JEE Advanced 2023 Paper 2 Q06 Physics Rotational Mechanics Rolling Motion Hard

JEE Advanced 2023 Paper 2 · Q06 · Rolling Motion

An annular disk of mass $M$, inner radius $a$ and outer radius $b$ is placed on a horizontal surface with coefficient of friction $\mu$, as shown in the figure. At some time, an impulse $J_0\hat{x}$ is applied at a height $h$ above the centre of the disk. If $h = h_m$ then the disk rolls without slipping along the $x$-axis. Which of the following statement(s) is(are) correct?

  1. A. For $\mu \ne 0$ and $a \to 0$, $h_m = b/2$.
  2. B. For $\mu \ne 0$ and $a \to b$, $h_m = b$.
  3. C. For $h = h_m$, the initial angular velocity does not depend on the inner radius $a$.
  4. D. For $\mu = 0$ and $h = 0$, the wheel always slides without rolling.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, B, C, D

Solution

Linear impulse: $J_0 = Mv_{cm}$. Angular impulse about CM: $J_0 h = I_{cm}\omega$. For pure rolling: $v_{cm} = b\omega$, so $\omega = J_0/(Mb)$ — independent of $a$, confirming (C). Then $h_m = I_{cm}/(Mb)$. For solid disk ($a\to 0$): $I = Mb^2/2$, $h_m = b/2$ — (A) TRUE. For ring ($a\to b$): $I = Mb^2$, $h_m = b$ — (B) TRUE. For $\mu=0, h=0$: no torque, so $\omega = 0$ but $v\ne 0$ — slides without rolling — (D) TRUE.

Want to drill deeper? Open this question inside SolveKar AI and tap "Why?" on any step — the AI Mentor reads your full solution context and explains until it clicks. Download SolveKar AI →