{M}_{\mathrm{Q}}}{\mathrm{M}_{\mathrm{P}}} & =\frac{3}{2} \times \frac{6}{4}=\frac{9}{4} \end{aligned}

and

\mathrm{M}_{\mathrm{P}}=20

(given)

So,

\frac{\mathrm{M}_{\mathrm{Q}}}{20}=\frac{9}{4}

\begin{aligned} & \mathrm{M}_{\mathrm{Q}}=\frac{9 \times 26^5}{4} \\ & \mathrm{M}_{\mathrm{Q}}=45 \end{aligned}

So, option (B) is correct.

(C) If empirical formula of compound 2 is PQ, So the molar ratio is

\frac{44.4}{1}: \frac{55.6}{1}=\frac{44.4}{55.6}=0.79 \sim 0.8

Empirical formula of compound 1 is

P_5 Q_4

so the molar ratio is

\frac{50}{5} : { } \frac{50}{4}

.

=\frac{50}{5} \times \frac{4}{50}=\frac{4}{5}=0.8

Since, molar ratio of both the compound is equal hence, state

(\mathrm{C})

is correct.

So, option (C) is correct.

(D)

\mathrm{M}_{\mathrm{P}}=70, \mathrm{M}_{\mathrm{Q}}=35

Molar ratio of compound 1 is

\begin{aligned} \frac{50}{M_{\mathrm{P}}}: \frac{50}{\mathrm{M}_{\mathrm{Q}}} & =\frac{50}{70}: \frac{50}{35} \\ & =\frac{50}{70} \times \frac{35}{50}=\frac{1}{2} \end{aligned}

Hence, empirical formula of compound

\mathrm{PQ}_2

.

So, option (D) is incorrect.

Q 27ChemistrySection B • Numericald and f block elements

+3 / -1

Thermal decomposition of

\mathrm{AgNO}_{3}

produces two paramagnetic gases. The total number of electrons present in the antibonding molecular orbitals of the gas that has the higher number of unpaired electrons is ____________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 28ChemistrySection B • Numericalionic equilibrium

+3 / -1

Concentration of $\mathrm{H}_{2} \mathrm{SO}_{4}$ and $\mathrm{Na}_{2} \mathrm{SO}_{4}$ in a solution is $1 \mathrm{M}$ and $1.8 \times 10^{-2} \mathrm{M}$ , respectively. Molar solubility of $\mathrm{PbSO}_{4}$ in the same solution is $\mathrm{X} \times 10^{-\mathrm{Y}} \mathrm{M}$ (expressed in scientific notation). The value of $Y$ is ________.

[Given: Solubility product of $\mathrm{PbSO}_{4}\left(K_{s p}\right)=1.6 \times 10^{-8}$ . For $\mathrm{H}_{2} \mathrm{SO}_{4}, K_{a l}$ is very large and $\left.K_{a 2}=1.2 \times 10^{-2}\right]$

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:6

Step-by-Step Solution:

Given,

Conentration of

\mathrm{H}_2 \mathrm{SO}_4=1 \mathrm{M}

Conentration of

\mathrm{Na}_{2} \mathrm{SO}_{4}

=

1.8 \times 10^{-2} \mathrm{M}

qImage-solubility-product-and-common-ion-effect-ionic-equilibrium-physical-chemistry-Jeemate

From the above two equations, we get

\left[\mathrm{H}^{+}\right]=1 \mathrm{M} \text { and }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2}

K_c=\frac{1.8 \times 10^{-2} \times 1}{1}=1.8 \times 10^{-2}

and it is given that

\mathrm{K}_{a_2}\left(\mathrm{Q}_c\right)=1.2 \times 10^{-2} \mathrm{M}

Since,

\mathrm{K}_{a_2}

(i.e.,

\mathrm{Q}_c

)

<\mathrm{K}_c

\therefore

Rather than dissociation of

\mathrm{HSO}_4^{-}

into

\mathrm{H}^{+}

and

\mathrm{SO}_4^{2-}

ions, association between already present

\mathrm{H}^{+}

and

\mathrm{SO}_4^{2-}

will take place.

