Analysis of Galvanometer Conversion to Voltmeter and Ammeter in Ohms Law

To determine the correct statement(s), let us analyze each device and its behavior in the Ohm's law experiment.

1. Galvanometer Parameters:

• Resistance of the galvanometer, $G = 10\ \Omega$
• Full scale deflection current, $I_g = 2\ \mu\text{A} = 2 \times 10^{-6}\ \text{A}$

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2. Analysis of Option A (Voltmeter Resistance):

A galvanometer is converted into a voltmeter of full-scale voltage $V = 100\text{ mV} = 0.1\text{ V}$ by connecting a high resistance $R_s$ in series with it. The total resistance of the voltmeter $R_V$ is: $R_V = \frac{V}{I_g} = \frac{100 \times 10^{-3}\text{ V}}{2 \times 10^{-6}\text{ A}} = 50000\ \Omega = 50\text{ k}\Omega$

Since $R_V = 50\text{ k}\Omega \neq 100\text{ k}\Omega$, Option A is incorrect.

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3. Analysis of Option B (Ammeter Resistance):

A galvanometer is converted into an ammeter of full-scale current $I = 1\text{ mA} = 10^{-3}\text{ A}$ by connecting a shunt resistance $S$ in parallel with the galvanometer.

The potential difference across the ammeter at full-scale deflection is: $V_A = I_g G = 2 \times 10^{-6}\text{ A} \times 10\ \Omega = 2 \times 10^{-5}\text{ V}$

The total effective resistance of the ammeter $R_A$ is: $R_A = \frac{V_A}{I} = \frac{2 \times 10^{-5}\text{ V}}{10^{-3}\text{ A}} = 2 \times 10^{-2}\ \Omega = 0.02\ \Omega$

Rounded off to two decimal places, $R_A = 0.02\ \Omega$. Thus, Option B is correct.

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4. Analysis of Option C (Measured Value of $R$):

In a standard Ohm's law experiment circuit:

• The voltmeter is connected in parallel with the unknown resistor $R = 1000\ \Omega$.
• The ammeter is connected in series with the parallel combination of $R$ and $R_V$.

The voltage measured by the voltmeter is $V_{\text{meas}} = V_{\text{parallel}}$, and the current measured by the ammeter is $I_{\text{meas}} = I_{\text{total}}$.

The measured value of resistance $R_{\text{meas}}$ is given by: $R_{\text{meas}} = \frac{V_{\text{meas}}}{I_{\text{meas}}} = R_{\text{eq, parallel}} = \frac{R \cdot R_V}{R + R_V}$

Substituting $R = 1000\ \Omega$ and $R_V = 50000\ \Omega$: $R_{\text{meas}} = \frac{1000 \times 50000}{1000 + 50000} = \frac{50000}{51} \approx 980.39\ \Omega$

Since $978\ \Omega < 980.39\ \Omega < 982\ \Omega$, Option C is correct.

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5. Analysis of Option D (Effect of Cell's Internal Resistance):

If the ideal cell is substituted with a cell having internal resistance $r = 5\ \Omega$: The total circuit resistance changes, which in turn changes both $V_{\text{meas}}$ and $I_{\text{meas}}$ proportionally. However, the ratio $\frac{V_{\text{meas}}}{I_{\text{meas}}}$ depends solely on the parallel combination of $R$ and $R_V$: $R_{\text{meas}} = \frac{R \cdot R_V}{R + R_V} \approx 980.39\ \Omega$

The measured value remains $\approx 980.39\ \Omega$, which does not exceed $1000\ \Omega$. Thus, Option D is incorrect.

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Conclusion:

The correct statements are B and C.