A hemispherical portion of radius -R is removed from the bottom of a cylinder of radius R. The volume of remaining cylinder is V and its mass is M. It is suspended by a string in a liquid of density ρ, where it stays vertical. The upper surface of the cylinder is at a depth h below the liquid surface. The force on the bottom of the cylinder by the liquid 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
M g
b
Mg−Vρg
c
Mg+πR2hρg
d
ρgV+πR2h
answer is D.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
The net upward force on the bottom of the cylinder = weight of the liquid displaced by cylinder + thrust on the upper surface of cyliner due to i column ofliquid =Vρg+hρg×πR2=ρgV+πR2h