Question

a wooden block floats in a glycerine in such a way that its 2/5th volume remains above surface.If relative density of wood is 0.78 ,calculate the relative density of glycerine

Answer 1

From the laws of flotation

Weight of the wooden block=Weight of the glycerine displaced by immersed portion

(Vwood)(g)(RDwood)=(Vglycerine)(g)(RDglycerine)

(Vwood)(RDwood)=(Vglycerine)(RDglycerine)

RDglycerine=(Vwood)*(RDwood)/(Vglycerine) =(V)*(0.78)/(1−2/5)V =(V)*(0.78)/(3/5)V =1.3

Answer 2

Volume of block = V
Volume of water displaced = V - 2V/5 = 3V/5 = 0.6V
Density of wood (d) = 0.78 × 1 g/cm³ = 0.78 g/cm³
Density of glycerine = d’

Weight of block = Buoyant Force
Vdg = 0.6Vd’g
d’ = d / 0.6
d’ = 0.78 g/cm³ / 0.6
d’ = 1.3 g/cm³

Relative density of glycerine = 1.3 g/cm³ / (1 g/cm³)
= 1.3

Relative density of glycerine is 1.3