P HYSICS F UNDAMENTAL F OR IIT-J EE
KEY CONCEPTS & PROBLEM SOLVING STRATEGY
tension of a liquid depends on temperature. It decreases with rise in temperature.
Excess of Pressure : Inside a soap bubble or a gas bubble inside a liquid, there must be pressure in excess of the outside pressure to balance the tendency of the liquid surface to contract due to surface tension.
p(excess of pressure) = T
+
2
1 r
1 r
1 in general
where T is surface tension of the liquid, and r1 and r2
are the principal radii of curvature of the bubble in two mutually perpendicular directions.
For a spherical soap bubble, r1 = r2 = r and there are two free surfaces of the liquid.
∴ p = r
T 4
For a gas bubble inside a liquid, r1 = r2 = r and there is only one surface.
∴ p = r
T 2
For a cylindrical surface r1 = r and r2 = ∞ and there are two surfaces.
∴ p = r
T 2
Angle of Contact : The angle made by the surface of a liquid with the solid surface inside of a liquid at the point of contact is called the angle of contact. It is at this angle that the surface tension acts on the wall of the container.
The angle of contact θ depends on the natures of the liquid and solid in contact. If the liquid wets the solid (e.g., water and glass), the angle of contact is zero. In most cases, θ is acute (figure i). In the special case of mercury on glass, θ is obtuse (figure ii).
θ fig. (i)
θ
fig. (ii)
Rise of Liquid in a Capillary Tube : In a thin (capacity) tube, the free surface of the liquid becomes curved. The forces of surface tension at the edges of the liquid surface then acquire a vertical component.
T θ θ T
θ θ
h
r meniscus
The upward force by which a liquid surface is pulled up in a capillary tube is 2πrTcos θ, and the downward force due to the gravitational pull on the mass of liquid in the tube is (πr2h + v)ρg, where v is the volume above the liquid meniscus. If θ = 0º, the meniscus is hemispherical in shape. Then v = difference between the volume of the cylinder of radius r and height r and the volume of the hemisphere of radius r
= πr3 –
32 πr3 = 31 πr3
When θ ≠ 0, we cannot calculate v which is generally very small and so it may be neglected. For equilibrium
(πr2h + v) ρg = 2πrT cos θ
When a glass capillary tube is dipper in mercury, the meniscus is convex, since the angle of contact is obtuse. The surface tension forces now acquire a downward component, and the level of mercury inside the tube the falls below the level outside it. the relation 2T cos θ = hρgr may be used to obtain the fall in the mercury level.
Problem Solving Strategy Bernoulli's Equations :
Bernoulli's equation is derived from the work-energy theorem, so it is not surprising that much of the problem-solving strategy suggested in W.E.P. also applicable here.
Step 1: Identify the relevant concepts : First ensure that the fluid flow is steady and that fluid is incompressible and has no internal friction. This case is an idealization, but it hold up surprisingly well for fluids flowing through sufficiently large pipes and for flows within bulk fluids (e.g., air flowing around an airplane or water flowing around a fish).
Step 2: Set up the problem using the following steps Always begin by identifying clearly the points 1
and 2 referred to in Bernoulli's equation.
Define your coordinate system, particular the level at which y = 0.
Make lists of the unknown and known quantities in Eq. p1 + ρgy1 +
21 ρv12 = p2 + ρgy2 + 1 ρv2 22
(Bernoulli's equation)
The variables are p1, p2, v1, v2, y1 and y2, and the constants are ρ and g. Decide which unknowns are your target variables.
Step 3: Execute the solutions as follows : Write Bernoulli's equation and solve for the unknowns. In some problems you will need to use the continuity equation, Eq. A1v1 = A2v2 (continuity equation, incompressible fluid), to get a relation between the two speeds in terms of cross-sectional areas of pipes
or containers. Or perhaps you will know both speeds and need to determine one of the areas. You may also need to use Eq.
dt
dV = Av (volume flow rate) to find the volume flow rate.
Step 4: Evaluate your answer : As always, verify that the results make physical sense. Double-check that you have used consistent units. In SI units, pressure is in pascals, density in kilograms per cubic meter, and speed in meters per second. Also note that the pressures must be either all absolute pressure or all gauge pressures.
