Physics, asked by rudhrahprasanth, 1 year ago

A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2. If the surface tension of water is
70 dyne/cm then the normal force required to separate the glass plates from each other will be

Answers

Answered by Anonymous
122

Given:

Surface tension (T) = 70

Water drop (d) = 0.05

Area (A) = 40

In the given content:

\huge\boxed{\sf{F =  \frac{(2T)}{d} A}}

So:

\sf{F =  \frac{2T}{V}  \times  {A}^{2}}

Therefore:

\implies \sf{F =  \frac{2 \times 70 \times (40) ^{2} }{0.05}}

\implies \sf{F =   \frac{140 \times (40)^{2} }{0.05} }

\implies \sf{F = 45 \times  {10}^{5}  \: dyne}

Hence:

The normal force required to separate the glass plates from each other will be 45 N.


Anonymous: Ha
meettrivedi991: na
cyrus01: oo
cyrus01: herione
cyrus01: ooo
Anonymous: Hi
cyrus01: pagal
Anonymous: Ha
kartik15345: hii...
Answered by joker32
5

Final answer is 45 N.

A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2. If the surface tension of water is

70 dyne/cm then the normal force required to separate the glass plates from each other will be


arik86: hlo
rudhrahprasanth: hi
Similar questions