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Q.

A rod of length L and mass “m” fixed at one end, is hanging vertically. The other end is now raised so that the rod makes an angle $30 °$ with horizontal then workdone in this process will be

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a

mgL/3

b

mgL

c

mgL/4

d

mgL/2

answer is D.

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Detailed Solution

Let OA = L, mass = m

Centre of mass is at $L/2$

Question Image

Now rod is raised through an angle $60°$

OC = L now ${OC}^{|} = \frac{L}{2} \cos 60 = L/4$

Workdone = Force × displacement of centre of mass

$W = mg \times L/4$

$∴ W = mgL/4$

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