A resistance of 15 Ω is connected in series with an unknown resistance R₁. This combination is connected to tests one gap of a meter bridge, while a resistance R₂ is connected in the other gap. The balance point is at 50 cm. When 15 Ω is removed, the balance point shifts by 10 cm. The value of R₁ is

30 Ω

In the first case balance point is 50cm.Hence $\frac{R_1 + 15}{R_2} = 1$

$\Rightarrow R_1 + 15 = R_2$

When $15\Omega$ is removed then balance point will be at 40cm.

$\Rightarrow \frac{R_1}{R_2} =\frac{40}{60} = \frac{2}{3}$

$\Rightarrow R_2 = \frac{3R_1}{2} = R_1 +15$

$\Rightarrow R_1 = 30\Omega$