Question

A solid body floating in a water has 1/5th of its volume immersed in it. what fraction of its volume will be immersed,if it floats in a liquid of gravity 1.2?​

Answer 1

Answer :-

1/6th of the volume of the body will be immersed in the given liquid.

Explanation :-

We have :-

→ Solid body immersed in water [Case 1]

→ Volume immersed in water = 1/5th

→ Specific gravity of water (ρ) = 1

→ Specific gravity of the liquid (ρ') = 1.2

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Let the volume of the solid be v .

For water :-

Volume of water displaced (v) = v/5

According to Archimedes principle :-

⇒ W = vρg

⇒ W = v/5 × 1 × g

⇒ W = vg/5

For the other liquid :-

Let the volume of the body immersed (volume of the liquid displaced) be v' .

According to Archimedes principle :-

⇒ W = v' × ρ' × g

⇒ vg/5 = v' × 1.2 × g

⇒ v' = vg/(5 × 1.2g)

⇒ v' = vg/6g

⇒ v' = v/6

Answer 2

Given :-

A solid body floating in a water has 1/5th of its volume immersed in it.

To Find :-

Volume immersed

Solution :-

Let the volume of body be V

Density of water  = $\sf\rho_{w}$

Density of solid = $\sf\rho_{s}$

Now According to the question

$\sf \dfrac{\rho_{s}}{\rho_{w}} = \dfrac{\dfrac{V}{5}}{V}$

$\sf\dfrac{\rho_{s}}{1.2 \times \rho_{w}} = \dfrac{1}{5}$(1)

Now

Let the density of liquid be l

$\sf \dfrac{\rho_{s}}{\rho_{l}} = \dfrac{\dfrac{V}{V}}{5}$

$\sf\dfrac{\rho_{s}}{1.2 \rho_{w}} = \dfrac{1}{n}$

$\sf\dfrac{1}{1.2} = \dfrac{1}{n}$

$\sf\dfrac{1}{1.2 \times 5} = n$

$\sf\dfrac{1}{6}=n$