Answer
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Hint:- Power of any engine or application is defined as the rate of doing work , or tells about the energy consumption or production per unit of time . For finding the power first we find the value of force and after that we multiply it to the velocity of flowing fluid.
Complete step-by-step solution:-
Power of an engine / pump is defined as how much water it sucks and with what speed it delivers .
Power of a pump is defined in terms of force and velocity as
\[ \Rightarrow P = F.v\]
Where F = force applied & v = velocity of flowing fluid / water.
We know that force is defined as change in momentum (p) per unit time , hence force in terms of mass (m), velocity (v) and time (t) is written as…
\[ \Rightarrow F = \dfrac{{\Delta p}}{{\Delta t}} = \frac{{mv}}{t}\]
Putting the value of force \[F = \dfrac{{mv}}{t}\] , in the the relation \[P = F.v\]
\[ \Rightarrow P = \dfrac{{m{v^2}}}{t}\]; now putting the values of m = 200kg , & v = 2m/s
\[ \Rightarrow P = \dfrac{{m{v^2}}}{t} = \dfrac{{2.100{{(2)}^2}}}{1}\]
Now further solving the relation we get ,
\[ \Rightarrow P = \dfrac{{200.(4)}}{1}\]
Simplifying for the power (P) of the engine ,
\[ \Rightarrow P = 800\;W\]
Since, the power of the engine is 800 W
Hence option (A) is the correct answer.
Note:- If in this case it is said that , engine sucks water from a well eight metre deep and then delivers with the velocity 2m/s and the same amount of water then the rise in potential energy of water is also accounted in power of the engine as this is the engine due to which there is a rise in potential energy of the water.
Complete step-by-step solution:-
Power of an engine / pump is defined as how much water it sucks and with what speed it delivers .
Power of a pump is defined in terms of force and velocity as
\[ \Rightarrow P = F.v\]
Where F = force applied & v = velocity of flowing fluid / water.
We know that force is defined as change in momentum (p) per unit time , hence force in terms of mass (m), velocity (v) and time (t) is written as…
\[ \Rightarrow F = \dfrac{{\Delta p}}{{\Delta t}} = \frac{{mv}}{t}\]
Putting the value of force \[F = \dfrac{{mv}}{t}\] , in the the relation \[P = F.v\]
\[ \Rightarrow P = \dfrac{{m{v^2}}}{t}\]; now putting the values of m = 200kg , & v = 2m/s
\[ \Rightarrow P = \dfrac{{m{v^2}}}{t} = \dfrac{{2.100{{(2)}^2}}}{1}\]
Now further solving the relation we get ,
\[ \Rightarrow P = \dfrac{{200.(4)}}{1}\]
Simplifying for the power (P) of the engine ,
\[ \Rightarrow P = 800\;W\]
Since, the power of the engine is 800 W
Hence option (A) is the correct answer.
Note:- If in this case it is said that , engine sucks water from a well eight metre deep and then delivers with the velocity 2m/s and the same amount of water then the rise in potential energy of water is also accounted in power of the engine as this is the engine due to which there is a rise in potential energy of the water.
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