Q.30. 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

Question

Q.30. 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  is moment of inertia of a sphere having radius r about an axis through its centre and  is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia  = ________

FirstInTest · Accepted Answer

Correct answer: B. Sol. For the original sphere: $M = ho 	imes \frac{4}{3} \pi R^3$. Smaller part mass $M_1 = M/8$. So, $M/8 = ho 	imes \frac{4}{3} \pi r^3 \Rightarrow \frac{1}{8} ho \frac{4}{3} \pi R^3 = ho \frac{4}{3} \pi r^3 \Rightarrow r^3 = R^3/8 \Rightarrow r = R/2$. Moment of inertia of the smaller sphere $I_1 = \frac{2}{5} M_1 r^2 = \frac{2}{5} (M/8) (R/2)^2 = \frac{2}{5} \frac{M}{8} \frac{R^2}{4} = \frac{M R^2}{80}$. Larger part mass $M_2 = M - M/8 = 7M/8$. This part is converted into a circular disc of radius $2R$. Moment of inertia of the disc about its diameter $I_2 = \frac{1}{4} M_2 (2R)^2 = \frac{1}{4} (7M/8) (4R^2) = \frac{7M R^2}{8}$. The ratio $\frac{I_2}{I_1} = \frac{7M R^2 / 8}{M R^2 / 80} = \frac{7/8}{1/80} = \frac{7}{8} 	imes 80 = 70$.

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