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Question

A rod of mass M and length L is hinged about a end to swing freely in a vertical plane. Its density increases linearly from hinge to the free end, doubling is value at free end. The moment of inertia of the rod about the hinge is

A
13ML2
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B
524ML2
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C
718ML2
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D
935ML2
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Solution

The correct option is C 718ML2
Let the density at the hinged bed and at extreme end is 2d density at the distance x from hinge =d+dxL take dx width element at distance x from hindge. The mass of element is dx.
dx=L0(d+dxL)dx
M=dL+dL22L=3dL2
d=2M3L
for moment of inertia the MI of the dx with element
dI=x2dx
dI=L0x2(d+dxL)dx
I=dL[L43+L44]=7dL312=7L312[2M3L]=7ML218

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