In a metre bridge experiment $S$ is a standard resistance. $R$ is a resistance wire. It is found that the balancing length, measured from the end of the bridge wire on the side of $R$, is $l = 25$ cm. If $R$ is replaced by a wire of half length and half diameter that of $R$ of same material, then the balancing distance $l'$ (in cm) will now be ______.
Numerical value type. Enter your answer.
Answer: 40
Balance condition: $\dfrac{R}{S} = \dfrac{l}{100 - l} = \dfrac{25}{75} = \dfrac{1}{3}$.
$R = \dfrac{\rho L}{A}$ with $A \propto d^2$. Halving the length halves $R$; halving the diameter makes the area one quarter, which multiplies $R$ by $4$. So $R' = \dfrac{1}{2}\times4\,R = 2R$.
$$\frac{l'}{100 - l'} = \frac{2R}{S} = \frac{2}{3} \Rightarrow l' = 40\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics