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 (g = 10 m/s²)

Since two-thirds of the wooden block is immersed in water, its density = 2$\rho$/3,where $\rho$ is the density of water.
Let the volume of the chip be V. Then its weight is (2/3)V$\rho$g. The upthrust acting on it inside the water is Vpg. Hence, the unbalanced upward force is $\mathrm{V\rho g}(1-2/3)=\mathrm{V\rho g}/3$ Hence, its acceleration a in the upward direction is

$\mathrm{a}=\frac{\mathrm{V\rho g}/3}{(2/3)\mathrm{V\rho }}=\frac{\mathrm{g}}{2}$

If t is the required time, then $1=0+\frac{1}{2}{\mathrm{at}}^2=\frac{1}{2}\frac{\mathrm{g}}{2}{\mathrm{t}}^2$

Hence, $$\begin{gathered} {\mathrm{t}}^2=\frac{4}{\mathrm{g}}=\frac{4}{10}=0.4 \\ t=0.63\mathrm{s} \end{gathered}$$