Question

# A pump is used to lift $500\mathrm{kg}$ of water from a depth of $80\mathrm{m}$ in $10\mathrm{s}$. Calculate the power rating of the pump if its efficiency is $40%$

A

$100\mathrm{kW}$

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B

$40\mathrm{kW}$

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C

$16\mathrm{kW}$

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D

None of the above

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Solution

## The correct option is A $100\mathrm{kW}$Step 1: Calculate the force exerted by the pumpGiven, mass of the water lifted is $500\mathrm{kg}$We know that $\mathrm{F}=\mathrm{m}g$, where $\mathrm{F}$ is the force exerted, $\mathrm{m}$ is the mass of the object and $g$ is the acceleration due to gravity. Assuming $g=10{\mathrm{ms}}^{-2}$,$\begin{array}{rcl}\mathrm{F}& =& 500×10\\ \therefore \mathrm{F}& =& 5000\mathrm{N}\end{array}$There the pump exerts $5000\mathrm{N}$ of force.Step 2: Calculate the work done by the pumpGiven, water is moved $80\mathrm{m}$ high.The formula for calculating work done is,$\mathrm{W}=\mathrm{F}·\mathrm{s}$Where, $\mathrm{W}$ is the work done, $\mathrm{F}$ is the force exerted and $\mathrm{s}$ is the displacement of the object.Thus, work done is,$\mathrm{W}=5000·80\phantom{\rule{0ex}{0ex}}\therefore \mathrm{W}=400000=4×{10}^{5}\mathrm{J}$Therefore, the pump does $4×{10}^{5}\mathrm{J}$ of work.Step 3: Calculate the power of the pumpGiven, the pump does the work in $10\mathrm{s}$The formula for power ($\mathrm{P}$) is given as,$\mathrm{P}=\frac{\mathrm{W}}{t}$Where, $t$ is the time taken to do the work.Thus,$\mathrm{P}=\frac{4×{10}^{5}}{10}=40\mathrm{kW}$Step 4: Calculate power ratingGiven, efficiency$=40%=0.4$We know that,$\mathrm{Power}\mathrm{rating}=\frac{\mathrm{Useful}\mathrm{power}}{\mathrm{efficieny}}$Thus, the power rating of the given pump is,$\mathrm{Power}\mathrm{rating}=\frac{40}{0.4}=100\mathrm{kW}$Hence, option A is correct.

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