JEE · Physics
JEE Main 2023 (10 Apr, Shift 2)
2023
·
20 questions
·
First 10 Solutions Free
Questions 1–10 have full step-by-step solutions visible.
Sign in to unlock all 20 →
Q31
mcq
The distance between two plates of a capacitor is $d$ and its capacitance is $C_1$, when air is the medium between the plates. If a metal sheet of thickness $\dfrac{2d}{3}$ and of the same area as plate is introduced between the plates, the capacitance of the capacitor becomes $C_2$. The ratio $\dfrac{C_2}{C_1}$ is
a.
1:1
b.
2:1
Correct Answer
c.
3:1
d.
4:1
Step-by-Step AI Explanation
Verified Method
For a metal slab inserted between capacitor plates, the electric field inside the metal is zero, so only the air gaps contribute to potential drop. If the metal thickness is $t$, the effective separation becomes $d-t$.
Here $t=\dfrac{2d}{3}$, so effective separation is
$$d-\frac{2d}{3}=\frac{d}{3}.$$
Thus
$$C_2=\frac{\varepsilon_0 A}{d/3}=3\frac{\varepsilon_0 A}{d}=3C_1.$$
So
$$\frac{C_2}{C_1}=3,$$
which corresponds to option C as a ratio of 3:1. However since the available choices are written as ratios, the correct match is $3:1$.
Q32
mcq
For a periodic motion represented by the equation $y=\sin \omega t+\cos \omega t$ the amplitude of the motion is
a.
0.5
b.
1
c.
$\sqrt{2}$
Correct Answer
d.
2
Step-by-Step AI Explanation
Verified Method
Write the sum as a single sinusoid:
$$y=\sin\omega t+\cos\omega t = \sqrt{2}\sin\left(\omega t+\frac{\pi}{4}\right).$$
The amplitude is the coefficient of the sine term, i.e. $\sqrt{2}$.
Q33
mcq
A gas mixture consists of 2 moles of oxygen and 4 moles of neon at temperature $T$. Neglecting all vibrational modes, the total internal energy of the system will be,
a.
$4RT$
b.
$8RT$
c.
$11RT$
Correct Answer
d.
$16RT$
Step-by-Step AI Explanation
Verified Method
For ideal gases, internal energy depends on degrees of freedom.
- Oxygen is diatomic and at ordinary temperature with vibrational modes neglected, so $f=5$ and
$$U_{\mathrm{O_2}}=n\frac{f}{2}RT=2\cdot\frac{5}{2}RT=5RT.$$
- Neon is monoatomic, so $f=3$ and
$$U_{\mathrm{Ne}}=4\cdot\frac{3}{2}RT=6RT.$$
Total internal energy:
$$U=5RT+6RT=11RT.$$
Q34
mcq
A gas is compressed adiabatically, which one of the following statement is NOT true.
a.
The temperature of the gas increases.
b.
The change in internal energy is equal to the work done on the gas.
c.
There is no change in internal energy
Correct Answer
d.
There is no heat supplied to the system
Step-by-Step AI Explanation
Verified Method
In an adiabatic process, $Q=0$, so no heat is exchanged. For compression, work is done on the gas, so its internal energy increases and hence temperature also increases for an ideal gas.
Using the first law,
$$\Delta U = Q + W_{\text{on}} = W_{\text{on}}$$
since $Q=0$.
Therefore statements A, B, and D are true, while statement C is not true.
Q35
mcq
Young's modulus of the material of wires A and B are in the ratio of 1:4, while its area of cross sections are in the ratio of 1:3. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires A and B will be in the ratio of
[Assume length of wires A and B same]
a.
1:12
b.
1:36
c.
12:1
d.
36:1
Correct Answer
Step-by-Step AI Explanation
Verified Method
For a wire, elongation is
$$\Delta L = \frac{FL}{AY}$$
where $F$ is load, $L$ is length, $A$ is area of cross-section, and $Y$ is Young's modulus.
Given same $F$ and same $L$, so
$$\Delta L \propto \frac{1}{AY}$$
Now,
$$Y_A:Y_B = 1:4, \quad A_A:A_B = 1:3$$
Hence
$$\frac{\Delta L_A}{\Delta L_B} = \frac{A_B Y_B}{A_A Y_A} = \frac{3\times 4}{1\times 1} = 12$$
Wait, check carefully: if $Y_A:Y_B=1:4$, then $Y_B/Y_A=4$, and $A_B/A_A=3$. So
$$\frac{\Delta L_A}{\Delta L_B}=3\times 4=12$$
Thus the ratio should be $12:1$. Therefore the correct option is actually option C.
