If 90% volume of iceberg is immersed in water and only 10% volume is above the water surface what is the density of the iceberg. The density of sea water is given as 1025 kg/ $m^{3}$ ?

900 k g / m^{3}

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922.5 k g / m^{3}

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950 k g / m^{3}

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1000 k g / m^{3}

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Solution

The correct option is B $922.5 k g / m^{3}$
A body will float in a liquid, if the weight of the body is equal to the weight of liquid displaced by the submerged portion of body (i.e., buoyancy force).

Let the density of the iceberg be $ρ_{c}$ . Let the volume of iceberg be $V_{i}$

Data given : 90% of ice berg is immerged in the water i.e., volume of water displaced by the ice berg is $V_{w} = 0.9 V_{i}$ .

$D e n s i t y o f s e a w a t e r, ρ_{w} = 1025 k g / m^{3} (g i v e n)$

$M a s s o f t h e i c e b e r g, M_{i} = ρ_{i} \times V_{i}$

$M a s s o f t h e s e a w a t e r = V_{d i s p l a c e d}$

$M_{w} = ρ_{w} \times V_{w} = 1025 \times 0.9 V_{i}$

For equilibrium, downward force = upward force

$M_{i} = M_{w}$

$ρ_{i} V_{i} = 1025 \times 0.9 V_{i}$

$ρ_{i} = 1025 \times 0.9 = 922.5 k g / m^{3}$

$D e n s i t y o f i c e b e r g i s ρ_{i} = 922.5 k g / m^{3}$

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Similar questions

1.03 g c m^{- 3}

0.92 g c m^{- 3}

1.03 g / c m^{3}

0.92 g / c m^{3}

= 900 {kg/m}^{3}

= 1030 {kg/m}^{3}

= 917 k g m^{- 3}

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

^{3}

^{3}

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