Density of ice is 900 kg/m $^{3}$ . A piece of ice is floating in water (of density 1000 kg/m $^{3}$ ). The fraction of volume of the piece of ice outside the water is:

0.9

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0.1

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0.7

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Solution

The correct option is A 0.1
The forces acting on the ice must balance each other.
Hence buoyant force acting upwards = Weight of the ice

$⟹ V_{i m m} ρ g = V d g$
$⟹ \frac{V_{i m m}}{V} = \frac{d}{ρ} = 0.9$

Hence, the fraction of ice outside water $= 1 - 0.9 = 0.1$

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An ice berg of density 900 Kg/ $m^{3}$ is floating in water of density 1000 Kg/ $m^{3}$ . The percentage of volume of ice-cube outside the water is:

917 k g / m^{3}

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