Water is pumped from a depth of 10, m and del | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The power $P$ required to pump water is given by the rate of change of potential energy plus the rate of change of kinetic energy.$P = \frac{dW}{d

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) The power $P$ required to pump water is given by the rate of change of potential energy plus the rate of change of kinetic energy.$P = \frac{dW}{dt} = \frac{d}{dt}(mgh + \frac{1}{2}mv^2) = \frac{dm}{dt}gh + \frac{1}{2}\frac{dm}{dt}v^2$.Given: Depth $h = 10\, m$,Area $A = 10^{-2}\, m^2$,Volume flow rate $\frac{dV}{dt} = 10^{-1}\, m^3/s$.Density of water $ho = 10^3\, kg/m^3$.Mass flow rate $\frac{dm}{dt} = ho \frac{dV}{dt} = 10^3 	imes 10^{-1} = 100\, kg/s$.Velocity of water $v = \frac{dV/dt}{A} = \frac{10^{-1}}{10^{-2}} = 10\, m/s$.Now,$P = (100)(9.8)(10) + \frac{1}{2}(100)(10)^2$.$P = 9800 + 5000 = 14800\, W = 14.8\, kW$.Rounding to the nearest provided option,the required power is approximately $15\, kW$.

Water is pumped from a depth of $10\, m$ and delivered through a pipe of cross section $10^{-2}\, m^2$. If it is needed to deliver a volume of $10^{-1}\, m^3$ per second,the power required will be: ............ $kW$

Similar Questions

$A$ $TV$ transmitter has a range of $50 \ km$. The height of the $TV$ transmitter is . . . . . . (Radius of the earth,$R_e = 6.4 \times 10^6 \ m$). (in $m$)

The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^2 = m^2 + n^2$ is

If $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$,then the equation of the tangent at one of the points of contact is

Two identical thin bar magnets,each of length $l$ and pole strength $m$,are placed at a right angle to each other with the north pole of one touching the south pole of the other. The magnetic moment of the system is:

If $\cos 2x = (\sqrt{2} + 1) \left(\cos x - \frac{1}{\sqrt{2}}\right)$ and $\cos x \neq \frac{1}{\sqrt{2}}$,then $x \in$

Vedclass Test Series

Exam Paper Generator

Online Exam Module