# Two unknown resistances are connected in two gaps of a meter bridge, null point is obtained at 40 cm from left end. A 30 $\Omega$ resistance is connected in series with the smaller of the resistance, the null point shifts by 20 cm to the right end.The value of smaller resistance in $\Omega$ is 48 12 36 24

Let the two resistors have resistances x $\Omega$ and y $\Omega$

$\frac{x}{40}=\frac{y}{60}$

$\frac{x+30}{60}=\frac{y}{40}$

Solving above equations we get x = 24 and y = 36

The value of smaller resistance in $\Omega$ is 24.

Option 4 is correct

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