Resistance R = X Resistance S = Y Balancing length l=60cm Balancing length l=50cm Balancing condition for meter bridge RS=l100−lXY=6040⇒32 ...(1) When a 20 Ω resistance is connected in parallel to X XY=l′100−l′20X20+XY=5050⇒12020+X32=1⇒30=20+X∴X=10Ω∴Y=203Ω

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Two unknown resistances X and Y are connected in the left and right gaps of a meter 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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$20, 30$

$10, \frac{20}{3}$

$\frac{10}{3}, 20$

$10, 20$

answer is A.

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Detailed Solution

Resistance R = X Resistance S = Y Balancing length $l = 60 cm$ Balancing length $l = 50 cm$ Balancing condition for meter bridge $\frac{R}{S} = \frac{l}{100 - l} \frac{X}{Y} = \frac{60}{40} \Rightarrow \frac{3}{2}$ ...(1)

When a 20 $Ω$ resistance is connected in parallel to X
$\frac{X}{Y} = \frac{l^{'}}{100 - l^{'}} \frac{\frac{20 X}{20 + X}}{Y} = \frac{50}{50} \Rightarrow 1 \frac{20}{20 + X} (\frac{3}{2}) = 1 \Rightarrow 30 = 20 + X ∴ X = 10 Ω ∴ Y = \frac{20}{3} Ω$