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The R.D. of ice is 0.92 and that of sea water is 1.025. Find the total volume of an iceberg which floats with its volume 800 cm3 above water. - Physics

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प्रश्न

The R.D. of ice is 0.92 and that of sea water is 1.025. Find the total volume of an iceberg which floats with its volume 800 cm3 above water.
संक्षेप में उत्तर
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उत्तर

Relative density of Ice = 0.92
Relative density of sea water = 1.025
Let the total volume of iceberg = X cm3.
The volume of the iceberg above water = 800 cm3.
The volume of the iceberg is submerged in the water = (X - 800) cm3.
Fraction of iceberg submerged = (X- 800)/X
Now we know that the fractional part submerged equals the ratio of the density of the material of the block to the density of the liquid.
(Density of ice / Density of sea water) = fraction submerged
0.92/1.025 = (X-800)/X
0.8975 X = X - 800
X - 0.8975 X = 800
0.1025 X = 800
X = 800/0.1025 = 7804.8 cm3.
Total volume of iceberg = 7804.8 cm3.

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Determination of Relative Density of a Solid Substance by Archimedes’ Principle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Fluids - Exercise 2 [पृष्ठ १७४]

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फ्रैंक Physics [English] Class 9 ICSE
अध्याय 4 Fluids
Exercise 2 | Q 26 | पृष्ठ १७४

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