Question

The length of a potentiometer wire is $l$. A cell of emf $E$ is balanced at a length $\frac{l}{5}$ from the positive end of the wire. If the length of the wire is increased by $\frac{l}{2}$, the distance of the balance point with the same cell is

• A. $\frac{2}{15}l$
• B. $\frac{3}{15}l$
• C. $\frac{3}{10}l$
• D. $\frac{4}{10}l$

Answer 1

Answer: (C)

$\frac{3}{10}l$

Here,

The e.m.f. of the cell is

$E=\frac{kl}{5}$, where $k$ is potential gradient.

Also $k=\frac{E_0}{l}$, where $E_0$ is the e.m.f. of the battery in potentiometer circuit.

$\therefore E=\frac{E_0}{5}$

When the length of potentiometer wire is increased by $\frac{l}{2}$ the potential gradient becomes

$k=\frac{E_0}{(l+\frac{l}{2})}$

$k=\frac{2E_0}{3l}$

If $l_n$ is new balancing length then e.m.f. of cell is

$E=k{l}_n=\left(\frac{2E_0}{3l}\right)l_n$

$\therefore E=\frac{E_0}{5}=\left(\frac{2E_0}{3l}\right)l_n$

$\therefore l_n=\frac{3l}{10}$