A motor engine pumps \(1800 \mathrm{~L}\) of water per minute from a well of depth \(30 \mathrm{~m}\) and…
A motor engine pumps \(1800 \mathrm{~L}\) of water per minute from a well of depth \(30 \mathrm{~m}\) and allows to pass through a pipe of cross-sectional area \(30 \mathrm{~cm}^2\). Then the power of the engine is (Acceleration due to gravity, \(g=10 \mathrm{~ms}^{-2}\) )
\(20.5 \mathrm{~kW}\)
\(15.5 \mathrm{~kW}\)
\(10.5 \mathrm{~kW}\)
\(9.5 \mathrm{~kW}\)
Solution
Given,
depth of a well, \(d=30 \mathrm{~m}\)
water quantity per minute \(=1800\) litre
cross-section area of pipe, \(A=30 \mathrm{~cm}^2\)
\(=30 \times 10^{-4} \mathrm{~m}^2\)
Now,
velocity of water a jet \(=\frac{1800 \times 10^{-3}}{\left(30 \times 10^{-4}\right) \times 60}=10 \mathrm{~m} / \mathrm{s}\)
and work done by engine \(=m g d+\frac{1}{2} m v^2\)
\(\begin{aligned}
& =1800 \times 10 \times 30+\frac{1}{2} \times 1800 \times(10)^2 \\
& =540000+90000=630000 \mathrm{~J}
\end{aligned}\)
Power of engine, \(P=\frac{W}{t}=\frac{630000}{60}\)
\(=10500 \mathrm{~W}=10.5 \mathrm{~kW}\)