Question

The length of a potentiometer wires is $l$. A cell of $emfE$ is balanced at a length $\left(\frac{l}{3}\right)$ from positive end of the wire. If the length of the wire is increased by $\left(\frac{l}{2}\right)$, the distance at which the same cell gives the balancing point is
(Cell in the primary is ideal and no series resistance is present in the primary circuit.)

• A. $\frac{21}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{6}$
• D. $\frac{41}{3}$

Answer 1

Answer: (B)

According to question, the figure is as

If $K$ be the potential gradient of the potentiometer
wire, then emf of the cell which gives balancing
length $\frac{1}{3}$ is given by
$E=K\cdot \frac{l}{3}\cdots (i)$
Where, $E=\frac{V}{l}\cdot \frac{l}{3}=\frac{V}{3}\cdots (ii)[\because K=\frac{V}{l}]$
When, length of potentiometer wire is increased 1
by $\frac{l}{2}$, then new length,
$l_1=l+\frac{l}{2}=\frac{3l}{2}$
$\therefore$ New potential gradient,
$K^{\prime }=\frac{V}{\frac{3l}{2}}\Rightarrow K^{\prime }=\frac{2V}{3l}$
If $l^{\prime }$ be the new balancing length,
then, $E=K^{\prime }{l}^{\prime }.....(iii)$
or $E=\frac{2V}{3l}{l}^{\prime }$
$\frac{2V}{3l}\cdot l^{\prime }=\frac{V}{3}\Rightarrow l^{\prime }=\frac{l}{2}[\text{ From Eq. (iii), }E=\frac{V}{3}]$