Q126 / 180 · KCET 2017 · Physics
A motor pump lifts \( 6 \) tonnes of water from a well of depth \( 25 \mathrm{~m} \) to the first floor of height \( 35 \mathrm{~m} \)
from the ground floor in \( 20 \) minutes. The power of the pump (in \( \mathrm{kW} \) ) is: \( \left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right] \)
from the ground floor in \( 20 \) minutes. The power of the pump (in \( \mathrm{kW} \) ) is: \( \left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right] \)
- Correct answerA\( 03 \)
- B\( 06 \)
- C\( 1.5 \)
- D\( 12 \)
Solution
Given, mass of water $=6$ tonnes $=6 \times 1000$; depth of well $=25 \mathrm{~m}$; height of first floor $=35 \mathrm{~m}$; time $=20$ minutes
$=20 \times 60 \mathrm{~s}$
Power of pump $=\frac{\text { work done }}{\text { time }}=\frac{\mathrm{mgh}}{\text { time }}$
$\Rightarrow$ Power
$=\frac{6 \times 1000 \times 10 \times(25+35)}{20 \times 60}=3000 \mathrm{~W}$
Therefore, power of pump is $3 \mathrm{~kW}$
Hence, the correct answer is option: 'A'.