A cubical block of wood of specific gravity `0.5` and chunk of concrete of specific gravity `2.5` are fastened together. the ratio of mass of wood to the mass of concrete which makes the combination to float with entire volume of the combination submerged in water is

Explanation

m=Mass of solid sphere
p=Density of solid sphere
m.=Mass of cavity
p.=Density of cavity
R=Radius of sphere
r=Radius of cavity
Masss of solid sphere `=4/3 pi (R^(3)-r^(3)) p=m`
Mass of cavity `=4/3 pir^(3) p.=m.`
`m/(m,)=4/3 pi (R^(3)-r^(3)p)/(4/3 pir^(3)p.)`
`=(R^(3)-r^(3))/(r^(3)) p/(p.) ......(i)`
As the sphere of volume V and radius R is submerged under water.
Therefore, using principle of floatation we can write
Weight of water displaced= Weight of solid sphere +Weight of sawdust
`vp_(w)g=mg+m.g`
`p_(w)="density of water"`
`vp_(w)=4/3 pi (R^(3)-r^(3))p+4/3 pi r^(3) p.`
`43 pi R^(3)p_(w)=4/3 pi (R^(3)-r^(3))p+4/3 pir^(3) p.`
`R^(3)=R^(3)-r^(3) p/p_(w)+r^(3) (p.)/(p_(w))`
As `p/p_(w)=` relative density of solid sphere `=sigma`
`(p.)/(p_(w))` =relative density of sawdust `=sigma.`
`R^(3)=(R^(3)-r^(3)) sigma+r^(3) sigma.`
`R^(3)-R^(3)sigma=r^(3)sigma.-r^(3)sigma`
`R^(3) (1-sigma)=r^(3) (sigma.-sigma)`
`R^(3)/r^(3)=(sigma.-sigma)/(1-sigma)`
`R^(3)/r^(3)-1=(sigma.-sigma)/(1-sigma)-1`
`(R^(3)-r^(3))/(r^(3))=(sigma.-sigma-1+sigma)/(1-sigma)=(sigma.-1)/(1-sigma)`
From (i) `m/(m.) =(R^(3)-r^(3))/(r^(3)) (p/p^.)`
As `p/(p.)=sigma/(sigma.)`
`m/(m.)=((sigma.-1)/(1-sigma)) (sigma/sigma^(.))`
As `sigma =4.8, sigma^(.)=0.6`
`m/(m.)=((0.6 -1)/(1-4.8)) ((4.8)/(0.6))`
`=(-0.4)/(3.8) xx (4.8)/(0.6)=0.8`