Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
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
APPEARS IN
संबंधित प्रश्न
A solid of density 5000 kg m-3 weighs 0.5 kgf in air. It is completely immersed in water of density 1000 kg m-3. Calculate the apparent weight of the solid in water.
Differentiate between density and relative density of a substance.
Relative density of a substance is expressed by comparing the density of that substance with the density of :
The unit of relative density is :
A solid weighs 120 gf in air and and 105 gf when it is completely immersed in water. Calculate the relative density of solid.
A body of volume 100 cm3 weighs 1 kgf in air. Find:
- Its weight in water and
- Its relative density.
A body of mass 70 kg, when completely immersed in water, displaces 20,000 cm3 of water. Find: (i) The weight of body in water and (ii) The relative density of material of the body.
A piece of wax floats in brine. What fraction of its volume will be immersed?
R.D. of wax = 0.95, R.D. of brine = 1.1.
A body of volume 100 cm3 weighs 1 kgf in air. Calculate its weight in water. What is its relative density?
