A block of wood is floating in water such that `1//2` of it is submerged in water when the same block is floated in alcohol, `1//3^(rd)` of it's volume is submerged Now a mixture of water and alcohol is made taking equal volume of both and block is floated in it. What is the % of it's volume that is now submerged?

Answer

`40`

Explanation

`mg=rho_(1)xx0.5 V_(0)g Rightarrow rho_(1)=(2m)/V_(0)`
`mg=rho_(2)xx(V_(0))/3 g Rightarrowrho_(2)=(3m)/V_(0)`
`rho=(rho_(1)V+rho_(2)V)/(2V)=(rho_(1)+rho_(2))/2`
`mg=rhoVg=((rho_(1)+rho_(2))Vg)/2 Rightarrow m=1/2((2m)/V_(0)+(3m)/V_(0))V`
`m=(5mV)/(2V_(0))`
`V=0.4 V_(0) Rightarrow 40%`