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In the meter bridge experiment, the length AB of the wire is 1 m. The resistors X and Y have values $5 Ω$ and $2 Ω$ respectively. When a shunt resistance S is connected to X, the balancing point is found to be 0.625 m from A. Then, the resistance of the shunt is

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$7.5 Ω$

$10 Ω$

$12.5 Ω$

$5 Ω$

answer is B.

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Detailed Solution

$\frac{R_{1}}{R_{2}} = \frac{L}{1 - L} [L is in meter]$

$\frac{(\frac{5 S}{5 + S})}{2} = \frac{0.625}{0.375} = \frac{5}{3}$

$\Rightarrow S = 10 Ω$

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