Q.40. Two resistance X and Y are connected in the two gaps of a meterbridge and the null points is obtained at 20 cm from zero end. When the resistance of 20Ω is connected in series with the smaller of the two resistance X and Y, the null point shifts to 40 cm from left end. The value of smaller resistance in ohm is
A. 6
B. 9
C. 12
D. 15
Answer :- C. 12
Explanation :-
For a meterbridge, \frac{\mathrm{X}}{\mathrm{Y}}=\frac{l}{100-l}
In the first case, l=20 \mathrm{~cm})
\frac{X}{Y} =\frac{20}{100-20}=\frac{20}{80}=\frac{1}{4} \therefore \quad 4 X =YIn the second case, Y(l^{\prime}=40 \mathrm{~cm})
\therefore \frac{X^{\prime}}{Y^{\prime}}=\frac{40}{100-40}=\frac{40}{60}=\frac{2}{3} \text { But, } X^{\prime}=X+20 \text { and } Y^{\prime}=Y \therefore \frac{X+20}{Y}=\frac{2}{3} \therefore \frac{X+20}{4 X}=\frac{2}{3} \therefore 8 X=3 X+60 \therefore X=12 \Omega