Question

Please help me to solve the above sum. This question is of 9th standard of science 1 chapter ' Work and energy '. ( In the above question = 1 HP means 1 Horse power , and g = 10 m/s means gravity = 10 m/s ).

Please help me to solve this question.​

Answer 1

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Electric motor used to life the water stored in tank at the ground level to overhead tank of height 20 m .If the power of motor is 1 HP how much time will be required to pump 746 kg of water?

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Let us write the given data first

So It will be easy to solve the sums.

$\pink{ \boxed{GIVEN: }}$

Height (h) = 20 mts

Mass/Weight (m) = 746 kg

Power (p) = 1 HP [Horse Power]

Gravity (g) = 10m/s

$\purple{ \boxed{TO\: FIND:}}$

Time (t) = ???

${\huge{\underline{\small{\mathbb{\blue{ SOLUTION:}}}}}}$

We Know the formula

$➠ Power\: = \dfrac{Work\: Done}{Time }$

$➠ P\: = \dfrac{W}{t}$

Here work done is stored in the form of Mechanical Energy so we should use the formula W = mgh & not 1/2 mv²

Another Hint which is given in this question is they given height so we should mgh formula only

Also we need to convert HP unit to watt unit [∵ S.I. Unit of Power is Watt ]

1 Horse Power = 745.7 watt ≈ 746 watt

Let us substitute in the formula,

$➠ P\: = \dfrac{mgh}{t}$

$➠ 746 \: = \dfrac{746\times 10\times 20}{t}$

$➠ t \: = \dfrac{746\times10\times20}{746}$

$➠ t\: = \dfrac{\cancel{740}\times 10\times20 }{\cancel{740} }$

$➠ t\: = 10 \times 20$

$➠ t\: = 200\: seconds = \dfrac{20\cancel0}{6\cancel0} =\dfrac{10}{3} = 3.33\:mintues$

Hence it will require 200s to weight the lift of 746kg

ℌope ℑt ℌelps

Thanks for asking :)