A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60\mathrm{cm}$. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is ______ $m{s}^{-1}$. (Given $g=10m{s}^{-2}$) ![JEE Main 2023 Physics, Rotational Motion — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Rotational_Motion/648b5a6e417cc3fb48d67f0b/question_1__q_648b5a6e417cc3fb48d67f0b__cdn-question-pool.getmarks.app__f1512277-f579-41e7-9318-318cf103cbfb-9896745__caf5864061_final_ppt_sync.png)

A solid cylinder is released… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Mechanics

2023integerhard

A solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60\mathrm{cm}$ . If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is ______ $m{s}^{-1}$ . (Given $g=10m{s}^{-2}$ )

JEE Main 2023 Physics, Rotational Motion — question figure

Official previous-year question

Held on 1 Feb 2023 · Verified 6 Jul 2026.

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Solution

Let speed of cylinder upon reaching the bottom of the inclined plane be $v$ .

Here, gain in kinetic energy $=$ loss in potential energy

So, $\frac{1}{2}I{\omega }^{2}+\frac{1}{2}m{v}^{2}=mgl\mathrm{sin}\theta$

$\Rightarrow \frac{1}{2}(\frac{m{r}^{2}}{2}){\omega }^{2}+\frac{1}{2}m{v}^{2}=mgl\mathrm{sin}30^{\circ}$

For pure rolling, $v=r\omega$

$\Rightarrow \frac{1}{2}(\frac{m{r}^{2}}{2}){(\frac{v}{r})}^{2}+\frac{1}{2}m{v}^{2}=mgl\mathrm{sin}30^{\circ}$

$\Rightarrow \frac{1}{4}m{v}^{2}+\frac{1}{2}m{v}^{2}=mgl\mathrm{sin}30^{\circ} \Rightarrow \frac{3}{4}m{v}^{2}=\frac{mgl}{2} \Rightarrow v=\sqrt{\frac{4gl}{6}}$

Putting the values, we have

$\Rightarrow v=\sqrt{\frac{4\times 10\times 0.6}{6}}=2m{s}^{-1}$

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Solution

Let speed of cylinder upon reaching the bottom of the inclined plane be $v$ .

Here, gain in kinetic energy $=$ loss in potential energy

So, $\frac{1}{2}I{\omega }^{2}+\frac{1}{2}m{v}^{2}=mgl\mathrm{sin}\theta$

$\Rightarrow \frac{1}{2}(\frac{m{r}^{2}}{2}){\omega }^{2}+\frac{1}{2}m{v}^{2}=mgl\mathrm{sin}30^{\circ}$

For pure rolling, $v=r\omega$

$\Rightarrow \frac{1}{2}(\frac{m{r}^{2}}{2}){(\frac{v}{r})}^{2}+\frac{1}{2}m{v}^{2}=mgl\mathrm{sin}30^{\circ}$

$\Rightarrow \frac{1}{4}m{v}^{2}+\frac{1}{2}m{v}^{2}=mgl\mathrm{sin}30^{\circ} \Rightarrow \frac{3}{4}m{v}^{2}=\frac{mgl}{2} \Rightarrow v=\sqrt{\frac{4gl}{6}}$

Putting the values, we have

$\Rightarrow v=\sqrt{\frac{4\times 10\times 0.6}{6}}=2m{s}^{-1}$

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Solution

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Solution

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JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

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Thermodynamics

Waves & Oscillations

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