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The density of ice is 0.918 g"cm"^(-3) a...

The density of ice is `0.918 g"cm"^(-3)` and that of water is `1.03 g"cm"^(-3).` An iceberg floats with a portion of `224 m^(3)` outside the surface of water. Find the total volume of the iceberg.

Text Solution

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The correct Answer is:
2060
Density of ice, `rho = 0 . 918 xx 10 ^(3) kgm ^(-3)`
Density of water ,` rho. = 1.03 xx 10 ^(3) kgm^(-3)`
Let volume of the iceberg ` = V m ^(3)`
Then volume of water displaced
`V. = (v - 224 ) m^(3)`
`:. ` Weight of iceberg = Weight of water displaced
or ` V rho g = v. rho. g`
or ` v xx 0 . 918 xx 10 ^(3) xx g = ( V - 224 ) xx 1 . 03 xx 10 ^(3) xx g`
or ` V (1 . 03 - 0.918) = 224 xx 1.03 `
or `V = ( 22 4 xx 1.03)/(0.112) = 2060 m^(3)`
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