Surface Tension: Energy and Work Done Numerical Class-11 Nootan ISC Physics Nootan 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: Energy and Work Done Numerical Class-11 Nootan ISC Physics Solutions
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Writer | Kumar and Mittal |
| Publication | Nageen Prakashan |
| Chapter-16 | Surface Tension |
| Topics | Numericals on Surface Tension, Surface Energy and Work Done |
| Academic Session | 2024-2025 |
Numerical on Surface Tension, Surface Energy and Work Done
Ch-16 Surface Tension: Energy and Work Done Numerical Class-11 Nootan ISC Nootan Solutions of Kumar and Mittal Physics, Nageen Prakashan
Que-1: How much work will be done in enlarging the surface area of a soap bubble by 1.0 cm²?
Ans- ΔA = 2 x 1 x 10^-4 m²
soap bubble has two surfaces
∴ W = T x ΔA
=> 3 x 10^-2 x 10^-4
=> 6.0 x 10^-6 J
Que-2: How much work will be done in making a soap bubble of diameter 2.0 cm?
Ans- ΔA = 2 x 4π x r² = 2 x 4π x (1 x 10^-2)²
=> 25.12 x 10^-4 m²
∴ W = T x ΔA
=> 3 x 10^-2 x 25.12 x 10^-4
=> 7.54 x 10^-5 J
Que-3: In increasing the area of a film of soap solution from 50 cm² to 100 cm², 3.0 x 10^-4 J of work is done. Calculate the value of surface tension of the soap solution.
Ans- ΔA = 2 x (100 – 50) x 10^-4 = 10^-2 m²
ΔW = 3 x 10^-4 J
T = ΔW / ΔA
=> 3 x 10^-4 / 10^-2
=> 3 x 10^-2 N/m
Que-4: The surface area of a soap-bubble is 2.0 x 10^-3 m². How much work will be done in blowing the bubble to twice its surface area?
Ans- ΔA = 2 x 10^-3 x 2 = 4 x 10^-3
W = T x ΔA
=> 3 x 10^-2 x 4 x 10^-3
=> 1.2 x 10^-4 J
Que-5: How much work will be done in increasing the diameter of a soap bubble from 2 cm to 5 cm ?
Ans- ΔA = 4 π {(2.5)² – (1)²} x 10^-4 x 2
=> 4 x 3.14 x 5.25 x 2 x 10^-4
=> 0.0132 m²
∴ W = T x ΔA
=> 3 x 10^-2 x 0.0132
=> 3.96 x 10^-4 J
Que-6: A liquid drop of radius 1 mm is broken into 1000 equal small drops. How much work will be done ? Surface tension of water = 0.07 N/m and π = 22/7.
Ans- ΔW = T . 4 π r² (n^(1/3) – 1)
=> 0.07 x 4 x 22/7 x (1 x 10^-3)² {1000^(1/3) – 1}
=> 0.01 x 4 x 22 x 10^-6 x 9
=> 7.92 x 10^-6 J
Que-7: A drop of water of diameter 0.2 cm is broken up into 27000 droplets of equal volume. How much work will be done against surface tension in the process? (Surface tension of water = 7 × 10^-2 N/m)
Ans- ΔW = T . 4 π r² (n^(1/3) – 1)
=> 4 x 3.14 x (0.1 x 10^-2)² (27000^(1/3) – 1)
=> 4 x 3.14 x 0.01 x 10^-4 x 29
=> 2.55 x 10^-5 J
Que-8: A mercury drop of radius 1.0 cm is sprayed into 10^6 droplets of equal size. Calculate the energy expanded.
Ans- ΔW = T . 4 π r² (n^(1/3) – 1)
=>4 x 3.14 x (1 x 10^-2)² x 46.5 x 10^-2 [(10^6)^(1/3) – 1]
=> 4 x 3.14 x 10^-4 x 46.5 x 10^-2 x 99
=> 5.78 x 10^-2 J
Que-9: Calculate the energy released when 1000 small water drops each of same radius 10^-7 m coalesce to form one large drop. The surface tension of water is 7.0 × 10^-2 N/m.
Ans- Energy released
=> ΔW = T . 4 π r² (n^(1/3) – 1)
=> 4 x 3.14 x (10^-7)² (1000^(1/3) – 1)
=> 4 x 3.14 x 10^-14 x 9
=> 7.9 x 10^-12 J
Que-10: A big drop is formed by combining 27 small droplets of water. What will be the change in the surface energy? What will be the ratio of the surface energy of big drop to the surface energy of 27 small droplets?
Ans- Energy of big drop = T x A
where A = 4 π R² (surface area of big drop)
now volume of big drop = volume of 27 small drops
=> 4/3 π R³ = 27 x 4/3 π r³ {r = radius of small drop}
=> R = 3r
new surface energy
=> 27 x 4 π R²/9 T
=> 4 π x 3R² T
∴ratio => 4 π x R² T : 4 π x 3R² T
=> 1 : 3
Que-11: A thin wire is bent in the form of a ring of diameter 3.0 cm. The ring is placed horizontally on the surface of soap solution and then raised up slowly. How much upward force is necessary to break the vertical film formed between the ring and the solution?
Ans- T = F / l
=> F = T x l
=> T x 2 π r x 2
=> 3 x 10^-2 x 2 x 3.14 x 1.5 x 10^-2 x 2
length is doubled so film has two free surfaces
=> 5.652 x 10^-3 N
Que-12: The length of a needle floating on water is 2.5 cm. How much minimum force, in addition to the weight of the needle, will be needed to lift the needle above the surface of water ? Surface tension of water = 7.2 × 10^4 N/cm.
Ans- F = T x 2l {film has two surface}
=> F = 7.2 x 10^-2 x 25 x 10^-2 x 2
=> 36 x 10^-4 N
—: end of Surface Tension: Energy and Work Done Numerical Class-11 Nootan ISC Nootan Solutions :—
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