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Question: A non-viscous liquid of constant density \[1000\,{\text{kg}}\,{{\text{m}}^{ - 3}}\] flows in a strea...

A non-viscous liquid of constant density 1000kgm31000\,{\text{kg}}\,{{\text{m}}^{ - 3}} flows in a streamlined 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 PP and QQ at heights of 2m2\,{\text{m}} and 5m5\,{\text{m}} are respectively 4×103m24 \times {10^{ - 3}}\,{{\text{m}}^2} and 8×103m28 \times {10^{ - 3}}\,{{\text{m}}^2} . The velocity of the liquid at point PP is 1ms11\,{\text{m}}\,{{\text{s}}^{ - 1}} . Then,

A. Work done by pressure per unit volume is 29625Jm3 - 29625\,{\text{J}}\,{{\text{m}}^{ - 3}}
B. Work done by pressure per unit volume is +29625Jm3 + 29625\,{\text{J}}\,{{\text{m}}^{ - 3}}
C. Work done by gravity is 30000Jm3 - 30000\,{\text{J}}\,{{\text{m}}^{ - 3}}
D. Work done by gravity is +30000Jm3 + 30000\,{\text{J}}\,{{\text{m}}^{ - 3}}

Explanation

Solution

First of all, we will use the equation of continuity to find the velocity at the tube QQ . After that we will use Bernoulli's theorem to find the work done by pressure. Lastly, we will use the expression for the work done by gravity to calculate the value. Multiple options may be correct.

Complete step by step answer:
In the given question, we are supplied the following information:
The density of the non-viscous liquid is 1000kgm31000\,{\text{kg}}\,{{\text{m}}^{ - 3}} .The area of the cross section of the tube PP is 4×103m24 \times {10^{ - 3}}\,{{\text{m}}^2} .The area of the cross section of the tube QQ is 8×103m28 \times {10^{ - 3}}\,{{\text{m}}^2} .The velocity of the liquid at point PP is 1ms11\,{\text{m}}\,{{\text{s}}^{ - 1}} .We are asked to find the amount of work done and its type.

To begin with, we will use the equation of continuity at the tube PP and QQ .
The equation of continuity is given below:
aP×vP=aQ×vQ{a_{\text{P}}} \times {v_{\text{P}}} = {a_{\text{Q}}} \times {v_{\text{Q}}} …… (1)
Where,
aP{a_{\text{P}}} indicates the area of the cross section of PP .
vP{v_{\text{P}}} indicates the velocity of fluid at PP .
aQ{a_{\text{Q}}} indicates the area of the cross section of QQ .
vQ{v_{\text{Q}}} indicates the velocity of fluid at QQ .
Now,
aP×vP=aQ×vQ vQ=aPaQ×vP vQ=4×103m28×103m2×1 vQ=0.5ms1{a_{\text{P}}} \times {v_{\text{P}}} = {a_{\text{Q}}} \times {v_{\text{Q}}} \\\ \Rightarrow {v_{\text{Q}}} = \dfrac{{{a_{\text{P}}}}}{{{a_{\text{Q}}}}} \times {v_{\text{P}}} \\\ \Rightarrow {v_{\text{Q}}} = \dfrac{{4 \times {{10}^{ - 3}}\,{{\text{m}}^2}}}{{8 \times {{10}^{ - 3}}\,{{\text{m}}^2}}} \times 1 \\\ \Rightarrow {v_{\text{Q}}} = 0.5\,{\text{m}}\,{{\text{s}}^{ - 1}}
Therefore, the velocity of the fluid at QQ is 0.5ms10.5\,{\text{m}}\,{{\text{s}}^{ - 1}} .

From Bernoulli's theorem at points PP and QQ .
pP+12ρvP2+ρghP=pQ+12ρvQ2+ρghQ{p_{\text{P}}} + \dfrac{1}{2}\rho v_{\text{P}}^2 + \rho g{h_{\text{P}}} = {p_{\text{Q}}} + \dfrac{1}{2}\rho v_{\text{Q}}^2 + \rho g{h_{\text{Q}}}
Expression for the work done by pressure is given below:
pPpQ=12ρ(vQ2vP2)+ρg(hQhP)\Rightarrow {p_{\text{P}}} - {p_{\text{Q}}} = \dfrac{1}{2}\rho \left( {v_{\text{Q}}^2 - v_{\text{P}}^2} \right) + \rho g\left( {{h_{\text{Q}}} - {h_{\text{P}}}} \right) …… (2)
Where,
ρ\rho indicates the density of the liquid.
gg indicates the acceleration due to gravity.
pp indicates pressure exerted by the liquid.

All the terms in the equation (2) represents energy per unit volume of the liquid.
Substituting the required values in the equation (2), we get:
pPpQ=12×1000×(0.5212)+1000×10×(52) pPpQ=500×(0.75)+10000×3 pPpQ=29.625×103Jm3{p_{\text{P}}} - {p_{\text{Q}}} = \dfrac{1}{2} \times 1000 \times \left( {{{0.5}^2} - {1^2}} \right) + 1000 \times 10 \times \left( {5 - 2} \right) \\\ \Rightarrow {p_{\text{P}}} - {p_{\text{Q}}} = 500 \times \left( { - 0.75} \right) + 10000 \times 3 \\\ \Rightarrow {p_{\text{P}}} - {p_{\text{Q}}} = 29.625 \times {10^3}\,{\text{J}}\,{{\text{m}}^{ - 3}}
Therefore, work done by pressure is 29.625×103Jm329.625 \times {10^3}\,{\text{J}}\,{{\text{m}}^{ - 3}} .

Now, we calculate the work done by gravity is:
W=ρghPρghQ W=ρg(hPhQ) W=1000×10×(25) W=3×104Jm3W = \rho g{h_{\text{P}}} - \rho g{h_{\text{Q}}} \\\ \Rightarrow W = \rho g\left( {{h_{\text{P}}} - {h_{\text{Q}}}} \right) \\\ \Rightarrow W = 1000 \times 10 \times \left( {2 - 5} \right) \\\ \therefore W = - 3 \times {10^4}{\text{J}}\,{{\text{m}}^{ - 3}}
Therefore, work done by gravity is 3×104Jm3 - 3 \times {10^4}{\text{J}}\,{{\text{m}}^{ - 3}} .

The correct options are B and C.

Note: In the given question, we must remember that the work done by the gravity is always negative when the substance or the object goes up. This is due to the fact that gravity tries to decrease the kinetic energy of the substance.