Question

A ship floats with 1/3 of its volume outside water and 2/3 of its volume in a liquid. What is the relative density of the liquid?

Answer 1

By Archimedes principle: ship weight=displaced liquid weight.

Let’s say ρ=specific Mass("density")ρ=specific mass("density")

we have:

ρ(ship).Vol(ship).g=ρ(liq).Vol(liq).g ρ(ship).Vol(ship).g=ρ(liq).Vol(liq).g

then

ρ(ship)/ρ(liq)=Vol(liq)/Vol(ship)ρ(ship)/ρ(liq)=Vol(liq)/Vol(ship)

As we know that Vol(liq)=2/3 Vol(ship), we have: rho(ship)/rho(liq)=2/3rho(ship)/rho(liq)=2/3

or: ρ(ship)=2/3ρ(liq)ρ(ship)=2/3ρ(liq)

The “density” of the ship is equal to 2/3 the density of the liquid. Or, more correctly, the specific mass of the ship is 2/3 the specific mass of the liquid


hope this helps!

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Answer 2

Answer:

8000kg/m^3

Explanation:

Let V be the volume of the ship.

Let D be the density of other liquid.

Let m be the mass of the ship.

bouyant force in water = volume submerged×density of water= m

m=1000×2/3V=====>1

bouyant force in other liquid = volume submerged×D= m

1/4VD=m ========>2

Combining equation 1&2

1/4vd=1000×2/3V

(8000/3) Kg/m^3=D

Converting to g/cm^3

8/3gm/cm^3=D

Relative Density=8/3

HOPE IT HELPS

PLEASE MARK BRAINLIEST