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JIPMER 2014
Exam
JIPMER
Year
2014
Subject
Physics
Chapter
Work, Power and Energy

JIPMER 2014 · Physics · Work, Power and Energy

A engine pumps up 100 kg of water through a height of 10 m in 5 s ....

A engine pumps up 100 kg of water through a height of 10 m in 55s . Given that, the efficiency of engine is 60%60 \%. If g=10 ms−2g=10 \mathrm{~ms}^{-2}, the power of this engine is

Options

  1. A3.33.3 kW
  2. B0.330.33 kW
  3. C0.0330.033 kW
  4. D3333 kW

Correct answer

A. 3.33.3 kW

Step-by-step solution

Efficiency of engine η=60%\eta=60 \%.  Thus, power = work / time η=10060×mght\begin{aligned} \text{ Thus, power } & =\frac{\text{ work } / \text{ time }}{\eta} \\ & =\frac{100}{60} \times \frac{m g h}{t} \end{aligned} Given, m=100kg,h=10m,t=5sm=100 \mathrm{ kg}, \quad h=10 \mathrm{ m}, \quad t=5 \mathrm{ s} and g=10 ms−2\mathrm{ ms}^{-2} Hence, power =10060×100×10×105&=3.3×103W=3.3kW\begin{aligned}\text{ Hence, power } & =\frac{100}{60} \times \frac{100 \times 10 \times 10}{5} \& =3.3 \times 10^3 \mathrm{ W}=3.3 \mathrm{ kW}\end{aligned}$$
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