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(C) Let $G = 100 \ \Omega$ be the resistance of the galvanometer and $I_g$ be the full-scale deflection current.For the first case,the total resistanc

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$1400$

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(C) Let $G = 100 \ \Omega$ be the resistance of the galvanometer and $I_g$ be the full-scale deflection current.For the first case,the total resistance is $(X + G)$. The voltage range is $V_1 = 15 \ V$.So,$V_1 = I_g(X + G) \implies 15 = I_g(X + 100) \quad ... (1)$For the second case,the range is doubled,so $V_2 = 2 	imes 15 = 30 \ V$. The total resistance is $(X + 1500 + G)$.So,$V_2 = I_g(X + 1500 + G) \implies 30 = I_g(X + 1500 + 100) \implies 30 = I_g(X + 1600) \quad ... (2)$Dividing equation $(2)$ by equation $(1)$:$\frac{30}{15} = \frac{I_g(X + 1600)}{I_g(X + 100)}$$2 = \frac{X + 1600}{X + 100}$$2(X + 100) = X + 1600$$2X + 200 = X + 1600$$X = 1400 \ \Omega$.

With a resistance $X$ connected in series with a galvanometer of resistance $100 \Omega$,it acts as a voltmeter of range $0-15 \ V$. To double the range,a resistance of $1500 \Omega$ is to be connected in series with $X$. The value of $X$ in ohm is:

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