 # An IceBerg Of Density 900 Kg/M3 Is Floating In Water Of Density 1000 Kg/M3. What Will Be The Percentage Of Volume Of Ice-cube Outside The Water?

Let the total volume of the iceberg be V.

Vsub= Volume of an iceberg submerged

$\rho_{b} = Density of iceberg = 900 Kg/m3$ $\rho_{w} = Density of water = 1000 Kg/m3$

For the flotation of the body, the weight of the body= the weight of water is displaced.

$\Rightarrow \rho_{b} V = \rho_{w} V_{sub}$ $\Rightarrow \frac{V_{sub}}{V} = \frac{\rho _{b}}{\rho _{w}}$

Now by subtracting both sides from 1, we get:

$\Rightarrow 1 -\frac{V_{sub}}{V} = 1 – \frac{\rho _{b}}{\rho _{w}}$ $\Rightarrow \frac{V – V_{sub}}{V} = \frac{\rho _{w}-\rho _{b}}{\rho _{w}}$

Now by converting it in percentage, we get:

$\Rightarrow \frac{V – V_{sub}}{V} * 100 = \frac{\rho _{w} – \rho _{b}}{\rho _{w}} * 100$

Now by substituting the values, we get

$\Rightarrow \frac{1000 – 900}{1000} * 100$ $\Rightarrow \frac{100}{1000} * 100$ $\Rightarrow$ 0.1 * 100

$\Rightarrow$ 10

Therefore, the percentage volume outside the water is 10%.

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