Assuming '

x

' mol/L of

\mathrm{SO}_4^{2-}

and

\mathrm{H}^{+}

combines to form

\mathrm{HSO}_4^{-}

\begin{aligned} & \mathrm{K}_{a 2}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{SO}_4^{2-}\right]}{\left[\mathrm{HSO}_4^{-}\right]} \\ & 1.2 \times 10^{-2}=\frac{(1-x)\left(1.8 \times 10^{-2}-x\right)}{(1+x)} \\ & \because x<<1, \text { so }(1+x) \approx 1 \text { and }(1-x) \approx 1 \\ & 1.2 \times 10^{-2}=1.8 \times 10^{-2}-x \\ & x=\left(1.8 \times 10^{-2}\right)-\left(1.2 \times 10^{-2}\right) \\ & x=0.6 \times 10^{-2} \mathrm{M} \\ & \text { So, }\left[\mathrm{SO}_4^{2-}\right]=1.8 \times 10^{-2}-x \\ & {\left[\mathrm{SO}_4^{2-}\right] }=\left(1.8 \times 10^{-2}\right)-\left(0.6 \times 10^{-2}\right) \\ & {\left[\mathrm{SO}_4^{2-}\right] }=1.2 \times 10^{-2} \mathrm{M}\end{aligned}

Given,

\quad \mathrm{K}_{s p}=1.6 \times 10^{-8}

\therefore

y\left(1.2 \times 10^{-2}+y\right)=1.6 \times 10^{-8}

Since,

y<<1

, So

1.2 \times 10^{-2}+y \approx 1.2 \times 10^{-2}

So,

y \times 1.2 \times 10^{-2}=1.6 \times 10^{-8}

\Rightarrow

y =\frac{1.6 \times 10^{-8}}{1.2 \times 10^{-2}}

\Rightarrow

y =1.33 \times 10^{-6}

$ X \times 10^{-Y} \mathrm{M} =1.33 \times 10^{-6} \mathrm{M} $

So,

\mathrm{Y}=6

Hence, the value of

Y

is 6 .

Q 29ChemistrySection B • Numericalhydrocarbons

+3 / -1

The number of isomeric tetraenes (NOT containing

s p

-hybridized carbon atoms) that can be formed from the following reaction sequence is ___________.

qImage-reaction-of-alkynes-hydrocarbons-organic-chemistry-Jeemate

qImage-reaction-of-alkynes-hydrocarbons-organic-chemistry-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 30ChemistrySection B • Numericalhydrocarbons

+3 / -1

The number of

-\mathrm{CH}_{2}-

(methylene) groups in the product formed from the following reaction sequence is _________.

qImage-reaction-of-alkenes-hydrocarbons-organic-chemistry-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 31ChemistrySection B • Numericalhydrocarbons

+4 / -2

Considering the following reaction sequence, the correct statement(s) is(are)

qImage-aromatic-hydrocarbons-hydrocarbons-organic-chemistry-Jeemate

qImage-aromatic-hydrocarbons-hydrocarbons-organic-chemistry-Jeemate

Q 32ChemistrySection B • Numericalisolation of elements

+4 / -2

The correct option(s) related to the extraction of iron from its ore in the blast furnace operating in the temperature range

900-1500 \mathrm{~K}

is(are)

Q 33ChemistrySection B • Numericald and f block elements

+3 / -1

The reaction of

\mathrm{Xe}

and

\mathrm{O}_{2}{F}_{2}

gives a

\mathrm{Xe}

compound

\mathbf{P}

. The number of moles of

\mathrm{HF}

produced by the complete hydrolysis of

1 \mathrm{~mol}

of

\mathbf{P}

is __________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 34ChemistrySection B • Numericalelectrochemistry

