Given Data:

• Two identical moving coil galvanometers.
• Galvanometer resistance, $G = 10 \, \Omega$.
• Full-scale deflection current, $I_g = 2 \, \mu A = 2 \times 10^{-6} \, A$.
• One is converted to a voltmeter with a full-scale reading, $V_{max} = 100 \, mV = 0.1 \, V$.
• The other is converted to an ammeter with a full-scale reading, $I_{max} = 1 \, mA = 1 \times 10^{-3} \, A$.
• These are used in an Ohm's law experiment with a resistor $R = 1000 \, \Omega$ and an ideal cell.

We will evaluate each statement step-by-step.

Step 1: Analyze Option A (Resistance of the voltmeter)

To convert a galvanometer into a voltmeter, a high resistance ($R_s$) is connected in series with it. The total resistance of the voltmeter is $R_V = G + R_s$. At full-scale deflection, the voltage across the voltmeter is $V_{max}$. According to Ohm's law for the voltmeter: $V_{max} = I_g \times R_V$ $R_V = \frac{V_{max}}{I_g}$ Substituting the given values: $R_V = \frac{0.1 \, V}{2 \times 10^{-6} \, A} = 0.05 \times 10^6 \, \Omega = 50 \times 10^3 \, \Omega = 50 \, k\Omega$ The resistance of the voltmeter is $50 \, k\Omega$. Option A states the resistance is $100 \, k\Omega$. Therefore, statement A is incorrect.

Step 2: Analyze Option B (Resistance of the ammeter)

To convert a galvanometer into an ammeter, a low resistance shunt ($S$) is connected in parallel with it. The total resistance of the ammeter is $R_A$. At full-scale deflection, the total current is $I_{max}$. The current through the galvanometer is $I_g$, and the current through the shunt is $I_s = I_{max} - I_g$. Since the galvanometer and shunt are in parallel, the voltage across them is the same: $V_g = V_s$ $I_g G = (I_{max} - I_g) S$ We can calculate the shunt resistance $S$: $S = \frac{I_g G}{I_{max} - I_g} = \frac{(2 \times 10^{-6} \, A) \times (10 \, \Omega)}{1 \times 10^{-3} \, A - 2 \times 10^{-6} \, A} = \frac{20 \times 10^{-6}}{998 \times 10^{-6}} \, \Omega = \frac{10}{499} \, \Omega$ The total resistance of the ammeter ($R_A$) is the equivalent resistance of $G$ and $S$ in parallel: $\frac{1}{R_A} = \frac{1}{G} + \frac{1}{S} = \frac{1}{10} + \frac{1}{10/499} = \frac{1}{10} + \frac{499}{10} = \frac{500}{10} = 50 \, \Omega^{-1}$ $R_A = \frac{1}{50} \, \Omega = 0.02 \, \Omega$ Alternatively, the voltage across the ammeter at full scale is $V_A = I_g G = (2 \times 10^{-6} \, A)(10 \, \Omega) = 2 \times 10^{-5} \, V$. The total current is $I_{max} = 1 \times 10^{-3} \, A$. The resistance of the ammeter is $R_A = V_A / I_{max} = (2 \times 10^{-5} \, V) / (1 \times 10^{-3} \, A) = 0.02 \, \Omega$. This value is exactly 0.02 $\Omega$. Option B is correct.

Step 3: Analyze Option D (Measured value of R)

In the Ohm's law experiment, the voltmeter is connected in parallel with the resistor $R$, and the ammeter is connected in series with this parallel combination.

• The voltmeter measures the voltage across the parallel combination of $R$ and $R_V$. Let this be $V_{measured}$.
• The ammeter measures the total current flowing into this parallel combination. Let this be $I_{measured}$.

The measured resistance, $R_{measured}$, is the ratio of the voltmeter reading to the ammeter reading: $R_{measured} = \frac{V_{measured}}{I_{measured}}$ This ratio is equal to the equivalent resistance of the parallel combination of the resistor $R$ and the voltmeter $R_V$. $R_{measured} = R_{eq,p} = \frac{R \times R_V}{R + R_V}$ Using the values $R = 1000 \, \Omega$ and $R_V = 50000 \, \Omega$: $R_{measured} = \frac{1000 \times 50000}{1000 + 50000} = \frac{50 \times 10^6}{51000} = \frac{50000}{51} \, \Omega$ Calculating the value: $R_{measured} \approx 980.39 \, \Omega$ Option D states that the measured value of R will be $978 \, \Omega < R < 982 \, \Omega$. Since $978 < 980.39 < 982$, statement D is correct.

Step 4: Analyze Option C (Effect of cell's internal resistance)

If the ideal cell is replaced by a cell with internal resistance $r = 5 \, \Omega$, the total resistance of the circuit increases. This will decrease the current drawn from the cell and the voltage across the resistor-voltmeter combination. However, the measured resistance is the ratio of the voltmeter and ammeter readings: $R_{measured} = \frac{V_{measured}}{I_{measured}} = R_{eq,p}$ This value depends only on the resistances of the components in the measurement part of the circuit ($R$ and $R_V$), not on the source (EMF or internal resistance). Therefore, the measured value of R remains unchanged. $R_{measured} \approx 980.39 \, \Omega$ This value is less than the true resistance $R = 1000 \, \Omega$. Option C states that the measured value will be more than 1000 $\Omega$. Therefore, statement C is incorrect.

Conclusion

• Statement A is incorrect.
• Statement B is correct.
• Statement C is incorrect.
• Statement D is correct.

The correct options are B and D.