Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

Two glass plates are separated by water. If surface tension of water is 75 dynes per cm and area of each plate wetted by water is 8 cm2 and the distance between the plates is 0.12 mm, then the force applied to separate the two plates is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

104 dynes

b

102 dynes

c

105 dynes

d

106 dynes

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

The shape of water layer between the two plates is shown in the figure.
Thickness d of the film = 0.12 mm = 0.012 cm.
Radius R of cylindrical face = d2 
F = T(2l) = P(l× 2R)

 

Question Image

P = TR
Pressure difference across the cylindrical surface = TR = 2Td
Area of each plate wetted by water = A.
Force F required to separate the two plates is given by
F= pressure difference × area =2TdA
=2×75×80.012=105dynes

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring