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

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 $R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^{2}}$
In material remains same $\rho =\rho ^{′}$
Given, $l^{′}=2l$
$ \, 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^{′}=8R$
Therefore, the new balancing point is expected to be 8 $l$ .