Advertisements
Advertisements
प्रश्न
The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle θ. If the acceleration is a ms–2, what will be the slope of the free surface?
Advertisements
उत्तर
The behaviour of a liquid contained in a horizontally accelerated vessel can be understood by understanding the behaviour of a pendulum suspended from the ceiling of a horizontally accelerated trolley.
Every fluid element attains an equilibrium position under the action of gravity and pseudo-force. The free surface of the liquid orients itself perpendicular to the direction of net effective gravity.
tan θ = a/g
Suppose a tanker accelerates along the x-axis with acceleration a, the free surface of the tanker will not be horizontal because pseudo force acts as shown in the diagram.
Consider an elementary particle of the oil of mass m.
The acting forces on the particle with respect to the tanker are shown in the figure alongside.
Now, balancing forces (as the particle is in equilibrium) along the inclined direction of the surface.
ma = pseudo force
mg = weight of small part of the oil.
Along the free surface,
Net force = 0
⇒ ma cos θ = mg sin θ
⇒ a = g tan θ
⇒ θ = tan-1(a/g)
APPEARS IN
संबंधित प्रश्न
Explain why The angle of contact of mercury with glass is obtuse, while that of water with glass is acute
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
Define surface tension and surface energy.
The free surface of a liquid resting in an inertial frame is horizontal. Does the normal to the free surface pass through the centre of the earth? Think separately if the liquid is (a) at the equator (b) at a pole (c) somewhere else.
A uniform vertical tube of circular cross section contains a liquid. The contact angle is 90°. Consider a diameter of the tube lying in the surface of the liquid. The surface to the right of this diameter pulls the surface on the left of it. What keeps the surface on the left in equilibrium?
When the size of a soap bubble is increased by pushing more air in it, the surface area increases. Does it mean that the average separation between the surface molecules is increased?
A liquid is contained in a vertical tube of semicircular cross section. The contact angle is zero. The force of surface tension on the curved part and on the flat part are in ratio
A capillary tube of radius 1 mm is kept vertical with the lower end in water. (a) Find the height of water raised in the capillary. (b) If the length of the capillary tube is half the answer of part , find the angle θ made by the water surface in the capillary with the wall.
Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = 0.075 Nm−1.
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 dyn/cm.
Define surface tension.
Obtain an expression for the capillary rise or fall using the forces method.
A spherical soap bubble A of radius 2 cm is formed inside another bubble B of radius 4 cm. Show that the radius of a single soap bubble which maintains the same pressure difference as inside the smaller and outside the larger soap bubble is lesser than the radius of both soap bubbles A and B.
The surface tension of the two liquids is respectively 20 and 60 dyne cm-1. The liquids drop from the ends of two tubes of the same radius. The ratio of the weights of the two drops is ______
Two small drops of mercury each of radius 'R' coalesce to form a large single drop. The ratio of the total surface energies before and after the change is ____________.
If the surface tension of a soap solution is 3 × 10-2 N/m then the work done in forming a soap film of 20 cm × 5 cm will be ______.
Surface tension is exhibited by liquids due to force of attraction between molecules of the liquid. The surface tension decreases with increase in temperature and vanishes at boiling point. Given that the latent heat of vaporisation for water Lv = 540 k cal kg–1, the mechanical equivalent of heat J = 4.2 J cal–1, density of water ρw = 103 kg l–1, Avagadro’s No NA = 6.0 × 1026 k mole–1 and the molecular weight of water MA = 18 kg for 1 k mole.
- Estimate the energy required for one molecule of water to evaporate.
- Show that the inter–molecular distance for water is `d = [M_A/N_A xx 1/ρ_w]^(1/3)` and find its value.
- 1 g of water in the vapor state at 1 atm occupies 1601 cm3. Estimate the intermolecular distance at boiling point, in the vapour state.
- During vaporisation a molecule overcomes a force F, assumed constant, to go from an inter-molecular distance d to d ′. Estimate the value of F.
- Calculate F/d, which is a measure of the surface tension.
In a U-tube, the radii of two columns are respectively r1 and r2. When a liquid of density ρ(θ = 0°) is filled in it, a level difference of h is observed on two arms, then the surface tension of the liquid is ______.
The surface tension of a soap solution is T. The work done in blowing a soap bubble of diameter d to that of a diameter 2d is ______.
Define angle of contact.
