Answer:
Explanation:
Given
No of logs [tex]n=14[/tex]
diameter of log [tex]d=42\ cm[/tex]
Length of log [tex]L=6.4\ m[/tex]
42 % of log volume(V) is above water when no one is on raft
so 58 % of log volume(V) is submerged in the water
Weight of 14 log
[tex]W=14\times \rho _{log}\times V\times g[/tex]
Buoyancy force on 14 logs [tex]F_b=14\times \rho _{water}\times 0.58V\times g[/tex]
as system is in equilibrium so
[tex]W=F_b[/tex]
[tex]14\times \rho _{log}\times V\times g=14\times \rho _{water}\times 0.58V\times g[/tex]
[tex]\rho _{log}=0.58\rho _{water}[/tex]
[tex]\frac{\rho _{log}}{\rho _{water}}=0.58[/tex]
Specific gravity of log [tex]=0.58[/tex]
