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  • A water drop of radius  is broken into  equal droplets. If surface tensio

A water drop of radius 1 \mathrm{~cm} is broken into 729 equal droplets. If surface tension of water is \mathrm{75 \text { dyne/cm, }} then the gain in surface energy upto first decimal place will be :

\mathrm{\text { (Given } \pi=3.14 \text { ) }}

Option: 1

\mathrm{8.5 \times 10^{-4} \mathrm{~J}}


Option: 2

\mathrm{8.2 \times 10^{-4} \mathrm{~J}}


Option: 3

\mathrm{7.5 \times 10^{-4} \mathrm{~J}}


Option: 4

\mathrm{5.3 \times 10^{-4} \mathrm{~J}}


Answers (1)

best_answer

\mathrm{T =75 \frac{\text { dyne }}{\mathrm{cm}} }

\mathrm{=75 \times \frac{10^{-5} \mathrm{~N}}{10^{-2} \mathrm{~m}} }

\mathrm{T =75 \times 10^{-3} \mathrm{~N} / \mathrm{m}}

\mathrm{1 \: Drop \rightarrow 729\: Droplets}

\mathrm{\frac{4}{3} \pi R^3 =729 \times\left(\frac{4 \pi r^3}{3}\right) }

\mathrm{R =9 r=1 \mathrm{~cm}=10^{-2} \mathrm{~m} }

\mathrm{r =\frac{1}{9} \mathrm{~cm}=\frac{10^{-2}}{9}}

The gain in surface energy \mathrm{=S_f-S_i}

                                        \mathrm{=T \Delta A}

\mathrm{\Delta S =T(\Delta A) }

\mathrm{=75 \times 10^{-3}\left(729 \times 4 \pi r^2-4 \pi R^2\right) }

\mathrm{=75 \times 10^{-3}(36 \pi-4 \pi) \times 10^{-4} }

\mathrm{=75 \times 32 \pi \times 10^{-7} }

\mathrm{\Delta S =7.5 \times 10^{-4} \mathrm{~J}}

Hence (3) is correct option    





 

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chirag

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