A drop of liquid of density `rho` is floating half-immersed in a liquid of density `d`. If `sigma` is the surface tension the diameter of the drop of the liquid is

Options

A. the speed of the particle when it returns to its equilibrium position is `u_(0)`
B. the time at which the particle passes through the equilibrium position for the first time is `t=pisqrt((m)/(K))`
C. the time at which the maximum compression of the spring occurs is `t=(4pi)(3)sqrt((m)/(k))`
D. the time at which the particle passes through the equilibrium position for the second time is the time at which the maximum compression of the spring occurs is `t=(5pi)(3)sqrt((m)/(k))`

Answer

AD

Explanation

`v=u_(0)cosomegat`
(suppose `t_(01)` is the time of collision) `(u_(0))/(2)=u_(0)cosomegat_(1)implies t_(1)=(pi)/(3omega)`
Now the particle returns to equilibrium position at time `t_(2)=2t_(1)`,i.e.,`(2pi)/(3omega)` with same mechanical energy, i.e., its speed will be `u_(0)` Let t be the time at which the particle passes through the equilibrium position for the second time.
`t_(3)=(T)/(2)+2t_(1)=(pi)/(omega)+(2pi)/(3omega)=(5pi)/(3omega)=(5pi)/(3)sqrt((m)/(k))`
Time at which maximum compression of spring occurs =`2t_(1)+(T)/(4)=(2pi)/(3omega)+(pi)/(2omega)=(7pi)/(6omega)`