Mechanical Properties of Fluids

Radius of the mercury droplet, r = 3.00 mm = 3 $*{10}^{-3}$ m

Surface tension, S = 4.65 $*{10}^{-1}$ N/m

Atmospheric pressure,  $P_o$ = 1.01 $*{10}^5$ Pa

Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure

= $\frac{2S}{r}$ + $P_o$ = $\frac{2*4.65\mathrm{}*{10}^{-1}}{3*{10}^{-3}}$ + 1.01 $*{10}^5$ = 1.01310 $*{10}^5$ Pa

Excess pressure inside mercury = $\frac{2S}{r}$ = $\frac{2*4.65\mathrm{}*{10}^{-1}}{3*{10}^{-3}}$ = 310 Pa

The length of the liquid film supported by the weight, l = 40 cm = 0.4 m

The weight supported by the film, W = 4.5 $*{10}^{-2}$ N

Since a liquid film has two free surfaces,

Surface tension = $\frac{W}{2l}$ = $\frac{4.5\mathrm{}*{10}^{-2}}{2*0.4}$ = 5.625 $*{10}^{-2}$ N/m

In all 3 figures, the liquid is the same, temperature is also the same. Hence the surface tension in (b) and (c) are also going to be the same, with the value of 5.625 $*{10}^{-2}$ N/m and weight supported in each case is also going to be the same, since length of the film is same.

The weight that the soap film supports, W = 1.5 $*{10}^{-2}$ N

Length of the slider, l = 30 cm = 0.3 m

A soap film has two free surfaces, hence total length = 2l = 0.6 m

Surface tension, S= $\frac{Forces/Weight}{2l}$ = $\frac{1.5\mathrm{}*{10}^{-2}}{2*0.3}$ = 2.5 $*{10}^{-2}$ N/m

Area of cross-section of the spray pump, $A_1$ = 8 ${cm}^2$ = 8 $*{10}^{-4}$ $m^2$

Number of holes, n = 40

Diameter of each hole, d = 1 mm = 1 $*{10}^{-3}$ m

Radius of each hole, r = d/2 = 0.5 $*{10}^{-3}$ m

Area of cross-section of each hole, a = $\pi r^2$ = $\pi ({0.5\mathrm{}*{10}^{-3})}^2$

Total area of 40 holes, $A_2$ = 40 $\pi *({0.5\mathrm{}*{10}^{-3})}^2$ = 3.14 $*{10}^{-5}$ $m^2$

Speed of liquid inside the tube, $V_1$ = 1.5 m/min = 0.025 m/s

Speed of ejection of liquid = $V_2$

According to law of continuity, we have $A_1{V}_1$ = $A_2{V}_2$ or $V_2=\frac{A_1{V}_1}{A_2}$ = $\frac{8*{10}^{-4}*0.025}{3.14*{10}^{-5}}$

= 0.637 m/s

Take the case given in figure (b)

$A_1$ = Area of pipe 1,  $A_2$ = Area of pipe 2,  $V_1$ = Speed of fluid in pipe 1,  $V_2$ = Speed of fluid in pipe 2

From the law of continuity, we have

$A_1{V}_1$ = $A_2{V}_2$

When the area of cross-section in the middle of the venturimeter is small, the speed of the flow of liquid through this part is more. According to Bernoulli's principle, if speed is more, the pressure is less. Pressure is directly proportional to height, hence the level of water in pipe 2 is less. Therefore, figure (a) is not possible.

It does not matter if one uses gauge pressure, instead of absolute pressure while applying Bernoulli's equation. There should be significantly different atmospheric pressures, where Bernoulli's equation is applied.

No, Bernoulli's equation cannot be used to describe the flow of water through a rapid in a river. In rapid, the flow is turbulent whereas Bernoulli's equation is applicable for laminar flow only.