1
JEE Advanced 2026 Paper 1 Online
MCQ (Single Correct Answer)
+4
-1

List-I shows four planar structures made of uniform solid rods each of mass $m$ and length $l$. In the List-II the possible moment of inertia of these structures about an axis $OCO'$, which lies in the plane of the structures, are given.

Choose the option that describes the correct match between the entries in List-I to those in List-II.

List-I List-II
(P)

JEE Advanced 2026 Paper 1 Online Physics - Rotational Motion Question 3 English 1
(1) $$\frac{5}{4}ml^2$$
(Q)

JEE Advanced 2026 Paper 1 Online Physics - Rotational Motion Question 3 English 2
(2) $$\frac{1}{6}ml^2$$
(R)

JEE Advanced 2026 Paper 1 Online Physics - Rotational Motion Question 3 English 3
(3) $$\frac{1}{12}ml^2$$
(S)

JEE Advanced 2026 Paper 1 Online Physics - Rotational Motion Question 3 English 4
(4) $$\frac{2}{3}ml^2$$
(5) $$\frac{1}{3}ml^2$$
A

P → 5, Q → 1, R → 4, S → 2

B

P → 1, Q → 3, R → 4, S → 2

C

P → 5, Q → 3, R → 2, S → 1

D

P → 5, Q → 4, R → 2, S → 1

2
JEE Advanced 2025 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
Change Language

The center of a disk of radius $r$ and mass $m$ is attached to a spring of spring constant $k$, inside a ring of radius $R>r$ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following the Hooke's law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $T=\frac{2 \pi}{\omega}$. The correct expression for $\omega$ is ( $g$ is the acceleration due to gravity):

JEE Advanced 2025 Paper 1 Online Physics - Rotational Motion Question 7 English

A

$ \sqrt{\frac{2}{3} \left( \frac{g}{R - r} + \frac{k}{m} \right)} $

B

$ \sqrt{\frac{2g}{3(R - r)} + \frac{k}{m}} $

C

$ \sqrt{\frac{1}{6} \left( \frac{g}{R - r} + \frac{k}{m} \right)} $

D

$ \sqrt{\frac{1}{4} \left( \frac{g}{R - r} + \frac{k}{m} \right)} $

3
JEE Advanced 2023 Paper 1 Online
MCQ (Single Correct Answer)
+3
-1
Change Language
A bar of mass $M=1.00 \mathrm{~kg}$ and length $L=0.20 \mathrm{~m}$ is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass $m=0.10 \mathrm{~kg}$ is moving on the same horizontal surface with $5.00 \mathrm{~m} \mathrm{~s}^{-1}$ speed on a path perpendicular to the bar. It hits the bar at a distance $L / 2$ from the pivoted end and returns back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity $\omega$.

Which of the following statement is correct?
A
$\omega=6.98 ~\mathrm{rad}~ \mathrm{s}^{-1}$ and $\mathrm{v}=4.30 \mathrm{~m} \mathrm{~s}^{-1}$
B
$\omega=3.75 ~\mathrm{rad} ~\mathrm{s}^{-1}$ and $\mathrm{v}=4.30 \mathrm{~m} \mathrm{~s}^{-1}$
C
$\omega=3.75 ~\mathrm{rad}~ \mathrm{s}^{-1}$ and $\mathrm{v}=10.0 \mathrm{~m} \mathrm{~s}^{-1}$
D
$\omega=6.80 ~\mathrm{rad} ~\mathrm{s}^{-1}$ and $\mathrm{v}=4.10 \mathrm{~m} \mathrm{~s}^{-1}$
4
JEE Advanced 2022 Paper 2 Online
MCQ (Single Correct Answer)
+3
-1
Change Language

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 $n$ ?

JEE Advanced 2022 Paper 2 Online Physics - Rotational Motion Question 31 English

A
2
B
5
C
$\frac{7}{2}$
D
$\frac{9}{2}$

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