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
The balance condition is given by,
$\frac{R_1}{R_2}=\frac{l_1}{l_2}$
where,
$R_1$ = Resistance in the first gap.
$R_2$ = Resistance in the second gap.
$l_1$ = Distance from the left end of the meter bridge
$l_2$ = Distance from the right end of the meter bridge
Case 1: (10 Ω)
$⇒\frac{R_1+10}{R_2}=\frac{50}{50}=1$
$⇒R_1+10=R_2$ ...(1)
Case 2: (without 10 Ω)
$\frac{R_1}{R_2}=\frac{2}{3}$ ...(2)
Using (1) and (2),
$\frac{R_1+10}{R_2}=\frac{2}{3}$
$⇒R_1=20 Ω$