In meter bridge experiment, two resistances X and Y in the two gaps give a null point dividing the wire in the ratio 2 : 3 . When each resistance is increased by 30 Ω , the null point divides the wire

Question

In meter bridge experiment, two resistances X and Y in the two gaps give a null point dividing the wire in the ratio 2 : 3 . When each resistance is increased by 30 Ω , the null point divides the wire in the ratio 5 : 6 . The resistance X and Y are respectively

TestHub · Accepted Answer

For a meter bridge, the ratio of resistances in the two gaps is equal to the ratio of the balancing lengths.

Initially, the null point divides the wire in the ratio $2 : 3$ .

$\frac{X}{Y} = \frac{2}{3}$

$\Rightarrow Y = \frac{3}{2} X$

When each resistance is increased by $30 Ω$ , the new ratio becomes $5 : 6$ .

$\frac{X + 30}{Y + 30} = \frac{5}{6}$

$\Rightarrow 6 X + 180 = 5 Y + 150$

Substituting $Y = \frac{3}{2} X$ into the equation:

$6 X + 180 = 5 (\frac{3}{2} X) + 150$

$6 X - \frac{15}{2} X = 150 - 180$

$- \frac{3}{2} X = - 30$

$X = 20 Ω$

Substituting the value of $X$ to find $Y$ :

$Y = \frac{3}{2} (20) = 30 Ω$

Answer: $20 Ω$ , $30 Ω$

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