JEE Main 2023 13th April Evening Shift Question Paper Question Paper with Solutions | JEE Main Mock Test | JeeMate

Q 1PhysicsSection A • MCQmotion

+4 / -1

The distance travelled by an object in time $t$ is given by $s=(2.5) t^{2}$ . The instantaneous speed of the object at $\mathrm{t}=5 \mathrm{~s}$ will be:

A

5 \mathrm{~ms}^{-1}

B

12.5 \mathrm{~ms}^{-1}

C

62.5 \mathrm{~ms}^{-1}

D

25 \mathrm{~ms}^{-1}

Q 2PhysicsSection B • Numericalheat and thermodynamics

+4 / -1

Two plates $\mathrm{A}$ and $\mathrm{B}$ have thermal conductivities $84 ~\mathrm{Wm}^{-1} \mathrm{~K}^{-1}$ and $126 ~\mathrm{Wm}^{-1} \mathrm{~K}^{-1}$ respectively. They have same surface area and same thickness. They are placed in contact along their surfaces. If the temperatures of the outer surfaces of $\mathrm{A}$ and $\mathrm{B}$ are kept at $100^{\circ} \mathrm{C}$ and $0{ }^{\circ} \mathrm{C}$ respectively, then the temperature of the surface of contact in steady state is _____________ ${ }^{\circ} \mathrm{C}$ .

Answer Format:Enter numerical value rounded to the nearest integer

Q 3PhysicsSection A • MCQcapacitor

+4 / -1

In the network shown below, the charge accumulated in the capacitor in steady state will be:

qImage-rc-circuit-capacitor-electricity-Jeemate

A

10.3

\mu

C

B

7.2

\mu

C

4.8

\mu

C

D

12

\mu

C

Q 4PhysicsSection A • MCQunits and measurements

+4 / -1

In the equation $\left[X+\frac{a}{Y^{2}}\right][Y-b]=\mathrm{R} T, X$ is pressure, $Y$ is volume, $\mathrm{R}$ is universal gas constant and $T$ is temperature. The physical quantity equivalent to the ratio $\frac{a}{b}$ is:

A

Impulse

B

Energy

C

Pressure gradient

D

Coefficient of viscosity

Q 5PhysicsSection A • MCQgravitation

+4 / -1

Given below are two statements:

Statement I : For a planet, if the ratio of mass of the planet to its radius increases, the escape velocity from the planet also increases.

Statement II : Escape velocity is independent of the radius of the planet.

In the light of above statements, choose the most appropriate answer form the options given below

A

Both Statement I and Statement II are correct

B

Statement I is correct but statement II is incorrect

C

Both Statement I and Statement II are incorrect

D

Statement I is incorrect but statement II is correct

Q 6PhysicsSection A • MCQheat and thermodynamics

+4 / -1

The initial pressure and volume of an ideal gas are P $_0$ and V $_0$ . The final pressure of the gas when the gas is suddenly compressed to volume $\frac{V_0}{4}$ will be :

(Given $\gamma$ = ratio of specific heats at constant pressure and at constant volume)

A

P

_0

(4)

^{\frac{1}{\gamma}}

B

P

_0

C

4P

_0

D

P

_0

(4)

^{\gamma}

Q 7PhysicsSection B • Numericalwork power and energy

+4 / -1

A car accelerates from rest to $u \mathrm{~m} / \mathrm{s}$ . The energy spent in this process is E J. The energy required to accelerate the car from $u \mathrm{~m} / \mathrm{s}$ to $2 \mathrm{u} \mathrm{m} / \mathrm{s}$ is $\mathrm{nE~J}$ . The value of $\mathrm{n}$ is ____________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 8PhysicsSection A • MCQwave optics

+4 / -1

In a Young's double slits experiment, the ratio of amplitude of light coming from slits is $2: 1$ . The ratio of the maximum to minimum intensity in the interference pattern is:

A

25 : 9

B

9 : 1

C

9 : 4

D

2 : 1

Q 9PhysicsSection B • Numericalray optics

+4 / -1

A bi convex lens of focal length $10 \mathrm{~cm}$ is cut in two identical parts along a plane perpendicular to the principal axis. The power of each lens after cut is ____________ D.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:5

Step-by-Step Solution:

When a biconvex lens is cut into two identical parts along a plane perpendicular to the principal axis, each part becomes a plano-convex lens. To find the new power of each plano-convex lens, we can use the lensmaker's formula:

\frac{1}{f} = (n - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)

For the original biconvex lens, both radii of curvature have the same magnitude but opposite signs, so let's denote them as ±R. The focal length of the original lens is given as 10 cm, and the refractive index (n) is constant for both the original lens and the new plano-convex lenses.

For the original lens, the lensmaker's formula becomes:

\frac{1}{f} = (n - 1) \left(\frac{1}{R} - \frac{1}{-R}\right)

Since the focal length is given as 10 cm, we can plug in the value:

\frac{1}{10} = (n - 1) \left(\frac{1}{R} + \frac{1}{R}\right)

\frac{1}{10} = (n - 1) \left(\frac{2}{R}\right)

For each plano-convex lens, one radius of curvature (R1) is infinite (the flat side), and the other radius (R2) is the same as the original lens (R). The lensmaker's formula for the plano-convex lens becomes:

\frac{1}{f'} = (n - 1) \left(\frac{1}{\infty} - \frac{1}{R}\right)

\frac{1}{f'} = (n - 1) \left(-\frac{1}{R}\right)

Now, we can substitute the expression for (n-1)(2/R) from the original lens equation:

\frac{1}{f'} = \frac{1}{10} \cdot \frac{1}{2}

\frac{1}{f'} = \frac{1}{20}

So, the focal length of each plano-convex lens (f') is 20 cm. To find the power of each lens, we can use the formula:

P (\text{in diopters}) = \frac{1}{f (\text{in meters})}

Converting the focal length to meters and calculating the power:

P = \frac{1}{0.2}

P = 5 \text{ D}

So, the power of each plano-convex lens after the cut is 5 diopters (D).

Q 10PhysicsSection A • MCQcircular motion

+4 / -1

A vehicle of mass $200 \mathrm{~kg}$ is moving along a levelled curved road of radius $70 \mathrm{~m}$ with angular velocity of $0.2 ~\mathrm{rad} / \mathrm{s}$ . The centripetal force acting on the vehicle is:

A

560 \mathrm{~N}

B

14 \mathrm{~N}

C

2800 \mathrm{~N}

D

2240 \mathrm{~N}

Q 11PhysicsSection A • MCQheat and thermodynamics

+4 / -1

The mean free path of molecules of a certain gas at STP is $1500 \mathrm{~d}$ , where $\mathrm{d}$ is the diameter of the gas molecules. While maintaining the standard pressure, the mean free path of the molecules at $373 \mathrm{~K}$ is approximately:

A

750 \mathrm{~d}

B

1500 \mathrm{~d}

C

\mathrm{2049~ d}

D

1098 \mathrm{~d}

Q 12PhysicsSection B • Numericaldual nature of radiation

+4 / -1

An atom absorbs a photon of wavelength $500 \mathrm{~nm}$ and emits another photon of wavelength $600 \mathrm{~nm}$ . The net energy absorbed by the atom in this process is $n \times 10^{-4} ~\mathrm{eV}$ . The value of n is __________. [Assume the atom to be stationary during the absorption and emission process] (Take $\mathrm{h}=6.6 \times 10^{-34} ~\mathrm{Js}$ and $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ )

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:4125

Step-by-Step Solution:

The energy

E

of a photon is related to its wavelength

\lambda

by the formula:

E=\frac{hc}{\lambda}

where

h

is Planck's constant and

c

is the speed of light. In this problem, we are given that an atom absorbs a photon of wavelength

\lambda_1=500~\mathrm{nm}

and emits another photon of wavelength

\lambda_2=600~\mathrm{nm}

. We can use the formula above to calculate the energy absorbed by the atom:

\mathrm{Energy~absorbed}=E_1-E_2=\frac{hc}{\lambda_1}-\frac{hc}{\lambda_2}=hc\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right)

Substituting the given values for

h

and

c

, we get:

\mathrm{Energy~absorbed}=6.6\times10^{-34}~\mathrm{J\cdot s}\cdot3\times10^8~\mathrm{m/s}\cdot\left(\frac{1}{500\times10^{-9}~\mathrm{m}}-\frac{1}{600\times10^{-9}~\mathrm{m}}\right)

Simplifying this expression, we get:

\mathrm{Energy~absorbed}=6.6\times10^{-20}~\mathrm{J}

We need to express this energy in electron volts (eV), which is a more convenient unit for atomic and molecular energies. To do this, we can divide the energy in joules by the charge of an electron:

\mathrm{Energy~absorbed~in~eV}=\frac{6.6\times10^{-20}~\mathrm{J}}{1.6\times10^{-19}~\mathrm{C/eV}}=0.4125~\mathrm{eV}

Finally, we can express the net energy absorbed in terms of the given value of

n

, as follows:

n\times10^{-4}~\mathrm{eV}=0.4125~\mathrm{eV}

Solving for

n

, we get:

n=\frac{0.4125}{10^{-4}}=4125

Therefore, the value of

n

is

\boxed{4125}

.

