Question

Water is being poured into a vessel at a constant rate $q{m}^2/s$. There is a small aperture of cross-section area 'a' at the bottom of the vessel. The maximum level of water in the vessel is proportional to:

• A. q
• B. $q^2$
• C. $\frac{1}{a}$
• D. $\frac{1}{a^2}$

Answer 1

The correct option is D $\frac{1}{a^2}$

Rate of flow is given by : $q=av$

where $v=\sqrt{2g{h}_{max}}$

so $q=a\sqrt{2g{h}_{max}}$

$⟹h_{max}=\frac{q^2}{2a^{2}g}$

$⟹h_{max}\propto \frac{1}{a^2}$

So option D is correct.