JEE · Physics
JEE Main 2021 (24 Feb, Shift 1)
2021
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19 questions
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Q1
mcq
The workdone by a gas molecule in an isolated system is given by, $W = \alpha \beta^2 e^{-\frac{x^2}{\alpha k T}}$, where $x$ is the displacement, $k$ is the Boltzmann constant and $\text{T}$ is the temperature. $\alpha$ and $\beta$ are constants. Then the dimensions of $\beta$ will be :
a.
[M^2 L T^2]
b.
[M^0 L T^0]
c.
[M L T^{-2}]
Correct Answer
d.
[M L^2 T^{-2}]
Step-by-Step AI Explanation
Verified Method
The exponent must be dimensionless, so $\left[\frac{x^2}{\alpha k T}\right] = 1$.
Since $[k T] = [\text{Energy}] = [\text{M L}^2\text{T}^{-2}]$ and $[x^2] = [\text{L}^2]$:
$[\alpha] = \frac{[x^2]}{[k T]} = \frac{[\text{L}^2]}{[\text{M L}^2\text{T}^{-2}]} = [\text{M}^{-1}\text{T}^2]$.
The dimension of work is $[W] = [\text{M L}^2\text{T}^{-2}] = [\alpha \beta^2]$.
Thus, $[\beta^2] = \frac{[W]}{[\alpha]} = \frac{[\text{M L}^2\text{T}^{-2}]}{[\text{M}^{-1}\text{T}^2]} = [\text{M}^2\text{L}^2\text{T}^{-4}] \implies [\beta] = [\text{M L T}^{-2}]$.
Q2
mcq
Two stars of masses $\text{m}$ and $2\text{m}$ at a distance $\text{d}$ rotate about their common centre of mass in free space. The period of revolution is :
a.
$\frac{1}{2\pi} \sqrt{\frac{3\text{Gm}}{\text{d}^3}}$
b.
$2\pi \sqrt{\frac{\text{d}^3}{3\text{Gm}}}$
Correct Answer
c.
$2\pi \sqrt{\frac{3\text{Gm}}{\text{d}^3}}$
d.
$\frac{1}{2\pi} \sqrt{\frac{\text{d}^3}{3\text{Gm}}}$
Step-by-Step AI Explanation
Verified Method
For a binary star system, both masses revolve around their common centre of mass with the same angular velocity $\omega$.
The gravitational force between them provides the necessary centripetal force:
$\frac{\text{G}(m)(2m)}{d^2} = m \omega^2 r_1 = (2m) \omega^2 r_2$, where $r_1 = \frac{2}{3}d$.
Thus, $\frac{2\text{G}m^2}{d^2} = m \omega^2 \left(\frac{2}{3}d\right) \implies \omega^2 = \frac{3\text{G}m}{d^3} \implies \omega = \sqrt{\frac{3\text{G}m}{d^3}}$.
Therefore, the period of revolution is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{d^3}{3\text{G}m}}$.
Q3
mcq
Four identical particles of equal masses $1\text{ kg}$ made to move along the circumference of a circle of radius $1\text{ m}$ under the action of their own mutual gravitational attraction. The speed of each particle will be :
a.
$\frac{\sqrt{(1+2\sqrt{2})\text{G}}}{2}$
Correct Answer
b.
$\sqrt{\frac{\text{G}}{2}(1+2\sqrt{2})}$
c.
$\sqrt{\text{G}(1+2\sqrt{2})}$
d.
$\sqrt{\frac{\text{G}}{2}(2\sqrt{2}-1)}$
Step-by-Step AI Explanation
Verified Method
The four particles are placed at the vertices of a square inscribed in a circle of radius $R = 1\text{ m}$.
The side of the square is $a = \sqrt{2}R = \sqrt{2}\text{ m}$, and the diagonal is $2R = 2\text{ m}$.
The net gravitational force on any particle directed toward the centre is:
$F_{\text{net}} = 2\left(\frac{\text{G}m^2}{a^2}\cos 45^\circ\right) + \frac{\text{G}m^2}{(2R)^2} = 2\left(\frac{\text{G}m^2}{2R^2}\frac{1}{\sqrt{2}}\right) + \frac{\text{G}m^2}{4R^2} = \frac{\text{G}m^2}{R^2}\left(\frac{1}{\sqrt{2}} + \frac{1}{4}\right) = \frac{\text{G}m^2}{4R^2}(1 + 2\sqrt{2})$.
Equating this to the centripetal force $\frac{m v^2}{R}$:
$\frac{m v^2}{R} = \frac{\text{G}m^2}{4R^2}(1 + 2\sqrt{2}) \implies v = \frac{\sqrt{(1+2\sqrt{2})\text{G}}}{2}$ for $m = 1\text{ kg}$ and $R = 1\text{ m}$.
Q4
mcq
Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as ;
$\text{I}_1 = \text{M.I. of thin circular ring about its diameter}$,
$\text{I}_2 = \text{M.I. of circular disc about an axis perpendicular to disc and going through the centre}$,
$\text{I}_3 = \text{M.I. of solid cylinder about its axis and}$
$\text{I}_4 = \text{M.I. of solid sphere about its diameter}$.
