Question

A 10 ohm thick wire is stretched so that its original length becomes 3 times. Assuming that there is no change in uts density on streching, calculate the resistance of the new wire.

Answer 1

We can assume the wire to be a cylinder so since volume will be constant then we have

π*r^2*l=π*R^2*3*l

r=√3*R

R=(1/√3)*r

So resistance previously was R1=(pl)/(π*r^2)

Present resistance will be

R2=(p*3*l)/(π*R^2)

=(p*3*l)/((π*r^2)/3)

=9*(pl)/(π*r^2)

=9*R1

=9*10=90ohms

Resistivity will remain unchanged since it is the property of a material and it itself does not depend on length or cross sectional area

Answer 2

Answer:- 90 ohms

Explanation:-

Initial length of wire ,L1 = l

Final length of wire L2= 3l

Initial resistance of wire is R1= 10 ohms (given)

Final resistance = ? (R2)

W.K.T,

R Is directly proportional to l^2

=> R2/R1=(L2/L1)^2 = (3l/l)^2 = 9

R2 = 9R1

=> 9*10 = 90 ohms.