Question

A wire of 1.1 m length and radius 7 multiply 10-5 m is connected across right gap of a meter bridge. What a resistance of 45 ohm is introduced in left gap from a resistance box connected across it, balance point is obtained at 0.6 m from left side. Find specific resistance of the material of wire

Answer 1

Given that,

Length l= 1.1 m = 110 cm

Radius $r=7\times10^{-5}\ m$

Resistance = 45 Ω

Length l'=0.6 m = 60 cm

We need to calculate the specific resistance of the material of the wire

Using formula of resistivity of the material

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

Where, A = area

l = length

R = resistance

Put the value into the formula

$\rho=\dfrac{45\times\pi\times(7\times10^{-5})^2}{1.1}$

$\rho=6.29\times 10^{-7}\ \Omega-m$

Hence, The specific resistance of the material of the wire is $6.29\times 10^{-7}\ \Omega-m$