In a metre 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 where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?

50 cm

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80 cm

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40 cm

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70 cm

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Solution

The correct option is A 50 cm
$\frac{X}{Y} = \frac{20}{100 - l} = \frac{20}{80} = \frac{1}{4} Y = 4 X \frac{4 X}{Y} = \frac{l^{'}}{100 - l^{'}}; l^{'} = 50 c m$

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Similar questions

Q. In a meter bridge experiment, null point is obtained at

20 cm

from one end of the wire when resistance

x

is balanced against resistance

y

. If

x < y

, then where will be the new position of the null point from the same end, if one decides to balance a resistance of

4 x

against

y

Q. In a metre 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 where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?

In a meter bridge experiment, a 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 and if one decides to balance a resistance of 4x against y, then the new position of the null point from the same end is,

In a metre bridge experiment, two unknown resistances X and Y are connected to left and right gaps respectively and the balancing point is obtained at 20 cm from right. What will be the new position of the null point from left if one decides to balance a resistance of 4X against Y? (Assume that length of wire of metre bridge is 120 cm.)

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In a metre 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 where will be the new position of the null point from the same end, if one decides to balance a resistance of 4X against Y?

The correct option is A 50 cmXY=20100−l=2080=14Y=4X4XY=l′100−l′;l′=50 cm

The correct option is A 50 cm
$\frac{X}{Y} = \frac{20}{100 - l} = \frac{20}{80} = \frac{1}{4} Y = 4 X \frac{4 X}{Y} = \frac{l^{'}}{100 - l^{'}}; l^{'} = 50 c m$