Question

A person decided to use his bath tub water to generate electric power to run a $40W$ bulb. The bath tub is located at a height of $h$ meters from ground and it holds $V$ litres of water. He installs a water driven wheel generator on ground. The rate in $m^3/s$ at which water should drain from bath tub to light the bulb, if efficiency of machine be $90\%$, is (Given density of water $=\rho$ and acceleration due to gravity $=g$)

• A. $\frac{11.11}{\rho gh}$
• B. $\frac{33.33}{\rho gh}$
• C. $\frac{44.44}{\rho gh}$
• D. $\frac{22.22}{\rho gh}$

Answer 1

Answer: (C)

$\frac{44.44}{\rho gh}$

$Power=P=\frac{dW}{dt}=\frac{d}{dt}\left(mhg\right)$

$=\frac{dV\rho gh}{dt}$

or $P=\frac{dV\rho gh}{dt}=pdh\frac{dV}{dt}$

Since efficiency $\eta =\frac{P_{output}}{P_{input}}$

$\therefore P_{output}=\eta P_{input}=\eta \rho gh\frac{dV}{dt}$

or $\frac{dV}{dt}\eta =\frac{P_{output}}{\eta \rho g{h}^2}$

$\frac{dV}{dt}=\frac{40}{0.9\rho gh}=\frac{44.44}{\rho gh}$