To measure the internal resistance of a battery, potentiometer is used. For $\mathrm{R}=10 \Omega$, the balance point is observed at $\ell=500 \mathrm{~cm}$ and for $\mathrm{R}=1 \Omega$ the balance point is observed at $\ell=400 \mathrm{~cm}$. The internal resistance of the battery is approximately :
JEE MAIN 2024, Diffcult
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Let potential gradient be $\lambda$.
$\therefore \mathrm{i} \times 10=\lambda \times 500=\varepsilon-\mathrm{ir}_{\mathrm{s}}$
$\Rightarrow 500 \lambda=\varepsilon-50 \lambda \mathrm{r}_{\mathrm{s}}$
Also,
$\mathrm{i}^{\prime} \times 1=\lambda \times 400=\varepsilon-\mathrm{i}^{\prime} \mathrm{r}_{\mathrm{s}}$
$\Rightarrow 400 \lambda=\varepsilon-400 \lambda \mathrm{r}_{\mathrm{s}}$
$\therefore 100 \lambda=350 \lambda \mathrm{r}_{\mathrm{s}} \Rightarrow \mathrm{r}_{\mathrm{s}}=\frac{10}{35} \approx 0.3 \Omega$
Hence option $(4)$
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