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 A large number of liquid drops each of radius r coalesce to from a single drop of radius R. The energy released in the process is converted into kinetic energy of the big drop so formed. The speed of the big drop is (given surface tension of liquid T,density \rho)

  • Option 1)

    \sqrt{\frac{T}{\rho }\left ( \frac{1}{r}-\frac{1}{R} \right )}

  • Option 2)

    \sqrt{\frac{2T}{\rho }\left ( \frac{1}{r}-\frac{1}{R} \right )}

  • Option 3)

    \sqrt{\frac{4T}{\rho }\left ( \frac{1}{r}-\frac{1}{R} \right )}

  • Option 4)

    \sqrt{\frac{6T}{\rho }\left ( \frac{1}{r}-\frac{1}{R} \right )}

 

Answers (1)

best_answer

As we have learned

Surface Energy -

W=T\times \Delta A

T= \frac{W}{\Delta A}

- wherein

\Delta A\rightarrow increase\: in\: area

 

 Let number of drop is n 

Initial volume = Final volume 

 \Rightarrow n \frac{4 \pi }{3}r^3 = \frac{4 \pi }{3}R^3 \: \: or\: \: n r^= R^3......(1)

Energy release = T \Delta A

T = surface tension 

\Delta A = Change in surface area 

\Rightarrow \Delta V = T \cdot [ n4 \pi r^2- 4 \pi R^2 ]

\Delta V =1/2 mv ^2 = 1/ 2 (\rho \cdot \frac{4 \pi }{3}R^3 )v^2

\Rightarrow \frac{2 \rho \pi R^3 }{3 }v^2 = T 4 \pi (r^2 \frac{R^3}{r^3}-R^2)

v^2= \sqrt{\frac{6T}{\rho }\cdot \left ( 1/r-1/R \right )}

 

 

 

 

 

 


Option 1)

\sqrt{\frac{T}{\rho }\left ( \frac{1}{r}-\frac{1}{R} \right )}

Option 2)

\sqrt{\frac{2T}{\rho }\left ( \frac{1}{r}-\frac{1}{R} \right )}

Option 3)

\sqrt{\frac{4T}{\rho }\left ( \frac{1}{r}-\frac{1}{R} \right )}

Option 4)

\sqrt{\frac{6T}{\rho }\left ( \frac{1}{r}-\frac{1}{R} \right )}

Posted by

SudhirSol

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