A moving-coil galvanometer has a rectangular coil of 60 turns, 2.0 cm wide and 2.5 cm high…

Step 1. Sensitivity (a). $NAB = 60 \times 5.0 \times 10^{-4}\ \text{m}^2 \times 0.25\ \text{T} = 7.50 \times 10^{-3}\ \text{N m/A}$. So $\dfrac{\theta}{I} = \dfrac{NAB}{k} = 1500\ \text{rad/A} = 85.9^\circ$ per mA, which is $28.6$ divisions per mA.

Step 2. Full scale (b). $30$ divisions need $I_g = \dfrac{30}{28.65}\ \text{mA} = 1.047\ \text{mA}$.

Step 3. Shunt (c). At full scale the coil takes $I_g$ and the shunt the rest, at the same voltage: $S = \dfrac{I_gG}{I - I_g} = \dfrac{1.047 \times 10^{-3}\ \text{A} \times 40\ \Omega}{5.0\ \text{A} - 1.047 \times 10^{-3}\ \text{A}} = 8.38\ \text{m}\Omega$.

Step 4. Series resistance (d). The whole voltmeter must have $R_V = \dfrac{V}{I_g} = 14.32\ \text{k}\Omega$, so the series part is $R_V - G = 14.28\ \text{k}\Omega$.

Step 5. The reading (e). The voltmeter and the $10\ \text{k}\Omega$ resistor in parallel make $5.89\ \text{k}\Omega$. In series with the other $10\ \text{k}\Omega$:

Step 6. Twice the turns (f). The current sensitivity $NAB/k$ doubles. The voltage sensitivity is the current sensitivity divided by the resistance, and the resistance doubles too. So the factor is $1.00$: no gain at all.