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 sphere of iron and another of wood, both of same radius are placed on the surface of water. State which of the two will sink? Give a reason for your answer.
How are the (i) Mass, (ii) Volume and (iii) Density of a metallic piece affected, if at all, with an increase in temperature?
The density of iron is 7.8 x 103 kg m-3. What is its relative density?
A stone of density 3000 kgm3 is lying submerged in water of density 1000 kgm3. If the mass of stone in air is 150 kg, calculate the force required to lift the stone. [g = 10 ms2]
A glass cylinder of length 12 x 10-2 m and area of crosssection 5 x 10-4 m2 has a density of 2500 kgm-3. It is immersed in a liquid of density 1500 kgm-3, such that 3/8. of its length is above the liquid. Find the apparent weight of glass cylinder in newtons.
A solid weighs 0.08 kgf in air and 0.065 kgf in water. Find
(1) R.D. of solid
(2) Density of solid in SI system. [Density of water = 1000 kgm3]
A solid of R.D. 4.2 is found to weigh 0.200 kgf in air. Find its apparent weight in water.
A solid weighs 105 kgf in air. When completely immersed in water, it displaces 30,000 cm3 of water, calculate relative density of solid.
