I am extremely puzzled in the situation mentioned below, which is a problem of Excessive Pressure in Liquid Drop.
This is my question:
A liquid drop exerts a pressure on walls of the drop, and the drop expands (consider Internal Pressure to be Higher than both the Pressure exerted by atmosphere on drop as well as the pressure due to the force of surface tension) . Therefore the work done by the force due to this internal pressure is Positive.
We know that Negative of Work Done by internal conservative forces is equal to Change in Potential Energy.
This would mean that Initial Potential Energy is Greater than the Final Potential Energy. This seems to contradict the fact that surface energy of liquid drop increases due to expansion of drops.
I tried to think about work done considering the force of surface Tension, but we can't only and only think of work done by Surface Tension but we also have to think about work done by the Internal Pressure. And since the Internal Pressure is larger than force of Surface Tension (because the liquid drop expands), The net work done is still positive. This contradicts the -W = ∆U equation for the Expanded Drop as its final surface energy should be more.
Answers
Answer:
We know that Negative of Work Done by internal conservative forces is equal to Change in Potential Energy.
This would mean that Initial Potential Energy is Greater than the Final Potential Energy. This seems to contradict the fact that surface energy of liquid drop increases due to expansion of drops.
I tried to think about work done considering the force of surface Tension, but we can't only and only think of work done by Surface Tension but we also have to think about work done by the Internal Pressure. And since the Internal Pressure is larger than force of Surface Tension (because the liquid drop expands), The net work done is still positive. This contradicts the −W=ΔU equation for the Expanded Drop as its final surface energy should be more.
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Explanation: