The resistance of a certain length of
wire having a diameter of 6 mm is 5
ohm. The wire is drawn such that the
diameter becomes 3 mm. The new
resistance will be
(A) 30 ohms.
B) 5 ohms.
(C) 60 ohms.
(D) 80 ohms.
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Answer:
The volume of the wire remains the same
Initial volume= Area × length = π× 3²×l1
Final volume= π× 1.5²×l2
As the volume is equal Final length =4l1
Resistance= Resistivity× length÷ area
Intial resistance = k ×l1÷ 9 =5 be equation 1
Final resistance = k× 4l1× 4÷ 9 be equation 2
dividing equation 1 by 2 5÷ final resistance = 1÷ 16
Final resistance = 16× 5 = 80 ohms
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