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:
Answer & explanation
Correct answer: option 3
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 -
$R_1'=R_1+15$
$∴\frac{R_1'}{R_2}=\frac{40}{100-40}=\frac{2}{3}$
$∴\frac{R_1+15}{R_2}=\frac{2}{3}$
$⇒\frac{R_2+60}{4R_2}=\frac{2}{3}$
$⇒R_2=36Ω$
$∴R_1=\frac{R_2}{4}=\frac{36}{4}$
$R_1=9Ω$