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(C) For a meter bridge,the balancing condition is given by $\frac{X}{Y} = \frac{l}{100-l}$.In the first case,$l = 20 	ext{ cm}$.$\frac{X}{Y} = \frac{

Vedclass · Accepted Answer

$12$

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(C) For a meter bridge,the balancing condition is given by $\frac{X}{Y} = \frac{l}{100-l}$.In the first case,$l = 20 	ext{ cm}$.$\frac{X}{Y} = \frac{20}{100-20} = \frac{20}{80} = \frac{1}{4}$.Therefore,$Y = 4X$.In the second case,the null point shifts to $l' = 40 	ext{ cm}$. Since $X So,$X' = X + 20$ and $Y' = Y = 4X$.The new balancing condition is $\frac{X+20}{4X} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$.Cross-multiplying gives $3(X + 20) = 2(4X)$.$3X + 60 = 8X$.$5X = 60$,which gives $X = 12 	ext{ } \Omega$.Since $Y = 4X = 48 	ext{ } \Omega$,the smaller resistance is $X = 12 	ext{ } \Omega$.

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