A drop of some liquid of volume 0.04 cm 3 is placed on the surface of… | MHT CET 2026 (14 May, Shift 1)

Q3MHT CET 2026 · 14 MAY · SHIFT 1Medium

Force due to Surface Tension

A drop of some liquid of volume $0.04$ cm $3$ is placed on the surface of the glass plate. Then another glass plate is placed on it such that the liquid forms a thin layer of area $20$ cm $2$ between the surfaces. To separate the plates a force of $16 \times 10^{5}$ dyne has to be applied normal to the surfaces. The surface tension of the liquid in dyne/cm is

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Answer: (c)

Thickness of layer $t = \frac{V}{A} = \frac{0.04}{20} = 0.002\text{ cm}$ .
Force to separate plates: $F = \dfrac{2AT}{t} \Rightarrow T = \dfrac{Ft}{2A}$ .
$T = \dfrac{16\times10^5 \times 0.002}{2 \times 20} = 80\text{ dyne/cm}$ . So the answer is (c).

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Force due to Surface Tension question from MHT CET 2026 (14 May, Shift 1)

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