
A pump is used to lift \[500kg\] of water from a depth of $80m$ in $10s$. Calculate the power rating of the pump if its efficiency is $40\% $:
$(A)100kW$
$(B)40kW$
$(C)16kW$
$(D)$ None of these
Answer
490.2k+ views
Hint: In order to calculate the power rating, first we will calculate the force, then with the help of force we will calculate the work done. With the help of work done, we will calculate the power consumed and hence, with the help of the power consumed, we will calculate the power rating.
Complete step by step solution:
Efficiency is basically a signification of the peak level of performance that can be obtained by using the least amount of inputs. Efficiency deals with the process of reducing the number of unnecessary resources which are being and it helps in providing a very high output.
Given,
Mass of the water which is to be lifted by the pump $m = 500kg$
Depth of the water $s = 80m$
Time taken $t = 10s$
Efficiency of the pump $ = 40\% $
We know that the expression for the gravitational force acting on the pump is given by,
$F = mg$
On putting the value of mass and the acceleration due to gravity, we get,
$F = 500 \times 10$
Here, we have taken the value of acceleration due to gravity as $10\dfrac{m}{{{s^2}}}$, in order to amke our calculations easy.
$F = 5000N$
Now, we need to find the work done by the motor,
$W = F \times s$
On putting the required values,
$W = 5000 \times 80$
$W = 4 \times {10^5}J$
We know that the expression of power is,
$P = \dfrac{W}{t}$
$P = \dfrac{{4 \times {{10}^5}}}{{10}}$
$P = 4 \times {10^4}W$
On converting the value of power from watt to kilowatt, we get,
$P = 40kW$
Now, we need to calculate the power rating of the pump. Power rating is defined as the ratio of useful power to efficiency. So,
Power rating $ = \dfrac{{40000}}{{0.4}}$
On solving this, we get,
Power rating $ = 100000W$
On converting it into kilowatt, we get,
Power rating $ = 100kW$
Thus, the power rating of the pump if its efficiency is $40\% $ is $100kW$.
So, the final answer is $(A)100kW$.
Note:
There is a simple relationship between work done and power. As an example, the work done against the force of gravity comes out to be equal to the change in the potential energy of the body and all the work which is done against the resistive forces comes out to be equal to the change in the total energy.
Complete step by step solution:
Efficiency is basically a signification of the peak level of performance that can be obtained by using the least amount of inputs. Efficiency deals with the process of reducing the number of unnecessary resources which are being and it helps in providing a very high output.
Given,
Mass of the water which is to be lifted by the pump $m = 500kg$
Depth of the water $s = 80m$
Time taken $t = 10s$
Efficiency of the pump $ = 40\% $
We know that the expression for the gravitational force acting on the pump is given by,
$F = mg$
On putting the value of mass and the acceleration due to gravity, we get,
$F = 500 \times 10$
Here, we have taken the value of acceleration due to gravity as $10\dfrac{m}{{{s^2}}}$, in order to amke our calculations easy.
$F = 5000N$
Now, we need to find the work done by the motor,
$W = F \times s$
On putting the required values,
$W = 5000 \times 80$
$W = 4 \times {10^5}J$
We know that the expression of power is,
$P = \dfrac{W}{t}$
$P = \dfrac{{4 \times {{10}^5}}}{{10}}$
$P = 4 \times {10^4}W$
On converting the value of power from watt to kilowatt, we get,
$P = 40kW$
Now, we need to calculate the power rating of the pump. Power rating is defined as the ratio of useful power to efficiency. So,
Power rating $ = \dfrac{{40000}}{{0.4}}$
On solving this, we get,
Power rating $ = 100000W$
On converting it into kilowatt, we get,
Power rating $ = 100kW$
Thus, the power rating of the pump if its efficiency is $40\% $ is $100kW$.
So, the final answer is $(A)100kW$.
Note:
There is a simple relationship between work done and power. As an example, the work done against the force of gravity comes out to be equal to the change in the potential energy of the body and all the work which is done against the resistive forces comes out to be equal to the change in the total energy.
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