Q 13PhysicsSection A • MCQatoms and nuclei

+4 / -1

Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$

Assertion A : The binding energy per nucleon is practically independent of the atomic number for nuclei of mass number in the range 30 to 170 .

Reason R : Nuclear force is short ranged.

In the light of the above statements, choose the correct answer from the options given below

A

\mathrm{A}

is false but

\mathbf{R}

is true

B

\mathrm{A}

is true but

\mathbf{R}

is false

C

Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

D

Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

Q 14PhysicsSection A • MCQmotion

+4 / -1

A passenger sitting in a train A moving at $90 \mathrm{~km} / \mathrm{h}$ observes another train $\mathrm{B}$ moving in the opposite direction for $8 \mathrm{~s}$ . If the velocity of the train B is $54 \mathrm{~km} / \mathrm{h}$ , then length of train B is:

A

80 m

B

200 m

C

120 m

D

320 m

Q 15PhysicsSection A • MCQcommunication systems

+4 / -1

To radiate EM signal of wavelength $\lambda$ with high efficiency, the antennas should have a minimum size equal to:

A

\frac{\lambda}{2}

B

\lambda

C

\frac{\lambda}{4}

D

2 \lambda

Q 16PhysicsSection B • Numericalelectrostatics

+4 / -1

Three point charges $\mathrm{q},-2 \mathrm{q}$ and $2 \mathrm{q}$ are placed on $x$ -axis at a distance $x=0, x=\frac{3}{4} R$ and $x=R$ respectively from origin as shown. If $\mathrm{q}=2 \times 10^{-6} \mathrm{C}$ and $\mathrm{R}=2 \mathrm{~cm}$ , the magnitude of net force experienced by the charge $-2 q$ is ___________ N.

qImage-electric-charges-and-coulombs-law-electrostatics-electricity-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 17PhysicsSection A • MCQelectrostatics

+4 / -1

A $10 ~\mu \mathrm{C}$ charge is divided into two parts and placed at $1 \mathrm{~cm}$ distance so that the repulsive force between them is maximum. The charges of the two parts are:

A

9 ~\mu\mathrm{C}, 1 ~\mu \mathrm{C}

B

5 ~\mu\mathrm{C}, 5 ~\mu \mathrm{C}

C

8 ~\mu\mathrm{C}, 2 ~\mu \mathrm{C}

D

7 ~\mu\mathrm{C}, 3 ~\mu \mathrm{C}

Q 18PhysicsSection B • Numericalelectromagnetic induction

+4 / -1

An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area $24 \mathrm{~cm}^{2}$ . The two ends of the wire are connected to a resistor. The total resistance in the circuit is $12 ~\Omega$ . If an externally applied uniform magnetic field in the core along its axis changes from $1.5 \mathrm{~T}$ in one direction to $1.5 ~\mathrm{T}$ in the opposite direction, the charge flowing through a point in the circuit during the change of magnetic field will be ___________ $\mathrm{mC}$ .

Answer Format:Enter numerical value rounded to the nearest integer

Q 19PhysicsSection A • MCQalternating current

+4 / -1

Given below are two statements:

Statement I : An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.

Statement II : An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.

In the light of above statements, choose the correct answer form the options given below:

A

Both Statement I and Statement II are false

B

Statement I is true but statement II is false

C

Statement I is false but statement II is true

D

Both Statement I and Statement II are true

Q 20PhysicsSection B • Numericalmagnetics

+4 / -1

A straight wire $\mathrm{AB}$ of mass $40 \mathrm{~g}$ and length $50 \mathrm{~cm}$ is suspended by a pair of flexible leads in uniform magnetic field of magnitude $0.40 \mathrm{~T}$ as shown in the figure. The magnitude of the current required in the wire to remove the tension in the supporting leads is ___________ A.

$\left(\right.$ Take $g=10 \mathrm{~ms}^{-2}$ ).

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

Answer Format:Enter numerical value rounded to the nearest integer

Q 21PhysicsSection A • MCQelectromagnetic waves

+4 / -1

In an electromagnetic wave, at an instant and at particular position, the electric field is along the negative $z$ -axis and magnetic field is along the positive $x$ -axis. Then the direction of propagation of electromagnetic wave is:

A

at

45^{\circ}

angle from positive y-axis

B

positive

y

-axis

C

negative

\mathrm{y}

-axis

D

positive z-axis

Q 22PhysicsSection B • Numericalrotational motion

+4 / -1

A light rope is wound around a hollow cylinder of mass 5 kg and radius 70 cm. The rope is pulled with a force of 52.5 N. The angular acceleration of the cylinder will be _________ rad s $^{-2}$ .

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:15

Step-by-Step Solution:

In this problem, the net force on the cylinder is the tension

T

in the rope, which is equal to the force applied to the rope:

F=T=52.5~\mathrm{N}

The force causes the cylinder to accelerate with an angular acceleration

\alpha

, which is related to its linear acceleration

a

and the radius of the cylinder

R

by the equation:

\alpha=\frac{a}{R}

The linear acceleration

a

of the cylinder can be found using the formula

F=ma

:

ma=F=52.5~\mathrm{N}

where

m=5~\mathrm{kg}

is the mass of the cylinder. Solving for

a

, we get:

a=\frac{F}{m}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}}=10.5~\mathrm{m/s^2}

Substituting this value of

a

into the equation for

\alpha

, we get:

\alpha=\frac{a}{R}=\frac{10.5~\mathrm{m/s^2}}{0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}

Therefore, the angular acceleration of the cylinder is

15~\mathrm{rad/s^2}

.

Alternate Method:

Let's first draw a free body diagram of the cylinder. The force

F

applied to the rope creates a tension in the rope, which in turn exerts a force on the cylinder in the opposite direction. This force is given by:

T=F

where

T

is the tension in the rope. The cylinder also experiences a torque due to the tension in the rope, which causes it to rotate. The torque is given by:

\tau=TR

where

R

is the radius of the cylinder.

The net torque on the cylinder is equal to the product of the moment of inertia

I

of the cylinder and its angular acceleration

\alpha

:

\tau=I\alpha

The moment of inertia of a hollow cylinder about its geometrical axis which is parallel to its length is given by:

I=MR^2

where

M

is the mass of the cylinder.

Substituting the given values, we get:

\tau=TR=I\alpha=MR^2\alpha

Solving for

\alpha

, we get:

\alpha=\frac{T}{MR}=\frac{F}{MR}=\frac{52.5~\mathrm{N}}{5~\mathrm{kg}\cdot0.7~\mathrm{m}}=\boxed{15~\mathrm{rad/s^2}}

Therefore, the angular acceleration of the cylinder is

15~\mathrm{rad/s^2}

.

