Question

Hey !


A sphere has a hollow portion which is one third of it's total volume . It floats in water with four fifth of it's volume immersed .The specific gravity (relative density )of it's material is ?

WITH EXPLANTION(STEPS )
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Answer 1

Answer:

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Case 1: Body in water

So if 1/3 of its volume stays outside of water, then 2/3 of it is submerged.

Buoyant Force = ρg(2v/3) = 2ρgv/3

This force must be equal to the gravitational force acting on the body

Buoyant Force = Gravitational Force

2ρgv/3 = mg

Case 2: Body in liquid

Let the density of this munchkin be ρ’

If 3/4 of its volume is outside, only 1/4 is submerged.

Buoyant Force = ρ’g(v/4) = ρ’gv/4

Again, this must be equal to Gravitational Force

ρ’gv/4 = mg

Considering both cases, we find that:

2ρgv/3 = mg and ρ’gv/4 = mg

thanks

Answer 2

Answer:

Okay I'll stop but tell me the "Second OHM s law ."

Show your guts young man .

I respect you too !