Question

Water flows along a horizontal pipe whose cross-section is not constant. The pressure is $1cm$ 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. $0.89cm$ of $Hg$
• B. $8.9cm$ of $Hg$
• C. $0.5cm$ of $Hg$
• D. $1cm$ 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, $p_1={\rho }_{m}g{h}_1=13600\times 9.8\times 10^{-2}$
$p_2=13600\times 9.8\times h$
$\rho =1000kg-m^{-3}$
$v_1=35\times 10^{-2}m{s}^{-1},v_2=65\times 10^{-2}m{s}^{-1}$
$\therefore$ From Eq. (i), 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, $h=0.89cm$ of $Hg$