JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Rotational Mechanics

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For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is:

(A) $1/2$
(B) $2/3$
(C) $1/4$ ✓ Correct
(D) $1/8$

Answer: (C)

Solution

For a flat lamina the moment of inertia about an axis through the centre and perpendicular to its plane is

I_{O} = \frac{M}{12} (ℓ_{1}^{2} + ℓ_{2}^{2})

, where

ℓ_{1}

and

ℓ_{2}

are the sides. The corner

O^{'}

is a distance

\frac{1}{2} ℓ_{1}^{2} + ℓ_{2}^{2}

from the centre, so by the parallel-axis theorem

I_{O^{'}} = \frac{M}{12} (ℓ_{1}^{2} + ℓ_{2}^{2}) + M \frac{ℓ _{1}^{2} + ℓ _{2}^{2}}{4} = \frac{M}{3} (ℓ_{1}^{2} + ℓ_{2}^{2})

. Their ratio is therefore

\frac{I _{O}}{I _{O^{'}}} = \frac{M /12}{M /3} = \frac{1}{4}

JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Single correct option

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JEE Main · 2020 · 04 Sep 2020 · Shift 2 · Rotational Mechanics

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