Physics - Current Electricity Question with Solution | TestHub

Let$\rho$be the resistance per unit length of the metre bridge wire.

From the condition of balance of Wheatstone bridge.

$\frac{P}{Q}=\frac{R}{S}$

$\Rightarrow \frac{6}{R}=\frac{30\rho }{\left(100-30\right)\rho }=\frac{30}{70}=\frac{3}{7}$

$\Rightarrow R=\frac{7\times 6}{3}$

$\Rightarrow R=14 \mathrm{\Omega }$