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Question

Two unknown resistances X and Y are placed on the left and right gaps of a meter bridge. The null point in the galvanometer is obtained at a distance of 80 cm from the left. A resistance of 100 Ω is now connected in parallel across X. The null point is then found by shifting the sliding contact left by 20 cm. Calculate X and Y. [Neglect the end corrections]

A

X=20050 Ω , Y=20050 Ω
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B

X=50030 Ω , Y=1253 Ω
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C

X=1003 Ω , Y=253 Ω
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D

X=2003 Ω , Y=253 Ω
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Solution

The correct option is B
X=50030 Ω , Y=1253 Ω
From the first null point:


XY=8020 .....(1)

Now, when 100 Ω is connected across X, then


(100X100+X)Y=6040 .....(2)

From eqs.(1) and (2), we get

(100×4Y100+4Y)Y=6040

3Y(100+4Y)=800Y

12Y2500Y=0

Y(12Y500)=0

Y=50012=1253 Ω

Using equation (1),

X=5003 Ω

Hence, option (b) is the correct answer.

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