Search any question & find its solution
Question:
Answered & Verified by Expert
Two glass plates are separated by water. If surface tension of water is 75 dyne/cm and area of each plate wetted by water is $8 \mathrm{~cm}^2$ and the distance between the plates is 0.12 mm , then the force applied to separate the two plates is
Options:
Solution:
1972 Upvotes
Verified Answer
The correct answer is:
$10^5$ dyne
The shape of water layer between the two plates is shown in the figure.
Thickness $d$ of the film $=0.12 \mathrm{~mm}$
$=0.012 \mathrm{~cm}$
Radius $R$ of the cylindrical face $=\frac{d}{2}$
Pressure difference across the surface
$=\frac{T}{R}=\frac{2 T}{d}$
Area of each plate wetted by water $=A$
Force $F$ required to separate the two plates is given by
$F=$ pressure difference $\times$ area
$=\frac{2 T}{d} A$
Putting the given values, we get
$F=\frac{2 \times 75 \times 8}{0.012}=10^5$ dyne
Thickness $d$ of the film $=0.12 \mathrm{~mm}$
$=0.012 \mathrm{~cm}$
Radius $R$ of the cylindrical face $=\frac{d}{2}$
Pressure difference across the surface
$=\frac{T}{R}=\frac{2 T}{d}$
Area of each plate wetted by water $=A$
Force $F$ required to separate the two plates is given by
$F=$ pressure difference $\times$ area
$=\frac{2 T}{d} A$
Putting the given values, we get
$F=\frac{2 \times 75 \times 8}{0.012}=10^5$ dyne
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.