If the galvanometer reading is zero,then the | vedclass

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(B) When the galvanometer reading is zero,no current flows through the branch containing the galvanometer. This means the potential difference across

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$100$

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(B) When the galvanometer reading is zero,no current flows through the branch containing the galvanometer. This means the potential difference across the resistor $X$ must be equal to the electromotive force $(EMF)$ of the battery in the right branch,which is $2 \ V$.Let the potential at the bottom wire $DC$ be $0 \ V$. Then the potential at the top junction point (where $500 \ \Omega$ and $X$ meet) must be $2 \ V$ for the galvanometer current to be zero.Using the voltage divider rule for the left loop:$V_X = \frac{X}{500 + X} 	imes 12 = 2$Solving for $X$:$12X = 2(500 + X)$$12X = 1000 + 2X$$10X = 1000$$X = 100 \ \Omega$

If the galvanometer reading is zero,then the resistance $X = \dots \Omega$.

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