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Q.

Two unknown resistances X and Y are connected to left and right gaps of a meter bridge and the balancing point is obtained at 80 cm from left. When a $10 Ω$ resistance is connected parallel to X, the balancing point is 50 cm from left. The values of X and Y respectively are

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a

$20 Ω, 6 Ω$

b

$40 Ω, 9 Ω$

c

$30 Ω, 7.5 Ω$

d

$10 Ω, 3 Ω$

answer is C.

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Detailed Solution

$\begin{aligned} \frac{X}{Y} = \frac{80}{(100 - 80)} (∵ \frac{X}{Y} = \frac{l}{(100 - l)}) \frac{X}{Y} = \frac{4}{1} & \frac{10 X}{(10 + X)} \times \frac{1}{Y} = \frac{50}{50} \\ ∴ \frac{10}{(10 + X)} \times \frac{4}{1} = 1 \Rightarrow 40 = 10 + X \Rightarrow X = 30 ΩY = \frac{X}{4} = \frac{30}{4} = 7 .5 Ω \end{aligned}$

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