A solid ball of density half that of water falls freely under gravity from a height of 19.6 m and then enters water. Up to what depth will the ball go. How much time will it take to come again to the water surface? Neglect air resistance and viscosity effects in water. (Take `g=9.8 m//S^(2))`.

Options

A. 4 s
B. 8s
C. 6s
D. 2s

Answer

A

Explanation

`v=sqrt(2gh)=sqrt(2xx9.8xx19.6)=19.6 ms^(-1)`
Let `rho` be the density of ball and `2rho` the density of water. Net retardation inside the water,

`a=("upthrust-weight")/("mass")`
`=(V(2rho)g-V(rho)(g))/(V(rho)g) " " (V="volume of ball")`
`=g=9.8 ms^(-2)`
Hence, the ball will go upto the same depth 19.6 m below the water surface. Further, time taken by the abll to come back to water surface is, `t=2((v)/(a))=2((19.6)/(9.8))=4s`