An iceberg is floating in water. The density of ice in the iceberg is $917 k g m^{- 3}$ and the density of water is $1024 k g m^{- 3}$ . What percentage fraction of the iceberg would be visible?

5 %

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11 %

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12 %

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8 %

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Solution

The correct option is B $11 %$
The buoyancing force due to water = Weight of iceberg

$V_{1} ρ_{1} g = m g = V_{2} ρ_{2} g$

$V_{1} \times 1024 = V_{2} \times 917$

$V_{1} = 0.89 V_{2}$

Iceberg visible $= V_{2} - V_{1}$

$= V_{2} - 0.89 V_{2}$

$= 0.11 V_{2}$

Fraction of iceberg visible $= \frac{0.11 V_{2}}{V_{2}} = 0.11$

% Fraction of iceberg visible $= 11$ %

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Similar questions

917 k g m^{- 3} .

1024 k g m^{- 3} .

(917 k g m^{- 3}) ?

0.92 g m / c m^{3}

1.03 g m / c m^{3}

= 917 k g m^{- 3}

= 1.024 \times 10^{3} k g m^{- 3}

1.03 g c m^{- 3}

0.92 g c m^{- 3}

= 900 k g / m^{3}

= 1030 k g / m^{3}

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