Question

A body is floating in water such that 6 out of 10 parts of its volume is under water. Its density is

Answer 1

Since Volume Under Water =6/10V(V=Total Volume )

Now Mg=Bouyant Force

D(body)×V×g=D(water)×6V/10×g

So

D(body)=1000×6/10=600kg/m^3

Answer 2

Given :

A body is floating in water such that 6 out of 10 parts of its volume is under water.

Therefore , ratio of water displaced and volume of body is , $\dfrac{V_w}{V_b}=\dfrac{6}{10}$ .

To Find :

The density of body .

Solution :

We know , Weight of body = Weight of water displaced

$\rho_bV_bg=\rho_wV_wg\\\\\rho_b=\rho_w\dfrac{V_w}{V_b}$

Here , $\rho_w$ is the density of water , $\rho_w=1000\ kg/m^3$ .

Putting all these in above equation .

We get ,

$\rho_b=1000\times \dfrac{6}{10}\\\\\rho_b=600\ kg/m^3$

Therefore , density of body is $600\ kg/m^3$ .

Learn More :

Fluids Mechanics

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