Question

A metallic wire of resistance $20\Omega$ is stretched such that its length becomes three times. The new resistance of the wire will be

• A. $6.67\Omega$
• B. $60.0\Omega$
• C. $120\Omega$
• D. $180\Omega$

Answer 1

Answer: (D)

$180\Omega$

The resistance of the wire is

$R=\frac{\rho L}{A}$

$\frac{\rho L}{A}=20\Omega$...................(1)...(given)

When the wire is stretched three times its length becomes

$L^{\prime }=3L$

but the volume remains same hence

$A^{\prime }{L}^{\prime }=AL$

where, A' is the new area o the wire.

$A^{\prime }=\frac{AL}{L^{\prime }}=\frac{A}{3}$

Therefore new resistance is

$R^{\prime }=\frac{\rho L^{\prime }}{A^{\prime }}$

$R^{\prime }=\frac{\rho 3L}{\frac{A}{3}}$

$R^{\prime }=9\frac{\rho L}{A}$

$R^{\prime }=9\times 20=180\Omega$ ................from(1)