Question

In the meter bridge experiment shown in the figure, the balance length $AC$ corresponding to null deflection of the galvanometer is $x$. What would be the balance length if the radius of the wire $AB$ is doubled?

• A. $4x$
• B. $2x$
• C. $x$
• D. $x/2$

Answer 1

Answer: (C)

According to Wheatstone bridge principle
$\frac{R_1}{R_2}=\frac{R_{AC}}{R_{CB}}=\frac{AC}{CB}...$ (i)
Now, radius of the wire $AB$ is doubled
$As$ $R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^2}$
If radius of wire is doubled
Then new resistance, $R^{\prime }=\frac{R}{4}$
Similarly $R_{AC}=\frac{R_{AC}}{4}\dots$ (ii)
and $R_{CB}=\frac{R_{CB}}{4}$.. (iii)
$\therefore$ From Eqs. (i), (ii) and (iii), we get
$\frac{R_1}{R_2}=\frac{4R_{AC}^{\prime }}{4R_{CB}^{\prime }}=\frac{R_{AC}^{\prime }}{R_{AC}^{\prime }}=\frac{R_{AC}}{R_{CB}}=\frac{AC}{CB}$
Thus, new balancing length $AC=x$ is same as before.