A drop of water of volume 0.05 cm^(3) is...

A drop of water of volume `0.05 cm^(3)` is pressed between two glass plates, as a consequence of which, it spreads and occupies an are of `40 cm^(2)`. If the surface tension of water is `70 "dyne"//cm`, find the normal force required to separate out the two glass plates is newton.

Similar Questions

Explore conceptually related problems

A drop of water volume 0.05 cm^(3) is pressed between two glass -paltes, as a consequence of which, it spreads and occupies an area of 40 cm^(2) . If the surface tension of water is 70 dyn e//cm , find the normal force requried to separte out the two glass plates in newton.

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 water drop of 0.05 cm^(3) is squeezed between two glass plates and spreads into area of 40 cm^(2) . If the surface tension of water is 70 dyne cm^(-1) , then the normal force required to separate the glass plates from each other will be

A drop of water of volume V is pressed between the two glass plates so as to spread to an area. A . If T is the surface tension, the normal force required to separate the glass plates is

A drop of water of volume `0.05 cm^(3)` is pressed between two glass plates, as a consequence of which, it spreads and occupies an are of `40 cm^(2)`. If the surface tension of water is `70 "dyne"//cm`, find the normal force required to separate out the two glass plates is newton.

Similar Questions

Recommended Questions