Question

Consider a water tank as shown in the figure. It’s cross-sectional area is $0.4m^2$. The tank has an opening $B$ near the bottom whose cross-sectional area is $1c{m}^2$. A load of $24kg$ is applied on the water at the top when the height of the water level is $40cm$ above the bottom, the velocity of water coming out the opening $B$ is $vm{s}^{-1}$. The value of $v$, to the nearest integer, is $\_\_\_\_.$

[Take value of $g$ to be $10m{s}^{-2}$]

Answer 1

Step 1: Given data,

Cross-sectional area $A=0.4m^2$

The cross-sectional area of $B$ $=1c{m}^2$,

Mass applied on the water at the top $m=24kg$, is

Height of the water level $h=40cm$,

Step 2: On applying Bernoulli’s theorem at points $A$ and $B$

$P_0+\frac{mg}{A}+\rho gh+\frac{1}{2}\rho V^2=P_0+\frac{1}{2}\rho v^2$

(Where, $P_0$ is pressure on both point $A$ and $B$, $\rho$ is the density of water, $V$ is the volume at point $A$ and $v$ is the volume at the point $B$)

Therefore, $V=0$ (At $A$)

Since,

$$\begin{gathered} \frac{mg}{A}+\rho gh=P_0+\frac{1}{2}\rho v^2 \\ \Rightarrow \frac{24\times 10}{0.4}+{10}^3\times 10\times 0.4=\frac{1}{2}\left({10}^3{v}^2\right) \\ \Rightarrow v=\simeq 3m{s}^{-1} \end{gathered}$$

Hence, The value of $v$, to the nearest integer, is $3m{s}^{-1}$.