Question:
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 :
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 :
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Use \(I_{cm} = \dfrac{M(a^2+b^2)}{12}\) and the parallel axis theorem with \(d^2 = \left(\dfrac a2\right)^2+\left(\dfrac b2\right)^2\).
Updated On: Nov 5, 2026
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{4}\)
- \(\dfrac{1}{8}\)
- \(\dfrac{2}{3}\)
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The Correct Option is B
Solution and Explanation
Sheet dimensions: \(a = 60\) cm, \(b = 80\) cm, mass \(M\).
About the perpendicular axis through the centre O: \(I_O = \dfrac{M(a^2+b^2)}{12} = \dfrac{M(3600+6400)}{12} = \dfrac{10000M}{12}\).
Distance from O to the corner O': \(d^2 = 30^2 + 40^2 = 2500\) cm2.
Parallel axis theorem: \(I_{O'} = I_O + Md^2 = \dfrac{10000M}{12} + 2500M = \dfrac{10000M}{3}\).
\(\dfrac{I_O}{I_{O'}} = \dfrac{10000/12}{10000/3} = \dfrac{3}{12} = \dfrac{1}{4}\).
Answer: (2) \(\dfrac{1}{4}\)
About the perpendicular axis through the centre O: \(I_O = \dfrac{M(a^2+b^2)}{12} = \dfrac{M(3600+6400)}{12} = \dfrac{10000M}{12}\).
Distance from O to the corner O': \(d^2 = 30^2 + 40^2 = 2500\) cm2.
Parallel axis theorem: \(I_{O'} = I_O + Md^2 = \dfrac{10000M}{12} + 2500M = \dfrac{10000M}{3}\).
\(\dfrac{I_O}{I_{O'}} = \dfrac{10000/12}{10000/3} = \dfrac{3}{12} = \dfrac{1}{4}\).
Answer: (2) \(\dfrac{1}{4}\)
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