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Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done is
- a)7.3 erg
- b)7.3 x 104 erg
- c)7.3 J
- d)0.73 J
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Surface tension of water is 73 dynes cm -1 at 20° C. If surface a...
Work done = surface tension×Change in area
Surface tension = 73 dyne cm-1 or 73 x 10-3 N/m
Area = 0.10 m2
So, work done = 73×10-3 x 0.10
= 73 x 10-4 J or 73×10-4 x 107 erg
= 7.3×104 erg
Surface tension = 73 dyne cm-1 or 73 x 10-3 N/m
Area = 0.10 m2
So, work done = 73×10-3 x 0.10
= 73 x 10-4 J or 73×10-4 x 107 erg
= 7.3×104 erg
Most Upvoted Answer
Surface tension of water is 73 dynes cm -1 at 20° C. If surface a...
Work done = surface tension×Change in area
Surface tension = 73 dyne cm-1 or 73 x 10-3 N/m
Area = 0.10 m2
So, work done = 73×10-3 x 0.10
= 73 x 10-4 J or 73×10-4 x 107 erg
= 7.3×104 erg
Surface tension = 73 dyne cm-1 or 73 x 10-3 N/m
Area = 0.10 m2
So, work done = 73×10-3 x 0.10
= 73 x 10-4 J or 73×10-4 x 107 erg
= 7.3×104 erg
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Community Answer
Surface tension of water is 73 dynes cm -1 at 20° C. If surface a...
Given:
Surface tension of water = 73 dynes cm^-1
Change in surface area = 0.10 m^2
To find:
Work done
Formula:
Work done = Surface tension * Change in surface area
Calculation:
1. Convert the given surface tension from dynes cm^-1 to N m^-1:
1 N = 10^5 dynes
1 m = 100 cm
Surface tension = 73 dynes cm^-1 * (1 N / 10^5 dynes) * (1 m / 100 cm)
= 0.073 N m^-1
2. Convert the change in surface area from m^2 to cm^2:
1 m^2 = 10^4 cm^2
Change in surface area = 0.10 m^2 * 10^4 cm^2 / 1 m^2
= 1000 cm^2
3. Calculate the work done:
Work done = Surface tension * Change in surface area
= 0.073 N m^-1 * 1000 cm^2
= 73 N m^-1 * cm^2
= 73 N cm
= 73 J (since 1 N m = 1 J)
Therefore, the work done is 73 J.
Answer:
The correct answer is option (c) 7.3 J.
Surface tension of water = 73 dynes cm^-1
Change in surface area = 0.10 m^2
To find:
Work done
Formula:
Work done = Surface tension * Change in surface area
Calculation:
1. Convert the given surface tension from dynes cm^-1 to N m^-1:
1 N = 10^5 dynes
1 m = 100 cm
Surface tension = 73 dynes cm^-1 * (1 N / 10^5 dynes) * (1 m / 100 cm)
= 0.073 N m^-1
2. Convert the change in surface area from m^2 to cm^2:
1 m^2 = 10^4 cm^2
Change in surface area = 0.10 m^2 * 10^4 cm^2 / 1 m^2
= 1000 cm^2
3. Calculate the work done:
Work done = Surface tension * Change in surface area
= 0.073 N m^-1 * 1000 cm^2
= 73 N m^-1 * cm^2
= 73 N cm
= 73 J (since 1 N m = 1 J)
Therefore, the work done is 73 J.
Answer:
The correct answer is option (c) 7.3 J.
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Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer?.
Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer?.
Solutions for Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for Class 11.
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Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of Surface tension of water is 73 dynes cm -1 at 20° C. If surface area is increased by 0.10 m2, work done isa)7.3 ergb)7.3 x 104ergc)7.3 Jd)0.73 JCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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