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Q.

Two ends of a copper wire are connected to the terminals of a battery. If the wire is stretched to twice its original length, then power dissipated in the wire. The thermal power dissipated in the wire.

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a

is doubled

b

is halved

c

becomes one-fourth

d

becomes four times

answer is C.

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Detailed Solution

When the wire is stretched to n times its original length, its resistance becomes n2R.Now, P=V2R.So  P'=V2R =V2n2R=Pn2=P22=P4.
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