A drop of water of volume 0.05cm^(3) is pressed between two glass plates, as a consequence of which it spreads and occupies an area of 40cm^(2). If the surface tension of water is 70 dyne/cm, then the normal force required to separate out the two glass plates will be in Newton

A drop of water of volume 0.05cm^(3) is pressed between two glass plat

1.	A drop of water of volume 0.05cm^(3) is pressed between two glass plates, as a consequence of which it spreads and occupies an area of 40cm^(2). If the surface tension of water is 70 dyne/cm, then the normal force required to separate out the two glass plates will be in Newton
Answer» <html><body><p>90<br/>45<br/>22.5<br/>450</p>Answer :<a href="https://interviewquestions.tuteehub.com/tag/b-387190" style="font-weight:bold;" target="_blank" title="Click to know more about B">B</a></body></html>