Q 23PhysicsSection A • MCQelectronic devices

+4 / -1

The output from NAND gate having inputs A and B given below will be,

qImage-digital-circuits-electronic-devices-modern-physics-Jeemate

A

B

C

D

Official Correct Answer:Option (B)

Step-by-Step Solution:

\begin{aligned} &\text {Truth table for NAND gate is }\\ &\begin{array}{|c|c|c|} \hline \mathrm{A} & \mathrm{B} & \mathrm{Y}=\overline{\mathrm{A} \cdot \mathrm{B}} \\ \hline 0 & 0 & 1 \\ \hline 0 & 1 & 1 \\ \hline 1 & 0 & 1 \\ \hline 1 & 1 & 0 \\ \hline \end{array} \end{aligned}

On the basis of given input

\mathrm{A}

and

\mathrm{B}

the truth table is

$\mathrm{A}$	$\mathrm{B}$	$\mathrm{Y}$
1	1	0
0	0	1
0	1	1
1	0	1
1	1	0
0	0	1
0	1	1

Q 24PhysicsSection A • MCQgravitation

+4 / -1

Two planets A and B of radii $\mathrm{R}$ and 1.5 R have densities $\rho$ and $\rho / 2$ respectively. The ratio of acceleration due to gravity at the surface of $\mathrm{B}$ to $\mathrm{A}$ is:

A

2 : 1

B

2 : 3

C

4 : 3

D

3 : 4

Q 25PhysicsSection B • Numericalwaves

+4 / -1

In an experiment with sonometer when a mass of $180 \mathrm{~g}$ is attached to the string, it vibrates with fundamental frequency of $30 \mathrm{~Hz}$ . When a mass $\mathrm{m}$ is attached, the string vibrates with fundamental frequency of $50 \mathrm{~Hz}$ . The value of $\mathrm{m}$ is ___________ g.

Answer Format:Enter numerical value rounded to the nearest integer

Q 26PhysicsSection A • MCQdual nature of radiation

+4 / -1

Given below are two statements:

Statement I : Out of microwaves, infrared rays and ultraviolet rays, ultraviolet rays are the most effective for the emission of electrons from a metallic surface.

Statement II : Above the threshold frequency, the maximum kinetic energy of photoelectrons is inversely proportional to the frequency of the incident light.

In the light of above statements, choose the correct answer form the options given below

A

Both Statement I and Statement II are true

B

Statement I is true but statement II is false

C

Statement I is false but statement II is true

D

Both Statement I and Statement II are false

Q 27PhysicsSection A • MCQproperties of matter

+4 / -1

Given below are two statements: one is labelled as Assertion $\mathbf{A}$ and the other is labelled as Reason $\mathbf{R}$

Assertion A : A spherical body of radius $(5 \pm 0.1) \mathrm{mm}$ having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is $4 \%$ .

Reason R : The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius.

In the light of the above statements, choose the correct answer from the options given below

A

A is false but

\mathbf{R}

is true

B

\mathrm{A}

is true but

\mathbf{R}

is false

C

Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

D

Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

Q 28PhysicsSection B • Numericalcapacitor

+4 / -1

In the circuit shown, the energy stored in the capacitor is $n ~\mu \mathrm{J}$ . The value of $n$ is __________

qImage-capacitor-in-circuit-capacitor-electricity-Jeemate

Answer Format:Enter numerical value rounded to the nearest integer

Q 29PhysicsSection A • MCQsimple harmonic motion

+4 / -1

A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is :

A

\frac{1}{\sqrt{2}} A

B

\frac{1}{2} A

C

2 \mathrm{~A}

D

\sqrt{2 A}

Q 30PhysicsSection A • MCQmagnetics

+4 / -1

An electron is moving along the positive $\mathrm{x}$ -axis. If the uniform magnetic field is applied parallel to the negative z-axis, then

A. The electron will experience magnetic force along positive y-axis

B. The electron will experience magnetic force along negative y-axis

C. The electron will not experience any force in magnetic field

D. The electron will continue to move along the positive $\mathrm{x}$ -axis

E. The electron will move along circular path in magnetic field

Choose the correct answer from the options given below:

A

A and E only

B

B and D only

C

B and E only

D

C and D only

Q 31ChemistrySection A • MCQs block elements

+4 / -1

Identify the correct order of standard enthalpy of formation of sodium halides :

A

\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaCl}<\mathrm{NaF}

B

\mathrm{NaI}<\mathrm{NaBr}<\mathrm{NaF}<\mathrm{NaCl}

C

\mathrm{NaF}<\mathrm{NaCl}<\mathrm{NaBr}<\mathrm{NaI}

D

\mathrm{NaCl}<\mathrm{NaF}<\mathrm{NaBr}<\mathrm{NaI}

Q 32ChemistrySection A • MCQhydrogen

+4 / -1

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Isotopes of hydrogen have almost same chemical properties, but difference in their rates of reaction.

Reason R : Isotopes of hydrogen have different enthalpy of bond dissociation.

In the light of the above statements, choose the most appropriate answer from the options given below:

A

A is not correct but R is correct

B

Both A and R are correct but R is NOT the correct explanation of A

C

A is correct but R is not correct

D

Both A and R are correct and R is the correct explanation of A

Q 33ChemistrySection A • MCQcoordination compounds

+4 / -1

The total number of stereoisomers for the complex $\left[\mathrm{Cr}(o x)_{2} \mathrm{ClBr}\right]^{3-}$ (where $o x=$ oxalate) is :

A

1

B

3

C

2

D

4

Q 34ChemistrySection A • MCQalcohols phenols and ethers

+4 / -1

Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Order of acidic nature of the following compounds is A > B > C.

qImage-properties-preparation-and-uses-of-alcohols-alcohols-phenols-and-ethers-organic-chemistry-Jeemate

Reason R : Fluoro is a stronger electron withdrawing group than Chloro group.

In the light of the above statements, choose the correct answer from the options given below:

A

Both A and R are correct but R is NOT the correct explanation of A

B

A is false but R is true

C

A is true but R is false

D

Both A and R are correct and R is the correct explanation of A

Q 35ChemistrySection A • MCQalcohols phenols and ethers

+4 / -1

The major product for the following reaction is:

qImage-properties-preparation-and-uses-of-alcohols-alcohols-phenols-and-ethers-organic-chemistry-Jeemate

A

B

C

D

Q 36ChemistrySection A • MCQisolation of elements

+4 / -1

Given below are two statements related to Ellingham diagram:

Statement I : Ellingham diagrams can be constructed for formation of oxides, sulfides and halides of metals.

Statement II : It consists of plots of $\Delta_{\mathrm{f}} \mathrm{H}^{0}$ vs $\mathrm{T}$ for formation of oxides of elements.

In the light of the above statements, choose the most appropriate answer from the options given below:

A

Both Statement I and Statement II are correct

B

Both Statement I and Statement II are incorrect

C

Statement I is correct but Statement II is incorrect

D

Statement I is incorrect but Statement II is correct

Q 37ChemistrySection A • MCQchemical bonding and molecular structure

+4 / -1

Given below are two statements :

Statement I : $\mathrm{SO}_{2}$ and $\mathrm{H}_{2} \mathrm{O}$ both possess V-shaped structure.

Statement II : The bond angle of $\mathrm{SO}_{2}$ is less than that of $\mathrm{H}_{2} \mathrm{O}$ .

In the light of the above statements, choose the most appropriate answer from the options given below:

A

Both Statement I and Statement II are incorrect

B

Both Statement I and Statement II are correct

C

Statement I is correct but Statement II is incorrect

D

Statement I is incorrect but Statement II is correct

Q 38ChemistrySection B • Numericalredox reactions

+4 / -1

See the following chemical reaction:

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{XH}^{+}+6 \mathrm{~F}_{e}^{2+} \rightarrow \mathrm{YCr}^{3+}+6 \mathrm{~F}_{e}^{3+}+\mathrm{Z} \mathrm{H}_{2} \mathrm{O}$

The sum of $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ is ___________

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:23

Step-by-Step Solution:

To find the sum of X, Y, and Z, we need to balance the given chemical reaction:

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{XH}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow \mathrm{YCr}^{3+} + 6 \mathrm{Fe}^{3+} + \mathrm{ZH}_{2} \mathrm{O}

Let's balance the reaction step by step:

1. Balance the chromium atoms:

There are 2 chromium atoms on the left side of the equation, so we need 2 chromium atoms on the right side. Thus, Y = 2.

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{XH}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + \mathrm{ZH}_{2} \mathrm{O}

2. Balance the oxygen atoms:

There are 7 oxygen atoms on the left side of the equation, so we need 7 oxygen atoms on the right side. Since each water molecule (H₂O) contains one oxygen atom, Z = 7.

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + \mathrm{XH}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}

3. Balance the hydrogen atoms and charges:

There are 14 hydrogen atoms on the right side of the equation, so we need 14 hydrogen atoms on the left side. Thus, X = 14.

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}

Now the chemical reaction is balanced:

\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-} + 14 \mathrm{H}^{+} + 6 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{Fe}^{3+} + 7 \mathrm{H}_{2} \mathrm{O}

The sum of X, Y, and Z is:

X + Y + Z = 14 + 2 + 7 = 23

So, the sum of X, Y, and Z is 23.

Q 39ChemistrySection B • Numericalchemical kinetics and nuclear chemistry

+4 / -1

A(g) $\to$ 2B(g) + C(g) is a first order reaction. The initial pressure of the system was found to be 800 mm Hg which increased to 1600 mm Hg after 10 min. The total pressure of the system after 30 min will be _________ mm Hg. (Nearest integer)

Answer Format:Enter numerical value rounded to the nearest integer

Q 40ChemistrySection A • MCQp block elements

+4 / -1

The covalency and oxidation state respectively of boron in $\left[\mathrm{BF}_{4}\right]^{-}$ , are :

A

3 and 4

B

4 and 3

C

4 and 4

D

3 and 5

Q 41ChemistrySection B • Numericalsolid state

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Sodium metal crystallizes in a body centred cubic lattice with unit cell edge length of $4~\mathop A\limits^o$ . The radius of sodium atom is __________ $\times ~10^{-1}$ $\mathop A\limits^o$ (Nearest integer)

Answer Format:Enter numerical value rounded to the nearest integer

Q 42ChemistrySection B • Numericalatomic structure

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The orbital angular momentum of an electron in $3 \mathrm{~s}$ orbital is $\frac{x h}{2 \pi}$ . The value of $x$ is ____________ (nearest integer)

Answer Format:Enter numerical value rounded to the nearest integer

Q 43ChemistrySection B • Numericalsome basic concepts of chemistry

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$1 \mathrm{~g}$ of a carbonate $\left(\mathrm{M}_{2} \mathrm{CO}_{3}\right)$ on treatment with excess $\mathrm{HCl}$ produces $0.01 \mathrm{~mol}$ of $\mathrm{CO}_{2}$ . The molar mass of $\mathrm{M}_{2} \mathrm{CO}_{3}$ is __________ $\mathrm{g} ~\mathrm{mol}^{-1}$ . (Nearest integer)

Answer Format:Enter numerical value rounded to the nearest integer

Q 44ChemistrySection B • Numericalionic equilibrium

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20 mL of $0.1 ~\mathrm{M} ~\mathrm{NaOH}$ is added to $50 \mathrm{~mL}$ of $0.1 ~\mathrm{M}$ acetic acid solution. The $\mathrm{pH}$ of the resulting solution is ___________ $\times 10^{-2}$ (Nearest integer)

Given : $\mathrm{pKa}\left(\mathrm{CH}_{3} \mathrm{COOH}\right)=4.76$

$\log 2=0.30$

$\log 3=0.48$

Answer Format:Enter numerical value rounded to the nearest integer

Q 45ChemistrySection A • MCQcompounds containing nitrogen

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Compound $\mathrm{A}$ from the following reaction sequence is:

qImage-aromatic-amines-compounds-containing-nitrogen-organic-chemistry-Jeemate

A

Benzoic acid

B

Phenol

C

Salicylic acid

D

Aniline

Q 46ChemistrySection B • Numericalpractical organic chemistry

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$0.400 \mathrm{~g}$ of an organic compound $(\mathrm{X})$ gave $0.376 \mathrm{~g}$ of $\mathrm{AgBr}$ in Carius method for estimation of bromine. $\%$ of bromine in the compound $(\mathrm{X})$ is ___________.

(Given: Molar mass $\mathrm{AgBr=188~g~mol^{-1}}$

$\mathrm{Br}=80 \mathrm{~g} \mathrm{~mol}^{-1}$ )

Answer Format:Enter numerical value rounded to the nearest integer

Q 47ChemistrySection B • Numericalp block elements

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If the formula of Borax is $\mathrm{Na_2B_4O_x(OH)_y~.~zH_2O}$ , then $x+y+z=$ ___________

Answer Format:Enter numerical value rounded to the nearest integer

Q 48ChemistrySection A • MCQenvironmental chemistry

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Which of the following are the Green house gases?

A. Water vapour

B. Ozone

C. I $_2$

D. Molecular hydrogen

Choose the most appropriate answer from the options given below :

A

A and D only

B

A and B only

C

B and C only

D

C and D only

Q 49ChemistrySection A • MCQpractical organic chemistry

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In the wet tests for detection of various cations by precipitation, $\mathrm{Ba}^{2+}$ cations are detected by obtaining precipitate of

A

\mathrm{BaSO}_{4}

B

\mathrm{Ba}(\mathrm{OAc})_{2}

C

\mathrm{BaCO}_{3}

D

\mathrm{Ba}(\mathrm{ox})

: Barium oxalate

Q 50ChemistrySection A • MCQbiomolecules

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The naturally occurring amino acid that contains only one basic functional group in its chemical structure is

A

asparagine

B

arginine

C

histidine

D

lysine

Q 51ChemistrySection A • MCQcoordination compounds

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Which of the following complexes will exhibit maximum attraction to an applied magnetic field?

A

\left[\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}

B

\left[\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}

C

\left[\mathrm{Co}(\mathrm{en})_{3}\right]^{3+}

D

\left[\mathrm{Zn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}

Q 52ChemistrySection A • MCQp block elements

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The correct group of halide ions which can be oxidised by oxygen in acidic medium is :

A

\mathrm{Br}^{-}

only

B

\mathrm{Cl}^{-}, \mathrm{Br}^{-}

and

\mathrm{I}^{-}

only

C

\mathrm{Br}^{-}

and

\mathrm{I}^{-}

only

D

\mathrm{I}^{-}

only

Q 53ChemistrySection A • MCQbasics of organic chemistry

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Given below are two statements :

Statement I : Tropolone is an aromatic compound and has $8 \pi$ electrons.

Statement II : $\pi$ electrons of $> \mathrm{C}=\mathrm{O}$ group in tropolone is involved in aromaticity.

In the light of the above statements, choose the correct answer from the options given below:

A

Statement I is false but Statement II is true

B

Both Statement I and Statement II are false

C

Statement I is true but Statement II is false

D

Both Statement I and Statement II are true

Q 54ChemistrySection B • Numericalsolutions

+4 / -1

Sea water contains $29.25 \% ~\mathrm{NaCl}$ and $19 \% ~\mathrm{MgCl}_{2}$ by weight of solution. The normal boiling point of the sea water is _____________ ${ }^{\circ} \mathrm{C}$ (Nearest integer)

Assume $100 \%$ ionization for both $\mathrm{NaCl}$ and $\mathrm{MgCl}_{2}$

Given : $\mathrm{K}_{\mathrm{b}}\left(\mathrm{H}_{2} \mathrm{O}\right)=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$

Molar mass of $\mathrm{NaCl}$ and $\mathrm{MgCl}_{2}$ is 58.5 and 95 $\mathrm{g} \mathrm{~mol}^{-1}$ respectively.

Answer Format:Enter numerical value rounded to the nearest integer

Q 55ChemistrySection A • MCQpolymers

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Match List I with List II

	LIST I		LIST II
A.	Weak intermolecular forces of attraction	I.	Hexamethylenediamine + adipic acid
B.	Hydrogen bonding	II.	$\mathrm{AlEt_3+TiCl_4}$
C.	Heavily branched polymer	III.	2 - chloro - 1, 3 - butadiene
D.	High density polymer	IV.	Phenol + formaldehyde

Choose the correct answer from the options given below:

A

A-III, B-I, C-IV, D-II

B

A-IV, B-I, C-III, D-II

C

A-II, B-IV, C-I, D-III

D

A-IV, B-II, C-III, D-I

Q 56ChemistrySection A • MCQsurface chemistry

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Given below are two statements, one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : The diameter of colloidal particles in solution should not be much smaller than wavelength of light to show Tyndall effect.

Reason R : The light scatters in all directions when the size of particles is large enough.

In the light of the above statements, choose the correct answer from the options given below:

A

Both A and R are correct and R is the correct explanation of A

B

A is true but R is false

C

Both A and R are correct but R is NOT the correct explanation of A

D

A is false but R is true

Q 57ChemistrySection B • Numericalelectrochemistry

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At $298 \mathrm{~K}$ , the standard reduction potential for $\mathrm{Cu}^{2+} / \mathrm{Cu}$ electrode is $0.34 \mathrm{~V}$ .

