Solved Examples for JEE: Liquids at Rest | NCERT Exemplar & Revision Notes for NEET PDF Download

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 = 2πr
The surface tension acting on the thin films between the glass plates is
γ = F/L
Substitute L = 2πr in 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 = 2πrγ
To obtain the normal force, substitute 12.0 cm for the radius r of the thin circular patch,0.072 N/m  for the surface tension of the water in the equation F = 2πrγ,
F = 2πrγ
= 2(3.14) (12.0 cm) (10^-2 m/1 cm) (0.072 N/m)
= 5.42592×10^-2 N
Rounding 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 = ρgh
Here, density of the liquid is ρ, acceleration due to gravity is g and height of the cylindrical barrel is h.
The hydrostatic force acting at the bottom of the barrel is
F = pA
Here, surface area of the barrel at the bottom is A.
The weight of the liquid is
W = ρgV
Here, 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 h in the equation p = ρgh gives
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
V₁ = (4.6 cm²) (1.8 m)
= (4.6 cm²) (10^-2 m/1 cm)² (1.8 m)
= 8.28×10^-4 m³
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
V₂ = πr² (1.8 m)
= (3.14) (0.6 m)² (1.8 m)
= 2.03472 m³
Now, the total volume of the cylindrical barrel is
V = V₁+V₂
= 8.28×10^-4 m³ + 2.03472 m³
= 2.035548 m³
The surface area of the barrel at the bottom is
A = πr²
= (3.14) (0.6 m)²
= 1.1304 m²
To obtain the hydrostatic force exerted by the liquid at the bottom, substitute ρg(3.6 m) for p  and 1.1304 m² for A in the equation F = pA gives
F = pA
= ρg (3.6 m) (1.1304 m²)
= ρg (4.06944 m³)
To obtain the weight of the liquid in the cylindrical barrel, substitute 2.035548 m³ for V in the equation W = ρgV gives,
W =ρgV
= ρg (2.035548 m³)
Now, the ratio of the hydrostatic force to the weight of the liquid is
F/W = ρg (4.06944 m³)/ ρg (2.035548 m³)
= 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.

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
A₁ = x₁d₁
Here, width of the first stream is x₁ and depth of the stream is d₁.
The volume flow rate of the first stream is
R₁ = A₁v₁
= x₁d₁v₁
Here, speed of the first stream is v₁.
Similarly, the volume rate flow of the second stream is
R₂ = x₂d₂v₂
Here, width of the second stream is x₂, speed of the water in the second stream is v₂ and depth of the second stream is d₂.
Now, the volume flow rate of the main river is
R₃ = x₃d₃v₃
Here, width of the main river is x₃, speed of the water in the main river is v₃ and depth of the main river is d₃.
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
R₃ = R₁+R₂
x₃d₃v₃ = x₁d₁v₁+ x₂d₂v₂
d₃ = [x₁d₁v₁+ x₂d₂v₂]/[ x₃v₃]
This gives the depth of the main river.
Substitute 8.2 m for x₁, 3.4 m for d₁, 2.3 m/s  for v₁, 6.8 m for x₂, 3.2 m for d₂, 2.6 m/s  for v₂, 10.7 m for x₃ and 2.9 m/s for v₃ in the above equation gives,
d₃ = [x₁d₁v₁+ x₂d₂v₂]/[ x₃v₃]
= [(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 m³/s) + (56.576 m³/s)]/[31.03 m²/s]
= 3.8898 m
Rounding off to two significant figures, the depth of the main river is 3.9 m.