1. A vertical U-tube of uniform cross-section contains mercury in both arms. A glycerine (relative density 1.3) column of length 10 cm is introduced into one of the arms. Oil of density 800 kg m–3 is poured into the other arm until the upper surface of the oil and glycerine are at the same horizontal level. Find the length of the oil column. Density of mercury is 13.6
× 103 kg m–3.
Sol. Draw a horizontal line through the mercury-glycerine surface. This is a horizontal line in the same liquid at rest namely, mercury. Therefore, pressure at the points A and B must be the same.
h
10 cm
(1 – h)
A B
Pressure at B
= p0 + 0.1 × (1.3 × 1000) × g Pressure at A
= p0 + h × 800 × g + (0.1 – h) × 13.6 × 1000g
∴ p0 + 0.1 × 1300 × g
= p0 + 800gh + 1360g – 13600 × g × h
⇒ 130 = 800h + 1360 – 13600h
⇒ h = 12800
1230 = 0.096 m = 9.6 cm
2. A liquid flows out of a broad vessel through a narrow vertical pipe. How are the pressure and the velocity of the liquid in the pipe distributed when the height of the liquid level in the vessel is H from the lower end of the length of the pipe is h ?
Sol. Let us consider three points 1, 2, 3 in the flow of water. The positions of the points are as shown in the figure.
H x h
•1
•2
•3 ρ0
p + v12 2
1 + gH = ρ p2
+ v22 2
1 + g (h – x)
= ρ0 p + v23
2 1 + 0 By continuity equation
v 1A1 = A2v 2 = A2v 3
Since A1 >> A2,v1 is negligible and v2 = v3 = n (say).
∴ ρ0 p + gH =
ρ2 p +
2
1v2 + g (h – x)
= ρ0 p +
2 1 v2
∴ v = 2gH (i)
and ρ0 p + gH =
ρ2
p + gH + g (h – x)
⇒ p0 + p2 + ρg (h – x)
⇒ p2 = p0 – ρg (h – x) (ii) Thus pressure varies with distance from the upper end of the pipe according to equation (ii) and velocity is a constant and is given by (i).
3. A rod of length 6m has a mass of 12 kg. It is hinged at one end at a distance of 3 m below the water surface.
(i) What weight must be attached to the other end so that 5 m of the rod is submerged ?
(ii) Find the magnitude and direction of the force exerted by the hinge on the rod. The specific gravity of the material of the rod is 0.5
Sol. Mass per unit length of the rod is 2 kg. Therefore, mass of the submerged portion of the rod is 10 kg and its volume =
500
10 m3 (using the simple formula,
volume = density
mass and density = specific gravity × 1000 kg m–3).
B W
θ C 12 kgf O´
N A H
3m
Fb
O
Solved Examples
Therefore, buoyant force Fb =
500
10 × 1000 = 20 kgf
Let N and H be the vertical downward and horizontal reactions of the hinge on the rod. Considering horizontal and vertical translational equilibrium of the rod, N + 12 + W = 20 (where W is the weight to be attached and H = 0).
N + W = 8 and H = 0 ..(i) Considering the rotational equilibrium of the rod about A immersed in water. Is mechanical energy conserved when the water rises in the tube ? The tube is sufficiently long. If not, calculate the energy change.
Sol. In the equilibrium position (θ = 0º for pure water and glass) U, potential energy of water in the tube
= (πr2hρ)gh/2; it is multiple by h/2 because the cg of the water in the capillary tube is at a height h/2.
⇒ U =
Thus it is seen that the mechanical energy is not conserved. This energy is converted into heat.
5. Calculate the difference in water levels in two communicating tubes of diameter d = 1 mm and d = 1.5 mm. Surface tension of water = 0.07 Nm–1 and angle of contact between glass and water = 0º.
Sol. Pressure at A = p0 – r2
cos T
2 θ
(Q pressure inside a curved surface is greater than that outside)
Pressure at B = p0 –
Let this pressure difference correspond to h units of the liquid.
Then 2T cos θ