Question

Water is flowing through a horizontal tube having cross-sectional areas of its two ends being $A$ and $A'$ such that the ratio $A/A'$ is $5$. If the pressure difference of water between the two ends is $3\times 10^5\, N \,m^{-2}$, the velocity of water with which it enters the tube will be (neglect gravity effects)

• A. $5\, ms^{-1}$
• B. $10 \,ms^{-1}$
• C. $25\, ms^{-1}$
• D. $50\sqrt{10}m\,s^{-1}$

Answer 1

Answer: (A)

According to Bernoulli’s theorem
$P_{1} +\frac{1}{2}\rho v^{2}_{1} = P_{2} +\frac{1}{2}\rho v^{2}_{2}\quad...\left(i\right)$
From question,
$P_{1} - P_{2} = 3\times10^{5}, \frac{A_{1}}{A_{2}} =5$
According to equation of continuity
$A_{1}v_{1}=A_{2}v_{2}$
or, $\frac{A_{1}}{A_{2}} = \frac{v_{2}}{v_{1} } = 5$
$\Rightarrow \quad v_{2} = 5v_{1}$
From equation $\left(i\right)$
$P_{1}-P_{2} = \frac{1}{2}\rho\left(v^{2}_{2}-v^{2}_{1}\quad\quad\right)$
or $3 \times 10^{5} = \frac{1}{2}\times1000\left(5v^{2}_{1}-v^{2}_{1}\quad\quad\right)$
$\Rightarrow \quad600 = 6v_{1} \times 4v_{1}$
$\Rightarrow \quad v^{2}_{1} = 25$
$\therefore \quad v_{1}= 5 \,m/s$