Question

In a potentiometer of wire length $l$, a cell of emf $V$ is balanced at a length $\frac{l}{3}$ from the positive end of the wire. For another cell of emf $1.5\,V$, the balancing length becomes

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

Answer 1

Answer: (B)

$\frac{l}{2}$

Answer 2

For balanced potentiometer,
$\frac{E_{1}}{E_{2}}=\frac{l_{1}}{l_{2}}$
Hence, $\frac{V}{1.5 V} =\frac{l / 3}{l_{2}}$
$\frac{1}{1.5} =\frac{l}{3 l_{2}}$
$l_{2} =\frac{1.5}{3} l$
$l_{2} =\frac{l}{2}$