
Water from a hose pipe of radius \[5cm\]strikes a wall normally at a speed of $5ms^{-1}$. The force exerted on the wall in newtons is:
\[\begin{align}
& A.13.5\pi \\
& B.6.25\pi \\
& C.62.5\pi \\
& D.27\pi \\
& E.125\pi \\
\end{align}\]
Answer
578.4k+ views
Hint: We need to find the average force exerted by the water on the wall. Since velocity is given instead of acceleration, we can use the relationship between force and momentum to solve the problem. Since mass is not given directly, we need to calculate mass from volume.
Formula used:
$F=\rho\times\pi r^{2}\times v^{2}$
Complete answer:
From newton’s second law, we know that $F=ma$ where $F$ is the force ,$m$ is the mass and $a$ is the acceleration. Or $F=\dfrac{dP}{dt}$ where $P$ is the momentum here, of the water flowing from the pipe and $t$ is the time it takes.
But, $F=\dfrac{dP}{dt}$ gives the instantaneous force. But we need to calculate the average force, which is given by $F_{avg}=\dfrac{\Delta P}{\Delta t}$
We also know that the momentum is given as, $P=mv$, where $v$ is the velocity of the flowing water.
It is given that, the velocity $v=5ms^{-1}$ and the radius of the hose is $r=5cm=5\times 10^{-2}m$
We know that, mass $m=\rho\times vol$, where $\rho$ is the density of the water which is $1000kg/m^{3}$ and \[vol\] its volume.
Also since volume is not given, we can take volume as, $vol= Av$ where $A$ is the area of the circular opening.
The, we get $vol=\pi r^{2}\times v$
Substituting in $F$ we get,
$F=\rho\times\pi r^{2}\times v\times v=\rho\times\pi r^{2}\times v^{2}$
Substituting, the values, we get,
$F= 10^{3}\times \pi \times (5\times 10^{-2})^{2} \times 5=62.5\pi$
Then, the water from the hose exerts a force of $62.5\pi N$ on the wall.
Hence the answer is $62.5\pi$
So, the correct answer is “Option C”.
Note:
This is an easy sum, provided one knows the basics.We have done this calculation for only $1$ second, when the water from the pipe hits the wall. Also, we have done the calculation for an average force. Also, note that, since volume is not given in the data, we are taking volume as the amount of water flowing from the pipe.
Formula used:
$F=\rho\times\pi r^{2}\times v^{2}$
Complete answer:
From newton’s second law, we know that $F=ma$ where $F$ is the force ,$m$ is the mass and $a$ is the acceleration. Or $F=\dfrac{dP}{dt}$ where $P$ is the momentum here, of the water flowing from the pipe and $t$ is the time it takes.
But, $F=\dfrac{dP}{dt}$ gives the instantaneous force. But we need to calculate the average force, which is given by $F_{avg}=\dfrac{\Delta P}{\Delta t}$
We also know that the momentum is given as, $P=mv$, where $v$ is the velocity of the flowing water.
It is given that, the velocity $v=5ms^{-1}$ and the radius of the hose is $r=5cm=5\times 10^{-2}m$
We know that, mass $m=\rho\times vol$, where $\rho$ is the density of the water which is $1000kg/m^{3}$ and \[vol\] its volume.
Also since volume is not given, we can take volume as, $vol= Av$ where $A$ is the area of the circular opening.
The, we get $vol=\pi r^{2}\times v$
Substituting in $F$ we get,
$F=\rho\times\pi r^{2}\times v\times v=\rho\times\pi r^{2}\times v^{2}$
Substituting, the values, we get,
$F= 10^{3}\times \pi \times (5\times 10^{-2})^{2} \times 5=62.5\pi$
Then, the water from the hose exerts a force of $62.5\pi N$ on the wall.
Hence the answer is $62.5\pi$
So, the correct answer is “Option C”.
Note:
This is an easy sum, provided one knows the basics.We have done this calculation for only $1$ second, when the water from the pipe hits the wall. Also, we have done the calculation for an average force. Also, note that, since volume is not given in the data, we are taking volume as the amount of water flowing from the pipe.
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