Power calculation for pump using gravitational work and efficiency

Steps:

• Mass of water: m = density × volume = 1000 kg/m³ × 0.5 m³ = 500 kg
• Work done against gravity: W = mgh = 500 kg × 9.8 m/s² × 30 m = 147,000 J
• Time interval: t = 1 min = 60 s; hydraulic power P_h = W/t = 147,000 J / 60 s = 2,450 W
• Engine output power: P = P_h / efficiency = 2,450 W / 0.7 = 3,500 W = 3.5 kW

Why B is correct:

• B equals the input power to the 70% efficient pump, derived from P = (mgh/t) / η, where η is pump efficiency.

Why the others are wrong:

• A is the hydraulic power without dividing by efficiency.
• C and D overestimate by factors of 40–60, likely from unit errors like minutes instead of seconds.

Final answer: B