Physics · Rotational Dynamics

JEE Main 2024 — 4 April, Shift 1 — Question 58

A solid sphere and a hollow cylinder roll up without slipping on same inclined plane with same initial speed $v$ . The sphere and the cylinder reaches upto maximum heights $h_{1}$ and $h_{2}$ , respectively, above the initial level. The ratio $h_{1}: h_{2}$ is $\frac{n}{10}$ . The value of $n$ is $\qquad$ .

Answer: 7

Numerical answer — enter this value.

Step-by-step solution

Gain in P.E. $=$ Loss in K.E.

$\mathrm{mgh}=\frac{1}{2} \mathrm{mv}^{2}\left(1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}\right)$

$\mathrm{h} \propto 1+\frac{\mathrm{K}^{2}}{\mathrm{R}^{2}}$

$\frac{\mathrm{h}_{1}}{\mathrm{~h}_{2}}=\frac{1+\frac{2}{5}}{1+1}=\frac{7}{5 \times 2}=\frac{7}{10}$

$\mathrm{n}=7$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 1
Subject: Physics
Chapter: Rotational Dynamics
Topic: Rolling Motion