JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 19)
Suppose you have taken a dilute solution of oleic acid in such a way that its concentration becomes 0.01 cm3 of oleic acid per cm3 of the solution. Then you make a thin film of this solution (monomolecular thickness) of area 4 cm2 by considering 100 spherical drops of radius $${\left( {{3 \over {40\pi }}} \right)^{{1 \over 3}}} \times {10^{ - 3}}$$ cm. Then the thickness of oleic acid layer will be x $$\times$$ 10$$-$$14 m. Where x is ____________.
Answer
25
Explanation
4tT = 100 $$ \times $$ $${4 \over 3}\pi {r^3}$$
= $$100 \times {4 \over 3}\pi \times {3 \over {40\pi }} \times {10^{ - 9}}$$
= 10-8 cm3
$$ \Rightarrow $$ tT = 25 $$ \times $$ 10-10 cm
= 25 $$ \times $$ 10-12 m
t0 = 0.01 tT = 25 $$ \times $$ 10-14 m
$$ \therefore $$ x = 25
= $$100 \times {4 \over 3}\pi \times {3 \over {40\pi }} \times {10^{ - 9}}$$
= 10-8 cm3
$$ \Rightarrow $$ tT = 25 $$ \times $$ 10-10 cm
= 25 $$ \times $$ 10-12 m
t0 = 0.01 tT = 25 $$ \times $$ 10-14 m
$$ \therefore $$ x = 25
Comments (0)
