nificant current from the circuit being measured, which can alter the voltage being measured. Therefore, out of the options given, we can immediately disregard the options mentioning a resistance in parallel with the galvanometer or resistances that are too low to make a sensible voltmeter.

The current sensitivity of the galvanometer is $$10 \mu \mathrm{A} / \text{div}$$, and it has 20 divisions. This means that the full-scale deflection current of the galvanometer is:

$20 \times 10 \mu \mathrm{A} = 200 \mu \mathrm{A} = 200 \times 10^{-6} \mathrm{A}$

To read up to $$1 \mathrm{V}$$, we require this full-scale current of $$200 \mu \mathrm{A}$$ to flow when a potential difference of $$1 \mathrm{V}$$ is applied across the modified galvanometer (now a voltmeter).

The internal resistance of the galvanometer, $R_g$, is $$100 \Omega$$. The voltage $V$ across the galvanometer and the series resistance $R_s$ together is given by Ohm's law, $V = I_t \times (R_g + R_s)$, where $I_t$ is the total current for full-scale deflection, which is $$200 \mu \mathrm{A} = 200 \times 10^{-6} \mathrm{A}\). Set this equal to 1 V to find $R_s$.

$1 = (200 \times 10^{-6}) \times (100 + R_s)$

$1 = (200 \times 10^{-6} \times 100) + (200 \times 10^{-6} \times R_s)$

$1 = 0.02 + (200 \times 10^{-6} \times R_s)$

Subtracting 0.02 from both sides yields:

$0.98 = 200 \times 10^{-6} \times R_s$

Solving for $R_s$ gives:

$R_s = \frac{0.98}{200 \times 10^{-6}} = \frac{0.98}{0.2 \times 10^{-3}} = 4900 \Omega $

Hence, the correct answer is a $$4900 \Omega$$ resistor in series with the galvanometer to convert it into a voltmeter that can read up to $$1 \mathrm{V}$$. This matches Option D: $4900 \Omega$ in series with the galvanometer.

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