A rolling wheel of 12 kg is on an inclined… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Mechanics

2022integermedium

A rolling wheel of $12\mathrm{kg}$ is on an inclined plane at position $P$ and connected to a mass of $3\mathrm{kg}$ through a string of fixed length and pulley as shown in figure.

Consider $PR$ as friction free surface.

The velocity of centre of mass of the wheel when it reaches at the bottom $Q$ of the inclined plane $PQ$ will be $\frac{1}{2}\sqrt{xgh}m{s}^{-1}$ . The value of $x$ (rounded off to the nearest integer) is _____.

JEE Main 2022 Physics, Work Power Energy — question figure

Official previous-year question

Held on 27 Jun 2022 · Verified 6 Jul 2026.

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Solution

If we take the wheel as a ring then its moment of inertia will be, $I=M{r}^{2}$ .

Using mechanical energy conservation,

$Mgh-mgh=\frac{1}{2}m{v}^{2}+\frac{1}{2}M{v}^{2}+\frac{1}{2}I{\omega }^{2} \Rightarrow 12gh-3gh=\frac{1}{2}3{v}^{2}+\frac{1}{2}12{v}^{2}+\frac{1}{2}12{r}^{2}\frac{{v}^{2}}{{r}^{2}}$

$\Rightarrow 9gh=\frac{15{v}^{2}+12{v}^{2}}{2}$

$\Rightarrow v=\sqrt{\frac{2}{3}gh}=\frac{1}{2}\sqrt{\frac{8}{3}gh}$

Therefore, $x=\frac{8}{3}\simeq 3$

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