Then :
a.
$\text{I}_1 + \text{I}_2 = \text{I}_3 + \frac{5}{2} \text{I}_4$
b.
$\text{I}_1 + \text{I}_3 < \text{I}_2 + \text{I}_4$
c.
$\text{I}_1 = \text{I}_2 = \text{I}_3 < \text{I}_4$
d.
$\text{I}_1 = \text{I}_2 = \text{I}_3 > \text{I}_4$
Correct Answer
Step-by-Step AI Explanation
Verified Method
Given that all four bodies have the same mass $M$ and radius $R$:
$\text{I}_1 = \frac{1}{2}MR^2 = 0.5 MR^2$
$\text{I}_2 = \frac{1}{2}MR^2 = 0.5 MR^2$
$\text{I}_3 = \frac{1}{2}MR^2 = 0.5 MR^2$
$\text{I}_4 = \frac{2}{5}MR^2 = 0.4 MR^2$
Therefore, $\text{I}_1 = \text{I}_2 = \text{I}_3 > \text{I}_4$.
Q5
mcq
Consider two satellites $\text{S}_1$ and $\text{S}_2$ with periods of revolution $1\text{ hr.}$ and $8\text{ hr.}$ respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite $\text{S}_1$ to the angular velocity of satellite $\text{S}_2$ is :
a.
$8 : 1$
Correct Answer
b.
$1 : 8$
c.
$2 : 1$
d.
$1 : 4$
Step-by-Step AI Explanation
Verified Method
The angular velocity $\omega$ of a satellite is related to its time period of revolution $T$ by $\omega = \frac{2\pi}{T}$.
Therefore, the ratio of their angular velocities is:
$\frac{\omega_1}{\omega_2} = \frac{T_2}{T_1} = \frac{8\text{ hr.}}{1\text{ hr.}} = \frac{8}{1} = 8 : 1$.
Q6
mcq
Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is '$\alpha$'. The metal sheet is heated uniformly, by a small temperature $\Delta\text{T}$, so that its new temperature is $\text{T} + \Delta\text{T}$. Calculate the increase in the volume of the metal box.
a.
$4\pi\text{a}^3\alpha\Delta\text{T}$
b.
$4\text{a}^3\alpha\Delta\text{T}$
c.
$\frac{4}{3}\pi\text{a}^3\alpha\Delta\text{T}$
d.
$3\text{a}^3\alpha\Delta\text{T}$
Correct Answer
Step-by-Step AI Explanation
Verified Method
The initial volume of the cubic box is $V = a^3$.
The coefficient of volumetric expansion is $\gamma = 3\alpha$.
The increase in volume is $\Delta V = V \gamma \Delta T = a^3 (3\alpha) \Delta T = 3a^3\alpha\Delta T$.
Q8
mcq
If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph ?
a.
Graph with a constant negative acceleration followed by a step change to a constant positive acceleration.
Correct Answer
b.
Graph showing a sinusoidal wave for acceleration versus time.
c.
Graph showing a periodic square wave for acceleration versus time.
d.
Graph showing a triangular wave for acceleration versus time.
Step-by-Step AI Explanation
Verified Method
Acceleration is the derivative of velocity with respect to time ($a = \frac{dv}{dt}$):
1. From point A to point M, the velocity-time curve is a straight line with a constant negative slope, meaning acceleration is constant and negative.
2. From point M to point B, the velocity-time curve is a straight line with a constant positive slope, meaning acceleration is constant and positive.
Thus, the acceleration-time graph consists of a negative horizontal line segment followed by a positive horizontal line segment.
Q9
mcq
n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.
$A \rightarrow B$ : Isothermal expansion at temperature T so that the volume is doubled from $V_1$ to $V_2 = 2V_1$ and pressure changes from $P_1$ to $P_2$.
$B \rightarrow C$ : Isobaric compression at pressure $P_2$ to initial volume $V_1$.
$C \rightarrow A$ : Isochoric change leading to change of pressure from $P_2$ to $P_1$.
Total workdone in the complete cycle ABCA is :
a.
0
b.
nRT ln 2
c.
nRT (ln 2 + 1/2)
d.
nRT (ln 2 - 1/2)
Correct Answer
Step-by-Step AI Explanation
Verified Method
Work done in isothermal expansion $A \rightarrow B$ is $W_{AB} = nRT \ln(V_2/V_1) = nRT \ln 2$.
Work done in isobaric compression $B \rightarrow C$ is $W_{BC} = P_2(V_1 - V_2) = P_2(V_1 - 2V_1) = -P_2 V_1 = -\frac{1}{2} P_1 V_1 = -\frac{1}{2} nRT$.
Work done in isochoric process $C \rightarrow A$ is $W_{CA} = 0$.
Total work done $W = W_{AB} + W_{BC} + W_{CA} = nRT \left(\ln 2 - \frac{1}{2}\right)$.
Q10
mcq
Match List I with List II.
