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Water Near the Bed of a Deep River is Quiet While that Near the Surface Flows. Give Reasons.

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प्रश्न

Water near the bed of a deep river is quiet while that near the surface flows. Give reasons.

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उत्तर

The motion of any liquid is dependent upon the amount of stress acting on it. The motion of one layer of liquid is resisted by the other due to the property of viscosity. A river bed remains in a static state. Therefore, any immediate layer of liquid in contact with the river bed will also remain static due to the frictional force. However, the next layer of liquid above this static layer will have a greater velocity due to lesser resistance offered by the static layer. Moving upwards, subsequent layers provide lesser and lesser resistance to the movement of the layers above it. Finally, the topmost layer acquires the maximum velocity. Therefore, for a river, the surface waters flow the fastest .

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अध्याय 14: Some Mechanical Properties of Matter - Short Answers [पृष्ठ २९८]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 14 Some Mechanical Properties of Matter
Short Answers | Q 18 | पृष्ठ २९८

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संबंधित प्रश्न

Define the angle of contact.


'n' droplets of equal size of radius r coalesce to form a bigger drop of radius R. The energy liberated is equal to...................

(T =Surface tension of water)

`(a) 4piR^2T[n^(1/3)-1]`

`(b) 4pir^2T[n^(1/3)-1]`

`(c) 4piR^2T[n^(2/3)-1]`

`(d)4 pir^2T[n^(2/3)-1]`


The surface tension of water at 0ºc is 75·5 dyne/cm. Find surface tension of water at 25°C. [ α for water = 0·0021/°C ]


Explain why Surface tension of a liquid is independent of the area of the surface


The total free surface energy of a liquid drop is `pisqrt2` times the surface tension of the liquid. Calculate the diameter of the drop in S.l. unit.


Show that the surface tension of a liquid is numerically equal to the surface energy per unit
area.


Air is pushed into a soap bubble of radius r to double its radius. If the surface tension of the soap solution in S, the work done in the process is 


Which of the following graphs may represent the relation between the capillary rise hand the radius r of the capillary?


The rise of a liquid in a capillary tube depends on

(a) the material
(b) the length
(c) the outer radius
(d) the inner radius of the tube


The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury. (a) Find the depression of mercury column in the capillary. (b) If the length dipped inside is half the answer of part (a), find the angle made by the mercury surface at the end of the capillary with the vertical. Surface tension of mercury = 0.465 N m−1 and the contact angle of mercury with glass −135 °.


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 solid sphere of radius 5 cm floats in water. If a maximum load of 0.1 kg can be put on it without wetting the load, find the specific gravity of the material of the sphere.


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


The wettability of a surface by a liquid depends primarily on


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 ______


The upward force of 105 dyne due to surface tension is balanced by the force due to the weight of the water column and 'h' is the height of water in the capillary. The inner circumference of the capillary is ______.

(surface tension of water = 7 × 10-2 N/m)


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.

  1. Estimate the energy required for one molecule of water to evaporate.
  2. Show that the inter–molecular distance for water is `d = [M_A/N_A xx 1/ρ_w]^(1/3)` and find its value.
  3. 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.
  4. 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.
  5. Calculate F/d, which is a measure of the surface tension.

Define angle of contact.


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]


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