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When two resistances $\frac{50}{3} Ω$ and unknown resistance are connected in the left and right gaps of a meter bridge, the balance length is 70cms. When resistance ‘x’ is connected in parallel to $\frac{50}{3} Ω$ ,the balance length becomes 35cms. The value of ‘x’ is

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3Ω

4Ω

5Ω

10Ω

answer is C.

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Detailed Solution

$\frac{\frac{50}{3}}{R} = \frac{70}{30} \frac{\frac{50 x}{50 + 3 x}}{R} = \frac{35}{65} D i v i d i n g t h e a b o v e t w o e q u a t i o n s \frac{50 + 3 x}{3 x} = \frac{13}{3} \Rightarrow 150 + 9 x = 39 x; x = 5 Ω$

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