+3 / -1

Consider the strong electrolytes

Z_{m} X_{n}, U_{m} Y_{p}

and

V_{m} X_{n}

. Limiting molar conductivity (

\Lambda^{0}

) of

\mathrm{U}_{\mathrm{m}} \mathrm{Y}_{\mathrm{p}}

and

\mathrm{V}_{\mathrm{m}} \mathrm{X}_{\mathrm{n}}

are 250 and

440 \mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}

, respectively. The value of

(\mathrm{m}+\mathrm{n}+\mathrm{p})

is

Given:

Ion	$\mathrm{Z}^{\mathrm{n}+}$	$\mathrm{U}^{\mathrm{p}+}$	$\mathrm{V}^{\mathrm{n}+}$	$\mathrm{X}^{\mathrm{m}-}$	$\mathrm{Y}^{\mathrm{m}-}$
$\lambda^{0}\left(\mathrm{~S} \mathrm{~cm}^{2} \mathrm{~mol}^{-1}\right)$	$50.0$	$25.0$	$100.0$	$80.0$	$100.0$

\lambda^{0}

is the limiting molar conductivity of ions

The plot of molar conductivity (

\Lambda

) of

\mathrm{Z}_{\mathrm{m}} \mathrm{X}_{\mathrm{n}} v s\, \mathrm{c}^{1 / 2}

is given below.

qImage-conductance-and-electrolysis-electrochemistry-physical-chemistry-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:7

Step-by-Step Solution:

Given,

\Lambda_{\mathrm{V}_{\mathrm{m}} \mathrm{Y}_{\mathrm{p}}}^{\circ}=250 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

\Lambda_{\mathrm{V}_m \mathrm{x}_n}=440 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

It is also given that

\begin{aligned} & \lambda^{\circ} \mathrm{Z}^{n+}=50 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{V} p+}^0=25.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{V}^{n+}}^{\circ}=100.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{X}^{m-}}^{\circ}=80.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \lambda_{\mathrm{Y}^{m-}}^{\circ}=100.0 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ \end{aligned}

Now,

\Lambda_{\mathrm{V}_m \mathrm{Y}_p}=m \lambda_{\mathrm{V}^{+}}^{\circ}+p \lambda^{\circ}{ }_{\mathrm{Y}}

250=25 m+100 p

10=m+4p

..........(1)

Also,

\Lambda_{\mathrm{V}_m \mathrm{x}_n}=m \lambda_{\mathrm{v}^{+}}^{\circ}+n \lambda^{\circ} \mathrm{X}^{-}

440=100 m+80 n

22=5 m+4 n

........(2)

In the question, a graph of

(\Lambda)

of

Z_m X_n

Vs

C^{1 / 2}

is given,

We know,

\lambda_m=\lambda_m^0-A \sqrt{C}

For electrolyte

Z_m X_n

and from given curve

\begin{aligned} & \lambda_m\left(Z_m X_n\right)=\lambda_m^0\left(Z_m X_n\right)-A \sqrt{C} \\ & -A=\frac{336-339}{0.04-0.01}=-\frac{3}{0.03} \\ & \Rightarrow A=100 \\ & \therefore \text { For } \lambda_m=336 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \Rightarrow 336=\lambda_m^{\circ}\left(Z_m X_n\right)-100 \times 0.04 \\ & \lambda_m^{\circ}=336+4=340 \mathrm{~S} \mathrm{~cm}{ }^2 \mathrm{~mol}^{-1} \end{aligned}

So,

\begin{aligned} \Lambda_{Z_m x_n} & =m \lambda_{\mathrm{Z}^{+}}+n \lambda_{\mathrm{X}^{-}} \\ 340 & =50 m+80 n \\ 34 & =5 m+8 n ........(3) \end{aligned}

