Question

A drop of water of mass 0.2g is placed between two glass plates. The distance between them is 0.01 cm.
Find the force of attraction between the plates if surface tension of water = 0.07 Nm-1:​

Answer 1

Answer:

The force of attraction between the plates is 0.011 N

Explanation:

Mass of the drop of water is given as

m = 0.2 g

now we have

$0.2 \times 10^{-3} = \rho V$

$0.2\times 10^{-3} = 1000 \times A \times d$

$0.2 \times 10^{-3} = 1000 \times A \times (0.01 \times 10^{-2})$

$A = 2 \times 10^{-3} m^2$

now we have

$A = \pi r^2$

$2 \times 10^{-3} = \pi r^2$

$r = 0.025m$

so force due to surface tension between the two plates is given as

$F = S(2\pi r)$

$F = 0.07(2\pi)(0.025)$

$F = 0.011 N$

#Learn

Topic : Surface tension

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