Question

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

In horizontal pipe
$P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$
Here, $P_1={\rho }_{m}g{h}_1=13600\times 9.8\times 10^{-2}$ $P^2=13600\times 9.8\times h$
$\rho =1000kg{m}^{-3}$
$v_1=35\times 10^{-2}m{s}^{-1}$
$v_2=65\times 10^{-2}m{s}^{-1}$
$\therefore$ From Eq. (i). $13600\times 9.8\times 10^{-2}+\frac{1}{2}\times 1000\times (0.35{)}^2$
$=13600\times 9.8\times h+\frac{1}{2}1000\times (0.65{)}^2$
After solving, 0.89 cm of Hg.