Thus inner diameter of hollow sperical iron shell is 57.3 cm.
How is density and diameter of sperical iron shell related?
We use the formula 4/3r3 to calculate the volume of a sphere. The volume of the hollow area must be subtracted from the total volume of the sphere in order to get the volume of a hollow sphere. Call the sphere's overall radius r1 and the radius of the hollowed-out area r2. As a result, we have 4/3π*r1^3 – 4/3π*r2^3.
When you know the radius, you may use the formula volume = (4/3) r3 to calculate the sphere's volume. The mass may then be calculated by multiplying the density and volume, as in mass = volume density, using the density formula density = mass/volume. The following formula might also be used: mass = (4/3) r3 density.
Estimate of V we interpret "almost completely submerged" to mean
V ≈ 4 πR^3 / 3
here R =60cm.
Thus, equilibrium of forces (on the iron sphere) leads to
F = m (iron).g
ρ (water)gV =ρ (iron) g( 4πR^3/ 3 - 4πr^3/ 3)
r is the inner radius (half the inner diameter).
Substituting in our estimate for V as well as the densities of water (1.0g/cm ^3) and iron(7.87g/cm ^3) we get the inner diameter:
[tex]2r = 2R(1 - \frac{1.0}{7.87})^{\frac{1}{3} }[/tex] =57.3cm.
Thus inner diameter is 57.3 cm.
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