Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes $0.01 \,cm ^{3}$ of oleic acid per $cm ^{3}$ of the solution. Then you make a thin film of this solution (monomolecular thickness) of area $4\, cm ^{2}$ by considering $100$ spherical drops of radius $\left(\frac{3}{40 \pi}\right)^{\frac{1}{3}} \times 10^{-3}\, cm .$ Then the thickness of oleic acid layer will be $x \times 10^{-14} \,m$. Where $x$ is ...... .
JEE MAIN 2021, Diffcult
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$4 t _{ T }=100 \times \frac{4}{3} \pi r ^{3}$
$=100 \times \frac{4 \pi}{3} \times \frac{3}{40 \pi} \times 10^{-9}=10^{-8}\, cm ^{3}$
$t _{ T }=25 \times 10^{-10}\, cm$
$=25 \times 10^{-12} \,m$
$t _{0}=0.01 t _{ T }=25 \times 10^{-14}\, m$
$=25$
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