Question

There is a small hole in the bottom of a fixed container containing a liquid up to height "h". The top of the liquid, as well as the hole at the bottom, are exposed to the atmosphere. As the liquid comes out of the hole: (Area of the hole is "a' and that of the top surface is "A')

• A. The top surface of the liquid accelerates with acceleration = g
• B. The top surface of the liquid accelerates with acceleration = $g\frac{a^2}{2A^2}$
• C. The top surface of the liquid accelerates with acceleration = $g\frac{a}{A}$
• D. The top surface of the liquid accelerates with acceleration = $\frac{g{a}^2}{A^2}$

Answer 1

The correct option is D The top surface of the liquid accelerates with acceleration = $\frac{g{a}^2}{A^2}$

The velocity of the fluid at the hole is $V_2=\sqrt{\frac{2gh}{1-\frac{a^2}{A^2}}}$

Using the continuity equation at the two Top surface and hole section:

$V_{1}A=V_{2}a$

$V_1=\frac{a}{A}{V}_2$

acceleration (of top surface)

$=-V_1\frac{d{V}_1}{dh}=\frac{-a}{A}{V}_2\frac{d}{dh}\left(\frac{a}{A}{V}_2\right)=\frac{-a^2}{A^2}{V}_2\frac{d{V}_2}{dh}=\frac{-a^2}{A^2}\sqrt{\frac{2gh}{1-\frac{a^2}{A^2}}}\sqrt{\frac{2g}{1-\frac{a^2}{A^2}}}\frac{1}{2\sqrt{h}}=\frac{-a^2}{A^2}\frac{2g\sqrt{h}}{1-\frac{a^2}{A^2}}\frac{1}{2\sqrt{h}}=\frac{-a^2}{A^2}\frac{g}{1-\frac{a^2}{A^2}}=\frac{-a^2}{A^2}\frac{g}{\frac{A^2-a^2}{A^2}}=\frac{-a^{2}g}{A^2-a^2}$

We know a <<< A, Hence $A^2-a^2\approx A^2$

Hence $\frac{g{a}^2}{A^2}$ is correct answer.