Question

A drop of water is squeezed between two glass
plates and is found to spread in the form of a circle
of radius R. The force needed to pull the plates apart
is found to be “F”. Now the plates are pressed
9 towards each other such that the water spreads in
the form of a circle of radius 2R. The force needed to
pull the plates apart now will be (assume separation
between the plates is very small compared to
R​

Answer 1

Answer:

प नतमहनब

Explanation:

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Answer 2

The force needed to pull the plates apart now will be 4F.

Force needed to separate two glass plates with a drop of water squeezed between them is,

F = P$_{excess$ × A

Where, A is the area

P$_{excess$ = T/R

T = Surface tension of the drop of water.

R = Radius.

In the first case,

F = TA/R

⇒ F = TπR²/R [As the drop of water is circular]

⇒ F = TπR

In the second case,

The radius of the circle is increased to 2R.

F' = Tπ4R²/R [As the drop is made of water in both the cases, so T remains same]

⇒ F' = 4TπR  

∴ F' = 4F

So, the force needed to pull the plates apart now will be 4F.