Learn

Saturday, 31 October

JEE Main 2021 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 17 March 2021, Shift 2, Physics Q30

Question 30 of 90 in this shift, Physics question 30 of 30, Section B.

Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm3\mathrm{cm^3} of oleic acid per cm3\mathrm{cm^3} of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm2\mathrm{cm^2} by considering 100 spherical drops of radius (340π)13×10−3\left(\frac{3}{40\pi}\right)^{\frac{1}{3}}\times10^{-3} cm. Then the thickness of oleic acid layer will be x×10−14x\times10^{-14} m. Where xx is ____________.

Official answer

25

NTA final key.

Chapter
Units and Measurements
Topic
No close match in our taxonomy
Idea tested
No close match in our taxonomy

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.