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Question

A uniform straight rod of mass 'm; kg and length L is hinged at one end. It is free to oscillate in vertical plane. A point of mass (m kg) is attached to it at a distance 'x' from the hinge. The value of x for which time period of oscillations will be minimum is (use: 73=1.5)

A
2L/3
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B
L/2
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C
L/4
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D
2L/7
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Solution

The correct option is C L/4
For a rigid body oscillating about horizontal axis passing through the point O the time period is given by T=2πlmg(lin)
cg=length of centre of gravity.
For real centre=l/2 from ringed point
For mass of x
Centre of mass of system c=mx+ml/22m
=(2l+l/2)2
Now,
Moment of inertia of system =13ml2+mx2
T=2π     13ml2+mx2(2mg)(x+l/22)
T should be minimum then 13ml2+mx2(2mg)(x+l/22) should be minimum
Let =13l2+x2(x3+l4)
Should be minimum
x=L4.

1223011_1264807_ans_8a85ea81a85b43ea94593a47d8b046ec.png

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