The force required to separate two glass plates of area 10^-2 m^2 with a film of water 0.05 mm thick between them, is (Surface tension = 70*10^-3) (a) 28 N (b) 14 N (c) 50 N (d) 38 N
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05 mm = 0.05 * 10^-3 m). Show more…
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982 The force required to separate two glass plates of area 10–2 m2 with a film of water 0.05 mm thick between them is (surface tension of water is 70 × 10–3 N/m) (A) 28 N (B) 14 N (C) 50 N (D) 38 N
Kajal R.
A drop of water of mass $m=0.4 \mathrm{~g}$ is placed between two clean glass plates, the distance between which is $0.01 \mathrm{~cm}$. Find the force of attraction between the plates. Surface tension of water $=0.01 \mathrm{~N} / \mathrm{m}$.
Two parallel glass plates having separation ' $d^{\prime}$ are dipped in water. Some water rise up in the gap between the plates. The surface tension of water is $S$, atmospheric pressure $=P_{0}$, pressure of water just below the water surface in the region between the plates is $P$, find the relation between them : (a) $P=P_{0}-\frac{2 S}{d}$ (b) $P=P_{0}+\frac{2 S}{d}$ (c) $P=P_{0}-\frac{4 S}{d}$ (d) $P=P_{0}+\frac{4 S}{d}$
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