Two resistances are connected in two gaps of a metre bridge. The balance point is 20 cm from the zero end. A resistance of $15 \Omega$ is connected in series with the smaller resistance, then the null point shifts to 40 cm. The value of the smaller resistance is:

$9 \Omega$

The correct answer is Option (3) → $9 \Omega$

Given,

Initial balance at $l_1=20cm$ ($R_1$ is connected)

After connecting 40 Ω resistance, balance shifts to $l_2=40 cm$

let,

$R_1$ be the smaller resistance

$R_2$ be the larger resistance in second gap

$∴\frac{R_1}{R_2}=\frac{l}{100-l}$

$⇒\frac{R_1}{R_2}=\frac{20}{100-20}⇒R_1=\frac{R_2}{4}$

Now, when a 15Ω resistor is connected in series -

$R_1'=R_1+15$

$∴\frac{R_1'}{R_2}=\frac{40}{100-40}=\frac{2}{3}$

$∴\frac{R_1+15}{R_2}=\frac{2}{3}$

$⇒\frac{R_2+60}{4R_2}=\frac{2}{3}$

$⇒R_2=36Ω$

$∴R_1=\frac{R_2}{4}=\frac{36}{4}$

$R_1=9Ω$