When a piece of aluminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will become:-
Two times
Four times
Eight times
Sixteen times
A
Two times
B
Eight times
C
Four times
D
Sixteen times
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Solution
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The correct option is D Sixteen times In stretching wire,
We know that, Rα1r4
R=K1r4
Where the diameter is rediced to 1/2
We have r1=(r12)
R=K1(r2)4R=K1r416R=16K1r4R=16(K1r4)
[d]∴ The resistance will be 16 times.
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