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Rotational Motion

JEE Main level, single correct, medium

8 of 280

Question 8 of 280

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 $2R$ . 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 $I_2/I_1 =$ _____.

Show answer and solution

Answer: (B) $70$

Let the mass of the original solid sphere be

M

and its radius be

R

.

Mass of the smaller sphere,

m_1 = \dfrac{M}{8}

.
Since the material remains the same, the density is constant. The volume is directly proportional to the mass.

\dfrac{4}{3}\pi r^3 = \dfrac{1}{8} \left(\dfrac{4}{3}\pi R^3\right) \Rightarrow r^3 = \dfrac{R^3}{8} \Rightarrow r = \dfrac{R}{2}

Moment of inertia of the smaller sphere about an axis through its centre is:

I_1 = \dfrac{2}{5} m_1 r^2 = \dfrac{2}{5} \left(\dfrac{M}{8}\right) \left(\dfrac{R}{2}\right)^2 = \dfrac{2}{5} \times \dfrac{M}{8} \times \dfrac{R^2}{4} = \dfrac{M R^2}{80}

Mass of the larger part (which is converted into a disc) is:

m_2 = M - \dfrac{M}{8} = \dfrac{7M}{8}

The radius of the disc is given as

2R

. The moment of inertia of a disc about its diameter is:

I_2 = \dfrac{1}{4} m_2 (\text{radius})^2 = \dfrac{1}{4} \left(\dfrac{7M}{8}\right) (2R)^2 = \dfrac{1}{4} \times \dfrac{7M}{8} \times 4R^2 = \dfrac{7M R^2}{8}

The ratio of their moments of inertia is:

\dfrac{I_2}{I_1} = \dfrac{\dfrac{7M R^2}{8}}{\dfrac{M R^2}{80}} = \dfrac{7}{8} \times 80 = 70

Answer:

70

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