Consider a water tank as shown in the figure. It's cross-sectional area is 0.4m² .The tank has an opening B near the bottom whose  cross-section area is 1cm² . A load of 24 kg is applied on the water at the top when the height of the water level is 40 cm, $\mathrm{v}$ to the nearest integer, is __________  $\times \mathrm{m}/\mathrm{s}$ [Take value of g to be 10ms^-2]

 

Given,

Cross-sectional area A $=0.4{\mathrm{m}}^2$

The cross-sectional area of B $=1{\mathrm{cm}}^2$

The top mass applied to the water is m = 24km.

When Bernoulli's theorem is applied at sites A and B, the height of the water level is determined to be h=40cm.

${\mathrm{P}}_0+\frac{\mathrm{mg}}{\mathrm{A}}+\mathrm{\rho gh}+\frac{1}{2}{\mathrm{\rho V}}^2={\mathrm{P}}_0+\frac{1}{2}{\mathrm{\rho v}}^2$

From the figure at A,  V=0 

Since

$\frac{\mathrm{mg}}{\mathrm{A}}+\mathrm{\rho gh}={\mathrm{P}}_0+\frac{1}{2}{\mathrm{\rho V}}^2$

$=\frac{24\times 10}{0.4}+{10}^3\times 10\times 0.4=\frac{1}{2}\left({10}^3\right){\mathrm{v}}^2=3\mathrm{m}/\mathrm{s}$

Hence the correct answer is 3.