Resistance \( R_1 \) is with 10 Ω resistance in series and is in the first gap of meter bridge with \( R_2 \) in the second gap. The balance point is at 50 cm. When the 10 Ω resistance is removed then the balance point shifts to 40 cm. What is the value of \( R_1 \)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 20 Ω
At the balance point of a meter bridge, the ratio of the resistances in the two gaps equals the ratio of the lengths of wire on the two sides: \( X/R = l/(100-l) \).
First balance point (50 cm), with the 10 Ω in series
\( \displaystyle \frac{R_1 + 10}{R_2} = \frac{50}{50} = 1 \Rightarrow R_2 = R_1 + 10 \)
Second balance point (40 cm), with the 10 Ω removed
\( \displaystyle \frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \)
Putting \( R_2 = R_1 + 10 \):
\( \displaystyle \frac{R_1}{R_1 + 10} = \frac{2}{3} \Rightarrow 3R_1 = 2R_1 + 20 \)
\( R_1 = 20\,\Omega \)