Question

A pump of 200 W power is lifting 2 kg water from an average depth of 10 m per second. Velocity of water delivered by the pump is $\left( g=9.8\dfrac{m}{s^2} \right)$
$A.\,3\dfrac{m}{s}$
$B.\,2\dfrac{m}{s}$
$C.\,4\dfrac{m}{s}$
$D.\,1\dfrac{m}{s}$

Answer 1

The formula for calculating the power in terms of the work done and time should be used to solve this problem. In this case, the work done will be the sum of the kinetic energy and the potential energy of the water and the power will be of the pump. Upon substituting the given values in this formula, we can find the value of the velocity of the water.

Formula used:

& P=\dfrac{W}{t} \\\ & PE=mgh \\\ & KE=\dfrac{1}{2}m{{v}^{2}} \\\ \end{aligned}$$ **Complete step by step answer:** The formulae used to carry out the calculations are as follows. The power is the work done per time. $$P=\dfrac{W}{t}$$ Where W is the work done and t is the time taken. The potential energy and the kinetic energy formulae are as follows. $$\begin{aligned} & PE=mgh \\\ & KE=\dfrac{1}{2}m{{v}^{2}} \\\ \end{aligned}$$ Where m is the mass of the body, v is the velocity of the body, g is the gravitational constant and h is the height. From the data, we have, The power of a pump, P = 200 W The mass of water, m = 2 kg The height from where the water is lifted, g = 10 m The time taken is, t = 10 m per sec In this case, the work done is the sum of the potential energy and the kinetic energy. So, we have, $$W=mgh+\dfrac{1}{2}m{{v}^{2}}$$ Substitute the given values in the above equation. $$\begin{aligned} & W=2\times 9.8\times 10+\dfrac{1}{2}\times 2\times {{v}^{2}} \\\ & \Rightarrow W=196+{{v}^{2}} \\\ \end{aligned}$$ Now consider the power equation. $$\begin{aligned} & 200=\dfrac{196+{{v}^{2}}}{1} \\\ & \Rightarrow 200=196+{{v}^{2}} \\\ \end{aligned}$$ Continue the further calculation. $$\begin{aligned} & 4={{v}^{2}} \\\ & \therefore v=2\dfrac{m}{s} \\\ \end{aligned}$$ As, the velocity with which the water is delivered by the pump equals $$2\dfrac{m}{s}$$. **Thus, the option (B) is correct.** **Note:** The units of the parameters should be taken care of. Other than the velocity, the parameters like mass, height or the power can be asked. So, the above-mentioned formulae should be known to solve such problems.