JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Rotational Mechanics

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

\frac{7 M}{8}

and is converted into a uniform disc of radius 2R. 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 of

I_{1} / I_{2}

is given by:

Answer: (D)

Solution

Split the mass as the problem asks. The disc takes

m_{1} = \frac{7 M}{8}

with radius

2 R

, so its moment of inertia about the central axis is

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

. The leftover mass

m_{2} = \frac{M}{8}

is recast as a small solid sphere, and because the density is unchanged,

\frac{m _{2}}{M} = \frac{r _{2}^{3}}{R ^{3}} = \frac{1}{8}

gives

r_{2} = \frac{R}{2}

. Its moment of inertia is

I_{2} = \frac{2}{5} m_{2} r_{2}^{2} = \frac{2}{5} \cdot \frac{M}{8} \cdot \frac{R ^{2}}{4} = \frac{M R ^{2}}{80}

. Dividing one by the other,

\frac{I _{1}}{I _{2}} = \frac{7 M R ^{2} /4}{M R ^{2} /80} = \frac{7 \cdot 80}{4} = 140

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option