Question

If we stretch a wire to 5 times its present length and reduce the diameter to half its present value.Then what will be the change in its resistance?

• A. 5 times
• B. 10 times
• C. 20 times
• D. 40 times

Answer 1

The correct option is C 20 times

$R=\rho \frac{l}{A}$
$\rho$ = Resistivity of material
I = length of the material
A = Area of the cross-section

Wire is assumed to be with circular cross section of radius r
$A=\pi r^2=\frac{\pi d^2}{4}$
$R=\frac{\rho l}{\pi r^2}=\frac{4\rho l}{\pi d^2}$

Here $R_1=\frac{4\rho l_1}{\pi d_1^2}$
On streching $l_2=5L_1$
$d_2=\frac{d_1}{2}$
$R_2=\frac{4\rho \times 5l_1\times 4}{\pi d_1^2}=20\frac{4\rho l_1}{\pi d_1^2}$
$R_2=20R_1$