A solid sphere of radius \( 4\, \text{cm} \) has a mass of \( 64\, \text{kg} \). What will be the mass of a shell of the same metal whose internal and external radii are \( 2\, \text{cm} \) and \( 3\, \text{cm} \) respectively?

Calculate the density of the solid sphere:
The volume \( V \) of a solid sphere is given by the formula:
\(V = \frac{4}{3} \pi r^3\)
For the solid sphere with radius \( r = 4\, \text{cm} \):
\(V = \frac{4}{3} \pi (4)^3 = \frac{4}{3} \pi (64) = \frac{256}{3} \pi \, \text{cm}^3\)
To find the density \( \rho \):
\(\rho = \frac{\text{mass}}{\text{volume}} = \frac{64\, \text{kg}}{\frac{256}{3} \pi \, \text{cm}^3}\)
Converting mass to grams (since \( 1\, \text{kg} = 1000\, \text{g} \)):
\(\rho = \frac{64000\, \text{g}}{\frac{256}{3} \pi} = \frac{64000 \times 3}{256 \pi} = \frac{192000}{256 \pi} = \frac{750}{\pi} \, \text{g/cm}^3\)

 Calculate the volume of the shell:
The volume \( V_{\text{shell}} \) of the shell is the difference between the volume of the outer sphere and the inner sphere.

Outer sphere = (radius \( 3\, \text{cm} \)):
\(V_{\text{outer}} = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36 \pi \, \text{cm}^3\)

Inner sphere = (radius \( 2\, \text{cm} \)):
\(V_{\text{inner}} = \frac{4}{3} \pi (2)^3 = \frac{4}{3} \pi (8) = \frac{32}{3} \pi \, \text{cm}^3\)

Calculate the volume of the shell:
\(V_{\text{shell}} = V_{\text{outer}} - V_{\text{inner}} = 36 \pi - \frac{32}{3} \pi\)
Converting \( 36 \) to a fraction with a denominator of 3:
\(V_{\text{outer}} = \frac{108}{3} \pi\)
Now subtract:
\(V_{\text{shell}} = \left(\frac{108}{3} \pi - \frac{32}{3} \pi\right) = \frac{76}{3} \pi \, \text{cm}^3\)

Calculate the mass of the shell:
Using the density \( \rho = \frac{750}{\pi} \, \text{g/cm}^3 \):
\(\text{Mass of shell} = \text{Density} \times \text{Volume} = \frac{750}{\pi} \times \frac{76}{3} \pi\)
The \( \pi \) cancels out: \(\text{Mass of shell} = \frac{750 \times 76}{3} = \frac{57000}{3} \approx 19000 \, \text{g} \approx 19 \, \text{kg}\)