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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. $(g=10m/s^2)$

\[\begin {array} {1 1} (a)\;1.53\;N \\ (b)\;0.94\;N \\ (c)\;2.5\;N \\ (d)\;1.7\;N \end {array}\]
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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
 

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