Rotational Motion

JEE Main level, single correct, medium

66 of 280

Question 66 of 280


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 OO (the centre of mass) and OO' (corner point) is:

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Answer: (B) 1/41/4

We will use Parallel Axes Theorem here,
Distance between OO′O{O}^{'} is,
OO′=12a2+b2=a24+b24O{O}^{'}=\dfrac{1}{2}\sqrt{{a}^{2}+{b}^{2}}=\sqrt{\dfrac{{a}^{2}}{4}+\dfrac{{b}^{2}}{4}}
So, according to parallel axis theorem we have,
IOIO′=M12(a2+b2)M12(a2+b2)+M(a24+b24)=M12(a2+b2)M3(a2+b2)=14\dfrac{{I}_{O}}{{I}_{{O}^{'}}}=\dfrac{\dfrac{M}{12}\left({a}^{2}+{b}^{2}\right)}{\dfrac{M}{12}\left({a}^{2}+{b}^{2}\right)+M\left(\dfrac{{a}^{2}}{4}+\dfrac{{b}^{2}}{4}\right)}=\dfrac{\dfrac{M}{12}\left({a}^{2}+{b}^{2}\right)}{\dfrac{M}{3}\left({a}^{2}+{b}^{2}\right)}=\dfrac{1}{4}

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