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A resistance $R_{1}$ is connected in the left gap of a metre bridge and a resistance of $9 Ω$ is connected the right gap in the metre bridge. The null point divides the bridge wire in the ratio 4:3. Now another resistance $x Ω$ is connected in series with $R_{1}$ in the left gap of te metre bridge and this time the null point divides the bridge wire in the ratio 4:1. Then the value of ‘x’ is

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$18 Ω$

$16 Ω$

$8 Ω$

$24 Ω$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{R_{1}}{R_{2}} = \frac{4}{3} ..... (1) \\ and \frac{R_{1} + x}{R_{2}} = \frac{4}{1} ..... (2) \end{array}$

$∴ \frac{R_{1}}{R_{2}} + \frac{x}{R_{2}} = 4 \Rightarrow \frac{4}{3} + \frac{x}{R_{2}} = 4 \Rightarrow \frac{x}{R_{2}} = 4 - \frac{4}{3} = \frac{8}{3}$

$∴ x = \frac{8}{3} R_{2} = \frac{8}{3} \times 9 Ω = 24 Ω$

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