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.