Density of ice is 900 kg//m^(3). A piece...

Density of ice is `900 kg//m^(3)`. A piece of ice is floating in water of density `1000kg//m^(3)`. Find the fraction of volume of the picec of ice outside the water.

Text Solution

Verified by Experts

Let V be the total volume and `V_(1)` the volume of ice piece immersed in water. For equilibrium of ice piece,
weight =upthrust
`therefore " " V rhoig=V_(i) rho_(w)g`
Here, `rho_(i)`=density of water`=900 kgm^(-3)`
and `rho_(w)`=density of water `=1000kgm^(-3)`
Substituting in Eq. (i), we get
`(V_(i))/(V)=(rho_(i))/(rho_(w))=(900)/(1000)=0.9`
i.e., the fraction of volume outside the water, f=1-0.9=0.1

Topper's Solved these Questions

FLUID MECHANICS
DC PANDEY|Exercise Example 13.13|1 Videos
FLUID MECHANICS
DC PANDEY|Exercise Example 13.14|1 Videos
FLUID MECHANICS
DC PANDEY|Exercise Example 13.11|1 Videos
EXPERIMENTS
DC PANDEY|Exercise Subjective|15 Videos
GENERAL PHYSICS
DC PANDEY|Exercise INTEGER_TYPE|2 Videos

An iceberg of density 900kg//m^(3) is floating in water of density 1000 kg//m^(3) . The percentage of volume of ice cube outside the water is

`20%`

`35%`

`10%`

`25%`

An ice -berg of density 900kgm^(-3) is floating in water of density 1000kgm^(-3) .The percentage of volume of ice -berg outside the water is

`20%`

`35%`

`10%`

`11%`

A cube of density 250 kg/ m^2 floats in water, then what part of total volume of the cube outside the water?

0.75

0.25

0.333

0.677