Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A tank of diameter $0.6 \;m$ and height $1\;m$ has a hole at of diameter $10\;mm$ at a depth of $0.6\;m$ from the free surface. What is the thrust force due to the water issuing from the hole. $(g=10m/s^2)$

\[\begin {array} {1 1} (a)\;1.53\;N \\ (b)\;0.94\;N \\ (c)\;2.5\;N \\ (d)\;1.7\;N \end {array}\]
Can you answer this question?

1 Answer

0 votes
Now velocity of water coming out of the hole $=v=\sqrt {2gh}$
$\qquad= {2 \times 10 \times 0.6}$
$\qquad= \sqrt {12}\;m/s$
If a is area of cross section of the hole.
$a \times v$= volume of water flowing per second.
$\therefore\; Thrust =\large\frac{Volume}{Second} $$ \times \rho \times velocity$
$Thrust = \rho av^2$
$\qquad= 10^3 \times \large\frac{\pi}{4}$$ (10^{-2})^2 (\sqrt {12})^2$
$\qquad= \large\frac{3.14 \times 10^{-1} \times 12}{4}$
$\qquad= 0.942\;N$
Hence b is the correct answer.


answered Oct 8, 2013 by meena.p
edited Feb 18, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App