1. Set up the relation between deflection and current

For a galvanometer, $\theta \propto I$ where $\theta$ is deflection in divisions and $I$ is current through the galvanometer.

Since the battery has no internal resistance, the circuit current is $I = \frac{E}{R+G}$ where:

• $E$ = battery emf,
• $R$ = resistance from resistance box,
• $G$ = galvanometer resistance.

Thus, $\theta = k\,\frac{E}{R+G}$ for some proportionality constant $k$.

2. Use the two given observations

When $R=2400\,\Omega$, deflection is $40$ divisions: $40 = k\frac{E}{2400+G}$

When $R=4900\,\Omega$, deflection is $20$ divisions: $20 = k\frac{E}{4900+G}$

Divide the first equation by the second: $\frac{40}{20} = \frac{4900+G}{2400+G}$ $2 = \frac{4900+G}{2400+G}$

So, $2(2400+G)=4900+G$ $4800+2G=4900+G$ $G=100\,\Omega$

Therefore, the galvanometer resistance is $\boxed{100\,\Omega}$

So Option A is false.

3. Find current sensitivity

Since deflection is proportional to current, let current for 40 divisions be $I_1$ and for 20 divisions be $I_2$.

Then, $I_1 = \frac{E}{2400+100} = \frac{E}{2500}$ $I_2 = \frac{E}{4900+100} = \frac{E}{5000}$

As expected, $I_1 = 2I_2$ corresponding to $40$ and $20$ divisions.

Let current per division be $x$. Then current for 40 divisions is $I_1 = 40x$

Current for 20 divisions is $I_2 = 20x$

Using the resistance values found: $\frac{I_1}{I_2} = \frac{E/2500}{E/5000}=2$ which is consistent.

Now check the possibility of current sensitivity from the options.

If current sensitivity is $20\,\mu A$/division, then full-scale current for 50 divisions is $I_g = 50\times 20\,\mu A = 1000\,\mu A = 1\,mA$

Then current for 40 divisions would be $40\times 20\,\mu A = 800\,\mu A$

This gives emf $E = I(R+G)=800\times 10^{-6}\times 2500=2\,V$

Check second case: $I=\frac{2}{5000}=400\,\mu A$ which corresponds to $\frac{400\,\mu A}{20\,\mu A/\text{div}}=20\text{ divisions}$

So Option C is perfectly consistent.

Thus current sensitivity is $\boxed{20\,\mu A/\text{division}}$

4. Check Option B

Full-scale current for 50 divisions using Option C is $I_g = 50\times 20\,\mu A = 1000\,\mu A = 1\,mA$

So full-scale deflection current is not $2\,mA$.

Hence Option B is false.

5. Check Option D

For 10 divisions, current needed is $I_{10} = 10\times 20\,\mu A = 200\,\mu A$

Using $E=2\,V$ and $G=100\,\Omega$, $I = \frac{E}{R+G}$ $200\times 10^{-6} = \frac{2}{R+100}$ $R+100 = \frac{2}{200\times 10^{-6}} = 10000$ $R = 9900\,\Omega$

So required resistance is $\boxed{9900\,\Omega}$ not $9800\,\Omega$.

Hence Option D is false.

6. Final conclusion

Only Option C is correct.

$\boxed{\text{C: Current sensitivity of galvanometer is }20\,\mu A/\text{division}}$