Home
Class 11
PHYSICS
Water falling from a 50 m high fall is t...

Water falling from a 50 m high fall is to be used for generating electric enegy. If `1.8xx10^5kg` of water falls per hour and half the gravitational potential energyh can be converted into electric energy, how many 100 W lamps can be it?

Text Solution

Verified by Experts

The correct Answer is:
A, B
`h=50m, m=1.8x10^5kg`
`P=100 watt`
P.E=mgh`
`=1.8xx10^5xx9.8xx50`
`=882xx10^5J`
Because half the potetnial energy is converted into electricital energy.
Electircal energy `=1/2P.E. =441xx10^5J/hr`
So, power in watt(J/sec)`=(441xx10^5)`
`:.` Number of 100 W lamps, that can be lit `
(441xx10^5)/(3600xx100)=122.5~~122`
Promotional Banner

Similar Questions

Explore conceptually related problems

A family uses 8 kW of power. Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW?

A piece of ice balls from a height 'h' so that it melts completely .Only one quarter of the heat produced is absorbed by the ice and all energy . Of ice gets converted in to heat during its fall. The value of h is : [" latent heat of ice is " 3.4 xx 10^(5) "J/kg and " g = 10 N/(kg) ]

Scientists are working hard to develop nuclear fusion reactor Nuclei of heavy hydrogen, _(1)^(2)H , known as deuteron and denoted by D , can be thought of as a candidate for fusion rector . The D-D reaction is _(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+ energy. In the core of fusion reactor, a gas of heavy hydrogen of _(1)^(2) H is fully ionized into deuteron nuclei and electrons. This collection of _1^2H nuclei and electrons is known as plasma . The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually , the temperature in the reactor core are too high and no material will can be used to confine the to plasma for a time t_(0) before the particles fly away from the core. If n is the density (number volume ) of deuterons , the product nt_(0) is called Lawson number. In one of the criteria , a reactor is termed successful if Lawson number is greater then 5 xx 10^(14) s//cm^(2) it may be helpfull to use the following boltzmann constant lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm Assume that two deuteron nuclei in the core of fusion reactor at temperature energy T are moving toward each other, each with kinectic energy 1.5 kT , when the seperation between them is large enough to neglect coulomb potential energy . Also neglate any interaction from other particle in the core . The minimum temperature T required for them to reach a separation of 4 xx 10^(-15) m is in the range

Scientists are working hard to develop nuclear fusion reactor Nuclei of heavy hydrogen, _(1)^(2)H , known as deuteron and denoted by D , can be thought of as a candidate for fusion rector . The D-D reaction is _(1)^(2) H + _(1)^(2) H rarr _(2)^(1) He + n+ energy. In the core of fusion reactor, a gas of heavy hydrogen of _(1)^(2) H is fully ionized into deuteron nuclei and electrons. This collection of _1^2H nuclei and electrons is known as plasma . The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually , the temperature in the reactor core are too high and no material will can be used to confine the to plasma for a time t_(0) before the particles fly away from the core. If n is the density (number volume ) of deuterons , the product nt_(0) is called Lawson number. In one of the criteria , a reactor is termed successful if Lawson number is greater then 5 xx 10^(14) s//cm^(2) it may be helpfull to use the following boltzmann constant lambda = 8.6 xx 10^(-5)eV//k, (e^(2))/(4 pi s_(0)) = 1.44 xx 10^(-9) eVm Assume that two deuteron nuclei in the core of fusion reactor at temperature energy T are moving toward each other, each with kinectic energy 1.5 kT , when the seperation between them is large enough to neglect coulomb potential energy . Also neglate any interaction from other particle in the core . The minimum temperature T required for them to reach a separation of 4 xx 10^(-15) m is in the range

A person trying to lose weight (dieter ) lifts a 10 kg mass , one thousand times , to a height of 0.5 m each time .Assume that the potential energy lost each time she lowers the mass is dissipated . (a) How much work does she do against the gravitational force ? (b) Fat supplies 3.8 xx10^(7) J of energy per kilogram which is converted to mechanical energy with a 20 % efficiency rate . How much fat will the dieter use up ?

The human ciculatory system can be thought of as a closed system of interconnecting pipes through which fluid is continuously circulated by two pumps the two pumps the right and left verticles of the heart, work as simple two-stroke force pumps. The muscles of the heart regulate the force by contracting and relaxing. the contraction (systole) lasts about 0.2s and a complete systole/diastole (contraction/relaxation) cycle lasts about 0.8s. For flood pressures and speeds in the normal range. the volume flow rate of blood through a blood vessel is directly proportional tot he pressure difference over a length of the vessel and to the fourth power of the radius of the vessel. The total mechanical energy per unit volume of blood just as it leaves the heart is E//V=rhogh+P+rhov^(2) Q. What is the gravitational potential energy of 8cm^(3) of blood in a 1.8 m tall man, in a blood vessel 0.3 m above his heart? (Note: The man's blood pressure is 1.3xx10^(4)N//m^(2) )

Two stars bound together by gravity orbit othe because of their mutual attraction. Such a pair of stars is referred to as a binary star system. One type of binary system is that of a black hole and a companion star. The black hole is a star that has cullapsed on itself and is so missive that not even light rays can escape its gravitational pull therefore when describing the relative motion of a black hole and companion star, the motion of the black hole can be assumed negligible compared to that of the companion. The orbit of the companion star is either elliptical with the black hole at one of the foci or circular with the black hole at the centre. The gravitational potential energy is given by U=-GmM//r where G is the universal gravitational constant, m is the mass of the companion star, M is the mass of the black hole, and r is the distance between the centre of the companion star and the centre of the black hole. Since the gravitational force is conservative. The companion star and the centre of the black hole, since the gravitational force is conservative the companion star's total mechanical energy is a constant. Because of the periodic nature of of orbit there is a simple relation between the average kinetic energy ltKgt of the companion star Two special points along the orbit are single out by astronomers. Parigee isthe point at which the companion star is closest to the black hole, and apogee is the point at which is the farthest from the black hole. Q. Which of the following prevents the companion star from leaving its orbit and falling the black hole?

A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 1.8xx10^(-8)kgm^(-1)s^(-1) what will be the terminal velocity of the drop. Density of air can be neglected.

When water is boiled at 2 atm pressure the latent heat of vaporization is 2.2xx10^(6) J//kg and the boiling point is 120^(@)C At 2 atm pressure 1 kg of water has a volume of 10^(-3)m^(-3) and 1 kg of steam has volume of 0.824m^(3) . The increase in internal energy of 1 kg of water when it is converted into steam at 2 atm pressure and 120^(@)C is [ 1 atm pressure =1.013xx10^(5)N//m^(2) ]

The width of a depletion region is 400 nm. The intensity of the electric field at the depletion region is 5xx10^(5) V/m. Then calculate the following quantities: (1) The value of the potential barrier. (2) The minimum energy required by an electron to move from the n-type to the p-type region of the diode.