Answer:
Step-by-step explanation:
The power rating of the battery isn't provided. But let us assume that it is one of the common batteries with ratings of 12 V and 50 A.h
Potential energy possessed by water at that height = mgh
m = mass of the water = ρV
ρ = density of water = 1000 kg/m³
V = volume of water = ?
g = acceleration due to gravity = 9.8 m/s²
h = height of water = 50 cm = 0.5 m
Potential energy = ρVgh = 1000 × V × 9.8 × 0.5 = (4900V) J
Energy of the battery = qV
q = 50 A.h = 50 × 3600 = 180,000 C
V = 12 V
qV = 180,000 × 12 = 2,160,000 J
Energy = 2,160,000 J
At a 100% conversion rate, the energy of the water totally powers the battery
(4900V) = (2,160,000)
4900V = 2,160,000
V = (2,160,000/4900)
V = 440.82 m³
Hence, with our assumed power ratings for the battery (12 V and 50 A.h), 440.82 m³ of water at the given height of 50 cm would power the battery.
Incase the power ratings of the battery in the complete question is different, this solution provides you with how to obtain the correct answer, given any battery power rating.
Hope this Helps!!!