JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Rotational Mechanics

Four identical solid spheres each of mass '

m

' and radius '

a

' are placed with their centres on the four corners of a square of side '

b

'. The moment of inertia of the system about side of square where the axis of rotation is parallel to the plane of the square is :

(A) $\frac{4}{5} m a^{2}$
(B) $\frac{4}{5} m a^{2} + 2 m b^{2}$
(C) $\frac{8}{5} m a^{2} + 2 m b^{2}$ ✓ Correct
(D) $\frac{8}{5} m a^{2} + m b^{2}$

Answer: (C)

Solution

Each sphere has moment of inertia

\frac{2}{5} m a^{2}

about any diameter. With the axis taken along one side of the square, the centres of two spheres lie on this axis and the centres of the other two lie at perpendicular distance

b

from it. The two spheres whose centres are on the axis contribute

2 \times \frac{2}{5} m a^{2} = \frac{4}{5} m a^{2}

, while for the other two the parallel axis theorem gives

2 (\frac{2}{5} m a^{2} + m b^{2}) = \frac{4}{5} m a^{2} + 2 m b^{2}

. Adding the four contributions,

I = \frac{8}{5} m a^{2} + 2 m b^{2}

so the moment of inertia of the system about a side of the square is

\frac{8}{5} m a^{2} + 2 m b^{2}

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Rotational Mechanics

Solution

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