Given : $\mathrm{K}_{\mathrm{sp}} \mathrm{Cu}(\mathrm{OH})_{2}=1 \times 10^{-20}$

Take $\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}$

The reduction potential at $\mathrm{pH}=14$ for the above couple is $(-) x \times 10^{-2} \mathrm{~V}$ . The value of $x$ is ___________

Answer Format:Enter numerical value rounded to the nearest integer

Q 58ChemistrySection A • MCQcompounds containing nitrogen

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Match List I with List II

1 - Bromopropane is reacted with reagents in List I to give product in List II

	LIST I - Reagent		LIST II - Product
A.	$\mathrm{KOH}$ (alc)	I.	Nitrile
B.	$\mathrm{KCN}$ (alc)	II.	Ester
C.	$\mathrm{AgNO_2}$	III.	Alkene
D.	$\mathrm{H_3CCOOAg}$	IV.	Nitroalkane

Choose the correct answer from the options given below:

A

A-I, B-II, C-III, D-IV

B

A-IV, B-III, C-II, D-I

C

A-III, B-I, C-IV, D-II

D

A-I, B-III, C-IV, D-II

Q 59ChemistrySection A • MCQs block elements

+4 / -1

Better method for preparation of $\mathrm{BeF}_{2}$ , among the following is :

A

\mathrm{BeH}_{2}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}

B

\left(\mathrm{NH}_{4}\right)_{2} \mathrm{BeF}_{4} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}

C

\mathrm{Be}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}

D

\mathrm{BeO}+\mathrm{C}+\mathrm{F}_{2} \stackrel{\Delta}{\longrightarrow} \mathrm{BeF}_{2}

Q 60ChemistrySection A • MCQthermodynamics

+4 / -1

What happens when methane undergoes combustion in systems A and B respectively?

qImage-fundamentals-of-thermodynamics-thermodynamics-physical-chemistry-Jeemate

A

System A	System B
Temperature falls	Temperature rises

B

System A	System B
Temperature rises	Temperature remains same

C

System A	System B
Temperature falls	Temperature remains same

D

System A	System B
Temperature remains same	Temperature rises

Q 61MathematicsSection B • Numericalhyperbola

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The foci of a hyperbola are $( \pm 2,0)$ and its eccentricity is $\frac{3}{2}$ . A tangent, perpendicular to the line $2 x+3 y=6$ , is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $\mathrm{x}$ - and $\mathrm{y}$ -axes are $\mathrm{a}$ and $\mathrm{b}$ respectively, then $|6 a|+|5 b|$ is equal to __________

Answer Format:Enter numerical value rounded to the nearest integer

Q 62MathematicsSection A • MCQ3d geometry

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The plane, passing through the points $(0,-1,2)$ and $(-1,2,1)$ and parallel to the line passing through $(5,1,-7)$ and $(1,-1,-1)$ , also passes through the point :

A

(0,5,-2)

B

(2,0,1)

C

(1,-2,1)

D

(-2,5,0)

Official Correct Answer:Option (D)

Step-by-Step Solution:

The first step is to find the normal vector to the desired plane. Since the plane is parallel to the line passing through the points (5, 1, -7) and (1, -1, -1), the direction vector of that line is also parallel to the plane. The direction vector is the difference between the coordinates of the two points, which is (5-1, 1-(-1), -7-(-1)) = (4, 2, -6).

Next, let's find another vector that is parallel to the plane. This vector can be obtained by taking the difference between the coordinates of the points (0, -1, 2) and (-1, 2, 1), which the plane passes through. This gives us a vector of (0-(-1), -1-2, 2-1) = (1, -3, 1).

The normal to the plane is perpendicular to both these vectors. It can be found by taking the cross product of the two vectors.

The cross product of vectors (4, 2, -6) and (1, -3, 1) is :

\begin{aligned} & \vec{n}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 2 & -6 \\ 1 & -3 & 1 \end{array}\right|=\hat{i}(-16)-\hat{j}(+10)+\hat{k}(-14) \\ & =-16 \hat{i}-10 \hat{j}-14 \hat{k} \end{aligned}

The equation of the plane can now be written in the form :

-16x - 10y - 14z = d

We can find the constant 'd' by substituting one of the points through which the plane passes, say (0, -1, 2) :

d = -16.0 -10.(-1) -14.2 = 10 - 28 = -18

So, the equation of the plane is -16x - 10y - 14z = -18.

Now, we substitute the given options into the equation to check which one satisfies it :

Option A: (-16.0 -10.5 -14.(-2) = -18) => -50 + 28 = -22 ≠ -18, so A is not correct.

Option B: (-16.2 -10.0 -14.1 = -18) => -32 - 14 = -46 ≠ -18, so B is not correct.

Option C: (-16.1 -10.(-2) -14.1 = -18) => -16 + 20 -14 = -10 ≠ -18, so C is not correct.

Option D: (-16.(-2) -10.5 -14.0 = -18) => 32 - 50 = -18, so D is correct.

So, the plane also passes through the point given in Option D, which is (-2, 5, 0).

Q 63MathematicsSection A • MCQcircle

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Let the centre of a circle C be $(\alpha, \beta)$ and its radius $r < 8$ . Let $3 x+4 y=24$ and $3 x-4 y=32$ be two tangents and $4 x+3 y=1$ be a normal to C. Then $(\alpha-\beta+r)$ is equal to :

A

7

B

9

C

5

D

6

Q 64MathematicsSection A • MCQcomplex numbers

+4 / -1

Let $S=\left\{z \in \mathbb{C}: \bar{z}=i\left(z^{2}+\operatorname{Re}(\bar{z})\right)\right\}$ . Then $\sum_\limits{z \in \mathrm{S}}|z|^{2}$ is equal to :

A

\frac{7}{2}

B

4

C

3

D

\frac{5}{2}

Q 65MathematicsSection A • MCQsequences and series

+4 / -1

Let a $_1$ , a $_2$ , a $_3$ , .... be a G.P. of increasing positive numbers. Let the sum of its 6^th and 8^th terms be 2 and the product of its 3^rd and 5^th terms be $\frac{1}{9}$ . Then $6(a_2+a_4)(a_4+a_6)$ is equal to

A

2

\sqrt2

B

2

C

3

\sqrt3

D

3

Q 66MathematicsSection B • Numericaldifferential equations

+4 / -1

If $y=y(x)$ is the solution of the differential equation

$\frac{d y}{d x}+\frac{4 x}{\left(x^{2}-1\right)} y=\frac{x+2}{\left(x^{2}-1\right)^{\frac{5}{2}}}, x > 1$ such that

$y(2)=\frac{2}{9} \log _{e}(2+\sqrt{3}) \text { and } y(\sqrt{2})=\alpha \log _{e}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma \in \mathbb{N} \text {, then } \alpha \beta \gamma \text { is equal to }$ :

Answer Format:Enter numerical value rounded to the nearest integer

Q 67MathematicsSection A • MCQvector algebra

+4 / -1

Let $|\vec{a}|=2,|\vec{b}|=3$ and the angle between the vectors $\vec{a}$ and $\vec{b}$ be $\frac{\pi}{4}$ . Then $|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^{2}$ is equal to :

A

441

B

482

C

841

D

882

Q 68MathematicsSection B • Numericaldefinite integration

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Let $f_{n}=\int_\limits{0}^{\frac{\pi}{2}}\left(\sum_\limits{k=1}^{n} \sin ^{k-1} x\right)\left(\sum_\limits{k=1}^{n}(2 k-1) \sin ^{k-1} x\right) \cos x d x, n \in \mathbb{N}$ . Then $f_{21}-f_{20}$ is equal to _________

Answer Format:Enter numerical value rounded to the nearest integer

Q 69MathematicsSection A • MCQlimits continuity and differentiability

+4 / -1

If $\lim_\limits{x \rightarrow 0} \frac{e^{a x}-\cos (b x)-\frac{cx e^{-c x}}{2}}{1-\cos (2 x)}=17$ , then $5 a^{2}+b^{2}$ is equal to

A

64

B

68

C

72

D

76

Official Correct Answer:Option (B)

Step-by-Step Solution:

The Taylor series for $e^x$ , $\cos x$ , and $e^{-x}$ around $x=0$ are:

$e^{ax} = 1 + ax + \frac{(ax)^2}{2!} + \ldots,$

$\cos(bx) = 1 - \frac{(bx)^2}{2!} + \ldots,$

$e^{-cx} = 1 - cx + \frac{(cx)^2}{2!} - \ldots.$

Substituting these into the limit and simplifying:

$\lim\limits_{x \rightarrow 0} \frac{(1+ax+\frac{(ax)^2}{2!})-(1-\frac{(bx)^2}{2!})-\frac{cx}{2}(1-(cx)+\frac{(cx)^2}{2!})}{1-\cos 2x}.$

This simplifies to:

$\lim\limits _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{\left(\frac{1-\cos 2 x}{(2 x)^2}\right) \times 4 x^2}.$

Let's evaluate this expression:

$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2.$

We know that $1 - \cos 2x$ can be rewritten using the double-angle formula for cosine, which is $\cos 2x = 1 - 2\sin^2x$ . So, $1 - \cos 2x = 2\sin^2x$ .

