Question

A 250 cm long wire has diameter 1mm. It is connected at the right gap of a slide wire bridge. When a $3\Omega$ resistance is connected to left gap, the null point is obtained at 60 cm. The specific resistance of the wire is

• A. $6.28\times {10}^{-6}\Omega m$
• B. $6.28\times {10}^{-5}\Omega m$
• C. $6.28\times {10}^{-8}\Omega m$
• D. $6.28\times {10}^{-7}\Omega m$

Answer 1

The correct option is D $6.28\times {10}^{-7}\Omega m$

Let,

Resistance in left gap is $R_1=3\Omega$,

Resistance in right gap is $R_2$,

Balancing length is $l=60cm$

Length of the wire is $l=250cm=2.5m$

Radius of the wire is $r=\frac{1}{2}mm=5\times {10}^{-4}m$

The balance condition for meter bridge is given by

$\frac{R_1}{R_2}=\frac{l}{100-l}$

$\frac{3}{R_2}=\frac{l}{100-l}$

$R_2=\frac{120}{60}$

$R_2=2\Omega$

The resistance of the wire is

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

where, $\rho$ is the specific resistance of the wire, A is area of cross

section of wire.

$\rho =\frac{RA}{l}$

$\rho =\frac{R\pi r^2}{l}$

$\rho =\frac{\left(2\right)\left(3.14\right)(5\times {10}^{-4}{)}^2}{2.5}$

$\rho =\frac{6.28\times 25\times {10}^{-8}}{2.5}$

$\rho =6.28\times 10\times {10}^{-8}$

$\rho =6.28\times {10}^{-7}$