The answer marked by computation is $12:1$.
Q36
mcq
Given below are two statements:
Statement I : Rotation of the earth shows effect on the value of acceleration due to gravity $(g)$
Statement II : The effect of rotation of the earth on the value of $g$ at the equator is minimum and that at the pole is maximum.
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
Correct Answer
d.
Statement I is false but statement II is true
Step-by-Step AI Explanation
Verified Method
Earth's rotation affects the effective acceleration due to gravity because of the centrifugal effect, so Statement I is true.
The centrifugal effect is maximum at the equator and zero at the poles. Therefore the reduction in $g$ is maximum at the equator and minimum at the poles. Statement II says the opposite, so it is false.
Hence, the correct option is statement I true, statement II false.
Q37
mcq
The time period of a satellite, revolving above earth's surface at a height equal to $R$ will be
(Given $g = \pi^2\,\mathrm{m/s^2}$, $R =$ radius of earth)
a.
$\sqrt{2R}$
b.
$\sqrt{4R}$
c.
$\sqrt{8R}$
d.
$\sqrt{32R}$
Correct Answer
Step-by-Step AI Explanation
Verified Method
For a circular orbit at height $h=R$, orbital radius is
$$r = R+h = 2R.$$
The time period is
$$T = 2\pi\sqrt{\frac{r^3}{GM}}.$$
Using $g=\frac{GM}{R^2}$, we get $GM=gR^2$, so
$$T = 2\pi\sqrt{\frac{(2R)^3}{gR^2}} = 2\pi\sqrt{\frac{8R}{g}}.$$
Given $g=\pi^2$,
$$T = 2\pi\sqrt{\frac{8R}{\pi^2}} = 2\sqrt{8R} = \sqrt{32R}.$$
So option D is correct.
Q38
mcq
Two projectiles are projected at $30^\circ$ and $60^\circ$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:
a.
1:3
Correct Answer
b.
1:$\sqrt{3}$
c.
2:$\sqrt{3}$
d.
$\sqrt{3}$:1
Step-by-Step AI Explanation
Verified Method
Maximum height of a projectile is
$$H = \frac{u^2\sin^2\theta}{2g}.$$
For the same speed, the ratio is
$$H_{30}:H_{60} = \sin^2 30^\circ : \sin^2 60^\circ = \left(\frac12\right)^2 : \left(\frac{\sqrt3}{2}\right)^2 = \frac14 : \frac34 = 1:3.$$
So option A is correct.
Q39
mcq
In an experiment with vernier calipers of least count 0.1 mm, when two jaws are joined together the zero of vernier scale lies right to the zero of the main scale and 6th division of vernier scale coincides with the main scale division. While measuring the diameter of a spherical bob, the zero of vernier scale lies in between 3.2 cm and 3.3 cm marks, and 4th division of vernier scale coincides with the main scale division. The diameter of bob is measured as
a.
3.26 cm
b.
3.25 cm
c.
3.22 cm
Correct Answer
d.
3.18 cm
Step-by-Step AI Explanation
Verified Method
The zero error is positive because, when jaws are closed, vernier zero lies to the right of main scale zero.
Given least count $=0.1\,\mathrm{mm}=0.01\,\mathrm{cm}$.
With 6th vernier coincidence, zero error is
$$+ (10-6)\times 0.01 = +0.04\,\mathrm{cm}.$$
So zero correction is $-0.04\,\mathrm{cm}$.
Observed reading for the bob:
Main scale reading $= 3.2\,\mathrm{cm}$.
Vernier reading $= 4\times 0.01 = 0.04\,\mathrm{cm}$.
So observed value
$$= 3.24\,\mathrm{cm}.$$
Applying correction:
$$3.24 - 0.04 = 3.20\,\mathrm{cm}.$$
But among the given options, the intended standard interpretation in many JEE questions is to take the vernier zero lying between $3.2$ and $3.3$ cm and 4th coincidence giving $3.24$ cm, then subtract $0.02$ cm? That would be inconsistent with the stated zero error.
Using the standard vernier formula and stated positive zero error, the corrected diameter is $3.20\,\mathrm{cm}$. Since that is not listed, the closest listed option is $3.22\,\mathrm{cm}$, but the mathematically consistent result is $3.20\,\mathrm{cm}$.
Q40
mcq
A person travels $x$ distance with velocity $v_1$ and then $x$ distance with velocity $v_2$ in the same direction. The average velocity of the person is $v$, then the relation between $v$, $v_1$ and $v_2$ will be.
a.
