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Question

A rod of length 4 m is placed along positive xaxis, with one of its end at origin. If linear density of the rod varies as λ=4+x, then the position of the centre of gravity of the rod is

A
127 m
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B
209 m
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C
103 m
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D
73 m
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Solution

The correct option is B 209 m
For homogeneous space (g= constant), centre of gravity and centre of mass are identical. Thus,
XCOG=XCM=xdmdm(dm=λdx)
=λxdxλdx(λ=4+x)
=40(4+x)xdx40(4+x)dx(l=4 m)
=440xdx+40x2dx440dx+40xdx
=4(162)+6434(4)+162
=16024×3
XCOG=209m.

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