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Two unknown resistances x and y are connected in the left and right gaps of a metre bridge and the balancing point is obtained at 60 cm from the left. When a $20 Ω$ resistance is connected in parallel to x, the balance point is 50 cm. Calculate x and y (in ohm)

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10, 20/3

20, 30

10/3, 20

10, 20

answer is A.

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Detailed Solution

$\Rightarrow \frac{X}{Y} = \frac{60}{40} = 1.5 \Rightarrow X = 1.5 Y$

After connecting the resistor

$\frac{\frac{20 X}{20 + X}}{Y} = \frac{50}{50} = 1$

$\Rightarrow \frac{20 X}{20 + X} = Y \Rightarrow \frac{20}{20 + X} (1.5) Y = Y$

$\Rightarrow 20 + X = 30 \Rightarrow X = 10 Ω$

$\Rightarrow Y = \frac{2}{3} x 10 = \frac{20}{3} Ω$

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