LinkedIn Insight Solutions for Exercises of Mechanical Properties of Fluids - Grad Plus

Solutions for Exercises of Mechanical Properties of Fluids

1) Multiple Choice Questions

 i) A hydraulic lift is designed to lift heavy objects of maximum mass 2000 kg. The area of cross section of piston carrying the load is 2.25 x 10-2 m2. What is the maximum pressure the smaller piston would have to bear?

(A) 8711 x 106 N/m2

(B) 5862 x 107 N/m2

(C) 4869 x 105 N/m2

(D) 3271 x 104 N/m2

ii) Two capillary tubes of radii 0.3 cm and 6 cm are dipped in the same liquid. The ratio of heights through which the liquid will rise in the tubes is

(A) 1:2

(B) 2:1

(C) 1:4

(D) 4:1

iii) The energy stored in a soap bubble of diameter 6 cm and T = 0.04 N/m is nearly

(A) 0.9 x 10-3 J

(B) 0.4 x 10-3J

(C) 0.7 x 10-3 J

(D) 0.5 x 10-3 J

iv) Two hail stones with radii in the ratio of 1:4 fall from a great height through the atmosphere. Then the ratio of their terminal velocities is

(A) 1:2

(B) 1:12

(C) 1:16

(D) 1:8

v) In Bernoulli’s theorem, which of the following is conserved?

(A) linear momentum

(A) angular momentum

(A) mass

(A) energy


2) Answer in brief.

i) Why is the surface tension of paints and lubricating oils kept low?

ii) How much amount of work is done in forming a soap bubble of radius r?

iii) What is the basis of the Bernoulli’s principle?

iv) Why is a low density liquid used as a manometric liquid in a physics laboratory?

v) What is an incompressible fluid?


3. Why two or more mercury drops form a single drop when brought in contact with each other?


4. Why does velocity increase when water flowing in broader pipe enters a narrow pipe?


5. Why does the speed of a liquid increase and its pressure decrease when a liquid passes through constriction in a horizontal pipe?


6. Derive an expression of excess pressure inside a liquid drop.

Consider a free liquid drop. As it is spherical in shape, the inside pressure (Pi) will be greater than the outside pressure (Po).

Hence, Excess Pressure = Pi – Po

Let the radius of the drop increases from r to (r + Δr), where Δr is very small so that inside pressure is almost constant.

Initial surface area, A1 = 4 π r2

Final surface area is given as,

A=  4π(r+∆r)2

A2 = 4π (r2 + 2r∆r + ∆r2 )

A2 = 4πr2 + 8πr∆r + 4π ∆r2

As ∆r is very small we can neglect 4π∆r2

∴ A2 = 4πr2 + 8πr∆r

∴ Increase in Surface area,

dA = A2– A1

∴ dA = 4πr2 + 8πr∆r – 4πr2

∴ dA  = 8πr∆r

Work done to increase the surface area 8πr∆r is equal to surface energy.

∴ dW = T dA = T (8πr∆r)     ……(1)

Also, Excess force = Excess Pressure × Area

∴ dF = (Pi – Po ) × 4πr2

But, work done is also equal to force multiplied distance.

∴ dW = dF × ∆r = (Pi – Po ) × 4πr2 × ∆r       ……(2)

From equations (1) and (2) we have,

T (8πr∆r) = (Pi – Po ) × 4πr2 × ∆r

∴ (Pi – Po ) = 2T/r

This is an expression of excess pressure inside a liquid drop.


7. Obtain an expression for conservation of mass starting from the equation of mass starting from the equation of continuity.


8. Explain the capillary action.


9. Derive an expression for capillary rise for a liquid having a concave meniscus.


10. Find the pressure 200 m below the surface of the ocean if pressure on the free surface of liquid is one atmosphere.(Density of sea water = 1060 kg/m3)

h = 200m, ρw

h = 1060kg/m3,

Patm = Po = 1atm

Patm  = 1.013 × 105N/m2, P = ?

P = P + hρg

∴ P = (1.013 × 105) + (200 × 1060 × 9.8)

∴ P = (1.013 × 105) + (20.77 × 105)

∴ P = 21.789 × 105N/m2


11. In a hydraulic lift, the input piston had surface area 30 cm2 and the output piston has surface area of 1500 cm2. If a force of 25 N is applied to the input piston, calculate weight on output piston.

A1 = 30cm2, A2 = 1500cm2, F1 = 25N, F2 =?

$latex P_1\;=\;P_2\;\;\;\;\;\;\;\;\;\;\therefore\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$latex \therefore\;F_2=\frac{F_1}{A_1}\times A_2$

$latex \therefore\;F_2=\frac{25}{30}\times1500$

$latex \therefore\;F_2=1250N$


12. Calculate the viscous force acting on a rain drop of diameter 1 mm, falling with a uniform velocity 2 m/s through The coefficient of viscosity of air is 1.8 x 10-5 Ns/m2.

D = 1mm              ∴ r = 0.5mm = 0.5 × 10-3m ,

v = 2 m/s, η = 1.8 × 10-5Nsm-2, F = ?

