Concept explainers
A hollow copper (ρCu = 8.92 × 103 kg/m3) spherical shell of mass m = 0.950 kg floats on water with its entire volume below the surface. a. What is the radius of the sphere? b. What is the thickness of the shell wall?
(a)
The radius of the sphere.
Answer to Problem 25PQ
The radius of the sphere is
Explanation of Solution
Since the sphere is in equilibrium condition so that weight of the spherical shell is equal to the buoyant force of the water.
Here,
Write the expression for force due to gravity.
Here,
Write the expression for buoyant force.
Here,
Write the expression for volume of the spherical shell.
Here,
Conclusion:
Substitute the equation (II) and (III) in the equation (I).
Substitute equation (IV) in the above equation to solve for .
Substitute
Therefore, the radius of the sphere is
(b)
The thickness of the shell wall.
Answer to Problem 25PQ
The thickness of the shell wall is
Explanation of Solution
Write the expression for mass of the ball.
Here,
Rearrange the above equation for volume of the inner and outer radius of the sphere.
Here,
Write the expression for thickness of the shell wall.
Here,
Conclusion:
Substitute
Solve the above equation to find
Substitute
Therefore, the thickness of the shell wall is
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