In an experiment with Meter Bridge, with unknown resistance R in the left gap, the null point is obtained at 40 cm from the left end. Now a resistance of 10 Ω is connected in series with resistance R, the null point is obtained at 70 cm from the left end. The value of resistance R is

4 Ω

The correct answer is Option (1) → 4 Ω

Given:

First balance point at 40 cm → $\frac{R}{S} = \frac{40}{60} = \frac{2}{3}$

Hence, $R = \frac{2}{3}S$

When 10 Ω is connected in series with R, new balance point at 70 cm →

$\frac{R + 10}{S} = \frac{70}{30} = \frac{7}{3}$

Substitute $R = \frac{2}{3}S$ into the second equation:

$\frac{\frac{2}{3}S + 10}{S} = \frac{7}{3}$

$\frac{2}{3} + \frac{10}{S} = \frac{7}{3}$

$\frac{10}{S} = \frac{7}{3} - \frac{2}{3} = \frac{5}{3}$

$S = \frac{10 \times 3}{5} = 6 \, \Omega$

Now $R = \frac{2}{3}S = \frac{2}{3} \times 6 = 4 \, \Omega$

Final Answer:

$R = 4 \, \Omega$