Question

Please solve q 169?

Answer 1

Let us consider that

1 the external volume of the concrete sphere be `V_e`

2.the volume of its cavity full of sawdust be `V_s`

3.the material volume of the sphere `V_m=V_e-V_s`

Given specific gravity of concrete `rho_c=2.4` and that of sawdust `rho_s=0.3`
It is obvious the specific gravities given here w . r.to water. So the specific gravity of water will be `rho_w=1`

By the condition of the problem the concrete sphere floats completely in water when it's cavity is filled with sawdust.

So by the condition of floatation we can write

Mass of concrete sphere + Mass of sawdust = Mass of water displaced by external volume of the sphere

So

`V_m*rho_c+V_s*rho_s=V_e*rho_w`

Inserting `V_e=V_m+V_c`
And `rho_c=2.4,rho_s=0.3, rho_w=1` we get

`V_m*2.4+V_s*0.3=(V_m+V_s)*1`

`=>V_m*1.4=V_s*0.7`

`=>V_m/V_s=0.7/1.4=1/2`

So the ratio of mass of concrete to that of sawdust

`=>V_m/V_s*rho_c/rho_s=1/2xx2.4/0.3=4/1`

So option (2) is correct.