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Q.

An insulating long light rod of length L pivoted at its centre O and balanced with a weight W at a distance X from the left end as shown in figure. Charges q and 2q are fixed to the ends of the rod. Exactly below each of these charges at a distance h a positive charge Q is fixed. Then x is

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a

QLq +ε02h2Lwh2W

b

QLq+ε0h2LWε0h2W

c

4QLq+ε0h2LW8πh2W

d

QLq+4πε0h2LW8πε0h2W

answer is D.

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Detailed Solution

By law of conservation of Moments about C,Total Clockwise CWMoments=Total counter Clockwise CCW Moments ⇒qQ4πε0h2L2+Wx-L2=2qQ4πε0h2L2 ⇒ Wx-L2=QqL8πε0h2⇒x=QqL+4πε0h2LW8πε0h2W

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