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The internal resistance of a cell of e.m.f. $12 \mathrm{~V}$ is $5 \times 10^{-2} \Omega$. It is connected across an unknown resistance. Voltage across the cell, when a current of $60 \mathrm{~A}$ is drawn from it, is
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$9 \mathrm{~V}$
$V=E-$ ir $=12-60 \times 5 \times 10^{-2}=9 \mathrm{~V}$.
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