A solid sphere of mass M and radius R is divided into two unequal… | JEE Main 2019 (10 Apr, Shift 2)

Q166JEE Main 2019 · 10 APR · SHIFT 2Difficult

Moment of Inertia & Radius of Gyration

A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of $\frac{7M}{8}$ and is converted into a uniform disc of radius 2R. The second part is converted into a uniform solid sphere. Let I1 be the moment of inertia of the disc about its axis and I2 be the moment of inertia of the new sphere about its axis. The ratio I1/I2 is given by:

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Answer: (b)

Disc: $I_1 = \frac{1}{2}\frac{7M}{8}(2R)^2 = \frac{7MR^2}{4}$
New sphere has mass $\frac{M}{8}$ , so radius $\frac{R}{2}$ : $I_2 = \frac{2}{5}\frac{M}{8}\frac{R^2}{4} = \frac{MR^2}{80}$
$\frac{I_1}{I_2} = \frac{7}{4} \times 80 = 140$
So the answer is (b).

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