Question

A non viscous liquid of constant density $1000kg/m^3$ flows in a streamline motion along a tube of variable cross section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross section of the tube at two points $P$ and $Q$ at heights of $2m$ and $5m$ are respectively $4\times {10}^{-3}{m}^2$ and $8\times {10}^{-3}{m}^3$. The velocity of the liquid at point $P$ is $1m/s$. Then,

• A. work done by pressure per unit volume is $-29625J/m^3$
• B. work done by pressure per unit volume is $+29625J/m^3$
• C. work done by gravity is $-30000J/m^3$
• D. work done by gravity is $+30000J/m^3$

Answer 1

Answer: (B), (C)

The correct options are
B work done by gravity is $-30000J/m^3$
C work done by pressure per unit volume is $+29625J/m^3$

From the equation of continuity at P and Q, we have
$a_a{v}_1=a_2{v}_2$
$\therefore v_2=\frac{a_1}{a_2}\times v_1=\frac{4\times {10}^{-3}}{8\times {10}^{-3}}\times 1=\frac{1}{2}m{s}^{-1}$

From Bernoulli's theorem at points P and Q,
$p_1+\frac{1}{2}\rho v_1^2+\rho g{h}_1=p_2+\frac{1}{2}\rho v_2^2+\rho g{h}_2$

Each term in the above eqn represents energy per unit volume.
Work done by the pressure per unit volume is:
$p_1-p_2=\frac{1}{2}\rho (v_2^2-v_1^2)+\rho g(h_2-h_1)$

$=\frac{1}{2}\times 1000({0.5}^2-1^2)+1000\times 10(5-2)$

$=\frac{1}{2}\times 1000\times (-0.75)+1000\times 10\times 3$

$=-0.375\times {10}^3+30\times {10}^3$

$=29.625\times {10}^{3}J{m}^{-3}$
Work done by gravity:
$W=\rho g{h}_1-\rho g{h}_2=-1000\times 10\times (5-2)=-3\times {10}^{4}J{m}^{-3}$