JEE Main 2026 · 4 Apr · Shift 1 · Physics Q30

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A solid sphere of mass $M$ and radius $R$ is divided into two unequal parts. The smaller part having mass $M/8$ is converted into a sphere of radius $r$ and the larger part is converted into a circular disc of thickness $t$ and radius $2R$ . If $I_1$ is moment of inertia of a sphere having radius $r$ about an axis through its centre and $I_2$ is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia $I_2/I_1$ = ________.

140

210

Answer and worked solution

Official answer: B

Small sphere. It has mass $\frac M8$ at the same density, so $r=\frac R2$ :
$I_1=\frac25\cdot\frac M8\cdot\frac{R^2}4=\frac{MR^2}{80}.$

Disc. Mass $\frac{7M}8$ , radius $2R$ , about a diameter:
$I_2=\frac14\cdot\frac{7M}8\cdot4R^2=\frac{7MR^2}8.$

Ratio $\frac{I_2}{I_1}=70$ .

(B)

Topics:Moment of inertia

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JEE Main 2026 · 4 Apr · Shift 1, Physics question 30