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A resistor $6 Ω$ in the right gap and unknown resistance in left gap of meter bridge are connected. The balancing length is 52cm, to shift the balancing length to 40cm, the resistance to be connected in series in the right gap is

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$9.75 Ω$

$16 Ω$

$2.5 Ω$

$3.75 Ω$

answer is C.

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Detailed Solution

$\frac{R_{1}}{R_{2}} = \frac{l}{100 - l}$

$\frac{x}{6} = \frac{52}{48} \to (1)$

$\frac{x}{6 + R} = \frac{40}{60} \to (2)$

$Solving, R = 3.75 Ω$

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