JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2021
- Hard
- Numerical
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 ...... .
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(25\)
Step-by-step Solution
Detailed explanation
\(4 t _{ T }=100 \times \frac{4}{3} \pi r ^{3}\)
\(=100 \times \frac{4 \pi}{3} \times \frac{3}{40 \pi} \times 10^{-9}=10^{-8}\, cm ^{3}\)
\(t _{ T }=25 \times 10^{-10}\, cm\)
\(=25 \times 10^{-12} \,m\)
\(t _{0}=0.01 t _{ T }=25 \times 10^{-14}\, m\)
\(=25\)
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