Question

$I$ is moment of inertia of a solid copper sphere of radius $R$ about diameter as axis. If a spherical cavity of radius $\frac{R}{2}$ is made in that sphere such that centre of cavity coincides with centre of sphere, then the moment of inertia of remainder about same axis is:

• A. $\frac{3I}{4}$
• B. $\frac{7I}{8}$
• C. $\frac{15I}{16}$
• D. $\frac{31I}{32}$

Answer 1

The correct option is D $\frac{31I}{32}$

Let the mass of Sphere be M ,

lets assume we have another sphere of same material with radius r/2 , so $M_1=M/8$

MI ,$I_1=MIofsolidsphere-MIofremovedcavity$

$I=\frac{2M{r}^2}{5}$

$I_1=\frac{2M{r}^2}{5}-\frac{\left(M/8\right)(r/2{)}^2}{5}=I-\frac{I}{32}$

So $I_1=\frac{31I}{32}$