The length of a potentiometer wire is 'L'. A cell of e.m.f. 'E' is balanced at a length…
The length of a potentiometer wire is 'L'. A cell of e.m.f. 'E' is balanced at a length $\frac{\mathrm{L}^{\prime}}{5}$ from the positive end of the wire. If the length of the wire is increased by $\frac{\mathrm{L}}{2}$,
at what distance will the same cell give a balance point? '
$\frac{5 \mathrm{~L}}{12}$
$\frac{2 \mathrm{~L}}{15}$
$\frac{4 \mathrm{~L}}{15}$
$\frac{3 \mathrm{~L}}{10}$
Solution
If the length of potentiometer wire is increased by $\frac{L}{2}$, the balancing length will also increase in the same proportion. Hence from $\frac{L}{5}$ it will increase by $\frac{1}{2}\left(\frac{L}{5}\right)$ and become $\frac{L}{5}+\frac{L}{10}=\frac{3 L}{10}$
.