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(B) Let $E$ be the electromotive force $(EMF)$ of the battery and $r$ be its internal resistance.The terminal voltage $V$ across a resistor $R$ is giv

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$\frac{20}{7}$

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(B) Let $E$ be the electromotive force $(EMF)$ of the battery and $r$ be its internal resistance.The terminal voltage $V$ across a resistor $R$ is given by $V = I R$,where $I = \frac{E}{R+r}$.Thus,$V = E \left( \frac{R}{R+r} \right)$,which can be rewritten as $E = V + I r = V + \left( \frac{V}{R} \right) r$.For the first case: $R_1 = 16 \ \Omega$,$V_1 = 12 \ V$.$E = 12 + \left( \frac{12}{16} \right) r = 12 + 0.75 r$  --- $(i)$For the second case: $R_2 = 10 \ \Omega$,$V_2 = 11 \ V$.$E = 11 + \left( \frac{11}{10} \right) r = 11 + 1.1 r$  --- (ii)Equating $(i)$ and (ii):$12 + 0.75 r = 11 + 1.1 r$$12 - 11 = 1.1 r - 0.75 r$$1 = 0.35 r$$r = \frac{1}{0.35} = \frac{100}{35} = \frac{20}{7} \ \Omega$.

When a battery is connected across a resistor of $16 \Omega$,the voltage across the resistor is $12 \ V$. When the same battery is connected across a resistor of $10 \Omega$,the voltage across it is $11 \ V$. The internal resistance of the battery in ohm is

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