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A drop of water volume 0.05 cm^3 is pressed between two glass-

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Hard

Solving time: 4 mins

A drop of water volume 0.05 $c m^{3}$ is pressed between two glass-plates, as a consequence of which, it spreads between the plates. The area of contact with each plate is 40 $c m^{2}$ . If the surface tension of water is 70 dyne/cm, the minimum normal force required to separate out the two glass plates in newton is approximately ______ $N$ .(assumed angle of contact is zero)

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A drop of water volume 0.05

c m^{3}

is pressed between two glass-plates, as a consequence of which, it spreads between the plates. The area of contact with each plate is 40

c m^{2}

. If the surface tension of water is 70 dyne/cm, the minimum normal force required to separate out the two glass plates in newton is approximately ______

N

.(assumed angle of contact is zero)

7 mins ago

Discuss this question LIVE

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Text solutionVerified

Volume

= 0.05 cm^{3}

Thickness

= \frac{0.05}{40} cm

Since contact angle is

0^{\circ}, d = 2 r

Excess pressure inside the liquid

= T (\frac{1}{r _{1}} + \frac{1}{r _{2}})

where,

r_{1}

and

r_{2}

are radius in two perpendicular directions Here,

F = P A = T (\frac{2}{d} + \frac{1}{\infty}) A [Another surface is flat] F = \frac{2 T A}{d} F = \frac{2 \times 70 \times 40 \times 40}{0.05 \times 1 0 ^{2}} \times 1 0^{- 7} N F = 44.8 N F = 45 N

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Question Text	A drop of water volume 0.05 $c m^{3}$ is pressed between two glass-plates, as a consequence of which, it spreads between the plates. The area of contact with each plate is 40 $c m^{2}$ . If the surface tension of water is 70 dyne/cm, the minimum normal force required to separate out the two glass plates in newton is approximately ______ $N$ .(assumed angle of contact is zero)
Topic	Mechanical Properties of Fluids
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	70

Text solutionVerified

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