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Q.

A uniform rod of length 2m and mass 5kg is lying on a horizontal surface. The work done in raising one end of the rod with the other end in contact with the surface until the rod makes an angle 30^o with the horizontal is, (g = 10ms^–2)

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a

25J

b

50J

c

25 $\sqrt{3}$ J

d

50 $\sqrt{3}$ J

answer is A.

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Detailed Solution

$w = mgh = mg \frac{l}{2} sinθ = 5 \times 10 \times \frac{2}{2} \times \frac{1}{2} = 25 J$

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