Question

Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1 cm$ of $Hg$ where the velocity is $35 cms ^{-1}$. At a point where the velocity is 65 $cms ^{-1},$ the pressure will be

• A. $0.89 cm$ of $Hg$
• B. $8.9 cm$ of $Hg$
• C. $0.5 cm$ of $Hg$
• D. $1 cm$ of $Hg$

Answer 1

Answer: (A)

Using Bernoulli's theorem,
$p_{1}+\frac{1}{2} \rho v_{1}^{2}=p_{2}+\frac{1}{2} \rho v_{2}^{2}\dots$(i)
Here, $p_{1}=\rho_{m} g h_{1}=13600 \times 9.8 \times 10^{-2}$
$p _{2}=13600 \times 9.8 \times h$
$\rho=1000 kg - m ^{-3} $
$v _{1}=35 \times 10^{-2} ms ^{-1}, v _{2}=65 \times 10^{-2} ms ^{-1}$
$\therefore $ From Eq. (i), we get
$\Rightarrow 13600 \times 9.8 \times 10^{-2}+\frac{1}{2} \times 1000 \times(0.35)^{2}$
$=13600 \times 9.8 \times h +\frac{1}{2} \times 1000 \times(0.65)^{2}$
After solving, $h =0.89 \,cm$ of $Hg$