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
$\frac{R_{1}}{R_{2}}=\frac{R_{A C}}{R_{C B}}=\frac{A C}{C B} ...$ (i)
Now, radius of the wire $A B$ is doubled
$A s$ $R=\frac{\rho l}{A}=\frac{\rho l}{\pi r^{2}}$
If radius of wire is doubled
Then new resistance, $R'=\frac{R}{4}$
Similarly $R_{A C}=\frac{R_{A C}}{4} \ldots$ (ii)
and $R_{C B}=\frac{R_{C B}}{4}$.. (iii)
$\therefore $ From Eqs. (i), (ii) and (iii), we get
$\frac{R_{1}}{R_{2}}=\frac{4 R'_{A C}}{4 R'_{C B}}=\frac{R'_{A C}}{R'_{A C}}=\frac{R_{A C}}{R_{C B}}=\frac{A C}{C B}$
Thus, new balancing length $A C=x$ is same as before.