Two resistances of 100 Ω and 200 Ω are connected in series across a 20 V battery as shown in the figure below. The reading in a 200 Ω voltmeter connected across the 200 Ω resistance is ________

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The correct option is 20/3 V.

To understand the solution, let us analyze the circuit components shown in the diagram:
1. A battery of voltage $V=20\text{V}$.
2. Two resistors connected in series: $R_1=100\text{Ω}$ and $R_2=200\text{Ω}$.
3. A voltmeter with internal resistance $R_V=200\text{Ω}$.

1. Analysis of the Unloaded Circuit (Voltage across the 100 Ω resistor)
In a simple series circuit without the voltmeter's loading effect, the voltage is divided across the two resistors. The voltage drop across the $100\text{Ω}$ resistor ($R_1$) is given by the voltage division formula:
$V_{100}=V\times \frac{R_1}{R_1+R_2}$
Substituting the given values:
$V_{100}=20\times \frac{100}{100+200}=20\times \frac{100}{300}=\frac{20}{3}\text{V}$
Thus, the voltage drop across the $100\text{Ω}$ resistance is exactly $\frac{20}{3}\text{V}$.

2. Analysis of the Loaded Circuit (Voltmeter connected across the 200 Ω resistor)
When the voltmeter of resistance $R_V=200\text{Ω}$ is connected across the $200\text{Ω}$ resistor:
The equivalent resistance $R_p$ of this parallel combination is:
$R_p=\frac{R_2\times R_V}{R_2+R_V}=\frac{200\times 200}{200+200}=100\text{Ω}$
The total resistance of the entire circuit $R_{\text{total}}$ becomes:
$R_{\text{total}}=R_1+R_p=100+100=200\text{Ω}$
The total current $I$ flowing through the circuit is:
$I=\frac{V}{R_{\text{total}}}=\frac{20}{200}=0.1\text{A}$
The reading of the voltmeter (which measures the voltage drop across the parallel combination $R_p$) is:
$V_{\text{reading}}=I\times R_p=0.1\times 100=10\text{V}$
In this state, the remaining voltage drop across the $100\text{Ω}$ resistor is also $10\text{V}$.