A resistor 6Ω 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 40 cm, the resistance to be connected in series in the right gap is

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2.5Ω

9.75Ω

16Ω

3.75Ω

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{x}{6} = \frac{52}{48} \Rightarrow x = \frac{13}{2}; \frac{\frac{13}{2}}{6 + y} = \frac{40}{60} \Rightarrow \frac{13}{2 (6 + y)} = \frac{2}{3} \\ 39 = 24 + 4 y; 4 y = 15 \Rightarrow y = \frac{15}{4} = 3 .75 Ω \end{array}$

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