List I
(a) Isothermal
(b) Isochoric
(c) Adiabatic
(d) Isobaric
List II
(i) Pressure constant
(ii) Temperature constant
(iii) Volume constant
(iv) Heat content is constant
Choose the correct answer from the options given below :
a.
(a) → (i), (b) → (iii), (c) → (ii), (d) → (iv)
b.
(a) → (iii), (b) → (ii), (c) → (i), (d) → (iv)
c.
(a) → (ii), (b) → (iv), (c) → (iii), (d) → (i)
d.
(a) → (ii), (b) → (iii), (c) → (iv), (d) → (i)
Correct Answer
Step-by-Step AI Explanation
Verified Method
In an isothermal process, temperature remains constant: (a) → (ii).
In an isochoric process, volume remains constant: (b) → (iii).
In an adiabatic process, there is no heat exchange, so heat content is constant: (c) → (iv).
In an isobaric process, pressure remains constant: (d) → (i).
Q11
mcq
In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be :
a.
A \sqrt{\frac{M+m}{M}}
b.
A \sqrt{\frac{M}{M+m}}
Correct Answer
c.
A \sqrt{\frac{M-m}{M}}
d.
A \sqrt{\frac{M}{M-m}}
Step-by-Step AI Explanation
Verified Method
At the equilibrium position, the velocity of mass $M$ is maximum: $v_0 = \omega A = \sqrt{\frac{k}{M}} A$.
When mass $m$ is gently placed on $M$, linear momentum is conserved:
$M v_0 = (M + m) v' \implies v' = \frac{M}{M+m} v_0$.
For the new combined system, the maximum velocity is $v' = \omega' A'$, where $\omega' = \sqrt{\frac{k}{M+m}}$.
Therefore, $\sqrt{\frac{k}{M+m}} A' = \frac{M}{M+m} \sqrt{\frac{k}{M}} A \implies A' = A \sqrt{\frac{M}{M+m}}$.
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Q12
mcq
A cell $\text{E}_1$ of emf 6 V and internal resistance $2\ \Omega$ is connected with another cell $\text{E}_2$ of emf 4 V and internal resistance $8\ \Omega$ (as shown in the figure). The potential difference across points X and Y is :
a.
2.0 V
b.
3.6 V
c.
5.6 V
d.
10.0 V
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Q13
mcq
A current through a wire depends on time as
$$i = \alpha_0 t + \beta t^2$$
where $\alpha_0 = 20\text{ A/s}$ and $\beta = 8\text{ As}^{-2}$. Find the charge crossed through a section of the wire in $15\text{ s}$.
a.
260 C
b.
2100 C
c.
11250 C
d.
2250 C
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Q14
mcq
Two equal capacitors are first connected in series and then in parallel. The ratio of the equivalent capacities in the two cases will be :
a.
1 : 2
b.
2 : 1
c.
4 : 1
d.
1 : 4
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Q15
mcq
A cube of side 'a' has point charges +Q located at each of its vertices except at the origin where the charge is -Q. The electric field at the centre of cube is :
a.
\frac{-Q}{3\sqrt{3} \pi \varepsilon_0 a^2} (\hat{x} + \hat{y} + \hat{z})
b.
\frac{Q}{3\sqrt{3} \pi \varepsilon_0 a^2} (\hat{x} + \hat{y} + \hat{z})
c.
\frac{-2Q}{3\sqrt{3} \pi \varepsilon_0 a^2} (\hat{x} + \hat{y} + \hat{z})
d.
\frac{2Q}{3\sqrt{3} \pi \varepsilon_0 a^2} (\hat{x} + \hat{y} + \hat{z})
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Q16
mcq
If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of $\beta$ will be :
a.
7
b.
0.875
c.
0.5
d.
3.5
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Q17
mcq
In the given figure, the energy levels of hydrogen atom have been shown along with some transitions marked A, B, C, D and E.
The transitions A, B and C respectively represent :
a.
The first member of the Lyman series, third member of Balmer series and second member of Paschen series.
b.
The ionization potential of hydrogen, second member of Balmer series and third member of Paschen series.
c.
The series limit of Lyman series, second member of Balmer series and second member of Paschen series.
d.
The series limit of Lyman series, third member of Balmer series and second member of Paschen series.
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Q18
mcq
Given below are two statements :
Statement I : Two photons having equal linear momenta have equal wavelengths.
Statement II : If the wavelength of photon is decreased, then the momentum and energy of a photon will also decrease.
In the light of the above statements, choose the correct answer from the options given below.
a.
Both Statement I and Statement II are true
b.
Both Statement I and Statement II are false
c.
Statement I is true but Statement II is false
d.
Statement I is false but Statement II is true
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Q19
mcq
The focal length $f$ is related to the radius of curvature $r$ of the spherical convex mirror by :
a.
$f = r$
b.
$f = -r$
c.
$f = -\frac{1}{2}r$
d.
$f = +\frac{1}{2}r$
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Q20
mcq
In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
a.
4 : 1
b.
2 : 1
c.
1 : 4
d.
3 : 1
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