Solving eqn (2) and eqn (3),

n=\frac{12}{4}=3

\Rightarrow

n=3

Substituting the value of

n

in eqn (2), we get,

\begin{aligned} 22 & =5 m+4(3) \\ \Rightarrow 22 & =5 m+12 \\ \Rightarrow 5 m & =22-12=10 \\ \Rightarrow m & =\frac{10}{5}=2 \\ \Rightarrow m & =2 \end{aligned}

Now, substituting the value of

m

is eqn

(1)

, we get

\begin{aligned} 10 & =m+4 p \\ \Rightarrow 10 & =2+4 p \\ \Rightarrow 8 & =4 p \\ \Rightarrow p & =\frac{8}{4}=2 \\ \Rightarrow p & =22 \end{aligned}

So,

m+n+p=2+3+2

\Rightarrow m+n+p=7

Hence, the required value of

m+n+p

is 7 .

Q 35ChemistrySection B • Numericalsolutions

+3 / -1

An aqueous solution is prepared by dissolving

0.1 \mathrm{~mol}

of an ionic salt in

1.8 \mathrm{~kg}

of water at

35^{\circ} \mathrm{C}

. The salt remains

90 \%

dissociated in the solution. The vapour pressure of the solution is

59.724 \mathrm{~mm}

of Hg. Vapor pressure of water at

35{ }^{\circ} \mathrm{C}

is

60.000 \mathrm{~mm}

of

\mathrm{Hg}

. The number of ions present per formula unit of the ionic salt is _________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 36ChemistrySection A • MCQbiomolecules

+3 / -1

Treatment of D-glucose with aqueous

\mathrm{NaOH}

results in a mixture of monosaccharides, which are

A

B

C

D

Q 37MathematicsSection A • MCQmatrices and determinants

+3 / -1

If

M=\left(\begin{array}{rr}\frac{5}{2} & \frac{3}{2} \\ -\frac{3}{2} & -\frac{1}{2}\end{array}\right)

, then which of the

following matrices is equal to

M^{2022} ?

A

\left(\begin{array}{rr}3034 & 3033 \\ -3033 & -3032\end{array}\right)

B

\left(\begin{array}{ll}3034 & -3033 \\ 3033 & -3032\end{array}\right)

C

\left(\begin{array}{rr}3033 & 3032 \\ -3032 & -3031\end{array}\right)

D

\left(\begin{array}{rr}3032 & 3031 \\ -3031 & -3030\end{array}\right)

Q 38MathematicsSection A • MCQprobability

+3 / -1

Suppose that

Box-I contains 8 red, 3 blue and 5 green balls,

Box-II contains 24 red, 9 blue and 15 green balls,

Box-III contains 1 blue, 12 green and 3 yellow balls,

Box-IV contains 10 green, 16 orange and 6 white balls.

A ball is chosen randomly from Box-I; call this ball $b$ . If $b$ is red then a ball is chosen randomly from Box-II, if $b$ is blue then a ball is chosen randomly from Box-III, and if $b$ is green then a ball is chosen randomly from Box-IV. The conditional probability of the event 'one of the chosen balls is white' given that the event 'at least one of the chosen balls is green' has happened, is equal to

A

\frac{15}{256}

B

\frac{3}{16}

C

\frac{5}{52}

D

\frac{1}{8}

Q 39MathematicsSection B • Numericalcircle

+4 / -2

Let

G

be a circle of radius

R>0

. Let

G_{1}, G_{2}, \ldots, G_{n}

be

n

circles of equal radius

r>0

. Suppose each of the

n

circles

G_{1}, G_{2}, \ldots, G_{n}

touches the circle

G

externally. Also, for

i=1,2, \ldots, n-1

, the circle

G_{i}

touches

G_{i+1}

externally, and

G_{n}

touches

G_{1}

externally. Then, which of the following statements is/are TRUE?

