In a meter bridge, the null point is found at a distance of 60 cm from end A. If a 5 Ω resistor is now connected in series with S, the null point occurs at 50 cm. The values of R and S are, respectively

15 Ω and 10 Ω

The correct answer is Option (1) → 15 Ω and 10 Ω

In a meter bridge, balance condition is:

$\frac{R}{S} = \frac{l}{100-l}$

Case 1: Null point at $l = 60 \, \text{cm}$

$\frac{R}{S} = \frac{60}{40} = \frac{3}{2}$

⟹ $R = \frac{3}{2} S$ …(1)

Case 2: A $5 \, \Omega$ resistor is connected in series with $S$. Effective resistance = $(S+5)$

Null point at $l = 50 \, \text{cm}$

$\frac{R}{S+5} = \frac{50}{50} = 1$

⟹ $R = S+5$ …(2)

From (1) and (2):

$\frac{3}{2} S = S+5$

$\frac{3S}{2} - S = 5$

$\frac{S}{2} = 5$

$S = 10 \, \Omega$

$R = S+5 = 15 \, \Omega$

Answer: $R = 15 \, \Omega, \; S = 10 \, \Omega$