Question

To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimen weighs twice as much in air as it does in water. The density of the solid part of the specimen is 5.0×103kg/m35.0×10

3
kg/m
3
. What fraction of the specimen's apparent volume is solid?

Answer 1

Fraction of the specimen's is 0.4.

Step-by-step explanation:

We know,

Mass = volume × density

Weigh= mass × g

= volume × density× g

= density× g × volume

`=\rho.g.V`

An object weighs less submerged due to buoyant force acting on it.

`\therefore W_(wet)= W_(dry)-B`

`B= W_(dry)-W_(wet)`

`=W_{\textrm{fluid displaced}}`

`=\rho_(fluid). g.V_(submerged)`

Given that, the weighs of the specimen in dry is twice of the weighs in air.

`W_(wet)=\frac 12W_(dry)`

Then ,

`B= W_(dry)-W_(wet)`

`= W_(dry)-\frac12W_(dry)`

`=\frac12W_(dry)`

`=\rho_(Rock). g.V_(Rock)`

Therefore,

`\rho_(fluid). g.V_(submerged)=\frac12\rho_(Rock). g.V_(Rock)`

`\Rightarrow \rho_(Rock). g.V_(Rock)=2\rho_(fluid). g.V_(submerged)`

`\Rightarrow (.V_(Rock))/(V_(submerged))=(2\rho_(fluid). g)/(\rho_(Rock).g)`

`\Rightarrow (.V_(Rock))/(V_(submerged))=(2\rho_(fluid))/(\rho_(Rock))`

`\Rightarrow (.V_(Rock))/(V_(submerged))=(2* 1.0 * 10^3\ kg /m^3)/(5.0* 10^3 \ kg/m^3)`

=0.4

Fraction of the specimen's is 0.4.