Find how hollow a rock is given its density and its weight in water

AI Thread Summary
A geologist suspects a rock is hollow, as it weighs twice as much in air compared to its weight in water. The density of the solid part of the rock is 2500 kg/m^3, while water has a density of 1000 kg/m^3. Using the relationship between weight, density, and volume, the calculations show that the average density of the rock is 2000 kg/m^3. This leads to the conclusion that 80% of the rock's volume is solid. The solution confirms the rock's hollow nature based on its density and buoyancy in water.
season89
Messages
3
Reaction score
0

Homework Statement



A geologist thinks a rock is hollow. The rock weighed in air is twice as much as when it's in water. Find the fraction of the rock's volume that is solid given the solid part of the rock is 2500kg/m^3 and water is 1000kg/m^3


Homework Equations



rho = mass/volume
Fb = rho*volume*gravity
F(normal) = Force gravity (rock in air) - rho(fwater)*V(displaced)*gravity

The Attempt at a Solution


 
Physics news on Phys.org
Can you please show your attempt at the solution as per PF guidelines.

Jared
 
Given that density of solid part of rock is 2500kg/m^3, water is 1000kg/m^3, air has neglibile density and the whole object weighs twice is much in air than in water:

Apparent weight = m(object in air)*g - m(fluid displaced)*g
1/2*m(object in air)*g = m(object in air)*g - m(fluid displaced)g
1/2* rho (object)*V*g = rho (object)*V*g - rho (fluid)*V*g
1/2* rho (object) = rho (object) - rho (fluid)
-1/2*rho (object) = - 1000kg/m^3
rho (object) = 2000kg/m^3

therefore
2000kg/m^3 = x*(solid part) + (1-x)*(air)
2000kg/m^3 = 2500kg/m^3*x
x = 2000/2500 = 0.8 or 80%
 
Your answer is correct.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Back
Top