Question

A hollow spherical iron shell floats almost completely submerged in water. The outer diameter is 58.2 cm, and the density of iron is 7.87 g/cm3. Find the inner diameter.

Answer 1

Answer:

0.556m

Explanation:

unit conversion

58.2 cm = 0.582 m -> the outer radius is 0.582/2 = 0.291 m

$7.87 g/cm^33 = 7.87 g/cm^33 * 1/1000 (kg/g) * 100^3 cm^3/m^3 = 7870 kg/m^3$

Let r be the inner diameter we can find the volume of the hollow part of the sphere using the following formula

$V_h = \frac{4}{3}\pi r^3$

The volume of the iron shell is the volume of the whole sphere subtracted by volume of the hollow part

$V_s = V - V_h = \frac{4}{3}\pi 0.291^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(0.0246 - r^3)$

Let water density $\rho_w = 1000 kg/m^3$, then the buoyant force is the weight of water displaced by the shell

$F_b = g\rho_w V = g1000\frac{4}{3}\pi(0.291^3) = g1000\frac{4}{3}\pi(0.0246)$

For the shell to almost completely submerged in water, its buoyant force equal to the weight of the shell

$F_b = W_s$

$g1000\frac{4}{3}\pi(0.0246) = g\rho_sV_s$

$g1000\frac{4}{3}\pi(0.0246) = g7870\frac{4}{3}\pi(0.0246 - r^3)$

$1000*0.0246 = 7870(0.0246 - r^3)$

$24.6 = 193.602 - 7870r^3$

$r^3 = (193.602 - 24.6)/7870 = 0.02147$

$r = \sqrt[3]{0.02147} = 0.278 m$

So the inner diameter is 0.278*2 = 0.556 m