JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 3)
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 one side of square where the axis of rotation is parallel to the plane of the square is :
$${4 \over 5}m{a^2}$$
$${8 \over 5}m{a^2} + m{b^2}$$
$${4 \over 5}m{a^2} + 2m{b^2}$$
$${8 \over 5}m{a^2} + 2m{b^2}$$
Explanation
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$$I = {2 \over 5}m{a^2} + {2 \over 5}m{a^2} + \left[ {{2 \over 5}m{a^2} + m{b^2}} \right] + [{2 \over 5}m{a^2} + m{b^2}]$$
$$I = 4 \times {2 \over 5}m{a^2} + 2m{b^2}$$
$$ = {8 \over 5}m{a^2} + 2m{b^2}$$
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