To convert a galvanometer into a voltmeter, a high resistance needs to be added in series with the galvanometer. This is because a voltmeter must have a high resistance to ensure that it does not draw significant 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
10µA∕ div, and it has 20 divisions. This means that the full-scale deflection current of the galvanometer is:
20×10µA=200µA=200×10−6ATo read up to
1V, we require this full-scale current of
200µA to flow when a potential difference of
1V is applied across the modified galvanometer (now a voltmeter).
The internal resistance of the galvanometer,
Rg, is
100Ω. The voltage
V across the galvanometer and the series resistance
Rs together is given by ohm's law,
V=It×(Rg+Rs), where
It is the total current for fullscale deflection, which is
$$200\mu\ mathrm
{A}=200 \imes
10∧{−6}\ mathrm
{A}\). Set this equal to
1V to find
‌Rs‌1=(200×10−6)×(100+Rs)‌1=(200×10−6×100)+(200×10−6×Rs)‌1=0.02+(200×10−6×Rs) Subtracting 0.02 from both sides yields:
0.98=200×10−6×RsSolving for
Rs gives:
Rs=‌=‌=4900ΩHence, the correct answer is a
4900Ω resistor in series with the galvanometer to convert it into a voltmeter that can read up to
1V. This matches Option
D:4900Ω in series with the galvanometer.