Substituting this into our expression we get:

$(\frac{2\sin^2x}{(2x)^2}) \times 4x^2.$

This simplifies to:

${1 \over 2}$ $(\frac{\sin x}{x})^2 \times 4x^2 = {1 \over 2} \times 4x^2 = 2x^2.$

So,

$(\frac{1-\cos 2x}{(2x)^2}) \times 4x^2 = 2x^2.$

$\therefore$ $\lim _{x \rightarrow 0} \frac{(a-\frac{c}{2})x + (\frac{a^2+b^2+c^2}{2})x^2 + \ldots}{2x^2} = 17.$

Now, for the limit to exist, the coefficient of $x$ in the numerator must be zero, as the limit would be undefined otherwise. This gives us the equation:

$a-\frac{c}{2} = 0 \Rightarrow c=2a.$

Then, for the limit to equal $17$ , the coefficient of $x^2$ in the numerator must equal $17 \times 2 = 34$ . This gives us the equation:

$\frac{a^2+b^2+c^2}{2} = 34 \Rightarrow a^2+b^2+c^2 = 68.$

Since $c=2a$ , we can substitute $c$ in the equation to get:

$a^2 + b^2 + (2a)^2 = 68 \Rightarrow 5a^2 + b^2 = 68.$

Q 70MathematicsSection A • MCQdefinite integration

+4 / -1

The value of ${{{e^{ - {\pi \over 4}}} + \int\limits_0^{{\pi \over 4}} {{e^{ - x}}{{\tan }^{50}}xdx} } \over {\int\limits_0^{{\pi \over 4}} {{e^{ - x}}({{\tan }^{49}}x + {{\tan }^{51}}x)dx} }}$ is

A

51

B

50

C

25

D

49

Official Correct Answer:Option (B)

Step-by-Step Solution:

We're given the expression:

$\frac{e^{-\pi/4} + \int_0^{\pi / 4} e^{-x} \tan ^{50} x dx}{\int_0^{\pi / 4} e^{-x}(\tan x)^{49} dx + \int_0^{\pi / 4} e^{-x}(\tan x)^{51} dx}$

Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x} \tan^n x$ from 0 to $\pi / 4$ , where $n$ is an integer. Let's denote this integral as $I(n)$ :

$I(n) = \int_0^{\pi / 4} e^{-x} \tan^n x dx$

We can then rewrite the original expression in terms of $I(n)$ :

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)}$

Now, we'll apply the method of integration by parts, which states that for two functions $u(x)$ and $v(x)$ :

$\int u dv = uv - \int v du$

We'll choose:

$u = \tan^n x, \quad dv = e^{-x} dx$

Then we get:

$du = n \tan^{n-1} x \sec^2 x dx, \quad v = -e^{-x}$

Applying integration by parts, we have:

$I(n) = -e^{-x} \tan^n x \Bigg|_0^{\pi / 4} + n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x \sec^2 x dx$

Since $\tan(\pi / 4) = 1$ , the first term evaluates to:

$-e^{-\pi / 4}$

The second term becomes:

$n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x (1 + \tan^2 x) dx = n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x dx + n \int_0^{\pi / 4} e^{-x} \tan^{n+1} x dx$

This is equal to:

$n(I(n-1) + I(n+1))$

So we have:

$I(n) = -e^{-\pi / 4} + n(I(n-1) + I(n+1))$

Now we can substitute $n = 50$ into this equation:

$I(50) = -e^{-\pi / 4} + 50(I(49) + I(51))$

So the original expression becomes:

$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)} = \frac{e^{-\pi/4} - e^{-\pi / 4} + 50(I(49) + I(51))}{I(49) + I(51)} = 50$

Q 71MathematicsSection B • Numericalstatistics

+4 / -1

The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is _________

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:269

Step-by-Step Solution:

The initial mean is given by:

$\bar{x}=50$

So, the total sum of the marks initially was:

$\sum x_i = \bar{x} \times n = 50 \times 10 = 500$

We later realize that two marks were incorrectly read as 45 and 50, when they should have been 20 and 25. Therefore, the corrected sum of the marks is:

$\sum x_{i{\text{correct}}} = \sum x_i - 45 - 50 + 20 + 25 = 500 - 45 - 50 + 20 + 25 = 450$

The initial variance is given as:

$\sigma^2 = 144$

We know that variance is calculated as the mean of the squares minus the square of the mean. Therefore, rearranging gives:

$\frac{\sum x_i^2}{n} = \sigma^2 + \bar{x}^2 = 144 + 50^2 = 2644$

Then, the sum of the squares of the initial marks is:

$\sum x_i^2 = n \times \frac{\sum x_i^2}{n} = 10 \times 2594 = 26440$

The corrected sum of the squares of the marks is calculated by subtracting the squares of the incorrect marks and adding the squares of the correct marks:

$\sum x_{i{\text{correct}}}^2 = \sum x_i^2 - 45^2 - 50^2 + 20^2 + 25^2 = 26400 - 45^2 - 50^2 + 20^2 + 25^2 = 22940$

Now we can calculate the corrected variance. The variance is the mean of the squares minus the square of the mean. Using the corrected values gives:

$\sigma_{\text{correct}}^2 = \frac{\sum x_{i{\text{correct}}}^2}{n} - \left(\frac{\sum x{i_{\text{correct}}}}{n}\right)^2 = \frac{22940}{10} - \left(\frac{450}{10}\right)^2 = 2294 - 45^2 = 2294 - 2025 = 269$

Therefore, the correct variance is 269.

Q 72MathematicsSection B • Numericalsets and relations

+4 / -1

Let $\mathrm{A}=\{-4,-3,-2,0,1,3,4\}$ and $\mathrm{R}=\left\{(a, b) \in \mathrm{A} \times \mathrm{A}: b=|a|\right.$ or $\left.b^{2}=a+1\right\}$ be a relation on $\mathrm{A}$ . Then the minimum number of elements, that must be added to the relation $\mathrm{R}$ so that it becomes reflexive and symmetric, is __________

Answer Format:Enter numerical value rounded to the nearest integer

Q 73MathematicsSection A • MCQ3d geometry

+4 / -1

The line, that is coplanar to the line $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$ , is :

A

\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{4}

B

\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}

C

\frac{x-1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}

D

\frac{x+1}{1}=\frac{y-2}{2}=\frac{z-5}{5}

Official Correct Answer:Option (B)

Step-by-Step Solution:

Given two lines:

$\frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$

and

$\frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$

These lines are coplanar if the determinant of the matrix

$\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}$ = 0

Now let's apply this condition to the given problem. The given line is :

$\frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$

So, the coordinates of any point on this line are $(-3, 1, 5)$ and the direction ratios are $(-3, 1, 5)$ .

Now, let's calculate the determinants for each option and check which one equals zero.

For Option A :

$\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{4}$

The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 4)$ .

The determinant is :

$\begin{vmatrix} -1 - (-3) & 2 - 1 & 5 - 5 \\ -3 & 1 & 5 \\ -1 & 2 & 4 \end{vmatrix} = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 4 \end{vmatrix}$

Applying the formula :

= $2(1 \times 4 - 5\times2) - 1((-3\times4) - (5\times-1)) + 0((-3\times2) - (1\times-1))$

= $2(4 - 10) - 1(-12 - (-5)) + 0(-6 - (-1))$

= $2(-6) - 1(-7) + 0(-5)$

= $-12 + 7 + 0$

= $-5$

The determinant for Option A is not equal to zero, so this line is not coplanar with the given line.

For Option B, we have :

$\frac{x + 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$

The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$ .

