In the given setup of meter bridge, the null point D is obtained at a distance of 40 cm from end A. If a resistance of 10Ω is connected in series with $R_1$, the null points shift by 20 cm. The values of $R_1$ and $R_2$ respectively are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8Ω and 12Ω
In a meter bridge, condition of balance:
$\frac{R_1}{R_2}=\frac{l}{100-l}$
Case 1: Null point at $l=40\,\text{cm}$
$\frac{R_1}{R_2}=\frac{40}{60}=\frac{2}{3}$ … (1)
Case 2: A $10\ \Omega$ is added in series with $R_1$, null point shifts by $20\,\text{cm}$ ⇒ new $l=60\,\text{cm}$
$\frac{R_1+10}{R_2}=\frac{60}{40}=\frac{3}{2}$ … (2)
From (1): $R_1=\frac{2}{3}R_2$
Substitute in (2): $\frac{\frac{2}{3}R_2+10}{R_2}=\frac{3}{2}$
$\frac{2}{3}+\frac{10}{R_2}=\frac{3}{2}$
$\frac{10}{R_2}=\frac{3}{2}-\frac{2}{3}=\frac{9-4}{6}=\frac{5}{6}$
$R_2=\frac{10 \cdot 6}{5}=12\ \Omega$
Then $R_1=\frac{2}{3} \cdot 12=8\ \Omega$
Answer: $R_1=8\ \Omega,\ R_2=12\ \Omega$