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# An electric pump on the ground floor of a building takes 10 min to fill a tank of volume $30{{m}^{3}}$ with water. If the tank is 60m above the ground and the efficiency of the pump is 30% then how much electric power is consumed by the pump in filling the tank? Take $g=10m{{s}^{-2}}$.$\text{A}\text{. }100KW$$\text{B}\text{. 150}KW$$\text{C}\text{. }\dfrac{100}{3}KW$$\text{D}\text{. 50}KW$

Last updated date: 13th Jun 2024
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Answer
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Hint: This question involves many concepts. Therefore, solve it gradually. It is asked to find the power consumed by the pump and power is the work done in one unit of time. Work done is the product of the force applied and the displacement of the body along the direction of the force.
Complete step by step answer:
There are many concepts involved in this problem, so we will solve it step by step. It is told that the electric pump pumps water from the ground to a height of 60m. We know that to take a mass from the ground to height, we have to apply a force against the force of gravity. The electric pump does this work. The pump applies a suction force on the water to suck it up. To lift the mass of the water, the pump has to apply a force equal to the force of gravity (mg). It is told that the pump fills the tank of $30{{m}^{3}}$ in 10 min. let the mass of the water that the pump sucked I 10 minutes be m. Therefore, the force applied to the pump will be equal to mg. Let us calculate the mass of the water-filled in the tank. Since we know the density of water i.e. $1000kg{{m}^{-3}}$, we can use the density formula i.e. $\text{density=}\dfrac{\text{mass}}{\text{volume occupied by the mass}}$. The volume occupied by the water is equal to the volume of the tank i.e. $30{{m}^{3}}$. Substitute the values of the density and the volume. Therefore,
$1000=\dfrac{m}{30}\Rightarrow m=30000kg$ and it is given that $g=10m{{s}^{-2}}$. Therefore, the force will be equal to $mg=30000\times 10=300000N$. This force displaces the water to a height of h=60m. Therefore, there will be some work done by the pump, which is equal to $\text{Force }\!\!\times\!\!\text{ displacement}=mgh=300000\times 60=18000000J$. That means the pump does a work of $18\times {{10}^{6}}J$ to lift water of mass 30000kg and this is done in 10min. Therefore, the work that has to be done in 1 second i.e. power, is $\dfrac{\text{work done}}{\text{time}}=\dfrac{18\times {{10}^{6}}J}{10\min }=\dfrac{18\times {{10}^{6}}J}{10\times 60\sec }=3\times {{10}^{4}}W$.
Therefore, to fill the tank completely, the pump needs a power of $3\times {{10}^{4}}W$. But, it is told that the pump is not fully efficient and it has an efficiency of 30%. This means that the pump can use only 30% of the total power supplied to it in filling the tank. Therefore, if we have to supply power which is more than$3\times {{10}^{4}}W$. Let that power be PT and out of PT, 30% will be used to fill the tank which is $3\times {{10}^{4}}W$. This implies that$\text{30 }\!\!\%\!\!\text{ of }{{\text{P}}_{\text{T}}}=3\times {{10}^{4}}W\Rightarrow \dfrac{30}{100}\times {{P}_{T}}=3\times {{10}^{4}}W$.
$\Rightarrow {{P}_{T}}=\dfrac{100}{30}\times 3\times {{10}^{4}}=100\times {{10}^{3}}W=100KW$
Therefore, the pump consumes a power of 100KW. Hence, the correct option is (A).
Note: One may say that to lift a body if we apply a force equal to the gravitational force than the two forces will balance each other and the body will be in equilibrium. Therefore, the body will not move up. To lift the body, we need to apply a force greater than mg. Even if we add an infinitesimally small force to mg, the body will move. Therefore, the net force will almost be equal to mg.