A null point is found at $200\,cm$ in potentiometer when cell in secondary circuit is shunted by $5\,\Omega$. When a resistance of $15\,\Omega$ is used for shunting null point moves to $300\,cm$. The internal resistance of the cell is $..............\,\Omega$.
JEE MAIN 2023, Medium
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$\frac{\varepsilon}{ r +5} \times 5=200\,x$
$\frac{\varepsilon \times 15}{ r +15}=300\,x$
$\Rightarrow r =5$
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