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Q.

When the terminals of a cell are connected to the ends of a metallic wire, power dissipated in the wire is found to be 12 W. Now the battery is removed and wire is stretched to three times its original length and the terminals of the same cell are connected to the ends of the stretched wire. Now power dissipated in the stretched wire will be

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a

9 W

b

8 W

c

43 W

d

34 W

answer is C.

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Detailed Solution

Before the wire stretched, its resistance isR=v212 Ω , v = emf of cell. After the wire is stretched, its resistance given by R'=Rl'l2=R x 3ll2=9RNew power P'=v2R'=v29R=v29v212 W⇒P'=43 W

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