91Ó°ÊÓ

i) Weight of each child,$\mathrm{W}=356\mathrm{N}$

ii) Diameter of log,$\mathrm{d}=0.30\mathrm{m}$

iii) Radius of log,$\mathrm{r}=0.15\mathrm{m}$

iv) Length of log,$\mathrm{l}=1.80\mathrm{m}$

v) Density of log,${\mathrm{ÒÏ}}_{\log }=800\mathrm{kg}/{\mathrm{m}}^3$

Using Archimedes' principle, we can state that for the object to float, the buoyant force should be equal to the weight of the object. We can find the weight of the water, which should be displaced by the logs for children to float using the total weight of children. Using the weight and density of the logs, we can find the number of logs.

Formula:

The buoyant force exerted on a body by the fluid, ${\mathrm{F}}_{\mathrm{b}}={\mathrm{ÒÏ}}_{\mathrm{W}}×\mathrm{g}×{\mathrm{V}}_{\text{submerged}}$(i)

The volume of log, $\mathrm{V}={\mathrm{Ï€°ù}}^2\mathrm{h}$(ii)

$\mathrm{N}$Let’s assume that there is N number of logs. So, the total required force of buoyancy for logs and three children to float is given by:

${\mathrm{F}}_{\mathrm{b}}=3\left(356\right)+\mathrm{N}\left({\mathrm{ÒÏ}}_{\text{wood}}\right)×\mathrm{g}×\mathrm{V}……\left(1\right)$

Using equation (ii), the volume of log is given as:

$\mathrm{V}=3.14(0.15{)}^2×(1.80)$

$V=0.127{\mathrm{m}}^3$

Comparing equation (i) and equation (1), using the formula of total submerged volume of logs, ${\mathrm{V}}_{\text{submerge}}=\mathrm{N}×\left(\text{volume of one single log}\right)$we get,

$3\left(356\right)+\mathrm{N}\left({\mathrm{ÒÏ}}_{\text{wood}}\right)\mathrm{gV}={\mathrm{ÒÏ}}_{\mathrm{w}}\mathrm{g}(\mathrm{N}×\mathrm{V})$

$1068+\mathrm{N}\left(800\right)×\mathrm{g}×\mathrm{V}=1000\mathrm{g}(\mathrm{N}×\mathrm{V})$

$\mathrm{N}=\frac{1068}{\mathrm{gV}(1000−800)}$

$=4.29$

It means $5$logs are needed.