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 $R_1$?

20 Ω

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),

$\frac{R_1+10}{R_2}=\frac{2}{3}$

$⇒R_1=20 Ω$