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(B) Let the current flowing through the $500 \, \Omega$ resistor be $I$. Since the galvanometer $G$ reads zero, no current flows through the branch co

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$100$

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(B) Let the current flowing through the $500 \, \Omega$ resistor be $I$. Since the galvanometer $G$ reads zero, no current flows through the branch containing $E_2$ and $G$.The potential difference across the resistance $X$ must be equal to the $e.m.f.$ of battery $E_2$, which is $2 \, V$. Therefore, the potential at the junction point above $X$ is $2 \, V$ relative to the bottom wire.The current $I$ flowing through the $500 \, \Omega$ resistor is given by the potential drop across it divided by its resistance:$I = \frac{E_1 - V_X}{500} = \frac{12 \, V - 2 \, V}{500 \, \Omega} = \frac{10 \, V}{500 \, \Omega} = 0.02 \, A$.Since no current flows through the galvanometer, all this current $I$ must flow through the resistance $X$. Using Ohm's law for resistance $X$:$V_X = I 	imes X$$2 \, V = 0.02 \, A 	imes X$$X = \frac{2}{0.02} \, \Omega = 100 \, \Omega$.Thus, the value of the resistance $X$ is $100 \, \Omega$.

In the adjoining circuit, the battery $E_1$ has an $e.m.f.$ of $12 \, V$ and zero internal resistance, while the battery $E_2$ has an $e.m.f.$ of $2 \, V$. If the galvanometer $G$ reads zero, then the value of the resistance $X$ in $ohms$ is

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