Question

A wide vessel with a small hole at the bottom is filled with water and kerosene. Find the approximate velocity of the water flow if thickness of water layer is $30\text{cm}$ and that of kerosene layer is $20\text{cm}$.
Specific gravity of kerosene $=0.8$.

• A. $300\text{m/s}$
• B. $30\text{m/s}$
• C. $3\text{m/s}$
• D. $1\text{m/s}$

Answer 1

The correct option is C $3\text{m/s}$

As water is heavier than kerosene, it will form the lower layer.


Hydrostatic pressure at the bottom $={\rho }_{1}g{h}_1+{\rho }_{2}g{h}_2$

Now, considering the combination of two liquid layers as a single layer of water of height $h_0$, then

$h_0{\rho }_{2}g=h_1{\rho }_{1}g+h_2{\rho }_{2}g........\left(1\right)$

Now, applying Bernoulli's theorem at the free (open) surface of liquid and at the hole,

$P_a+\frac{1}{2}{\rho }_1{v}_1^2=P_a+\frac{1}{2}{\rho }_2{v}_2^2+{\rho }_{2}g{h}_0$

Now, let area of the hole $=a$ and area of the open surface $=A$

Therefore, applying continuity equation:

$a{v}_2=A{v}_1\Rightarrow v_1=\frac{a{v}_2}{A}$

As $a<<A$, we can almost neglect $v_1$ comparing with $v_2$

Hence, using the above condition, from $\left(2\right)$

$v_2=\sqrt{2g{h}_0}$

From $\left(1\right)$ : $h_0=h_2+\frac{{\rho }_1}{{\rho }_2}{h}_1=0.3+\left(0.2\right)\left(0.8\right)$

$\Rightarrow h_0=0.3+0.16=0.46\text{m}$.

Hence, $v_2=\sqrt{\left(2\right)\left(10\right)\left(0.46\right)}=\sqrt{9.2}\approx 3\text{m/s}$

Therefore, required velocity $=3\text{m/s}$

Hence, option $\left(\text{C}\right)$ is correct.