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JEE Advanced 2020 Concept Recapitulation Test 2 Paper 1
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© examsnet.com
Question : 1
Total: 54
Two glass plates are separated by water. If surface tension of water is 75 dynes per
c
m
and area of each plate wetted by water is
8
c
m
2
and the distance between the plates is
0.12
m
m
, then the force applied to separate the two plates is
10
2
dynes
10
4
dynes
10
5
dynes
10
6
dynes
Validate
Solution:
The shape of water layer between the two plates is shown in the figure.
Thickness
d
of the film
=
0.12
m
m
=
0.012
c
m
.
Radius
R
of cylindrical face
=
d
2
.
F
=
T
(
21
)
=
P
(
I
×
2
R
)
P
=
T
R
Pressure difference across the cylindrical surface
=
T
R
=
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
×
area
=
2
T
d
A
=
2
×
75
×
8
0.012
=
10
5
dynes
© examsnet.com
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