The moment of inertia of a big drop is I. If | vedclass

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(A) The moment of inertia of a solid sphere (big drop) is $I = \frac{2}{5} M R^{2}$.When the big drop is divided into $n = 8$ small droplets,the total

Vedclass · Accepted Answer

$\frac{I}{32}$

Vedclass · Answer

(A) The moment of inertia of a solid sphere (big drop) is $I = \frac{2}{5} M R^{2}$.When the big drop is divided into $n = 8$ small droplets,the total volume remains constant.$n \left( \frac{4}{3} \pi r^{3} ight) = \frac{4}{3} \pi R^{3}$$8 r^{3} = R^{3} \Rightarrow 2r = R \Rightarrow r = \frac{R}{2}$.The mass of each small droplet is $m = \frac{M}{n} = \frac{M}{8}$.The moment of inertia of each small droplet $i$ is:$i = \frac{2}{5} m r^{2}$$i = \frac{2}{5} \left( \frac{M}{8} ight) \left( \frac{R}{2} ight)^{2}$$i = \frac{1}{8} 	imes \frac{1}{4} 	imes \left( \frac{2}{5} M R^{2} ight)$$i = \frac{1}{32} I$.

The moment of inertia of a big drop is $I$. If $8$ droplets are formed from the big drop,then the moment of inertia of a small droplet is:

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