Question

Water is flowing through a horizontal tube having cross-sectional areas of its two ends being $A$ and $A^{\prime }$ such that the ratio $A/A^{\prime }$ 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. $5m{s}^{-1}$
• B. $10m{s}^{-1}$
• C. $25m{s}^{-1}$
• D. $50\sqrt{10}m{s}^{-1}$

Answer 1

Answer: (A)

According to Bernoulli’s theorem
$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2...\left(i\right)$
From question,
$P_1-P_2=3\times 10^5,\frac{A_1}{A_2}=5$
According to equation of continuity
$A_1{v}_1=A_2{v}_2$
or, $\frac{A_1}{A_2}=\frac{v_2}{v_1}=5$
$\Rightarrow v_2=5v_1$
From equation $\left(i\right)$
$P_1-P_2=\frac{1}{2}\rho \left(v_2^2-v_1^2\right)$
or $3\times 10^5=\frac{1}{2}\times 1000\left(5v_1^2-v_1^2\right)$
$\Rightarrow 600=6v_1\times 4v_1$
$\Rightarrow v_1^2=25$
$\therefore v_1=5m/s$