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 I₁ is moment of inertia of a sphere having radius r about an axis through its centre and I₂ is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia I₂/I₁ = (a) 35 (b) 70 (c) 140 (d) 210

Answer: (b). Mass M/8 has 1/8 the volume, so r = R/2: I₁ = 2/5M/8R²/4 = (MR²)/(80) Disc of mass 7M/8, radius 2R, about a diameter (thin): I₂ = 1/47M/8(2R)² = (7MR²)/(8) (I₂)/(I₁) = 7/8 × 80 = 70 So the answer is (b) 70.

A solid sphere of mass M and radius R is divided into two unequal… | JEE Main 2026 (4 Apr, Shift 1)

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Moment of Inertia & Radius of Gyration

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 $I_2/I_1 = \underline{\qquad}$

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Answer: (b)

Mass $\frac{M}{8}$ has $\frac{1}{8}$ the volume, so $r = \frac{R}{2}$ : $I_1 = \frac{2}{5}\frac{M}{8}\frac{R^2}{4} = \frac{MR^2}{80}$
Disc of mass $\frac{7M}{8}$ , radius $2R$ , about a diameter (thin): $I_2 = \frac{1}{4}\frac{7M}{8}(2R)^2 = \frac{7MR^2}{8}$
$\frac{I_2}{I_1} = \frac{7}{8} \times 80 = 70$
So the answer is (b) 70.

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