Two resistance of $$100\Omega$$ and $$200\Omega$$ are connected in series with a battery of 4 V and negligible internal resistance. A voltmeter is used to measure voltage across $$100\Omega$$ resistance, which gives reading as 1 V. The resistance of voltmeter must be _______ $$\Omega$$.

$$100\Omega$$ and $$200\Omega$$ in series, battery 4V. Voltmeter across $$100\Omega$$ reads 1V.

The voltmeter (resistance $$R_V$$) is in parallel with $$100\Omega$$. Equivalent = $$\frac{100 R_V}{100 + R_V}$$.

Total resistance = $$\frac{100R_V}{100+R_V} + 200$$.

Current = $$\frac{4}{\frac{100R_V}{100+R_V} + 200}$$.

Voltage across parallel combination = 1V.

So voltage across $$200\Omega$$ = 3V. Current = $$\frac{3}{200} = 0.015$$ A.

For the parallel combination: $$V = IR_{eq}$$: $$1 = 0.015 \times \frac{100R_V}{100+R_V}$$

$$\frac{100R_V}{100+R_V} = \frac{1}{0.015} = \frac{200}{3}$$

$$300R_V = 200(100 + R_V) = 20000 + 200R_V$$

$$100R_V = 20000 \Rightarrow R_V = 200\Omega$$

Therefore, the answer is $$\boxed{200}$$.