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  • Help me solve this - Properties of Solids and Liquids - BITSAT

Assume that a drop of liquid evaporates by decrease in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible ? The surface tension is T,density of liquid is \rho and L is its latent heat of vaporization.

 

  • Option 1)

    2T/\rho L\;

  • Option 2)

    \rho L/T\; \;

  • Option 3)

    \sqrt{T/\rho L}

  • Option 4)

    T/\rho L\;

 

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

 

 Surface energy of the drop = T 4 \pi R^2

If sm mass of water evaporated than sm L = 8 \pi RT......(1)

m = \frac{4 \pi }{3} R^3 \rho

\Rightarrow \frac{8 m }{8 R }= \rho 4 \pi R^2 \Rightarrow \rho 4 \pi R^2 L = 8 \pi RT

R = \frac{2T }{L \rho }

 

 

 

 

 


Option 1)

2T/\rho L\;

Option 2)

\rho L/T\; \;

Option 3)

\sqrt{T/\rho L}

Option 4)

T/\rho L\;

Posted by

SudhirSol

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