A container of dimension \[4m\,\,\times \,\,3m\,\,\times \,\,2m\] starts to move with uniform acceleration \[a=1.25\text{ }m/{{s}^{2}}\] at t = 0. The volume of liquid in vessel is \[18\text{ }{{m}^{3}}.\] The speed of liquid coming out from a very small orifice made at bottom of right side wall just after acceleration of container - |
A) zero
B) \[\sqrt{30}\,\,m/sec\]
C) 5 m/s
D) 10 m/s
Correct Answer: C
Solution :
When vessel is accelerating surface of fluid get inclined at angle \[\theta \] where is \[\tan \theta \] is given by \[\frac{a}{g}\] |
\[\tan \theta =\frac{a}{g}=\frac{{{y}_{1}}-{{y}_{2}}}{4}\] |
\[{{y}_{1}}-{{y}_{2}}=0.5\] ...(1) |
Volume of water in container |
\[4\,\,\left( \frac{{{y}_{1}}+{{y}_{2}}}{2} \right)\,\,\times \,\,3=18\] |
From (1) and (2) |
\[{{y}_{1}}+{{y}_{2}}=3\] |
\[{{y}_{1}}-{{y}_{2}}=0.5\] |
\[{{y}_{2}}=\frac{2.5}{2}\] |
Velocity of water coming out from small orifice \[V=\sqrt{2g{{y}_{2}}}\](this result is from Bernoulli equation) |
\[v=\sqrt{2\,\,\times \,\,10\,\,\times \,\,\frac{2.5}{2}}=5\] |
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