Question

In a hydroelectric power station water is allowed to fall at the rate of $2000$ kg per second on a turbine kept $100$ m below the water level. Find kinetic energy of this water when it falls through a height of $25$ m.

• A. $0.5\times {10}^{7}J$
• B. $0.5\times {10}^{8}J$
• C. $0.5\times {10}^6{J}^3$
• D. $0.5\times {10}^{6}J$

Answer 1

The correct option is D $0.5\times {10}^{6}J$

(ii) When water falls through a height of 25 m its potential energy decreases but the kinetic energy (K.E.) increases.
Total energy= P.E + K.E
$\therefore P_1=P.Eat{h}_1={mgh}_1=2000\times 10\times 75=150\times {10}^4=15\times {10}^{5}J=1.5\times {10}^{6}J$
From law of conservation of energy, total energy is constant.
$\therefore E_1=P_1+K.E{E}_1=2\times {10}^{6}J{P}_1=1.5\times {10}^{6}J2\times {10}^6=1.5\times {10}^6+K.EK.E=2\times {10}^6-1.5\times {10}^6=0.5\times {10}^{6}J$