Question

A siphon tube is discharging a liquid of specific gravity $0.9$ from a reservoir as shown in the figure. The velocity of the liquid coming out through the siphon is
(Take $g=10{\text{m/s}}^2$)

• A. $7\text{m/s}$
• B. $8\text{m/s}$
• C. $9\text{m/s}$
• D. $10\text{m/s}$

Answer 1

The correct option is D $10\text{m/s}$


From Bernoulli's theorem, we know that,

$P+\frac{1}{2}\rho v^2+\rho gh=\text{constant}$

where $P$ is the pressure, $\rho$ is the density of the liquid, $v$ is the velocity, $g$ is acceleration due to gravity and $h$ is vertical height.

On applying Bernoulli's theorem between points $\text{A}$ and $\text{B}$ and taking reference at point $\text{B}$ we obtain,

$P_A+\frac{1}{2}\rho v_A^2+\rho g{h}_A=P_B+\frac{1}{2}\rho v_B^2+\rho g{h}_B$

Here $P_A$ and $P_B$ are both equal to atmospheric pressure.

$P_A=P_B=P_0$

Also, the velocity of liquid at point $\text{A}$ can be assumed to be $0$.
[since area ratio, $\frac{a_B}{a_A}$ is small, therefore using continuity equation it can be said that $\frac{V_A}{V_B}$ will also be very small]

$\Rightarrow P_0+\frac{1}{2}\rho \times (0{)}^2+\rho g\times 5=P_0+\frac{1}{2}\rho \times v^2+\rho g\times 0$

$\Rightarrow \frac{1}{2}\rho v^2=5\rho g$

The density on both sides are equal as the liquid is the same and they cancel out.

$\Rightarrow v^2=2\times 5\times g$

$\Rightarrow v=\sqrt{2\times 5\times 10}$

$\Rightarrow v=10\text{m/s}$