Grade upto college level Mechanics

Two mercury drops (each of radius ‘r’ ) merge to form bigger drop. The surface energy of the bigger drop, if T is the surface tension, is:
(a) 4πr²T
(b) 2πr²T
(c) 2^8/3πr²T
(d) 2^5/3πr²T

Amit Saxena

12 Years agoGrade upto college level

2 Answers

Navjyot Kalra

12 Years ago

2.4/3πr³ = 4/3πR³ $\Rightarrow$ R = 2^1/3r

Surface energy = T.4πR²

= T4π2 r² = T.2^8/3 πr²

Kushagra Madhukar

5 Years ago

Dear student,

Please find your answer below.

Sum of volumes of 2 smaller drops = Volume of the bigger drop

2.4/3πr³ = 4/3πR³ $\Rightarrow$ R = 2^1/3r

Surface energy = T.4πR²

= T4π2 r² = 2^8/3 πr²T

Hence option c) is correct.

Regards,

Kushagra

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Two mercury drops (each of radius ‘r’ ) merge to form bigger drop. The surface energy of the bigger drop, if T is the surface tension, is:(a) 4πr2T(b) 2πr2T(c) 28/3πr2T(d) 25/3πr2T

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Two mercury drops (each of radius ‘r’ ) merge to form bigger drop. The surface energy of the bigger drop, if T is the surface tension, is:
(a) 4πr²T
(b) 2πr²T
(c) 2^8/3πr²T
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