Question

Consider a water tank as shown in the figure. Its cross-sectional area is $0.4{\text{m}}^2$. The tank has an opening $B$ near the bottom whose cross-sectional area is $1{\text{cm}}^2$. A load of $24\text{kg}$ is applied on the water at the top when the height of the water level is $40\text{cm}$ above the bottom, the velocity of water coming out from the opening $B$ is $v{\text{ms}}^{-1}$. The value of $v$, to the nearest integer, is _______.
$(g=10{\text{m/s}}^2)$

Answer 1

On applying Bernoulli's theorem at points $1$ and $2$, we get,
$p_0+\frac{mg}{A}+\rho gH+\frac{1}{2}\rho v_1^2=p_0+\frac{1}{2}\rho v^2$

As $v_1=0\left(\text{at A}\right)$

$\because \frac{mg}{A}+\rho gH=\frac{1}{2}\rho v^2$

$\Rightarrow \frac{24\times 10}{0.4}+{10}^3\times 10\times 0.4=\frac{1}{2}\times {10}^3\times v^2$

$\Rightarrow v\approx 3\text{m/s}$