MHT CET · Physics · Mechanical Properties of Fluids
A water drop of \(0.01 \mathrm{~cm}^3\) is squeezed between two glass plates and spreads in to area of \(10 \mathrm{~cm}^2\). If surface tension of water is 70 dyne \(/ \mathrm{cm}\) then the normal force required to separate glass plates from each other will be
- A 12 N
- B 14 N
- C 16 N
- D 28 N
Answer & Solution
Correct Answer
(B) 14 N
Step-by-step Solution
Detailed explanation
\(t = \frac{V}{A} = \frac{0.01 \mathrm{~cm}^3}{10 \mathrm{~cm}^2} = 0.001 \mathrm{~cm}\) \(F = \frac{2 \gamma A}{t} = \frac{2 \times 70 \mathrm{~dyne/cm} \times 10 \mathrm{~cm}^2}{0.001 \mathrm{~cm}} = 1400000 \mathrm{~dyne}\)
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