$\begin{vmatrix} -1 - (-3) & 2 - 1 & 5 - 5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$

Applying the formula, we get :

= $2(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$

= $2(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$

= $2(-5) - 1(-10) + 0(-5)$

= $-10 + 10 + 0$

= $0$

So, the determinant for Option B equals zero, which confirms that the line in Option B is coplanar with the given line.

For Option C :

$\frac{x - 1}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$

The coordinates of any point on this line are $(1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$ .

The determinant is:

$\begin{vmatrix} 1 - (-3) & 2 - 1 & 5 - 5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix} = \begin{vmatrix} 4 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}$

Applying the formula :

= $4(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$

= $4(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$

= $4(-5) - 1(-10) + 0(-5)$

= $-20 + 10 + 0$

= $-10$

The determinant for Option C is not equal to zero, so this line is not coplanar with the given line.

For Option D :

$\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z - 5}{5}$

The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(1, 2, 5)$ .

The determinant is :

$\begin{vmatrix} -1 - (-3) & 2 - 1 & 5 - 5 \\ -3 & 1 & 5 \\ 1 & 2 & 5 \end{vmatrix} = \begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ 1 & 2 & 5 \end{vmatrix}$

Applying the formula :

= $2(1*5 - 5*2) - 1((-3*5) - (5*1)) + 0((-3*2) - (1*1))$

= $2(5 - 10) - 1(-15 - 5) + 0(-6 - 1)$

= $2(-5) - 1(-20) + 0(-7)$

= $-10 + 20 + 0$

= $10$

So the determinant for Option D is not equal to zero, which means the line in Option D is not coplanar with the given line.

Q 74MathematicsSection A • MCQmathematical reasoning

+4 / -1

The statement $(p \wedge(\sim q)) \vee((\sim p) \wedge q) \vee((\sim p) \wedge(\sim q))$ is equivalent to _________.

A

(\sim p) \vee(\sim q)

B

p \vee(\sim q)

C

\mathrm{p} \vee \mathrm{q}

D

(\sim p) \vee q

Q 75MathematicsSection A • MCQvector algebra

+4 / -1

Let for a triangle $\mathrm{ABC}$ ,

$\overrightarrow{\mathrm{AB}}=-2 \hat{i}+\hat{j}+3 \hat{k}$

$\overrightarrow{\mathrm{CB}}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$

$\overrightarrow{\mathrm{CA}}=4 \hat{i}+3 \hat{j}+\delta \hat{k}$

If $\delta > 0$ and the area of the triangle $\mathrm{ABC}$ is $5 \sqrt{6}$ , then $\overrightarrow{C B} \cdot \overrightarrow{C A}$ is equal to

A

60

B

54

C

120

D

108

Q 76MathematicsSection A • MCQquadratic equation and inequalities

+4 / -1

Let $\alpha, \beta$ be the roots of the equation $x^{2}-\sqrt{2} x+2=0$ . Then $\alpha^{14}+\beta^{14}$ is equal to

A

-64

B

-64 \sqrt{2}

C

-128 \sqrt{2}

D

-128

Official Correct Answer:Option (D)

Step-by-Step Solution:

Find the roots of the quadratic equation:

The quadratic formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . For the quadratic equation $x^2 - \sqrt{2}x + 2 = 0$ , we have $a = 1, b = -\sqrt{2}, c = 2$ . Plugging these into the quadratic formula gives:

$x = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2 - 4(1)(2)}}{2(1)} = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm \sqrt{6}i}{2}.$

So, we have two roots, $\alpha$ and $\beta$ , which are:

$\alpha = \frac{\sqrt{2} + \sqrt{6}i}{2},$

$\beta = \frac{\sqrt{2} - \sqrt{6}i}{2}.$
Express the roots in exponential form:

We can express complex numbers in the form $re^{i\theta}$ . For $\alpha$ and $\beta$ , we find the magnitude $r = \sqrt{2}$ and the arguments $\theta = \frac{\pi}{3}, -\frac{\pi}{3}$ respectively. So, we have:

$\alpha = \sqrt{2}e^{i\frac{\pi}{3}},$

$\beta = \sqrt{2}e^{-i\frac{\pi}{3}}.$
Calculate the 14th power of the roots:

To find $\alpha^{14}$ and $\beta^{14}$ , we use the property of exponents which says that $(a^m)^n = a^{mn}$ . So, we have:

$\alpha^{14} = (\sqrt{2}e^{i\frac{\pi}{3}})^{14} = 2^7e^{i\frac{14\pi}{3}} = 128e^{i\frac{2\pi}{3}},$

$\beta^{14} = (\sqrt{2}e^{-i\frac{\pi}{3}})^{14} = 2^7e^{-i\frac{14\pi}{3}} = 128e^{-i\frac{2\pi}{3}}.$
Add the 14th powers of the roots:

We want to find the real part of $\alpha^{14} + \beta^{14}$ . To do this, we use the property that $e^{ix} = \cos(x) + i\sin(x)$ . We have:

$\alpha^{14} + \beta^{14} = 128e^{i\frac{2\pi}{3}} + 128e^{-i\frac{2\pi}{3}} = 128(2)\cos\left(\frac{2\pi}{3}\right) = -128.$

Q 77MathematicsSection A • MCQfunctions

+4 / -1

The range of $f(x)=4 \sin ^{-1}\left(\frac{x^{2}}{x^{2}+1}\right)$ is

A

[0,2 \pi]

B

[0,2 \pi)

C

[0, \pi)

D

[0, \pi]

Q 78MathematicsSection A • MCQarea under the curves

+4 / -1

The area of the region $\left\{(x, y): x^{2} \leq y \leq\left|x^{2}-4\right|, y \geq 1\right\}$ is

A

\frac{4}{3}(4 \sqrt{2}+1)

B

\frac{3}{4}(4 \sqrt{2}+1)

C

\frac{4}{3}(4 \sqrt{2}-1)

D

\frac{3}{4}(4 \sqrt{2}-1)

Q 79MathematicsSection B • Numericalbinomial theorem

+4 / -1

The remainder, when $7^{103}$ is divided by 17, is __________

Answer Format:Enter numerical value rounded to the nearest integer

Q 80MathematicsSection A • MCQmatrices and determinants

+4 / -1

If the system of equations

$2 x+y-z=5$

$2 x-5 y+\lambda z=\mu$

$x+2 y-5 z=7$

has infinitely many solutions, then $(\lambda+\mu)^{2}+(\lambda-\mu)^{2}$ is equal to

A

916

B

912

C

920

D

904

Q 81MathematicsSection A • MCQbinomial theorem

+4 / -1

The coefficient of $x^{5}$ in the expansion of $\left(2 x^{3}-\frac{1}{3 x^{2}}\right)^{5}$ is :

A

\frac{26}{3}

B

\frac{80}{9}

C

9

D

8

Q 82MathematicsSection A • MCQpermutations and combinations

+4 / -1

All words, with or without meaning, are made using all the letters of the word MONDAY. These words are written as in a dictionary with serial numbers. The serial number of the word MONDAY is :

A

324

B

328

C

326

D

327

Q 83MathematicsSection A • MCQ3d geometry

+4 / -1

Let $\mathrm{N}$ be the foot of perpendicular from the point $\mathrm{P}(1,-2,3)$ on the line passing through the points $(4,5,8)$ and $(1,-7,5)$ . Then the distance of $N$ from the plane $2 x-2 y+z+5=0$ is :

A

7

B

6

C

9

D

8

Q 84MathematicsSection B • Numericaldifferentiation

+4 / -1

Let $f(x)=\sum_\limits{k=1}^{10} k x^{k}, x \in \mathbb{R}$ . If $2 f(2)+f^{\prime}(2)=119(2)^{\mathrm{n}}+1$ then $\mathrm{n}$ is equal to ___________

Answer Format:Enter numerical value rounded to the nearest integer

Q 85MathematicsSection A • MCQstraight lines and pair of straight lines

+4 / -1

Let $(\alpha, \beta)$ be the centroid of the triangle formed by the lines $15 x-y=82,6 x-5 y=-4$ and $9 x+4 y=17$ . Then $\alpha+2 \beta$ and $2 \alpha-\beta$ are the roots of the equation :

A

x^{2}-7 x+12=0

B

x^{2}-13 x+42=0

C

x^{2}-14 x+48=0

D

x^{2}-10 x+25=0

Official Correct Answer:Option (B)

Step-by-Step Solution:

Solve the equations $15x - y = 82$ and $6x - 5y = -4$

Multiply the first equation by 5 and the second by 1 and then subtract the second from the first:

$75x - 5y = 410$

$6x - 5y = -4$

Subtracting these gives $69x = 414$ which leads to $x = 6$

Substitute $x = 6$ into the first equation to get $y = 8$

So, the first vertex is $(6,8)$ .