Viscous force is given as,

⇒F = 6πηrv

∴ F = 6 × 3.14× 1.8 × 10-5 × 0.5 × 10-3 × 2

∴ F = 33.928 × 10-8

∴ F = 3.3928 × 10-7N


13. A horizontal force of 1 N is required to move a metal plate of area 10-2 m2 with a velocity of 2 x 10-2 m/s, when it rests on a layer of oil 1.5 x 10-3 m thick. Find the coefficient of viscosity of oil.

A = 10-2m2, dv = 2 × 10-2m/s, F = 1 N

dx = 1.5 × 10-3m, η = ?

$latex F=\eta\;\times\;A\;\times\;\frac{dv}{dx}$

$latex \therefore\;\eta=\frac FA\times\frac{dx}{dv}$

$latex =\frac1{10^{-2}}\times\frac{1.5\times10^{-3}}{2\times10^{-2}}$

$latex =7.5\;Ns/m^2$


14. With what terminal velocity will an air bubble 0.4 mm in diameter rise in a liquid of viscosity 0.1 Ns/m2 and specific gravity 0.9? Density of air is 1.29 kg/m3.

ρ = 1.29kg/m3, σ = 0.9 × 1000 = 900kg/m3,

η = 0.1Nsm-2,

D = 0.4mm ∴ r = 0.2mm = 0.2 × 10-3m ,

vT =?

$latex v_T=\frac29\frac{r^2g}\eta(\rho-\sigma)$

$latex \therefore v_T=\frac29\times\frac{0.2\times10^{-3}\times10}{0.1}\times(1.29-900)$

$latex \therefore v_T=-0.782\times10^{-3}m/s$


15. The speed of water is 2m/s through a pipe of internal diameter 10 cm. What should be the internal diameter of nozzle of the pipe if the speed of water at nozzle is 4 m/s?

v1 = 2m/s, d1 = 10cm = 10 × 10-2m = 10m-1,

d2 =?, v2 = 4m/s

According to equation of continuity,

A1v1 = A2v2

$latex \therefore\pi\left(\frac{d_1}2\right)^2V_1=\pi\left(\frac{d_2}2\right)^2V_2$

$latex \therefore d_2^2=\frac{d_1^2v_1}{v_2}$

$latex =\frac{\left(10^{-1}\right)^2\times2}4$

$latex =0.5\times10^{-2}$

$latex \therefore\;d_2=\sqrt{0.5\times10^{-2}}$

$latex \therefore\;d_2\;=\;0.707\times10^{-1}$

$latex \therefore\;d_2=\sqrt{0.5\times10^{-2}}$

$latex =7.07\times10^{-2}m$


16. With what velocity does water flow out of an orifice in a tank with gauge pressure 4 x 105 N/m2 before the flow starts? (Density of water = 1000 kg/m3).

P = 4 × 105N/m2, dw = 103kg/m3, v = ?

We know that,

$latex P=\rho hg\;\;\;\;\;\;\;\;\;\;\;\therefore h=\frac P{\rho g}$

$latex \therefore\;h=\frac{4\times10^5}{10^3\times10}=40m$

Hence velocity of water is given as,

$latex v=\sqrt{2gh}$

$latex \therefore\;v=\sqrt{2\times10\times40}$

$latex \sqrt{800}=28.28\;m/s$


17. The pressure of water inside the closed pipe is 3 x 105 N/m2. This pressure reduces to 2 x 105 N/m2 on opening thevalue of the pipe. Calculate the speed of water flowing through the pipe. (Density of water = 1000 kg/m3).

P1 = 3 × 105N/m2 , P2 = 2 × 105N/m2,

ρw = 103kg/m3, assume h1 = h2 = 0,

v1= 0 (valve is closed),

v2 =?(after opening valve)

According to Bernoulli’s theorem,

we have, $latex P=\rho gh+\frac12pv^2=cons\tan t$

$latex \therefore P_1+\rho gh_1+\frac12\rho v_1^2=P_2+\rho gh_2+\frac12\rho v_2^2$

$latex 3\times10^5+0+0=2\times10^5+0+\frac12\times10^3\times v_2^2$

$latex \therefore v_2^2=\frac{2\times(3\times10^5-2\times10^5)}{10^3}$

$latex =\frac{2\times10^5}{10^3}=2\times10^2=200$

$latex \therefore v_2\;=\;\sqrt{200}$

$latex \therefore v_2\;=\;14.14m/s$


18. Calculate the rise of water inside a clean glass capillary tube of radius 1 mm, when immersed in water of surface tension 7 x 10-2 N/m. The angle of contact between water and glass is zero, density of water = 1000 kg/m3, g = 9.8 m/s2.

r = 0.1mm = 0.1 × 10-3m, T = 7 × 10-2N/m

θ = 0°, ρw = 103kg/m3, h = ?

