Question

A perosn decides to use his bath tub water to generate electric power to run a 40W bulb. The bath tub is located at a height of 10m from the ground and it holds 200 litres of water. He instals a water driven wheel generator on the ground. At what rate should the water drain from the bath tub to light the bulb? How long can he keep the bulb on , if bath tub was full initially ? Efficiency of generator is `90%`. Take `g=9.8m//s^(2)`

Answer 1

Here, `P_(0)=40 "watt", h=10m, V=200litres`
`(dv)/(dt)=?eta=90%, g=9.8 m//s^(2), t=?`
Work done when m kg of wter falls from a height h is
`W=mgh=(Vrho)gh` , where V is volume of water and `rho` is density of water.
Input power `=P_(1)=` rate of doing work.
`=rhogh((dV)/(dt))` ...(i)
As `eta=(P_(0))/(P_(i))`
`:. P_(i)=(P_(0))/(eta)=(40)/(90//1000)=(400)/(9)`
From (i) , `rho gh ((dV)/(dt))=(400)/(9)`
`(dV)/(dt)=(400)/(9rhogh)`
`=(400)/(9(10)^(3)xx9.8xx10)m^(30//s`
`=(400xx10^(3))/(9xx10^(3)xx9.8xx10) "litre"//"sec"`
`(dV)/(dt)=0.453 "litres"//"sec"`
Now, the time for which bulb can be lighted
`=("total volume of water")/(dV//dt)`
`t=(200)/(0.453)=441.5sec`