A cylindrical plastic bottle of negligible mass is filled with 500ml of water. On  slightly pressing it downwards and releasing, it starts executing simple harmonic  motion. If radius of the bottle is 5cm, then its angular frequency of oscillation is  close to ($densityofwater\rho ={10}^{3}kg/m^3$)

detailed solution

Correct option is B

The restoring force responsible for executing simple harmonic motion in this case  is the buoyant force.Using this, we can equate  FB=ρVg to the restoring force F=kx .Comparing the two equations,  kx=ρVgHere  V=Ax,Substituting this we get  kx=ρAxgHence,  k=ρAg For a body executing simple harmonic motion, its angular frequency can be written  as   ω=kmSubstituting k=ρAg in  ω=km, we get ω=ρAgm .Now, the mass of water in the bottle  m=ρAhHence,   ω=ρAgρAh=gh………………………(1)In the question, the volume of water is given as 500 ml. Using the conversion that  1l=10−3 m3,  we get  1ml=10−6 m3,Hence, volume of 500ml of water is   5×10−4 m3Again using  V=Ah,  5×10−4 =AhTherefore  h=5×10−4 A=5×10−4 πr2. Given the radius if the base is 5 cm,h=5×10−4 π×5×10−22 = 1 5πh=115.70 .Substituting this in equation (1), ω=gh=9.8115.70ω=12.7  rad/s