A cylindrical body of cross-sectional area A, height H and density ρs′ is immersed to a depth h in a liquid of density ρ, and tied to the bottom with a string. The tension in the string is
A
ρghA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(ρs−ρ)ghA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
(ρ−ρs)ghA
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(ρh−ρsH)gA
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D(ρh−ρsH)gA Let T = Tension in string acting in the downward direction
Weight of body,
W = Mg(↓)
=ρsAHg
Buoyant force, FB=ρg∀(↑)
=ρgAh
For equilibrium,
Weight of cylindrical body + Tension in string
= Buoyant force
W + T = FB
or T = FB−W
=ρgAh−ρsAHg
=(ρh−ρsH)gA