A brass cube of side a and density σ is floating in mercury of density ρ. If the cube is displaced a bit vertically, it executes S.H.M. Its time period will be
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a
2πσ aρ g
b
2πρ aσ g
c
2πρ gσ a
d
2πσ gρ a
answer is A.
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Detailed Solution
As a is the side of cube σ is its density. Mass of cube is a2σ, its weight =a3σgLet h be the height of cube immersed in liquid of density ρ in equilibrium then, F=a2h ρ g=Mg=a3σ gIf it is pushed down by y then the buoyant force F'=a2(h+y)ρ gRestoring force is ΔF=F'−F=a2(h+y)σ g−a2h σ g =a2y ρ gRestoring acceleration =ΔFM=−a2yρ gM=−a2ρ ga3σyMotion is S.H.M. ⇒T=2πa3σa2ρ g=2πaσρ g