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(B) For the galvanometer $G$ to read zero,no current flows through the branch containing $G$ and $E$.This implies that the potential difference across

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$100$

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(B) For the galvanometer $G$ to read zero,no current flows through the branch containing $G$ and $E$.This implies that the potential difference across the resistance $X$ must be equal to the $e.m.f.$ of the battery $E$.Let $I$ be the current flowing through the loop containing $E_1$,$500 \, \Omega$,and $X$.Using Ohm's law,the current $I$ is given by $I = \frac{E_1}{500 + X}$.The potential difference across $X$ is $V_X = I \cdot X = \left( \frac{E_1}{500 + X} \right) X$.Since $V_X = E$,we have $\left( \frac{12}{500 + X} \right) X = 2$.Dividing both sides by $2$,we get $\left( \frac{6}{500 + X} \right) X = 1$.$6X = 500 + X$.$5X = 500$.$X = 100 \, \Omega$.

In the adjoining circuit,the battery $E_1$ has an $e.m.f.$ of $12 \, V$ and zero internal resistance,while the battery $E$ has an $e.m.f.$ of $2 \, V$. If the galvanometer $G$ reads zero,then the value of the resistance $X$ in $\Omega$ is

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