Q. 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) $\frac{1}{8l}$
B) $\frac{1}{4l}$
C) $8l$
D) $16l$
Explanation
(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, $l^{'} = 2l$
$\ \ \ \ \ \ \ \ \ \ \ \ \ r^{'} = \frac{r}{2}$
$\therefore\ \ \ R^{'} = \frac{\rho l^{'}}{A^{'}} = \frac{\rho 2l}{\pi\left( \frac{r}{2} \right)^{2}} = \frac{8\rho l}{\pi r^{2}}$
$\ \ \ \ \ \ R^{'} = 8R$
Therefore, the new balancing point is expected to be 8$l$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.