Current Electricity and Electrical InstrumentHard
Question
wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10Ω its new resistance would be:
Options
A.40 Ω
B.80 Ω
C.120 Ω
D.160 Ω
Solution
Let the new length be l1, keeping volume constant,
πr2L = π
⇒ l1 = 4L
Now,
R = 60
πr2L = π
⇒ l1 = 4L
Now,
R = 60
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