To measure the internal resistance of a battery, potentiometer is used. For $R=10 \Omega$, the balan

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To measure the internal resistance of a battery, potentiometer is used. For $R=10 \Omega$, the balance point is observed at $l=500 \mathrm{~cm}$ and for $R=1 \Omega$ the balance point is observed at $l=400 \mathrm{~cm}$. The internal resistance of the battery is approximately\

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(c) : Given : $R=10 \Omega, l=500 \mathrm{~cm}$, $l^{\prime}=400 \mathrm{~cm}, R^{\prime}=1 \Omega$\ Let the potential gradient of the potentiometer be $K$.\ So, $i 	imes 10=K 	imes 500$\ Now, $i=50 K ; E-i r=500 K$ $$\begin{equation*} E-50 K r=500 K 	ag{i} \end{equation*}$$ Now, $i^{\prime} 	imes 1=K 	imes 400 \Rightarrow i^{\prime}=400 K$ $$\begin{equation*} E-i^{\prime} r=400 K ; E-400 K r=400 K 	ag{ii} \end{equation*}$$ From equation (i) and (ii), we get\ $E-50 K r-E+400 K r=100 K$\ $350 \mathrm{Kr}=100 \mathrm{~K} ; r=\frac{100}{350}=\frac{10}{35} \approx 0.3 \Omega$\