Question

In the meter bridge circuit shown, the null point is obtained at $C(AC=x)$ on the wire AB. If the diameter of the wire AB is doubled, the position of the new null point C' will correspond to.

• A. AC'$=x/2$
• B. AC'$=x/4$
• C. AC'$=4x$
• D. AC'$=$x

Answer 1

The correct option is B AC'$=x/4$

$\frac{AC}{x}=\frac{BC}{l-x}$

$\frac{R}{x}=\frac{R^1}{(l-x)}$

$\frac{\rho \frac{l}{A}}{x}=\frac{\rho \frac{l}{A^1}}{x^1}\frac{1}{Ax}=\frac{A^1}{x^1}Ax=A^1{x}^1\pi (\frac{2r}{2}{)}^2=\pi r^2\times x^1$

Let initial radius $\left(\frac{r}{2}\right)$

$\pi \frac{r^2}{4}=\pi r^2{x}^1{x}^1=\frac{x}{4}$ new AC will be $\frac{x}{4}$