Question

A 10 ohm thick wire is stretched so that its length becomes three times . Assuming that there is no change in density on stretching. Calculate the resistance of the new wire.

Answer 1

Answer:

90 ohm.

Explanation:

The total volume of wire remains the same.

l - previous length

l' - new length= 3l

A- previous area

A'=new area.

l*A=l'*A'. (As volume is constant)

or, l *A= 3l*A

or, A'=A/3

R = k*(l/A). (k is the resistivity)

R'(new resistance) = k*(l'/A').

Dividing the two equations,

R'/R= (l'/A')*(A/l)

A= A'/3

l' = 3l

R' : R = (3l/A)*(A/l)

or, R' : R = 3 * 3

or, R' = 9 * 10 ohm. (R = 10 ohm)

or, R' = 90 ohm.

Answer 2

answer:

The total volume of wire remains the same.

l - previous length

l' - new length= 3l

A- previous area

A'=new area.

l*A=l'*A'. (As volume is constant)

or, l *A= 3l*A

or, A'=A/3

R = k*(l/A). (k is the resistivity)

R'(new resistance) = k*(l'/A').

Dividing the two equations,

R'/R= (l'/A')*(A/l)

A= A'/3

l' = 3l

R' : R = (3l/A)*(A/l)

R' = 90 ohm.