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

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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?

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

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

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

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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 (P_i) will be greater than the outside pressure (P_o).

Hence, Excess Pressure = P_i – P_o

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,

A₂  =  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 = (P_i – P_o ) × 4πr²

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

∴ dW = dF × ∆r = (P_i – P_o ) × 4πr² × ∆r       ……(2)

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

T (8πr∆r) = (P_i – P_o ) × 4πr² × ∆r

∴ (P_i – P_o ) = 2T/r

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

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

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8. Explain the capillary action.

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

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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, ρ_w

h = 1060kg/m³,

P_atm = P_o = 1atm

P_atm  = 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

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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 =?

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

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

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

$latex \therefore\;F_2=1250N$

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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

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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 = ?

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

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

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

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

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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 ,

v_T =?

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

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

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

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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

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

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

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

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

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

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

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

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

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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, d_w = 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,

$latex v=\sqrt{2gh}$

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

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

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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,

ρ_w = 10³kg/m³, assume h₁ = h₂ = 0,

v₁= 0 (valve is closed),

v₂ =?(after opening valve)

According to Bernoulli’s theorem,

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

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

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

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

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

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

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

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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°, ρ_w = 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

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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 = ?

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

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

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

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

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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,

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

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

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

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

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

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

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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

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

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

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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

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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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