Solved Examples on Liquid at Rest:-

Question 1:-A thin film of water of thickness 80.0 pm is sandwiched between two glass plates and forms a circular patch of radius 12.0 cm. Calculate the normal force needed to separate the plates if the surface tension of water is 0.072 N/m.

Concept:-Surface tension is the surface force per unit length over which it acts.

The circumference of the thin films of circular patch of radius r between the glass plates is

L= 2πrThe surface tension acting on the thin films between the glass plates is

γ=F/LSubstitute

L= 2πrin the above equation, the surface tension becomes

γ=F/2πr

Solution:-From the above equation, the normal force needed to separate the plates is

F= 2πrγTo obtain the normal force, substitute 12.0 cm for the radius

rof the thin circular patch,0.072 N/m for the surface tension of the water in the equationF= 2πrγ,

F= 2πrγ= 2(3.14) (12.0 cm) (10

^{-2}m/1 cm) (0.072 N/m)= 5.42592×10

^{-2}NRounding off to three significant figures, the normal force needed to separate the plates is 5.43×10

^{-2}N._________________________________________________________________________________

Question 2:-A cylindrical barrel has a narrow tube fixed to the top, as shown with dimensions in below figure. The vessel is filled with water to the top of the tube. Calculate the ratio of the hydrostatic force exerted on the bottom of the barrel to the weight of the water contained inside. Why is the ratio not equal to one? (Ignore the presence of the atmosphere.)

Concept:-The pressure at the bottom of the cylindrical barrel is

p=ρghHere, density of the liquid is

ρ, acceleration due to gravity isgand height of the cylindrical barrel ish.The hydrostatic force acting at the bottom of the barrel is

F=pAHere, surface area of the barrel at the bottom is

A.The weight of the liquid is

W=ρgVHere, total volume of the cylindrical barrel is

V.

Solution:-Now, the total height of the cylindrical barrel is

h= 1.8 m + 1.8 m= 3.6 m

From above figure given in the problem, substitute 3.6 m for

hin the equationp=ρghgives

p=ρgh=

ρg(3.6 m)This gives the pressure at the bottom of the barrel.

The volume of the thin barrel at the top is

V_{1}= (4.6 cm^{2}) (1.8 m)= (4.6 cm

^{2}) (10^{-2}m/1 cm)^{2}(1.8 m)= 8.28×10

^{-4}m^{3}The radius of the barrel at the lower part is

r= 1.2 m/2= 0.6 m

The volume of the barrel at the lower part is

V_{2}= πr^{2}(1.8 m)= (3.14) (0.6 m)

^{2}(1.8 m)= 2.03472 m

^{3}Now, the total volume of the cylindrical barrel is

V=V_{1}+V_{2}= 8.28×10

^{-4}m^{3}+ 2.03472 m^{3}= 2.035548 m

^{3}The surface area of the barrel at the bottom is

A= πr^{2}= (3.14) (0.6 m)

^{2}= 1.1304 m

^{2}To obtain the hydrostatic force exerted by the liquid at the bottom, substitute

ρg(3.6 m) forpand 1.1304 m^{2}forAin the equationF=pAgives

F=pA=

ρg(3.6 m) (1.1304 m^{2})=

ρg(4.06944 m^{3})To obtain the weight of the liquid in the cylindrical barrel, substitute 2.035548 m

^{3}forVin the equationW=ρgVgives,

W=ρgV=

ρg(2.035548 m^{3})Now, the ratio of the hydrostatic force to the weight of the liquid is

F/W=ρg(4.06944 m^{3})/ρg(2.035548 m^{3})= 1.9992

Rounding off to two significant figures, the ratio of the hydrostatic force to the weight of the liquid is 2.0.

The hydrostatic pressure is depending on the height of the liquid column. So, the same amount of liquid when taken in different volume of containers having different heights will not be the same. Weight is the volume times the density of the liquid. Therefore, the ratio is not equal to one.

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Question 3:-Below figure shows the confluence of two streams to form a river. One stream has a width of 8.2 m, depth of 3.4 m, and current speed of 2.3 m/s. The other stream is 6.8 m wide, 3.2 m deep, and flows at 2.6 m/s. The width of the river is 10.7 m and the current speed is 2.9 m/s. What is its depth?

Concept:-The volume flow rate of the main river is equal to the sum of the volume flow rates of the two streams.

Assume that the cross-sectional area of the streams and the main river is rectangular in shape. The area of the streams is equal to the width of the streams times the depth of the streams. The area of the main river is equal to the width of the main river times the depth of the main river.

Solution:-The cross sectional area of the first stream is

A_{1}=x_{1}d_{1}Here, width of the first stream is

x_{1}and depth of the stream isd_{1}.The volume flow rate of the first stream is

R_{1}=A_{1}v_{1}=

x_{1}d_{1}v_{1}Here, speed of the first stream is

v_{1}.Similarly, the volume rate flow of the second stream is

R_{2}=x_{2}d_{2}v_{2}Here, width of the second stream is

x_{2}, speed of the water in the second stream isv_{2}and depth of the second stream isd_{2}.Now, the volume flow rate of the main river is

R_{3}=x_{3}d_{3}v_{3}Here, width of the main river is

x_{3}, speed of the water in the main river isv_{3}and depth of the main river isd_{3}.From the equation of continuity, the volume flow rate of the main river is equal to the sum of the volume flow rates of the two streams. It is given as

R_{3}=R_{1}+R_{2}

x_{3}d_{3}v_{3}=x_{1}d_{1}v_{1}+x_{2}d_{2}v_{2}

d_{3}= [x_{1}d_{1}v_{1}+x_{2}d_{2}v_{2}]/[x_{3}v_{3}]This gives the depth of the main river.

Substitute 8.2 m for

x_{1}, 3.4 m ford_{1}, 2.3 m/s forv_{1}, 6.8 m forx_{2}, 3.2 m ford_{2}, 2.6 m/s forv_{2}, 10.7 m forx_{3}and 2.9 m/s forv_{3}in the above equation gives,

d_{3}= [x_{1}d_{1}v_{1}+x_{2}d_{2}v_{2}]/[x_{3}v_{3}]= [(8.2 m) (3.4 m) (2.3 m/s) + (6.8 m) (3.2 m) (2.6 m/s)]/[(10.7 m) (2.9 m/s)]

= [(64.124 m

^{3}/s) + (56.576 m^{3}/s)]/[31.03 m^{2}/s]= 3.8898 m

Rounding off to two significant figures, the depth of the main river is 3.9 m.