$v = \dfrac{v_1 + v_2}{2}$
b.
$v = v_1 + v_2$
c.
$\dfrac{1}{v} = \dfrac{1}{v_1} + \dfrac{1}{v_2}$
d.
$\dfrac{2}{v} = \dfrac{1}{v_1} + \dfrac{1}{v_2}$
Correct Answer
Step-by-Step AI Explanation
Verified Method
Average velocity is total displacement divided by total time.
Total displacement:
$$2x$$
Total time:
$$\frac{x}{v_1}+\frac{x}{v_2} = x\left(\frac{1}{v_1}+\frac{1}{v_2}\right).$$
Thus,
$$v = \frac{2x}{x\left(\frac{1}{v_1}+\frac{1}{v_2}\right)} = \frac{2}{\frac{1}{v_1}+\frac{1}{v_2}}.$$
Therefore,
$$\frac{2}{v} = \frac{1}{v_1}+\frac{1}{v_2}.$$
So option D is correct.
🎉 10 Free Sample Solutions Completed
Unlock All 20 Questions & Solutions
Sign up free in seconds to unlock step-by-step AI answers for the remaining 10 questions.
Q41
mcq
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R
Assertion A : An electric fan continues to rotate for some time after the current is switched off.
Reason R : Fan continues to rotate due to inertia of motion.
In the light of above statements, choose the most appropriate answer from the options given below.
a.
Both A and R are correct and R is the correct explanation of A
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.
A is not correct but R is correct
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q42
mcq
A message signal of frequency $3\,\mathrm{kHz}$ is used to modulate a carrier signal of frequency $1.5\,\mathrm{MHz}$. The bandwidth of the amplitude modulated wave is
a.
6 MHz
b.
3 MHz
c.
3 kHz
d.
6 kHz
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q43
mcq
If each diode has a forward bias resistance of $25\ \Omega$ in the below circuit, which of the following options is correct:
a.
$\dfrac{I_1}{I_2} = 1$
b.
$\dfrac{I_2}{I_3} = 1$
c.
$\dfrac{I_3}{I_4} = 1$
d.
$\dfrac{I_1}{I_2} = 2$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q44
mcq
The half life of a radioactive substance is $T$. The time taken, for disintegrating $\dfrac{7}{8}$th part of its original mass will be:
a.
$T$
b.
$2T$
c.
$3T$
d.
$8T$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q45
mcq
The variation of stopping potential $(V_0)$ as a function of the frequency $(\nu)$ of the incident light for a metal is shown in figure. The work function of the surface is
a.
$1.36\ \text{eV}$
b.
$2.98\ \text{eV}$
c.
$2.07\ \text{eV}$
d.
$18.6\ \text{eV}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q46
mcq
A bar magnet is released from rest along the axis of a very long vertical copper tube. After some time the magnet will
a.
move down with an acceleration equal to $g$
b.
move down with an acceleration greater than $g$
c.
move down with almost constant speed
d.
oscillate inside the tube
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q47
mcq
The ratio of intensities at two points $P$ and $Q$ on the screen in a Young's double slit experiment, where phase difference between two waves of same amplitude are $\pi/3$ and $\pi/2$, respectively are
a.
$3:2$
b.
$2:3$
c.
$1:3$
d.
$3:1$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q48
mcq
The amplitude of magnetic field in an electromagnetic wave propagating along $y$-axis is $6.0\times 10^{-7}\ \text{T}$. The maximum value of electric field in the electromagnetic wave is
a.
$6.0\times 10^{-7}\ \text{V m}^{-1}$
b.
$180\ \text{V m}^{-1}$
c.
$2\times 10^{15}\ \text{V m}^{-1}$
d.
$5\times 10^{14}\ \text{V m}^{-1}$
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q49
mcq
Given below are two statements:
Statement I : For diamagnetic substance, $-1 \le \chi < 0$, where $\chi$ is the magnetic susceptibility.
Statement II : Diamagnetic substances when placed in an external magnetic field, tend to move from stronger to weaker part of the field.
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 correct but Statement II is false
d.
Statement I is incorrect but Statement II is true
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.
Q50
mcq
In a metallic conductor, under the effect of applied electric field, the free electrons of the conductor
a.
drift from higher potential to lower potential.
b.
move in the straight line paths in the same direction
c.
move with the uniform velocity throughout from lower potential to higher potential
d.
move in the curved paths from lower potential to higher potential
Correct Answer & Step-by-Step Solution Locked
Sign in free to reveal the correct answer and AI explanation for this question.