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(B) Let the full-scale deflection current of the galvanometer be $I_g$.When a resistance of $100 \Omega$ is connected in series,the total resistance i

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$800$

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(B) Let the full-scale deflection current of the galvanometer be $I_g$.When a resistance of $100 \Omega$ is connected in series,the total resistance is $(100 + R)$. The voltage range $V$ is given by $V = I_g(100 + R)$  --- $(i)$When a resistance of $1000 \Omega$ is connected in series,the new range is $2V$. The total resistance is $(1000 + R)$. Thus,$2V = I_g(1000 + R)$  --- (ii)Dividing equation (ii) by equation $(i)$:$\frac{2V}{V} = \frac{I_g(1000 + R)}{I_g(100 + R)}$$2 = \frac{1000 + R}{100 + R}$$2(100 + R) = 1000 + R$$200 + 2R = 1000 + R$$2R - R = 1000 - 200$$R = 800 \Omega$

When a resistance of $100 \Omega$ is connected in series with a galvanometer of resistance $R$,its range is $V$. To double its range,a resistance of $1000 \Omega$ is connected in series. Find $R$. (in $\Omega$)

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