A circular disc reaches from… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Mechanics

2021integerhard

A circular disc reaches from top to bottom of an inclined plane of length $L.$ When it slips down the plane, it takes time ${t}_{1}$ . When it rolls down the plane, it takes time ${t}_{2}$ . The value of $\frac{{t}_{2}}{{t}_{1}}$ is $\sqrt{\frac{3}{x}}$ . The value of $x$ will be

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Held on 20 Jul 2021 · Verified 6 Jul 2026.

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Solution

JEE Main 2021 Physics, Rotational Motion — solution figure

If disk slips on inclined plane, then its acceleration, ${a}_{1}=g\mathrm{sin}\theta$

$L=\frac{1}{2}{a}_{1}{t}_{1}^{2}$

$\Rightarrow {t}_{1}=\sqrt{\frac{2L}{{a}_{1}}}\ldots (i)$

If disk rolls on inclined plane, its acceleration,

${a}_{2}=\frac{g\mathrm{sin}\theta }{1+\frac{I}{m{R}^{2}}}$

${a}_{2}=\frac{g\mathrm{sin}\theta }{1+\frac{m{R}^{2}}{2m{R}^{2}}}$

${a}_{2}=\frac{2}{3}g\mathrm{sin}\theta$

Now $L=\frac{1}{2}{a}_{2}\cdot {t}_{2}^{2}$

$\Rightarrow {t}_{2}=\sqrt{\frac{2L}{{a}_{2}}}\ldots (\mathrm{ii})$

Now, $\frac{{t}_{2}}{{t}_{1}}=\sqrt{\frac{{a}_{1}}{{a}_{2}}}=\sqrt{\frac{3}{2}}$

$\Rightarrow x=2$

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Solution

JEE Main 2021 Physics, Rotational Motion — solution figure

If disk slips on inclined plane, then its acceleration, ${a}_{1}=g\mathrm{sin}\theta$

$L=\frac{1}{2}{a}_{1}{t}_{1}^{2}$

$\Rightarrow {t}_{1}=\sqrt{\frac{2L}{{a}_{1}}}\ldots (i)$

If disk rolls on inclined plane, its acceleration,

${a}_{2}=\frac{g\mathrm{sin}\theta }{1+\frac{I}{m{R}^{2}}}$

${a}_{2}=\frac{g\mathrm{sin}\theta }{1+\frac{m{R}^{2}}{2m{R}^{2}}}$

${a}_{2}=\frac{2}{3}g\mathrm{sin}\theta$

Now $L=\frac{1}{2}{a}_{2}\cdot {t}_{2}^{2}$

$\Rightarrow {t}_{2}=\sqrt{\frac{2L}{{a}_{2}}}\ldots (\mathrm{ii})$

Now, $\frac{{t}_{2}}{{t}_{1}}=\sqrt{\frac{{a}_{1}}{{a}_{2}}}=\sqrt{\frac{3}{2}}$

$\Rightarrow x=2$

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Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics