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Question
An iceberg floats in sea water of density 1.17 g cm 3, such that 2/9 of its volume is above sea water. Find the density of iceberg.
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Solution
Density of sea water =ρw = 1.17 g cm-3
Density of solid ice berg =ρi = ?
`2/9`th volume of an iceberg is above the seawater
∴ Volume of iceberg inside water = `"V"_"i" = (1 - 2/9)"V"`
Where V = Total volume of an iceberg
`=> "V"_"i" = (9 - 2)/9"V" = 7/9"V"`
⇒ Volume of sea water displaced by immersed part of the iceberg
`= "V"_"w" = 7/9"V"`
By law of floatation:
Weight of iceberg = Weight of sea water displaced by iceberg
`"V"_"i" xx rho_"i" xx "g" = "V"_"w" xx rho_"w" xx "g"`
`"V" xx rho_"i" = 7/9 "V" xx 1.17`
`=> rho_"i" = 7/9 xx 1.17 = 0.91 "gcm"^-3`
⇒ Density of iceburg = 0.91 g cm-3
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