Question

A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point length $l$. If the wire is replaced by another wire of same material but with double the length and half the thickness, the balancing point is expected to be

• A. $\frac{1}{8\ell}$
• B. $\frac{1}{4\ell}$
• C. ${8\ell}$
• D. ${16\ell}$

Answer 1

Answer: (C)

In a meter bridge the ratio of two resistances is $\frac{R}{R^{'}}=\frac{l}{l^{'}}$
where $l$ and $l^{'}$ are balancing lengths Resistance $R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^{2}}$
If material remains same $\rho=\rho^{'}$
Given $l^{'}=2 l$
$r^{'} =\frac{r}{2} $
$\therefore R^{'} =\frac{\rho l^{'}}{A^{'}}$
$=\frac{\rho 2 l}{\pi\left(\frac{r}{2}\right)^{2}}$
$= \frac{8 \rho l}{\pi r^{2}} $
$R^{'} =8 R$
Therefore, the new balancing point is expected to be $8 l$.