Respuesta :
Answer:
3/4 th fraction of the volume of block is submerged in oil.
Explanation:
We know that
density of the block, ρ[tex]_{b}[/tex] =SG[tex]\times[/tex]density of water
= 0.6[tex]\times[/tex] 1000
= 600 kg/[tex]m^{3}[/tex]
density of the oil, ρ[tex]_{o}[/tex] =SG [tex]\times[/tex] denity of water
= 0.8[tex]\times[/tex] 1000
= 800 kg/[tex]m^{3}[/tex]
Let acceleration due to gravity, g = 9.81 m/[tex]s^{2}[/tex]
Volume of block submerged in oil is [tex]V_{1}[/tex]
Volume of block above the oil surface is [tex]V_{2}[/tex]
The total volume of the block is V = [tex]V_{1}[/tex]+[tex]V_{2}[/tex]
Therefore for a partially submerged body, we know that
Buoyant force = total weight of the block
and
total weight of the block = Weight of the fluid displaced by the block
ρ[tex]_{b}[/tex]\timesV\timesg = ρ[tex]_{o}[/tex]\times[tex]V_{1}[/tex]\times g
600([tex]V_{1}[/tex]+[tex]V_{2}[/tex]) = 800 [tex]V_{1}[/tex]
600[tex]V_{1}[/tex] + 600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex]
600[tex]V_{2}[/tex] = 800 [tex]V_{1}[/tex] - 600[tex]V_{1}[/tex]
600[tex]V_{2}[/tex] = 200[tex]V_{1}[/tex]
[tex]\frac{V_{1}}{V_{2}}[/tex] = [tex]\frac{600}{200}[/tex]
Thus [tex]V_{1}[/tex] = 600
[tex]V_{2}[/tex] = 200
V = 600+200
= 800
Therefore fraction of volume of the block submerged in oil is,
[tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{600}{800}[/tex]
[tex]\frac{V_{1}}{V}[/tex] = [tex]\frac{3}{4}[/tex]
