question_answer

A person decides to use his bath tub water to generate electric power to run a 40 W bulb. The bath tub is located at a height of 10 m from the ground and it holds 200 litres of water. He installs a water driven wheel generator on the ground. At what rate shold 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 = 90%. Take  \[g=9.8\,m/{{s}^{2}}\].

Answer:

                Work done when m kg of water falls from a height h is \[W=mgh=\left( V\rho  \right)gh\]                 Where V is volume of water and \[\rho \]is density of water Rate of doing work = Input power \[{{P}_{i}}=\frac{dW}{dt}=\rho gh\left( \frac{dV}{dt} \right)\] By definition, \[{{P}_{0}}=40\,watt\] As \[\eta =\frac{{{P}_{0}}}{{{P}_{i}}}\]   \[\therefore \]   \[{{P}_{i}}=\frac{{{P}_{0}}}{\eta }=\frac{40}{90/100}=\frac{400}{9}\] \[\rho gh\left( \frac{dV}{dt} \right)=\frac{400}{9}\] \[\frac{dV}{dt}=\frac{400}{9\left( \rho gh \right)}=\frac{400}{9\left( {{10}^{3}}\times 9.8\times 10 \right)}{{m}^{3}}/s\] \[\frac{dV}{dt}=\frac{400\times {{10}^{3}}}{9\times {{10}^{3}}\times 9.8\times 10}litre/s=0.453\,litre/\sec \] Time for which bulb can be lighted \[t=\frac{total\,volume\,of\,water}{dV/dt}=\frac{200}{0.453}=411s\]