Case 1 : In the first case, it is given that, when the key \(K_{1}\) is closed but \(K_{2}\) is open then deflection in the galvanometer equals \(\theta_{0}\).
Let the current flowing in the galvanometer is \(i_{g}\) and the deflection in the galvanometer is given by \(\theta_{0}\). We know that the current in the galvanometer is proportional to the deflection in the galvanometer.
Thus, \(i_{g} \propto \theta_{0}\)
After removing proportional sign we get a constant C,
\(i_{g}=C \theta_{0}\)......(1)
Let us assume the emf of the battery is \(E\).
We know that the current flowing in the circuit is equal to the ratio of emf of the battery to the total resistance in the circuit.
So, \(i_{g}=\frac{E}{220+R_{g}}\)......(2)
From equation (1) and (2), we get
\(\frac{E}{220+R_{g}}=C \theta_{0}\)......(3)
Now, according to the second condition when \(K_{2}\) is closed and adjusting \(R_{2}\) to \(5 \Omega\), the deflection in the galvanometer becomes \(\frac{\theta_{0}}{5}\).
We know that the current in the galvanometer is proportional to the deflection in the galvanometer. Here deflection is\(\frac{\theta_{0}}{5} \)
\(i_{g}=C \frac{\theta_{0}}{5}\)........(4)
We know that the current flowing in the circuit is equal to the ratio of emf of the battery to the total resistance in the circuit.
\(i_{g}=\left(\frac{E}{220+\frac{5 R_{g}}{5+R_{g}}}\right) \times\left(\frac{5}{R_{g}+5}\right)\)...........(5)
From equation (4) and (5), we get
\(\left(\frac{E}{220+\frac{5 R_{g}}{5+R_{g}}}\right) \times\left(\frac{5}{R_{g}+5}\right)=C \frac{\theta_{0}}{5}\)
On further solving this, we get
\(\Rightarrow \frac{5 E}{225 R_{g}+1100}=\frac{C \theta_{0}}{5}\).......(6)
Now finally solving equation (3) and (6) from case 1 and 2 respectively, we get,
\(\Rightarrow \frac{225 R_{g}+1100}{1100+5 R_{g}}=5\)
On further solving, we get
\(\Rightarrow 5500+25 R_{g}=225 R_{g}+1100\)
\(\Rightarrow 200 R_{g}=4400\)
On finally solving this, we get
\(R_{g}=22 \Omega\)
Thus, the resistance of the galvanometer is \(22 \Omega\).