Question

Water is filled to a height of $2.5m$ in a container lying at rest on a horizontal surface as shown in the figure. The bottom of the container is square shaped of side $\sqrt{3}m$ and its top is open. There is a very small hole in the vertical wall of the container near the bottom. A small cap closes the hole. The hole gets opened when pressure near the bottom becomes more than $4\times {10}^4$ pascal.
(a) What minimum acceleration can be given to the container in the positive $x-$ direction so that the cap will came out of the hole? Assume water dose not spill out of container when the container accelerates.
(b) Find the velocity of efflux as soon as the container starts moving with the acceleration calculate in part (a) of the question.

Answer 1

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a.

$P_0=(l+h)\rho g$

$\tan \theta =\frac{l}{\sqrt{3}}$

$l=\sqrt{3}\tan \theta$

$l=\sqrt{3}.\frac{a}{g}$

$P_0=\sqrt{3}\frac{a}{g}.\rho g+h\rho g$

$P_0=\sqrt{3}a\rho +h\rho g[{\rho }_{water}=1000㎏/㎥]$

$4\times {10}^4=\sqrt{3}a\rho +h\rho g$

Comparing water volume:$\sqrt{3}\times 2.5=\sqrt{3}h+\frac{1}{2}\sqrt{3}.l$

$\frac{25}{10}=h+\frac{\sqrt{3}a}{2g}\Rightarrow h=2.5-\frac{\sqrt{3}}{2}\frac{a}{g}$

$4\times {10}^4=\sqrt{3}a\times {10}^3+(2.5-\frac{\sqrt{3}a}{2g}){10}^3\times g$

$40=\sqrt{3}a+2.5g-\frac{\sqrt{3}}{2}a\Rightarrow 40=2.5g+\frac{\sqrt{3}}{2}a$

$a=\frac{30}{\sqrt{3}}\Rightarrow a=10\sqrt{3}m/s^2$

b.

Velocity of efflux

According to Bernouli's principle,

$P_0+\frac{1}{2}\rho v_2^2=P_0+0+\frac{1}{2}\rho v_1^2$

$4\times {10}^4=\frac{1}{2}\rho (v_1^2-v_2^2)$

By equation of continuity, $A_1{v}_1=A_2{v}_2$

$4\times {10}^4=\frac{1}{2}\times 1000(v_1^2-\frac{A_1^2{v}_1^2}{A_2^2})\Rightarrow 80=v_1^2\left(\frac{A_1^2}{A_2^2}\right)$

$80=v_1^2[\frac{A_1}{A_2}\approx 0][\because (A_1<<A_2)]$

$v_1=\sqrt{80}=4\sqrt{5}m/s$