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In the meter bridge shown in figure, distance of the null point N from end A is 40 cm. When a $20 Ω$ resistor is connected in series with P, the null point is displaced to the right by 20 cm. Then value of Q is

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Correct option is A

Initially, PQ=40100−40⇒PQ=23. . . 1Finally, P+20Q=40+20100−40+20⇒PQ+20Q=32 . . . 2Solving, Q=65×20 Ω=24 Ω

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In the meter bridge shown in figure, distance of the null point N from end A is 40 cm. When a 20Ω resistor is connected in series with P, the null point is displaced to the right by 20 cm. Then value of Q is

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In the meter bridge shown in figure, distance of the null point N from end A is 40 cm. When a $20 Ω$ resistor is connected in series with P, the null point is displaced to the right by 20 cm. Then value of Q is