Consider a badminton racket with length scales as shown in the figure. ![JEE Main 2021 Physics, Rotational Motion — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Rotational_Motion/648b5a6b417cc3fb48d673cc/question_1__q_648b5a6b417cc3fb48d673cc__cdn-question-pool.getmarks.app__b61ab3d4-c125-4d30-9ee0-74102bcbce5a-image__a40e30215d_final_ppt_sync.png) If the mass of the linear and circular portions of the badminton racket are same ($M$) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{r}{2}$ distance from the end $A$ of the handle will be ______$M{r}^{2}$.

Consider a badminton racket with length… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Mechanics

2021integermedium

Consider a badminton racket with length scales as shown in the figure.

JEE Main 2021 Physics, Rotational Motion — question figure

If the mass of the linear and circular portions of the badminton racket are same ( $M$ ) and the mass of the threads are negligible, the moment of inertia of the racket about an axis perpendicular to the handle and in the plane of the ring at, $\frac{r}{2}$ distance from the end $A$ of the handle will be ______ $M{r}^{2}$ .

Official previous-year question

Held on 26 Aug 2021 · Verified 6 Jul 2026.

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Solution

${I}_{A}=[\frac{M{r}^{2}}{2}+M{(7r-\frac{r}{2})}^{2}]+[\frac{M(6r{)}^{2}}{12}+(M{(3r-\frac{r}{2})}^{2})]$

$={\frac{M{r}^{2}}{2}+M\frac{(169){r}^{2}}{4}}+{3M{r}^{2}+\frac{25M{r}^{2}}{4}} =\frac{171M{r}^{2}}{4}+\frac{37M{r}^{2}}{4} =\frac{208}{4}M{r}^{2} =52M{r}^{2}$

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