Question

In the diagram shown below, the difference between the tubes of the manometer is 5 cm, the cross-sections at A and B are 6 sq mm and 10 sq mm respectively. The rate at which water flows through the tube is;($g=10m{\mathrm{s}}^{-2}$)



(A) 7.5 cc/sec
(B) 8.0 cc/sec.
(C) 10.0 cc/sec.
(D) 12.5 cc/sec.

Answer 1

Hint: This problem uses Bernoulli's theorem. Bernoulli's theorem is the pressure version of energy conservation. It is also similar to conservation of momentum.

Complete solution:
Bernoulli’s principle: In fluid dynamics, Bernoulli's principle states that an increase in the speed of a fluid occurs simultaneously with a decrease in static pressure or a decrease in the fluid's potential energy. Bernoulli's principle can be applied to various types of fluid flow, resulting in various forms of Bernoulli's equation; there are different forms of Bernoulli's equation for different types of flow.
The simplest one being;
$P+{\displaystyle \frac{1}{2}}\rho v^2=C$.
Here, $P$ is the pressure at the surface of the liquid due to air, $\rho$ is the density of the fluid and v is the velocity of the fluid and C is constant.
 Bernoulli's theorem also describes the relationship between a fluid's velocity and cross-sectional area. Mathematically it is written as $A_1{v}_1=A_2{v}_2$.
Now in this case, the difference between the tubes is given viz. 5cm= 0.05 meters.
So first we find the velocity relation in the two tubes.
Hence by applying Bernoulli’s theorem we get;
$A_1{v}_1=A_2{v}_2$ here, $A_1=6m{m}^2$ and $A_2=10m{m}^2$ ;
Thus, $6v_1=10v_2$
Thus $v_1={\displaystyle \frac{5}{3}}{v}_2$ (equation:1)
Now apply Bernoulli’s theorem of conservation of pressure;
$P_1+{\displaystyle \frac{1}{2}}\rho {v_1}^2=P_2+{\displaystyle \frac{1}{2}}\rho {v_2}^2$
Solving the equation further we get;
$P_1-P_2={\displaystyle \frac{1}{2}}\rho \left({v_2}^2-{v_1}^2\right)$
$P_1-P_2={\displaystyle \frac{1}{2}}\rho \left({v_2}^2-{\displaystyle \frac{25}{9}}{v_2}^2\right)$ (from equation: 1)
Now we know that, $P_2-P_1=\rho g(h_2-h_1)=0.05\rho g$ (given)
$0.05\rho g={\displaystyle \frac{1}{2}}\rho \left({\displaystyle \frac{16}{9}}{v_2}^2\right)$
Solving the above equation we get,
$v_2=0.75m{s}^{-2}$
Therefore, rate of flow is mathematically Area X velocity;
Rate=$A_2{v}_2=75\times {10}^{-1}c{m}^3{\sec }^{-1}$

Hence option A is correct.

Note: (1) The units must be converted very carefully
(2) Av is the rate of flow of fluid so Bernoulli’s principle states that rate of flow is constant.