A pump draws water from a tank which is 2.5; | vedclass

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

Question

(B) The power $P$ required by the pump is given by the rate of change of mechanical energy: $P = \left(gh + \frac{v^{2}}{2}\right) \frac{dm}{dt}$.Firs

Vedclass · Accepted Answer

$185$

Vedclass · Answer

(B) The power $P$ required by the pump is given by the rate of change of mechanical energy: $P = \left(gh + \frac{v^{2}}{2}ight) \frac{dm}{dt}$.First,calculate the mass flow rate $\frac{dm}{dt} = ho A v$.Given $ho = 1000\; kg/m^{3}$,$A = 10\; cm^{2} = 10 	imes 10^{-4}\; m^{2}$,and $v = 5\; m/s$:$\frac{dm}{dt} = 1000 	imes 10 	imes 10^{-4} 	imes 5 = 5\; kg/s$.Now,substitute the values into the power equation using $g = 10\; m/s^{2}$,$h = 2.5\; m$,and $v = 5\; m/s$:$P = \left(10 	imes 2.5 + \frac{5^{2}}{2}ight) 	imes 5$.$P = (25 + 12.5) 	imes 5$.$P = 37.5 	imes 5 = 187.5\; W$.Rounding to the nearest provided option,the correct value is $185\; W$.

Similar Questions

An automobile of mass $m$ accelerates starting from the origin and is initially at rest,while the engine supplies constant power $P$. The position $x$ as a function of time $t$ is given by:

$A$ man is riding a cycle with a velocity of $7.2 \text{ km/hr}$ up a hill having a slope of $1$ in $20$. The total mass of the man and the cycle is $100 \text{ kg}$. The power of the man is ............ $\text{W}$. (in $\text{ W}$)

$A$ body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time $t$ is proportional to

$A$ particle of mass $m$ is driven by a machine that delivers a constant power $k$ watts. If the particle starts from rest,the force on the particle at time $t$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module