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In a meter bridge experiment, null point is obtained at $20 cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$ . If $X < Y$ , then the new position of the null point from the same end, if one decides to balance a resistance of $4 X$ against $Y$ will be at

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70 cm

50 cm

80 cm

40 cm

answer is A.

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Detailed Solution

$\frac{X}{Y} = \frac{20}{80} = \frac{1}{4}$

Y=4 X

Since, $\frac{4 X}{Y} = \frac{l}{100 - 1}$

$\frac{4 X}{4 X} = \frac{1}{100 - 1}$

$2 l = 100$

$l = 50 c m$

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