Question

A pump on the ground floor of a building can pump up water to fill the tank of $30m^3$ in $15min$. If the tank is $40m$ above the ground, and the efficiency of the pump is $30$, the power consumed by the pump is $(g=10m{s}^{-2})$

• A. 4.4 kW
• B. 44 kW
• C. 440 kW
• D. 0.44 kW

Answer 1

Answer: (B)

Here, Capacity of the tank = $30m^3$
Time taken by the pump to fill the tank,
$t=15min=15\times 60s=900s$
Height through which water is lifted, $h=40m$
Efficiency of the pump, $\eta =30\%=\frac{30}{100}=0.3$
As density of water $=10^{3}kg{m}^{-3}$
$\therefore$ Mass of water pumped
$m$ = volwne $\times$ density
$m=30m^3\times 10^{3}kg{m}^{-3}=3\times 10^{4}kg$
Work done by the pump,
$W=mgh=(3\times 10^{4}kg)(10m{s}^{-2})(40m)$
$=12\times 10^{6}J$
$\therefore$ Output power of the pump
$=\frac{W}{t}=\frac{12\times 10^{6}J}{900s}=\frac{4}{3}\times 10^{4}W=1.33\times 10^{4}W$
As , Input power = $\frac{Output}{\nu }$
$=\frac{1.33\times 10^{4}W}{0.3}=4.4\times 10^{4}W=44kW$