Surface Tension Numerical on Ascent Descent Formula Angle of Contact Class-11 Nootan ISC Physics Solutions Ch-16. Step by step Solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Surface Tension Numerical on Ascent Descent Formula Angle of Contact Class-11 Nootan ISC Physics Ch-16
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Writer | Kumar and Mittal |
| Publication | Nageen Prakashan |
| Chapter-16 | Surface Tension |
| Topics | Numericals on Ascent/Descent Formula and Angle of Contact |
| Academic Session | 2024-2025 |
Numericals on Ascent/Descent Formula and Angle of Contact
Que-18: The diameter of a capillary tube is 0.4 x 10^-3 m. It is held vertically, in a liquid whose density is 0.8 x 10³ kg/m³ and the surface tension is 9.8 x 10^-2 N/m. Determine the height to which the liquid will rise in the tube. (Angle of contact is zero)
Ans- h = (2 T cos θ) / (r ρ g)
=> (2 x 9.8 x 10^-2 x cos 0°) / (0.2 x 10^-3 x 0.8 x 10³ x 9.8)
=> 12.5 x 10^-2
=> 12.5 cm
Que-19: (a) The radius of a capillary tube is 0.025 mm. It is held vertically in a liquid whose density is 0.8 x 10³ kg/m³, surface tension is 3.0 x 10^-2 N m and for which the cosine of the angle of contact is 0.3. Determine the height up to which the liquid will rise in the tube. Take g = 10 m/s².
(b) If the capillary is taken down in water slowly until its upper end comes in level of water, will the water come out from this end?
Ans- (a) h = (2 T cos θ) / (r ρ g)
=> (2 x 3 x 10^-2 x 0.3) / (0.025 x 10^-3 x 0.8 x 10³ x 10)
=> 9 x 10^-2
=> 9 cm
(b) Liquid never comes out of capillary tube only radius of meniscus changes
Que-20: Water rises in a capillary tube to a height of 5.0 cm. If surface tension of water is 9.8 x 10^-2 N/m, then find out the diameter of the capillary tube.
Ans- h = (2 T cos θ) / (r ρ g)
=> (2 T cos θ x 2) / (D ρ g) {D = diameter}
=> D = (4 T cos θ ) / (ρ g h)
=> (4 x 9.8 x 10^-2 x 1) / (1 x 10³ x 9.8 x 5 x 10^-2)
=> 0.08 x 10^-2
=> 0.08 cm
Que-21. The radius of a capillary tube is 0.8 mm and the water rises in it up to a height of 2 cm. Calculate the surface tension of water. Take g = 10 m/s².
Ans- h = (2 T cos θ) / (r ρ g)
=> T = (h r ρ g) / (2 cos θ)
=> (2 x 10^-2 x 0.8 x 10^-3 x 10³ x 10) / (2 x 1)
=> 8 x 10^-2 N/m
Que-22: A liquid rises to a height of 7.0 cm in a capillary tube of radius 0.1 mm. The density of the liquid is 0.8 x 10³ kg/m³. If the angle of contact between the liquid and the surface of the tube be zero, calculate the surface tension of the liquid. Take g = 10 m/s²
Ans- h = (2 T cos θ) / (r ρ g)
=> T = (h r ρ g) / (2 cos θ)
=> (7 x 10^-2 x 0.1 x 10^-3 x 0.8 x 10³ x 10) / (2 x 1)
=> 2.8 x 10^-2 N/m
Que-23: Water rises up in a glass capillary up to a height of 9.0 cm, while mercury falls down by 3.4 cm in the same capillary. Assume angles of contact for water-glass and mercury-glass 0° and 135° respectively. Determine the ratio of surface tensions of mercury and water (cos 135° = -0.71).
Ans- h = (2 T cos θ) / (r ρ g)
=> T = (h r ρ g) / (2 cos θ)
=> (T1 / T2) = (ρ1 / ρ2) x (h1 / h2) x (cos θ2 / cos θ1)
=> (Thg / Tw) = (13.6 x 10³) / (1 x 10³) x (-3.4 / 9) x (1 / – 0.71)
{negative sign for h is for hg as it remain down of free level}
=> 7.2 :1
Que-24: Water rises up to 2.0 cm in a capillary tube of length 20.0 cm held vertically. If the capillary is bent by 30° from this vertical position, find the length of liquid risen in it.
Ans-
cos 30° = 2 / l
=> √3 / 2 = 2 / l
=> l = 4 / √3
=> 4 √3 / 3
=> 2.31 cm
Que-25: A capillary tube of radius 0.4 mm is dipped vertically in water. Find up to what height the water will rise in the capillary. If the capillary is inclined at an angle of 60° with the vertical, how much length of the capillary is occupied by water ? Surface tension of water is 7.0 × 10^-2 N/m.
Ans- h = (2 T cos θ) / (r ρ g)
=> (2 x 7 x 10^-2 x 1) / (0.4 x 10^-3 x 10³ x 9.8)
=> 3.57 x 10^-2 m
=> 3.57 cm
again
sec 60° = l / 3.57
=> 2 = l / 3.57
=> l = 3.57 x 2
=> 7.14 cm
Que-26 Water rises in a capillary up to a height of 8.0 cm. If the capillary is inclined at an angle of 45° with the vertical, then determine the vertical height of water. How much length of the capillary will be occupied by water? If the length of the capillary is reduced to 4.0 cm and it is held vertically in water, then what will be the position of water?
Ans-26
cos 45° = 8.0 / l
=> 1 / √2 = 8 / l
=> l = 8 √2
=>11.3 cm
again in case of insufficient length of tube the liquid rises up to mouth of tube but does not fall.
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