Water falling from a 50 m high fall is to be used for generating electric energy. If $1.8 \times 10^{\circ}$ kg of water falls per hour and half the gravitational potential energy can be converted into electric energy,how many 100 2 lamps can be lit ?

Open in App

Solution

$h = 50 m, m = 1.8 \times 10^{5} k g,$

$P = 100 w a l t$

$P . E = m g h$

$= 1.8 \times 10^{5} \times 9.8 \times 50$

$= 882 \times 10^{5} J$

Because, half the potential energy is converted into electricital energy,

Electrical energy $= \frac{1}{2} P . E = 441 \times 10^{5} J / h r$

So, power in watt $(J / s e c) = (441 \times 10^{5})$

$∴$ Number of 100 W lamps, that can be lit

$\frac{441 \times 10^{5}}{3600 \times 100} = 122.5 \approx 122$

Suggest Corrections

Similar questions

Water failing from a 50 m high fall is to be used for generating electric energy. If 1.8 $\times$ $10^{5}$ Kg of water falls per hour and half the gravitational potential energy can be converted into electric energy, how many 100 W lamps can be lit?Take g=10 $m / s^{2}$

Q. Question 13
In a hydroelectric power plant more electrical power can be generated if water falls from a greater height because

(a) its temperature increases
(b) larger amount of potential energy is converted into kinetic energy
(c) the electricity content of water increases with height
(d) more water molecules dissociate into ions

Q. A water fall is

126

meters high. If half of the potential energy of the falling water gets converted to heat , the rise in temperature of water will be

Q. Water is falling at the rate of

1000 k g

per second. The height of the fall is about

100 m

. The potential energy of water when it falls on the blades will be conveyed to them and P.E. is converted into K.E. of blades. Estimate the power that could be generated, if energy is not lost otherwise (Take the value of '

g

' as

10 m s^{- 2}

In a hydroelectric power plant, water falls at a rate of $1000 k g s^{- 1}$ from a height of $100 m$ . Assuming that 50% of the energy of falling water is converted into electrical energy, calculate the power generated. (Take $g = 10 m s^{- 2}$ )

Join BYJU'S Learning Program

Water falling from a 50 m high fall is to be used for generating electric energy. If 1.8×10∘ kg of water falls per hour and half the gravitational potential energy can be converted into electric energy,how many 100 2 lamps can be lit ?

h=50 m,m=1.8×105 kg, P=100 walt P.E=mgh =1.8×105×9.8×50 =882×105J Because, half the potential energy is converted into electricital energy, Electrical energy =12P.E=441×105J/hr So, power in watt (J/sec)=(441×105) ∴ Number of 100 W lamps, that can be lit 441×1053600×100=122.5≈122

Water falling from a 50 m high fall is to be used for generating electric energy. If $1.8 \times 10^{\circ}$ kg of water falls per hour and half the gravitational potential energy can be converted into electric energy,how many 100 2 lamps can be lit ?