A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7 M 8 and is converted into uniform disc of radius   2 R . The second part is converted into a uniform solid sphere. Let I 1 be the moment of inertia of the disc about its axis and I 2 be the…

Question

A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7 M 8 and is converted into uniform disc of radius &#160; 2 R . The second part is converted into a uniform solid sphere. Let I 1 be the moment of inertia of the disc about its axis and I 2 be the moment of inertia of the new sphere about its axis. The ratio &#160; I 1 / I 2 is given by:

Aicharya · Accepted Answer

Moment of inertia of disc about the axis passing its centre and perpendicular to its surface,

I_{1} = \frac{(\frac{7 M}{8}) {(2 R)}^{2}}{2} = \frac{7 M R^{2}}{4}

.
Radius of small sphere

r

is related as

\frac{M}{\frac{4}{3} π R^{3}} = \frac{\frac{M}{8}}{\frac{4}{3} π r^{3}}

r = \frac{R}{2}

Moment of inertia of sphere about its axis passing through its centre

∴ I_{2} = \frac{2}{5} (\frac{M}{8}) {(\frac{R}{2})}^{2} = \frac{M R^{2}}{80}

\frac{I_{1}}{I_{2}} = \frac{7}{4} \times \frac{80}{1} = 140

A solid sphere of mass M and radius R is divided into two unequal parts. The first part has a mass of 7 M 8…

Solution