A wooden block in the form of a uniform cylinder floats with one-third length above the water surface. A small chip of this block is held at rest at the bottom of a container containing water to a height of `1 m` and is then released. The time in which it will rise to the surface of water is `=10 m//s^(2)`)

Options

A. `0.84s`
B. `0.63s`
C. `0.5s`
D. `1s`

Answer

B

Explanation

Since two thirds of the wooden block is immersed in water, its densityy `=2rho//3` where `rho` is the density of water.
Let the volume of the chip be `V`. Then its weight is `(2//3)Vrhog`.
The upthrust acting on it inside the water is `Vrhog`. Hence, the unbalanced upward forces is `Vrhog(1-2//3)=Vrhog/3`.
Hence its aceleratioin a in the upward direction is
`a=(Vrhog//3)/((2//3)Vrho)=g/2`
If `t` is the required time time `1=01/2at^(2)=1/2g/2t^(2)`
Hence `t^(2)=4/g=4/10=0.4=0.63s`