The density ρ of a liquid varies with depth h from the free surface as ρ=kh. A small body of density ρ1 is released from the surface of the liquid. The body executes SHM with amplitude
A
2ρ1k
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
ρ1k
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
√2ρ1k
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
√3ρ1k
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Bρ1k Let V be the volume of body which is released on the surface of liquid. When body reaches at depth h from the surface of liquid, then upward thrust on body due to liquid =V(kh)g
Weight of body =Vρ1g
At equilibrium, `V(kx)g=Vrho_(1)g or x=rho_(1)//k`