Question

The length of a potentiometer wire is $l$. A cell of emf $E$ is balanced at a length $\frac{l}{3}$ from the positive end of the wire. If the length of the wire is increased by $\frac{l}{2}$. At what distance from the initial balance point will the same cell give a new balance point.

• A. $\frac{2l}{3}$
• B. $\frac{l}{2}$
• C. $\frac{l}{6}$
• D. $\frac{4l}{3}$

Answer 1

Answer: (C)

$\frac{l}{6}$

When potentiometer is balanced, we get $E=kl$
$E=\frac{IR}{L}l$
where, all the variables have their usual meanings.
Since, $E$ is same and $I$ and $R$ are constant hence,
$L\propto l$
$\Rightarrow l\propto \frac{l}{3}$ and $(l+\frac{l}{2})\propto l^{{}^{\prime }}$
where, $l^{{}^{\prime }}$ is the balancing length obtained when the length of the wire is increased by $\frac{l}{2}$.
Therefore we can write

$\frac{l}{(l+\frac{l}{2})}=\frac{\frac{l}{3}}{l^{{}^{\prime }}}$

$l{l}^{\prime }=\frac{3l^2}{6}$

$\therefore l^{\prime }=\frac{l}{2}$

Hence, the distance of balance point is $\frac{l}{2}-\frac{l}{3}=\frac{l}{6}$