Vertical drainage: Relies on deep tubewells or pumping wells to extract excess groundwater from deep semi-confined or unconfined aquifers and lower waterlogged water tables -> (IV).
Stuffing box: A mechanical component on the shaft of a centrifugal pump containing packing rings to prevent liquid leakage from pump casing -> (III).
Diffuser: A stationary casing ring with expanding guide vanes fitted in deep-well turbine pumps to efficiently convert the kinetic energy of water leaving the impeller into pressure head -> (II).
Groundwater recharge: The natural or artificial replenishment of groundwater through deep percolation of surface water -> (I).

Step-by-step hydraulic calculation:
Discharge $Q = 450\text{ L/min} = \frac{450}{60} = 7.5\text{ L/s}$
Total head $H = 50\text{ m}$
Pump efficiency $\eta_p = 65\% = 0.65$
Metric Water Horse Power (WHP):

$$\text{WHP} = \frac{Q (\text{L/s}) \times H (\text{m})}{75} = \frac{7.5 \times 50}{75} = 5.0\text{ hp}$$

Shaft Horse Power (SHP) required:

$$\text{SHP} = \frac{\text{WHP}}{\eta_p} = \frac{5.0}{0.65} = 7.692\text{ hp}$$

Selecting standard commercial electric motor sizing among the given options gives 7.50 hp (Option 2).

Mechanical Power and Efficiency Principles in Irrigation Pumps:

• Statement (A) is True: Pump efficiency is defined as $\eta_p = \frac{\text{Water Horsepower (WHP)}}{\text{Brake/Shaft Horsepower (BHP)}} \times 100$.
• Statement (B) is True: By the Second Law of Thermodynamics, real fluid machines cannot achieve $100\%$ efficiency; hence no pump converts all mechanical input into hydraulic power.
• Statement (C) is True: Mechanical energy losses inevitably arise from bearing and packing friction, disk friction on the impeller, shock/turbulence at blade entries, and leakage recirculation.
• Statement (D) is True: Because of these mechanical and hydraulic friction losses, input power (BHP) must always exceed useful hydraulic output (WHP), ensuring $\eta_p \lt 1.0$.

Condition for Maximum Efficiency in a DC Shunt Machine:

• Power Loss Balance: In a DC shunt generator or motor, total losses comprise constant iron, mechanical, and shunt field losses ($P_i$ or $P_c$) and variable armature copper losses ($I_a^2 R_a$).
• Maximum Efficiency Condition: Efficiency is maximized when the variable loss equals the constant loss:
  $$I_a^2 R_a = P_i$$
• Solving for Armature Current ($I_a$):
  $$I_a^2 = \frac{P_i}{R_a} \implies I_a = \sqrt{\frac{P_i}{R_a}}$$
  where $P_i$ is constant loss and $R_a$ is armature circuit resistance.