Question

A 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 was 10 Ω  its new resistance would be

(a) 40 Ω 

(b) 80 Ω 

(c) 120 Ω 

(d) 160 Ω 

Answer 1

Correct option (d) 160 Ω 

Explanation:

Let the original diameter of the  wire be D.

Therefor the new. diameter is D/2.

Original area of cross-section is πD²/4

and the  final area of cross-section is πD²/16.

The new length of the wire is given by

Answer 2

Option (d) is correct

Explanation

Let the diameter of the wire of length L be D. .Again let the length and  diameter of the string after drawing be l and d respectively.

Given D/d=2

Now the volumes of the wires will be same So πD^2/4×L=πd^2×l

So l/L=D^2/d^2=4

Now resistance of 1st wire

R1=p×L/(πD^2/4)

The resistance of 2nd stretched wire will be

R2=p×l{/(πd^2/4)

So R2/R1=D^2/d^2×l/L=4×4=16

R2=16×R1=16×10=160 ohm