A metal piece of mass $120 g m$ is stretched into a thin plane rectangular sheet of area of cross-section $0.54 m^{2}$ . If the length and breadth of this sheet are in the ratio 1:6, the moment of inertia about an axis passing through its centre and perpendicular to its plane is

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$6.6 \times 10^{- 2} k g m^{2}$

$3.3 k g m^{2}$

$6.6 k g m^{2}$

$3.3 \times 10^{- 2} k g m^{2}$

answer is B.

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Detailed Solution

$I = \frac{m}{12} (ℓ^{2} + b^{2})$

$A = ℓ b$ and

$\frac{ℓ}{b} = \frac{1}{6} b = 6 ℓ A = ℓ b = 6 ℓ^{2} 0.54 = 6 ℓ^{2} ℓ = 0.3 I = \frac{M (ℓ^{2} + b^{2})}{12} = \frac{0.12 [{(0.3)}^{2} + {(1.8)}^{2}]}{12} = 3.3 \times 10^{- 2} k g m^{2}$ .

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