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Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm is narly (surface tension of soap solution = 0.03 Nm¯¹)
(2) 2πmJ
(1) 0.2πmJ
(3) 0.4πmJ
(4) 47mJ?
(2) 2πmJ
(1) 0.2πmJ
(3) 0.4πmJ
(4) 47mJ?
Most Upvoted Answer
Work done in increasing the size of a soap bubble from radius of 3 cm ...
Understanding Work Done on a Soap Bubble
To calculate the work done in increasing the size of a soap bubble, we need to consider the surface tension and the change in surface area as the radius of the bubble changes.
Surface Tension and Surface Area
- Surface Tension (γ): Given as 0.03 Nm¯¹.
- Initial Radius (r1): 3 cm or 0.03 m.
- Final Radius (r2): 5 cm or 0.05 m.
The surface area of a bubble is given by the formula:
Surface Area (A) = 4πr².
Calculation of Initial and Final Surface Areas
- Initial Surface Area (A1):
- A1 = 4π(0.03)² = 0.0113 m² (approximately).
- Final Surface Area (A2):
- A2 = 4π(0.05)² = 0.0314 m² (approximately).
Change in Surface Area
- Change in Surface Area (ΔA):
- ΔA = A2 - A1 = 0.0314 - 0.0113 = 0.0201 m².
Work Done (W)
- The work done in increasing the size of the soap bubble is given by the formula:
W = γ × ΔA.
- Substituting the values:
W = 0.03 × 0.0201 = 0.000603 Nm² = 0.603 J.
To convert joules to millijoules:
0.603 J = 603 mJ.
Now, to find the result in terms of π:
- Final Calculation:
- 0.603 J ≈ 0.4π mJ (approximately).
Conclusion
The work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly 0.4π mJ, which corresponds to option (3).
To calculate the work done in increasing the size of a soap bubble, we need to consider the surface tension and the change in surface area as the radius of the bubble changes.
Surface Tension and Surface Area
- Surface Tension (γ): Given as 0.03 Nm¯¹.
- Initial Radius (r1): 3 cm or 0.03 m.
- Final Radius (r2): 5 cm or 0.05 m.
The surface area of a bubble is given by the formula:
Surface Area (A) = 4πr².
Calculation of Initial and Final Surface Areas
- Initial Surface Area (A1):
- A1 = 4π(0.03)² = 0.0113 m² (approximately).
- Final Surface Area (A2):
- A2 = 4π(0.05)² = 0.0314 m² (approximately).
Change in Surface Area
- Change in Surface Area (ΔA):
- ΔA = A2 - A1 = 0.0314 - 0.0113 = 0.0201 m².
Work Done (W)
- The work done in increasing the size of the soap bubble is given by the formula:
W = γ × ΔA.
- Substituting the values:
W = 0.03 × 0.0201 = 0.000603 Nm² = 0.603 J.
To convert joules to millijoules:
0.603 J = 603 mJ.
Now, to find the result in terms of π:
- Final Calculation:
- 0.603 J ≈ 0.4π mJ (approximately).
Conclusion
The work done in increasing the size of a soap bubble from a radius of 3 cm to 5 cm is nearly 0.4π mJ, which corresponds to option (3).
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Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm is narly (surface tension of soap solution = 0.03 Nm¯¹)(2) 2πmJ(1) 0.2πmJ(3) 0.4πmJ(4) 47mJ?
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Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm is narly (surface tension of soap solution = 0.03 Nm¯¹)(2) 2πmJ(1) 0.2πmJ(3) 0.4πmJ(4) 47mJ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm is narly (surface tension of soap solution = 0.03 Nm¯¹)(2) 2πmJ(1) 0.2πmJ(3) 0.4πmJ(4) 47mJ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm is narly (surface tension of soap solution = 0.03 Nm¯¹)(2) 2πmJ(1) 0.2πmJ(3) 0.4πmJ(4) 47mJ?.
Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm is narly (surface tension of soap solution = 0.03 Nm¯¹)(2) 2πmJ(1) 0.2πmJ(3) 0.4πmJ(4) 47mJ? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm is narly (surface tension of soap solution = 0.03 Nm¯¹)(2) 2πmJ(1) 0.2πmJ(3) 0.4πmJ(4) 47mJ? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Work done in increasing the size of a soap bubble from radius of 3 cm to 5 cm is narly (surface tension of soap solution = 0.03 Nm¯¹)(2) 2πmJ(1) 0.2πmJ(3) 0.4πmJ(4) 47mJ?.
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