Explain: "Increasing the current sensitivity may not necessarily increase the voltage sensitivity".

(N/A) Current sensitivity of a galvanometer is defined as the deflection per unit current, given by:
$\frac{\phi}{I} = \frac{NAB}{k} \quad \dots (1)$
where $N$ is the number of turns, $A$ is the area, $B$ is the magnetic field, and $k$ is the torsional constant.
If we double the number of turns $(N \rightarrow 2N)$, the current sensitivity becomes:
$\left(\frac{\phi}{I}\right)' = \frac{(2N)AB}{k} = 2 \left(\frac{\phi}{I}\right)$
Thus, the current sensitivity doubles.
However, the resistance $R$ of the galvanometer coil is proportional to the length of the wire. Since the length of the wire is proportional to the number of turns, doubling the turns also doubles the resistance $(R \rightarrow 2R)$.
Voltage sensitivity is defined as the deflection per unit voltage, given by:
$\frac{\phi}{V} = \frac{\phi}{IR} = \left(\frac{NAB}{k}\right) \frac{1}{R}$
If $N$ is doubled, $R$ also doubles. Substituting these into the voltage sensitivity formula:
$\left(\frac{\phi}{V}\right)' = \frac{(2N)AB}{k(2R)} = \frac{NAB}{kR} = \frac{\phi}{V}$
Therefore, the voltage sensitivity remains unchanged. This proves that increasing current sensitivity does not necessarily increase voltage sensitivity.