Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Water is flowing at a speed of $1.5m{{s}^{-1}}$ through a horizontal tube of cross-sectional area ${{10}^{-2}}{{m}^{2}}$ and you are trying to stop the flow by your palm. Assuming that the water stops immediately after hitting the palm, the minimum force that you must exert should be (density of water $={{10}^{3}}kg{{m}^{-3}}$).
A. 15N
B. 22.5N
C. 33.7N
D. 45N

seo-qna
Last updated date: 17th Jun 2024
Total views: 403.2k
Views today: 10.03k
Answer
VerifiedVerified
403.2k+ views
Hint: Here, to find the force needed to exert on the tube by the palm, first we need to find the force of the water coming out of the tube. To find the force we will use Newton's second law of motion. Taking the time of flow of water inside the tube as 1 second we can find the force required by using Newton's second law of motion.

Complete answer:
The speed of flowing water is $v=1.5m{{s}^{-1}}$
The water is flowing through a horizontal tube. The cross-sectional area of the tube is, $A={{10}^{-2}}{{m}^{2}}$
We are trying to stop the water flow by opposing the flow of water by the palm. So, the force we need to exert on the tube should be equal to the force of water flowing through the tube to stop the water flow.
The density of the water is given as, $\rho ={{10}^{3}}kg{{m}^{-3}}$
From Newton’s second law of motion, the force on an object can be defined as the rate of change of the linear momentum of the object.
$F=\dfrac{dp}{dt}$
Again, the linear momentum of an object can be defined as the product of the mass and velocity of the object.
$p=mv$
So, we can write,
$F=\dfrac{d\left( mv \right)}{dt}$
Since, the mass of the object is constant, we can write,
$F=m\dfrac{dv}{dt}$
Now, let the water flow through the pipe for $t=1s$.
So, the length of the pipe will be, $l=vt=1.5\times 1=1.5m$
So, the mass of the water inside the tube will be,
$\begin{align}
  & m=\rho Al \\
 & m={{10}^{3}}\times {{10}^{-2}}\times 1.5 \\
 & m=15kg \\
\end{align}$
Again, the final velocity of the water will be zero, so, the change in velocity will be,
$dv=1.5-0=1.5m{{s}^{-1}}$
So, the force of the water will be,
$\begin{align}
  & F=m\dfrac{dv}{dt} \\
 & F=15\times \dfrac{1.5}{1} \\
 & F=22.5N \\
\end{align}$
So, the minimum force we need to exert on the pipe to stop the flow of water will be 22.5 N.

The correct option is (B).

Note:
We can also use the equation of continuity to find the force needed to stop the flow of water. Find the change in momentum of the water when flowing out of the tube to find the force of the water and from this we can find the force needed to stop the flow of water.