Give the expression for the excess pressure in a liquid drop.
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Consider a liquid drop of radius 'r' . The molecules on the surface of the drop experience a resultant force acting inwards due to surface tension. Therefore, pressure inside the drop will be greater then pressure outside the drop. It provides a force acting outwards perpendicular to the surface, to balance the resultant force due to surface tension. Let the drop be divided into two equal halves.
Considering the equilibrium of the upper hemisphere of the drop, the upward force due to excess pressure will be P π r² , where P is the pressure.
If 'T' is the surface tension then force acting downward along the circumference of the circle due to surface tension is T2πr .
At equilibrium,
P π r² = T 2 π r
P = 2T/r
which is the required pressure.
Hopefully this answer helps.
MARK AS BRAINLIEST
Thanks.
Considering the equilibrium of the upper hemisphere of the drop, the upward force due to excess pressure will be P π r² , where P is the pressure.
If 'T' is the surface tension then force acting downward along the circumference of the circle due to surface tension is T2πr .
At equilibrium,
P π r² = T 2 π r
P = 2T/r
which is the required pressure.
Hopefully this answer helps.
MARK AS BRAINLIEST
Thanks.
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