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Water is pumped from a depth of $10 $ $m$ and delivered through a pipe of cross section $10^{-2}$ $m^2$. If it is needed to deliver a volume of $10^{-1} $ $m^3$ per second the power required will be ........ $kW$
$10$
$15$
$9.8$
$4.9$
Solution
The potential energy obtained by the water when it is raised from that depth is;
$\text { P.E. }=m g h$
$=m \times 9.8 \times 10$
$=(98 \times m) J$
Also, the volume of water pumped up in time, $t$, is given as:
$V=r \times t$
$V=10^{-1} \times t m ^3$
So, the mass of the water pumped up is given by, $h$
$m=\rho \times V$
$m=10^3 kgm ^{-3} \times 10^{-1} \times tm ^3$
$m=100 \times t kg$
Now, power can be expressed as energy consumed per unit time. So, the power required to pump up the water is,
$P=\frac{P . E}{t}$
On substituting the values in the above expression, we get,
$P=\frac{98 \times 100 \times t}{t}$
$P=9800 Js ^{-1}$
$P=9800 W$
$P=9.8 kW$
