Question

Water flows along a horizontal pipe of non-uniform cross-section. The pressure is $1\text{cm}$ of $Hg$ where the velocity is $35\text{cm/s}$. At a point where the velocity is $65\text{cm/s}$, the pressure will be

• A. $0.89\text{cm}$ of $Hg$
• B. $0.62\text{cm}$ of $Hg$
• C. $0.5\text{cm}$ of $Hg$
• D. $1\text{cm}$ of $Hg$

Answer 1

The correct option is A $0.89\text{cm}$ of $Hg$

Applying the Bernoulli's equation between given points,
$p_1+\frac{1}{2}\rho v_1^2=p_2+\frac{1}{2}\rho v_2^2...\left(i\right)$
$p_1={\rho }_{Hg}gh=13600\times 1\times (10{)}^{-2}\times 10{\text{N/m}}^2$
$v_1=35\times {10}^{-2}\text{m/s}$
$v_2=65\times {10}^{-2}\text{m/s}$
Since pipe is horizontal (constant elevation), hence the potential energy terms will cancel out from both sides.

From Eq.$\left(i\right)$,
$(13.6\times {10}^3\times 10\times {10}^{-2})+\frac{1}{2}\times {10}^3\times (0.35{)}^2=p_2+\frac{1}{2}\times {10}^3\times (0.65{)}^2$
$\Rightarrow p_2=1210{\text{N/m}}^2$
Hence, corresponding reading of mercury column,
$h=\frac{p_2}{{\rho }_{Hg}g}$
$\Rightarrow h=\frac{1210}{13.6\times {10}^3\times 10}$
$\Rightarrow h=8.89\times {10}^{-3}\text{m}$
$\therefore h=0.89\text{cm of}Hg$