# The balancing length for a cell is

Question:

The balancing length for a cell is $560 \mathrm{~cm}$ in a potentiometer experiment. When an external resistance of $10 \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \mathrm{~cm}$. If the internal resistance of the cell is

$\frac{N}{10} \Omega$, where $\mathrm{N}$ is an integer then value of $\mathrm{N}$ is _______

Solution:

(12) We know that

$E \propto \ell$ where $l$ is the balancing length

$\therefore \mathrm{E}=k(560)$     ....(1)

When the balancing length changes by $60 \mathrm{~cm}$

$\frac{E}{r+10} 10=k(500)$    ....(2

Dividing (i) by (ii) we get

$\Rightarrow \quad \frac{r+10}{10}=\frac{56}{50}$

$\Rightarrow \quad 50 r+500=560$

$\Rightarrow \quad r=\frac{6}{5} \Omega=\frac{N}{10} \Omega$

$\Rightarrow \quad N=12$