Question

A conical groove was made in an iron cylinder and was filled with gold. If the base and height of both the cylinder and cone are the same and are equal to $10\text{cm}$ and $10\text{cm}$ respectively, find the mass of the combination. (density of gold = $19.3{\text{g cm}}^{-3}$ and density of iron = $7.8{\text{g cm}}^{-3}$

• A. Cannot be determined
• B. $11.1\text{g}$
• C. $11.1\text{kg}$
• D. $11,100\text{kg}$

Answer 1

The correct option is C $11.1\text{kg}$

Volume of the cone,
${\text{v}}_{\text{cone}}=\frac{1}{3}\pi r^{2}h=\frac{1000}{3}\pi$
Volume of the cylinder,
${\text{v}}_{\text{cy}}=\pi r^{2}h=1000\pi$

volume of gold = volume of the cone

volume of iron = volume of cylinder - volume of the cone
$\Rightarrow {\text{v}}_i={\text{v}}_{\text{cy}}-{\text{v}}_{\text{cone}}=1000\pi -\frac{1000}{3}\pi$
$=\frac{2000}{3}\pi$

Mass of gold = volume of gold $\times$ density of gold
${\text{m}}_{\text{G}}={\text{v}}_{\text{cone}}\times 19.3=\frac{1000}{3}\pi \times 19.3$
$=6143.3\text{g}$

Mass of iron = volume of iron $\times$ density of iron
${\text{m}}_{\text{i}}={\text{v}}_{\text{i}}\times 7.8=\frac{2000}{3}\pi \times 7.8$
$=4965.6\text{g}$

Total mass of the combination = ${\text{m}}_{\text{i}}+{\text{m}}_{\text{G}}=6143.3+4965.6$
$=11108.9\text{g}\approx 11.1\text{kg}$