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A mercury drop of radius 1 cm is broken into 106 droplets of equal size. The work done is (S = 35 × 10−2 Nm−1)
  • a)
    4.35 × 10−2 J
  • b)
    4.35 × 10−3 J
  • c)
    4.35 × 10−6 J
  • d)
    4.35 × 10−8 J
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A mercury drop of radius 1 cm is broken into 106droplets of equal size...
If r is the radius of small droplet and R is the radius of big drop, then according to question,

Work done = surface tension × increase in area
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Community Answer
A mercury drop of radius 1 cm is broken into 106droplets of equal size...
Given data:
Radius of mercury drop (r) = 1 cm = 0.01 m
Number of droplets formed (n) = 106
Surface tension (S) = 35 x 10^-2 Nm^-1

Calculating work done:
- The initial volume of the mercury drop can be calculated using the formula for the volume of a sphere: V = (4/3)πr^3
- Initial volume (V1) = (4/3) x π x (0.01)^3 = 4.188 x 10^-6 m^3
- Final volume (V2) of each droplet = V1 / n = 4.188 x 10^-6 / 106 = 3.95 x 10^-8 m^3

Work done in breaking the drop into droplets:
- The work done in breaking the drop into droplets can be calculated using the formula: W = 2S x A
- Surface area of each droplet (A) = 4πr^2 = 4 x π x (0.01)^2 = 1.256 x 10^-3 m^2
- Work done (W) = 2 x 35 x 10^-2 x 1.256 x 10^-3 = 4.43 x 10^-2 J
Therefore, the correct answer is option 'A': 4.35 x 10^-2 J.
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A mercury drop of radius 1 cm is broken into 106droplets of equal size. The work done is (S = 35 × 10−2 Nm−1)a)4.35 × 10−2Jb)4.35 × 10−3Jc)4.35 × 10−6Jd)4.35 × 10−8JCorrect answer is option 'A'. Can you explain this answer?
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A mercury drop of radius 1 cm is broken into 106droplets of equal size. The work done is (S = 35 × 10−2 Nm−1)a)4.35 × 10−2Jb)4.35 × 10−3Jc)4.35 × 10−6Jd)4.35 × 10−8JCorrect answer is option 'A'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A mercury drop of radius 1 cm is broken into 106droplets of equal size. The work done is (S = 35 × 10−2 Nm−1)a)4.35 × 10−2Jb)4.35 × 10−3Jc)4.35 × 10−6Jd)4.35 × 10−8JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A mercury drop of radius 1 cm is broken into 106droplets of equal size. The work done is (S = 35 × 10−2 Nm−1)a)4.35 × 10−2Jb)4.35 × 10−3Jc)4.35 × 10−6Jd)4.35 × 10−8JCorrect answer is option 'A'. Can you explain this answer?.
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