According to Capillary action

we have, $latex h=\frac{2T}{r\rho g}$

$latex h=\frac{2\times7\times10^{-2}}{0.1\times10^{-3}\times10^3\times9.8}$

$latex h=\frac{14\times10^{-2}}{0.98}$

∴ h = 14.28 × 10-2m

∴ h = 0.1428m


19. An air bubble of radius 0.2 mm is situated just below the water surface. Calculate the gauge pressure. Surface tension of water = 7.2 x 10-2 N/m.

r = 0.2mm = 0.2 × 10-3m, T = 7.2 × 10-2N/m,

Gauge Pressure, P = ?

$latex P=\frac{2T}r$

$latex P=\frac{2\times7.2\times10^{-2}}{0.2\times10^{-3}}$

$latex P\;=\;72\times10$

$latex P\;=\;720\;N/m^2$


20. Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single Find the change in surface energy. Surface tension of water is 0.072 N/m.

r = 0.1mm = 0.1 × 10-3m,

T = 0.072 N/m, ΔW = ?

Let R be the radius of single drop.

As drops coalesce into a single drop, we have volume of big drop equal to volume of 27 small drops.

$latex \therefore\;\frac43\pi R^3=27\times\frac43\pi^3$

$latex \therefore\;R^3=27r^3$

$latex \therefore\;R\;=\;3r$

$latex \therefore\;R\;=\;3\;\times\;0.1\times10^{-3}$

$latex \therefore\;R\;=\;0.3\times10^{-3}m$

Now we have,

Surface Energy = (Surface Tension) × (Surface Area)

Hence surface energy of 27 drops is given as,

$latex W_1=T\times4\pi r^2$

$latex \therefore W_1=0.072\times4\times3.14\times{(0.1\times10^{-3})}^2$

$latex \therefore W_1=2.44\times10^{-7}J$

And surface energy of single big drop is given as,

$latex \therefore W_2=T\times4\pi R^2$

$latex \therefore W_2=0.072\times4\times3.14\times{(0.3\times10^{-3})}^2$

$latex \therefore W_2=0.81\times10^{-7}J$

$latex \therefore\triangle W=\left|W_2-W_1\right|$

$latex \therefore\triangle W=\left|0.81-2.44\right|\times10^{-7}$

$latex \therefore\triangle W=1.63\times10^{-7}J$


21. A drop of mercury of radius 0.2 cm is broken into 8 droplets of the same size. Find the work done if the surface tension of mercury is 435.5 dyne/cm.

R = 0.2cm = 0.2 × 10-2m, n = 8, W = ?

$latex T\;=\;435.5\frac{dyne}{cm}$

$latex T\;=\;435.5\times\frac{10^{-5}}{10^{-2}}\frac Nm$

$latex T\;=\;435.5\times10^{-3}\frac Nm$

Let r be the radius of small droplet. We have volume of big drop equal to volume of 8 small droplets.

$latex \therefore\frac43\pi R^3=8\times\frac43\pi r^3\;\;\;\;\;\;\therefore R^3=8r^3$

$latex \therefore R=2r\;\;\;\;\;\;\;\;\therefore r=\frac R2$

∴ r = 0.1cm = 0.1 × 10-2m = 10-3m

Change in surface area is the difference between total surface area of 8 droplets & surface area of big drop.

$latex \therefore\triangle A=(8\times4\pi r^2)-(4\pi R^2)$

$latex =4\pi(8r^2-R^2)$

$latex \therefore\triangle A=4\pi\left[8\left(\frac R2\right)^2-R^2\right]$

$latex \therefore\triangle A\;=\;4\pi\left[2R^2-R^2\right]$

$latex \therefore\triangle A\;=\;4\pi R^2$

$latex \therefore\triangle A=4\times3.14\times{(0.2\times10^{-2})}^2$

$latex \therefore\triangle A\;=\;0.5\times10^{-4}$

Hence work done is given as,

W = T × ∆A

$latex W=435.5\times10^{-3}\times0.5\times10^{-4}$

$latex W=218.5\times10^{-7}\times2.18\times10^{-50}J$


22. How much work is required to form a bubble of 2 cm radius from the soap solution having surface tension 0.07 N/m.

r = 2cm = 2 × 10-2m, T = 0.07N/m, W = ?

Work done is given as,

⇒W = 2 × T × ΔA

∴ W = 2 × 0.07× 4 × π × (2 × 10-2)2

∴ W = 1.76 × 4 × 10-4

∴ W = 7.033 × 10-4

∴ W = 0.7033 × 10-3J


23. A rectangular wire frame of size2 cm x 2 cm, is dipped in a soap solution and taken out. A soap film is formed, if the size of the film is changed to 3 cm x 3 cm, calculate the work done in the process. The surface tension of soap film is 3 x 10-2 N/m.

A1 = 2cm × 2cm = 4cm2 = 4 × 10-4m2,

A = 3cm × 3cm = 9cm2 = 9 × 10-4m2,

T = 3 × 10-2N/m, W = ?

ΔA = A2 − A1 = (9 − 4) × 10-4m2

ΔA = 5 × 10-4m2

W = 2 × T × ΔA

∴ W = 2 × 3 × 10-2 × 5 × 10-4

∴ W = 3 × 10-5J


 

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