a hollow spherical iron shell floats just completely submerged in water.the outer diameter is 50 cm and the density of iron is 7.9cm cubed .find he inner diameter of the shell
volume of iron = (4/3)pi (Rout^3-Rin^3)
mass of iron = (1 g/cm^3)(4/3) pi Rout^3
so
density of iron = Rout^3/(Rout^3-Rin^3)
To find the inner diameter of the hollow spherical iron shell, we need to consider the principle of buoyancy. The net buoyant force acting on the shell equals the weight of the water displaced by the shell.
Step 1: Calculate the volume of the hollow spherical shell.
The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius of the sphere.
The outer diameter of the shell is 50 cm, so the outer radius (R) is half the diameter, which is 25 cm or 0.25 m.
Let the inner radius be denoted by r.
The volume of the hollow shell is V_shell = (4/3)π((R^3) - (r^3)).
Step 2: Determine the weight of the hollow shell.
The weight of an object can be calculated using the formula W = mg, where m is the mass of the object and g is the acceleration due to gravity. In this case, since the density of iron is mentioned, we can use the density formula to find the mass.
The density (ρ) is given by the equation ρ = m/V, where m is the mass and V is the volume.
Since the density of iron (ρ_iron) is given as 7.9 g/cm^3, we convert it to kg/m^3, which gives us 7900 kg/m^3.
The mass (m_shell) of the shell can be determined as m_shell = ρ_iron * V_shell.
Step 3: Calculate the weight of the water displaced.
The weight of the water displaced is equal to the weight of the shell. This is given by the equation W_displaced = mg, where m is the mass of the water displaced and g is the acceleration due to gravity.
The mass of the water displaced (m_water_displaced) can be calculated as m_water_displaced = ρ_water * V_shell.
The density of water (ρ_water) is approximately 1000 kg/m^3.
Step 4: Equate the weight of the shell to the weight of the water displaced.
Since the shell floats just completely submerged in water, the net buoyant force is zero.
Therefore, equating the weight of the shell to the weight of the water displaced, we have m_shell * g = m_water_displaced * g.
Step 5: Solve the equation to find the inner radius.
By cancelling out the acceleration due to gravity on both sides of the equation, we have ρ_iron * V_shell = ρ_water * V_shell.
Now substitute the values of densities and the expression for the volume of the hollow shell.
(ρ_iron * (4/3)π((R^3) - (r^3))) = (ρ_water * (4/3)π((R^3) - (r^3))).
Cancel out the common terms, and we are left with ρ_iron * (R^3) - ρ_iron * (r^3) = ρ_water * (R^3) - ρ_water * (r^3).
Simplifying further, we have (ρ_iron - ρ_water) * (R^3) = (ρ_water - ρ_iron) * (r^3).
Divide both sides by (ρ_water - ρ_iron), and we have (R^3) = (r^3).
Take the cube root of both sides, and we get R = r.
Since the outer radius (R) is 0.25 m, the inner radius (r) is also 0.25 m.
Finally, the inner diameter of the shell is twice the inner radius, so the inner diameter is 2 * 0.25 = 0.5 m or 50 cm.