JEE Main 2026 — Apr 4, Shift 1 — Physics solution

JEE Main2026Apr 4, Shift 1Physics

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 $I_{2} / I_{1} =$ _______.

A
$35$
B
$70$
C
$140$
D
$210$

Solution

Density is conserved (same material). Original sphere: $M = ρ \cdot \frac{4 π}{3} R^{3}$ . Smaller sphere of mass $M /8$ :

$\frac{M}{8} = ρ \cdot \frac{4 π}{3} r^{3} \Rightarrow r^{3} = \frac{R ^{3}}{8} \Rightarrow r = \frac{R}{2} .$

$I_{1} = \frac{2}{5} \cdot \frac{M}{8} \cdot r^{2} = \frac{2}{5} \cdot \frac{M}{8} \cdot \frac{R ^{2}}{4} = \frac{M R ^{2}}{80} .$

Disc (mass $7 M /8$ , radius $2 R$ ), moment of inertia about a diameter:

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

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

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