question_answer

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)

\[v=\sqrt{2\,\,\times \,\,10\,\,\times \,\,\frac{2.5}{2}}=5\]