JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Rotational Mechanics

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

2 R

. If

I_{1}

is moment of inertia of a sphere having radius r about an axis through its centre and

I_{2}

is the moment of inertia of a disc about its diameter, the ratio of their moment of inertia

I_{2} / I_{1} =

______.

Answer: (B)

Solution

Equal density fixes the small sphere's radius:

\frac{M}{8} = ρ \frac{4 π}{3} r^{3}

with

M = ρ \frac{4 π}{3} R^{3}

give

r = \frac{R}{2}

. Its moment of inertia about a central axis is

I_{1} = \frac{2}{5} (\frac{M}{8}) (\frac{R}{2})^{2} = \frac{M R ^{2}}{80}

. The disc of mass

\frac{7 M}{8}

and radius

2 R

has, about a diameter,

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

. Hence

\frac{I _{2}}{I _{1}} = \frac{7 M R ^{2} /8}{M R ^{2} /80} = 70

JEE Main · 2026 · 04 Apr 2026 · Shift 1 · Single correct option