Solve the equations $15x - y = 82$ and $9x + 4y = 17$

Multiply the first equation by 4 and the second by 1 and then add:

$60x - 4y = 328$

$9x + 4y = 17$

Adding these gives $69x = 345$ which leads to $x = 5$

Substitute $x = 5$ into the second equation to get $y = -7$

So, the second vertex is $(5,-7)$ .

Solve the equations $6x - 5y = -4$ and $9x + 4y = 17$

Multiply the first equation by 4 and the second by 5 and then add:

$24x - 20y = -16$

$45x + 20y = 85$

Adding these gives $69x = 69$ which leads to $x = 1$

Substitute $x = 1$ into the first equation to get $y = 2$

So, the third vertex is $(1,2)$ .

So, the vertices of the triangle are $(6,8)$ , $(5,-7)$ , and $(1,2)$ .

Now that we have the vertices of the triangle, we can find the centroid. The centroid $(\alpha, \beta)$ of a triangle with vertices $(x_1, y_1)$ , $(x_2, y_2)$ , and $(x_3, y_3)$ is given by :

$\alpha=\frac{x_1+x_2+x_3}{3}$ ,

$\beta=\frac{y_1+y_2+y_3}{3}$

Substituting the coordinates of the vertices into these equations, we get:

$\alpha=\frac{6+5+1}{3}=4$ ,

$\beta=\frac{8-7+2}{3}=1$

Then, we find $\alpha+2\beta$ and $2\alpha-\beta$ :

$\alpha+2\beta=4+2(1)=6 \ 2\alpha-\beta=2(4)-1=7$

So, $\alpha+2\beta=6$ and $2\alpha-\beta=7$ are the roots of the quadratic equation.

We can write the quadratic equation as

$x^2-(\alpha+2\beta+2\alpha-\beta)x+(\alpha+2\beta)(2\alpha-\beta)=0$

Substituting $\alpha=4$ , $\beta=1$ gives us:

$x^2-13x+42=0$

Q 86MathematicsSection A • MCQprobability

+4 / -1

The random variable $\mathrm{X}$ follows binomial distribution $\mathrm{B}(\mathrm{n}, \mathrm{p})$ , for which the difference of the mean and the variance is 1 . If $2 \mathrm{P}(\mathrm{X}=2)=3 \mathrm{P}(\mathrm{X}=1)$ , then $n^{2} \mathrm{P}(\mathrm{X}>1)$ is equal to :

A

15

B

12

C

11

D

16

Q 87MathematicsSection A • MCQmatrices and determinants

+4 / -1

Let for $A = \left[ {\matrix{ 1 & 2 & 3 \cr \alpha & 3 & 1 \cr 1 & 1 & 2 \cr } } \right],|A| = 2$ . If $\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}$ , then $3n + \alpha$ is equal to

A

11

B

9

C

12

D

10

Official Correct Answer:Option (A)

Step-by-Step Solution:

\begin{aligned} & A=\left[\begin{array}{lll} 1 & 2 & 3 \\ \alpha & 3 & 1 \\ 1 & 1 & 2 \end{array}\right] \\ & |A|=2 \end{aligned}

\begin{aligned} \Rightarrow&1(6-1)-2(2 \alpha-1)+3(\alpha-3)=2\\ \Rightarrow&5-4 \alpha+2+3 \alpha-9=2\\ \Rightarrow&-\alpha-4=0\\ \Rightarrow&\alpha=-4 \end{aligned}

Now,

\mathrm{|2\,adj\,(2\,adj\,(2A))| = {32^n}}

\Rightarrow

2^3|\operatorname{adj}(2 \operatorname{adj}(2 A))|

=

{32^n}

\begin{aligned} \Rightarrow & 8|\operatorname{Adj}(2 \operatorname{Adj}(2 \mathrm{~A}))| = {32^n}\\ \Rightarrow & 8\left|\operatorname{Adj}\left(2 \times 2^2 \operatorname{Adj}(\mathrm{A})\right)\right| = {32^n} \\ \Rightarrow & 8\left|\operatorname{Adj}\left(2^3 \operatorname{AdjA}\right)\right| = {32^n} \\ \Rightarrow & 8\left|2^6 \operatorname{Adj}(\operatorname{AdjA})\right| = {32^n} \\ \Rightarrow & 2^3\left(2^6\right)^3|\operatorname{Adj}(\operatorname{Adj})| = {32^n} \\ \Rightarrow & 2^3 \cdot 2^{18}|\mathrm{~A}|^4 = {32^n} \\ \Rightarrow& 2^{21} \cdot 2^4 = {32^n}\\ \Rightarrow& 2^{25} = {32^n}\\ \Rightarrow& \left(2^5\right)^5 = {32^n}\\ \Rightarrow& (32)^5 = {32^n} \end{aligned}

\begin{array}{ll} \therefore n=5 \\ \Rightarrow 3 n+\alpha=15-4=11 \end{array}

Q 88MathematicsSection B • Numericalinverse trigonometric functions

+4 / -1

For $x \in(-1,1]$ , the number of solutions of the equation $\sin ^{-1} x=2 \tan ^{-1} x$ is equal to __________.

Answer Format:Enter numerical value rounded to the nearest integer

Official Correct Answer:2

Step-by-Step Solution:

We're given the equation $\sin^{-1}x = 2\tan^{-1}x$ for $x$ in the interval $(-1, 1]$ . We want to find the number of solutions.

Step 1: Apply the sine and tangent functions to both sides :

We can rewrite the equation by applying the sine function to both sides :

$\sin(\sin^{-1}x) = \sin(2\tan^{-1}x).$

This simplifies to:

$x = \sin(2\tan^{-1}x).$

Step 2: Use the double-angle identity for sine :

Recall that $\sin(2y) = 2\sin(y)\cos(y)$ . Applying this identity to the right-hand side gives :

$x = 2\sin(\tan^{-1}x)\cos(\tan^{-1}x).$

Step 3: Use the identities for sine and cosine of an inverse tangent :

Recall that $\sin(\tan^{-1}x) = \frac{x}{\sqrt{1 + x^2}}$ and $\cos(\tan^{-1}x) = \frac{1}{\sqrt{1 + x^2}}$ . Substituting these into the equation gives :

$x = 2 \cdot \frac{x}{\sqrt{1 + x^2}} \cdot \frac{1}{\sqrt{1 + x^2}}.$

This simplifies to :

$x = \frac{2x}{1 + x^2}.$

Step 4: Solve for $x$ :

We have :

$x = \frac{2x}{1 + x^2}.$

Cross-multiplying gives :

$x(1 + x^2) = 2x.$

This simplifies to :

$x^3 + x - 2x = 0.$

Rearranging terms gives :

$x^3 - x = 0.$

This factors to:

$x(x^2 - 1) = 0.$

Setting each factor equal to zero gives the solutions $x = 0$ , $x = -1$ , and $x = 1$ .

However, we are given that $x \in (-1, 1]$ . Therefore, the only solutions in this interval are $x = 0$ and $x = 1$ .

So there are 2 solutions to the equation $\sin ^{-1} x=2 \tan ^{-1} x$ in the interval $x \in(-1,1]$ .

Q 89MathematicsSection B • Numericalpermutations and combinations

+4 / -1

Total numbers of 3-digit numbers that are divisible by 6 and can be formed by using the digits $1,2,3,4,5$ with repetition, is _________.

Answer Format:Enter numerical value rounded to the nearest integer

Q 90MathematicsSection B • Numericalquadratic equation and inequalities

+4 / -1

Let $[\alpha]$ denote the greatest integer $\leq \alpha$ . Then $[\sqrt{1}]+[\sqrt{2}]+[\sqrt{3}]+\ldots+[\sqrt{120}]$ is equal to __________

Answer Format:Enter numerical value rounded to the nearest integer

JEE Main 2023 13th April Evening Shift Question Paper with Solutions