Q 40MathematicsSection A • MCQlimits continuity and differentiability

+3 / -1

For positive integer

n

, define

f(n)=n+\frac{16+5 n-3 n^{2}}{4 n+3 n^{2}}+\frac{32+n-3 n^{2}}{8 n+3 n^{2}}+\frac{48-3 n-3 n^{2}}{12 n+3 n^{2}}+\cdots+\frac{25 n-7 n^{2}}{7 n^{2}} .

Then, the value of

\mathop {\lim }\limits_{n \to \infty } f\left( n \right)

is equal to :

A

3+\frac{4}{3} \log _{e} 7

B

4-\frac{3}{4} \log _{e}\left(\frac{7}{3}\right)

C

4-\frac{4}{3} \log _{e}\left(\frac{7}{3}\right)

D

3+\frac{3}{4} \log _{e} 7

Q 41MathematicsSection B • Numericalcomplex numbers

+4 / -2

Let

\bar{z}

denote the complex conjugate of a complex number

z

. If

z

is a non-zero complex number for which both real and imaginary parts of

(\bar{z})^{2}+\frac{1}{z^{2}}

are integers, then which of the following is/are possible value(s) of

|z|

?

Q 42MathematicsSection B • Numericalvector algebra

+4 / -2

Let

\hat{\imath}, \hat{\jmath}

and

\hat{k}

be the unit vectors along the three positive coordinate axes. Let

\begin{aligned} & \vec{a}=3 \hat{\imath}+\hat{\jmath}-\hat{k} \text {, } \\ & \vec{b}=\hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}, \quad b_{2}, b_{3} \in \mathbb{R} \text {, } \\ & \vec{c}=c_{1} \hat{\imath}+c_{2} \hat{\jmath}+c_{3} \hat{k}, \quad c_{1}, c_{2}, c_{3} \in \mathbb{R} \end{aligned}

be three vectors such that

b_{2} b_{3}>0, \vec{a} \cdot \vec{b}=0

and

\left(\begin{array}{ccc} 0 & -c_{3} & c_{2} \\ c_{3} & 0 & -c_{1} \\ -c_{2} & c_{1} & 0 \end{array}\right)\left(\begin{array}{l} 1 \\ b_{2} \\ b_{3} \end{array}\right)=\left(\begin{array}{r} 3-c_{1} \\ 1-c_{2} \\ -1-c_{3} \end{array}\right) .

Then, which of the following is/are TRUE?

Q 43MathematicsSection B • Numericaltrigonometric functions and equations

+3 / -1

Let

\alpha

and

\beta

be real numbers such that

-\frac{\pi}{4}<\beta<0<\alpha<\frac{\pi}{4}

.

If

\sin (\alpha+\beta)=\frac{1}{3}

and

\cos (\alpha-\beta)=\frac{2}{3}

, then the greatest integer less than or equal to

\left(\frac{\sin \alpha}{\cos \beta}+\frac{\cos \beta}{\sin \alpha}+\frac{\cos \alpha}{\sin \beta}+\frac{\sin \beta}{\cos \alpha}\right)^{2}

is

Answer Format:Enter numerical value rounded to the nearest integer

Q 44MathematicsSection A • MCQpermutations and combinations

+3 / -1

Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen ?

A

21816

B

85536

C

12096

D

156816

Q 45MathematicsSection B • Numericaltrigonometric functions and equations

+4 / -2

Let

P Q R S

be a quadrilateral in a plane, where

Q R=1, \angle P Q R=\angle Q R S=70^{\circ}, \angle P Q S=15^{\circ}

and

\angle P R S=40^{\circ}

.

If

\angle R P S=\theta^{\circ}, P Q=\alpha

and

P S=\beta

, then the interval(s) that contain(s) the value of

4 \alpha \beta \sin \theta^{\circ}

is/are

Q 46MathematicsSection B • Numericaldifferential equations

+4 / -2

For

x \in \mathbb{R}

, let the function

y(x)

be the solution of the differential equation

\frac{d y}{d x}+12 y=\cos \left(\frac{\pi}{12} x\right), \quad y(0)=0

Then, which of the following statements is/are TRUE ?

