Question

In fig., $AB$ is a potentiometer wire of length $10\text{m}$ and resistance $2\Omega$. With key $\left(K\right)$ open the balancing length is $5.5\text{m}$. However, on closing key $\left(K\right)$ the balancing length reduces to $5\text{m}$. The internal resistance of the cell $\left(E_1\right)$ is –

• A. $0.01\Omega$
• B. $0.1\Omega$
• C. $0.2\Omega$
• D. $1\Omega$

Answer 1

The correct option is B $0.1\Omega$

When $K$ is open we are balancing the emf $E_1$, hence $E_1=\frac{V}{L}\times 5.5=\frac{2}{10}\times \mathrm{5.5.......}\left(1\right)$
When $K$ is closed we are balancing the potential across the discharging battery which is $\frac{r}{r+r_1}{E}_1$ where $r_1$ is the internal resistance of the battery.
Hence,
$\frac{E_1}{r+r_1}=\frac{2}{10}\times \mathrm{5.....}\left(2\right)$
On dividing $\left(1\right)$ and $\left(2\right)$, we get
$r_1+1=1.1\Rightarrow r_1=0.1\Omega$