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(D) Let $G$ be the resistance of the galvanometer and $I_g = 0.006 \ A$ be the current for full scale deflection.For a voltmeter of range $V = 30 \ V$

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$5$

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(D) Let $G$ be the resistance of the galvanometer and $I_g = 0.006 \ A$ be the current for full scale deflection.For a voltmeter of range $V = 30 \ V$ with a series resistance $R = 4990 \ \Omega$:$V = I_g(G + R)$$30 = 0.006(G + 4990)$$G + 4990 = \frac{30}{0.006} = 5000$$G = 5000 - 4990 = 10 \ \Omega$For an ammeter of range $I = 1.5 \ A$ with a shunt resistance $S = \frac{2n}{249} \ \Omega$:$I_g G = (I - I_g) S$$0.006 	imes 10 = (1.5 - 0.006) 	imes S$$0.06 = 1.494 	imes S$$S = \frac{0.06}{1.494} = \frac{60}{1494} = \frac{10}{249} \ \Omega$Equating the shunt resistance:$\frac{2n}{249} = \frac{10}{249}$$2n = 10 \Rightarrow n = 5$

$A$ galvanometer gives full scale deflection with $0.006 \ A$ current. By connecting it to a $4990 \ \Omega$ resistance,it can be converted into a voltmeter of range $0-30 \ V$. If connected to a $\frac{2n}{249} \ \Omega$ resistance,it becomes an ammeter of range $0-1.5 \ A$. The value of $n$ is:

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