Advertisements
Advertisements
Question
If a mosquito is dipped into water and released, it is not able to fly till it is dry again. Explain
Advertisements
Solution
A mosquito thrown into water has its wings wetted. Now, wet wing surfaces tend to stick together because of the surface tension of water. This does not let the mosquito fly.
APPEARS IN
RELATED QUESTIONS
The energy of the free surface of a liquid drop is 5π times the surface tension of the liquid. Find the diameter of the drop in C.G.S. system.
Derive an expression for excess pressure inside a drop of liquid.
Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 m s–2)
Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. Surface tension of mercury, soap solution and water are 0.465 N m−1, 0.03 N m−1 and 0.076 N m−1 respectively.
Consider a small surface area of 1 mm2 at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area (a) by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure = 1.0 × 105 Pa and surface tension of mercury = 0.465 N m−1. Neglect the effect of gravity. Assume all numbers to be exact.
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 wire forming a loop is dipped into soap solution and taken out so that a film of soap solution is formed. A loop of 6.28 cm long thread is gently put on the film and the film is pricked with a needle inside the loop. The thread loop takes the shape of a circle. Find the tension the the thread. Surface tension of soap solution = 0.030 N m−1.
A cubical metal block of edge 12 cm floats in mercury with one fifth of the height inside the mercury. Water in it. Find the height of the water column to be poured.
Specific gravity of mercury = 13.6.
Twenty-seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m.
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.
The water droplets are spherical in free fall due to ______
A certain number of spherical drops of a liquid of radius R coalesce to form a single drop of radius R and volume V. If T is the surface tension of the liquid, then
Obtain an expression for the surface tension of a liquid by the capillary rise 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.
For a surface molecule ______.
- the net force on it is zero.
- there is a net downward force.
- the potential energy is less than that of a molecule inside.
- the potential energy is more than that of a molecule inside.
If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius R, break into N small droplets each of radius r. Estimate the drop in temperature.
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.
A liquid drop of density ρ is floating half immersed in a liquid of density d. The diameter of the liquid drop is ______.
(ρ > d, g = acceleration due to gravity, T = surface tension)
A light metal disc of radius ‘r’ floats on water surface and bends the surface downwards along the perimeter making an angle ‘θ’ with the vertical edge of the disc. If the weight of water displaced by the disc is ‘W’, the weight of the metal disc is ______.
[T = surface tension of water]
