Had the voltmeter been an ideal one, what would have been its reading?

7, 2 V

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1.8 V

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0.5 V

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3.24 V

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Solution

The correct option is D $3.24 V$
Let $R_{1}$ and $R_{2}$ be the resistances of the ammeter and the voltmeter, respectively.
Let the external resistance be denoted by $R$ and the internal resistance of battery be $r$ .
The equivalent resistance of the parallel combination of $R$ and $R_{2}$ is given by: $R^{'} = \frac{R R_{2}}{R + R_{2}}$
The total resistance $R_{T}$ of circuit then becomes: $R T = R_{1} + r + \frac{R R_{2}}{R + R_{2}}$
The current in the circuit is given by: $I = \frac{E}{R_{1} + r + \frac{R R_{2}}{R + R_{2}}}$
This must be equal to $0.04 A$ , the reading indicated by the ammeter.
$\frac{3.4}{2 + 3 + \frac{100 R_{2}}{100 + R_{2}}} = 0.04$ $⟹ R_{2} = 400 Ω$
In case of an ideal voltmeter, not current flows through it. In that case, current in the circuit is
$I^{'} = \frac{3.4}{2 + 3 + 100} = \frac{3.4}{105} = 0.0324 A$
Potential drop across the resistance $R$ would be $100 \times 0.0324 = 3.24 V$ .
This should be the reading indicated by an ideal voltmeter.

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