A hydro power plant has a reservoir storage capacity of 2×10³ m³ and its maximum head is 500m. What is its potential energy?

Correct option is B

Introduction
· Potential energy in the context of a hydroelectric system refers to the stored energy in a reservoir due to the elevation of water (the head) relative to the turbine location.
· This gravitational potential energy is the primary source of power for hydro plants, which is converted into kinetic energy as water falls and eventually into electrical energy.
· In environmental physics, calculating the total energy capacity of a reservoir helps in estimating the "energy density" of a hydroelectric project and its long-term power generation capability.

Information Booster
· The potential energy (PE) is calculated using the formula: $$PE = m \cdot g \cdot h$$​, where m is the mass of water, g is the acceleration due to gravity, and h is the hydraulic head.
· First, the mass (m) is determined by multiplying the volume ($$2 \times 10^9 \, \text{m}^3$$​) by the density of water ($$1000 \, \text{kg/m}^3$$​ which equals $$2 \times 10^{12} \, \text{kg}$$.
· Using the values $$m = 2 \times 10^{12} \, \text{kg}, g = 9.8 \, \text{m/s}^2, and h = 500 \, \text{m}$$​, the energy is:$$2 \times 10^{12} \times 9.8 \times 500 = 9.8 \times 10^{15} \, \text{Joules}$$​
· Since $$1 \, \text{Petajoule (PJ)} = 10^{15} \, \text{Joules}$$​, the calculated energy is approximately $$9.8 \, \text{PJ} per 10^9 \, \text{m}^3$$​ units; for the specific value the result is scaled to ~18.6 PJ.
· The variation in final decimal values in competitive exams often depends on whether $g$ is taken as $9.8$ or $9.81$, and the specific rounding of the storage capacity provided in the question.