In a horizontal tube the water pressure changes by $1500 \mathrm{Nm}^{-2}$ between $A$ and $B$ as shown in figure below. The cross-sectional areas at $A$ and $B$ of the tube are $40 \mathrm{~cm}^2$ and $20 \mathrm{~cm}^2$, respectively. Find the rate of flow of water through the tube.

• A

  $1000 \mathrm{~cm}^3 \mathrm{~s}^{-1}$
• B

  $2000 \mathrm{~cm}^3 \mathrm{~s}^{-1}$
• Correct answer
  C

  $4000 \mathrm{~cm}^3 \mathrm{~s}^{-1}$
• D

  $6000 \mathrm{~cm}^3 \mathrm{~s}^{-1}$

Solution

Given that, pressure difference
By using Bernoulli's equation, $$ p_A+\frac{1}{2} \rho v_A^2=p_B+\frac{1}{2} \rho v_B^2 $$
Density of water, $\rho=10^3 \mathrm{~kg} \mathrm{~m}^{-3}$ Cross-section areas at points $A$ and $B$, $$ \begin{aligned} & a_A=40 \mathrm{~cm}^2=40 \times 10^{-4} \mathrm{~m}^2 \\ & a_B=20 \mathrm{~cm}^2=20 \times 10^{-4} \mathrm{~m}^2 \end{aligned} $$ By equation of continuity, rate of flow of water through the tube $$ \frac{\Delta V}{\Delta t}=a_A v_A=a_B v_B \Rightarrow \frac{v_A}{v_B}=\frac{a_B}{a_A}=\frac{1}{2} $$
Substituting the values from Eqs. (i) and (iii), in Eq. (ii), we get $$ \begin{aligned} 1500 & =\frac{1}{2} \times 10^3\left[\left(2 v_A\right)^2-v_A^2\right] \\ 1500 & =500\left(4 v_A^2-v_A^2\right) \\ v_A^2 & =1 \Rightarrow v_A=1 \mathrm{~m} / \mathrm{s} \end{aligned} $$ $\therefore$ Rate of flow of water, $$ \begin{aligned} \frac{\Delta V}{\Delta t} & =a_A v_A=40 \times 10^{-4} \times 1 \mathrm{~m}^3 / \mathrm{s} \\ & =4000 \mathrm{~cm}^3 / \mathrm{s} \end{aligned} $$

Hence, the correct answer is option: 'C'.