Question

Two unknown resistances are connected in two gaps of a meter-bridge. The null point is obtained at $40cm$ from left end. A $30\Omega$ resistance is connected in series with the smaller of the two resistances, the null point shifts by $20cm$ to the right end. The value of smaller resistance in $\Omega$ is

• A. $12$
• B. $24$
• C. $36$
• D. $48$

Answer 1

The correct option is B $24$

let the reistance at right be $R$ and at left be $r$

$\frac{r}{R}=\frac{l_x}{l_R}=\frac{40}{60}$ , $r$ is smaller

$\frac{r+30}{R}=\frac{60}{40}$

From above equation we get $r=24\Omega$