Q 47MathematicsSection B • Numericaldefinite integration

+3 / -1

The greatest integer less than or equal to

\int_{1}^{2} \log _{2}\left(x^{3}+1\right) d x+\int_{1}^{\log _{2} 9}\left(2^{x}-1\right)^{\frac{1}{3}} d x

is ___________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 48MathematicsSection B • Numericaldifferential equations

+3 / -1

If

y(x)

is the solution of the differential equation

x d y-\left(y^{2}-4 y\right) d x=0 \text { for } x > 0, y(1)=2,

and the slope of the curve

y=y(x)

is never zero, then the value of

10 y(\sqrt{2})

is

Answer Format:Enter numerical value rounded to the nearest integer

Q 49MathematicsSection B • Numericalquadratic equation and inequalities

+3 / -1

The product of all positive real values of

x

satisfying the equation

x^{\left(16\left(\log _{5} x\right)^{3}-68 \log _{5} x\right)}=5^{-16}

is __________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 50MathematicsSection B • Numericallimits continuity and differentiability

+3 / -1

If

\beta=\lim \limits_{x \to 0} \frac{e^{x^{3}}-\left(1-x^{3}\right)^{\frac{1}{3}}+\left(\left(1-x^{2}\right)^{\frac{1}{2}}-1\right) \sin x}{x \sin ^{2} x},

then the value of

6 \beta

is ___________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 51MathematicsSection B • Numericalapplication of integration

+3 / -1

Consider the functions

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

defined by

f(x)=x^{2}+\frac{5}{12} \quad \text { and } \quad g(x)= \begin{cases}2\left(1-\frac{4|x|}{3}\right), & |x| \leq \frac{3}{4} \\ 0, & |x|>\frac{3}{4}\end{cases}

If

\alpha

is the area of the region

\left\{(x, y) \in \mathbb{R} \times \mathbb{R}:|x| \leq \frac{3}{4}, 0 \leq y \leq \min \{f(x), g(x)\}\right\},

then the value of

9 \alpha

is

Answer Format:Enter numerical value rounded to the nearest integer

Q 52MathematicsSection B • Numericalapplication of derivatives

+4 / -2

Let

\alpha=\sum\limits_{k = 1}^\infty {{{\sin }^{2k}}\left( {{\pi \over 6}} \right)}

Let

g:[0,1] \rightarrow \mathbb{R}

be the function defined by

g(x)=2^{\alpha x}+2^{\alpha(1-x)} .

Then, which of the following statements is/are TRUE ?

Q 53MathematicsSection B • Numericalmatrices and determinants

+3 / -1

Let

\beta

be a real number. Consider the matrix

A=\left(\begin{array}{ccc} \beta & 0 & 1 \\ 2 & 1 & -2 \\ 3 & 1 & -2 \end{array}\right)

If

A^{7}-(\beta-1) A^{6}-\beta A^{5}

is a singular matrix, then the value of

9 \beta

is _________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 54MathematicsSection B • Numericalhyperbola

+3 / -1

Consider the hyperbola

\frac{x^{2}}{100}-\frac{y^{2}}{64}=1

with foci at

S

and

S_{1}

, where

S

lies on the positive

x

-axis. Let

P

be a point on the hyperbola, in the first quadrant. Let

\angle S P S_{1}=\alpha

, with

\alpha<\frac{\pi}{2}

. The straight line passing through the point

S

and having the same slope as that of the tangent at

P

to the hyperbola, intersects the straight line

S_{1} P

at

P_{1}

. Let

\delta

be the distance of

P

from the straight line

S P_{1}

, and

\beta=S_{1} P

. Then the greatest integer less than or equal to

\frac{\beta \delta}{9} \sin \frac{\alpha}{2}

is ________.

Answer Format:Enter numerical value rounded to the nearest integer

JEE Advanced 2022 Paper 2 Question Paper with Solutions