Question

A block of an unknown metal has a mass of 227.8 kg and a volume of 0.1189 cubic meters. If the object were a sphere, what would be its radius? (The value for pi ≈ 3.14159)

Answer 1

Explanation:

The given data is as follows.

          mass = 227.8 kg,      volume = 0.1189 $m^{3}$

It is known that formula for volume of a sphere is $\frac{4}{3} \pi r^{3}$ and density is as follows.

             Density = $\frac{mass}{volume}$ ........ (1)

                          = $\frac{227.8 kg}{0.1189 m^{3}}$

                          = 1915.895 $kg/m^{3}$

Hence, putting the given values into equation (1) as follows.

            Density = $\frac{mass}{volume}$

   1915.895 $kg/m^{3}$ = $\frac{227.8 kg}{\frac{4}{3} \pi r^{3}}$        

          $8021.213 r^{3}$ = 227.8

                      r = 0.305 $m^{3}$

Thus, we can conclude that radius of the sphere is 0.305 $m^{3}$.