Physics - Current Electricity Question with Solution | TestHub

Let the unknown resistance in the left gap be $R$ and the resistance in the right gap be $S=10\Omega$.

At the balance point of the meter bridge, the ratio of resistances is equal to the ratio of the lengths of the bridge wire segments:

$\frac{R}{S}=\frac{l_1}{l_2}$

Given that the balance point divides the bridge wire in the ratio $3:2$, we have:

$\frac{R}{10}=\frac{3}{2}$

$R=10\times \frac{3}{2}=15\Omega$

The length of the unknown resistance wire is $L=1.5\text{m}$.

The resistance per unit length of the wire is:

$\frac{R}{L}=\frac{15}{1.5}=10.0\Omega /\text{m}$

Answer: $10.0\Omega /\text{m}$