Question

An object floats in water (at 4°C) with 20% of its
volume outside the water surface. What percentage
of volume of the same body will be outside the
liquid surface whose relative density is 5?​

Answer 1

Answer:

Outside part = 20%

= 1/5

Immersed part in water = 1-1/5

= 4/5 or, 80%

According to the Floatation principle,

v/V = density of substance/density of liquid

4/5 = x/1

= 0.8 g/cc = x = Density of the substance

If R.D of substance is 5 then in g/cc it would also be 5,

v/V = density of substance/density of liquid

= 0.8/5

= 4/25

So, 4/25 of teh volume of that solid is immersed in that liquid. So, the volume outside must be =

1-4/25

= 21/25

So, 21/25 of the volume of the body is outside the liquid.

Hope this helps.

Answer 2

Answer:

Let the volume be V and density of the object be d

Mass is Vd, weight Vdg

In water

Vd g= 0.80V * 1 * g

Or d= 0.8 of the object.

Now it is floating in liquid of density 5

So

V*0.8*g= v * 5 * g where v is the volm of object in water

v= 0.8V/ 5 .

So object volm outside the water is

V-- 0.8V/ 5= 3.2V/5 answer= 32/50V= 16/25;V outside the water.

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