A liquid is contained in a vertical tube of semicircular cross section. The contact angle is zero. The forces of surface tension on the curved part and on the flat part are in ratio.
A
1:1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1:2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
π:2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2:π
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C
π:2
The surface is in contact with curved and the flat part.Here is the top view of the container,
Arrow represents the way surface is trying to pull the contact part of container. Let F1= Force due to surface tension on the curved part Let S = surface tension of the liquid r = radius of semicircular part ⇒F1=S{πr} F2= Force due to surface tension on the flat part F2=S×2r ⇒F1F2=SπrS2r=π2 Hence (c)