Question

A toy pyramid is made from poly (methyl methacrylate), better known by its trade term Lucite. The toy pyramid has a regular hexagonal base of $15c{{m}^{2}}$ and a height of $4$cm. There is a semi spherical indentation of diameter $2$ cm at the base of the pyramid. If the weight of the pyramid is $21.129$ grams, then find the density of Lucite. (Density equals mass divided by Volume).
A. $1.06g/c{{m}^{3}}$
B. $0.365g/c{{m}^{3}}$
C. $2.09g/c{{m}^{3}}$
D. $6.51g/c{{m}^{3}}$

Answer 1

Hint: In the problem we have the base area of the pyramid and height of the pyramid. From this data we will calculate the volume of the pyramid without any indentation by multiplying the base area with the height of the cone. Now we have a semi spherical indentation of given diameter $2$ cm, indentation means we need to subtract that shape from the given toy i.e., we need to subtract the volume of the sphere from the volume of the pyramid to get the perfect volume of the toy. In the problem they have mentioned the weight of the pyramid and the density of the pyramid can be calculated by the dividing mass with volume. For this problem we can use the volume of sphere formula $\dfrac{2}{3}\pi {{r}^{3}}$ .

Complete step by step solution:
Given that, A toy is in pyramid shape with a rectangular base of area $15c{{m}^{2}}$ and height $4$cm.
Now the volume of the pyramid is given by
$\begin{align}
  & V=15c{{m}^{2}}\times 4cm \\
 & V=60c{{m}^{3}} \\
\end{align}$

The above calculated volume of the pyramid has no indentation. But in the problem, they have mentioned that the pyramid has semi spherical indentation having diameter $2$cm. Now the volume of the indentation is equal to the volume of hemisphere i.e.
$\begin{align}
  & {{V}_{i}}=\dfrac{2}{3}\pi {{\left( \dfrac{2}{2} \right)}^{3}}c{{m}^{3}} \\
 & \Rightarrow {{V}_{i}}=2.095c{{m}^{3}} \\
\end{align}$
Now the true volume of the toy is given by subtracting the volume of indentation from the volume of the pyramid without any indentation. Mathematically
$\begin{align}
  & {{V}_{t}}=V-{{V}_{i}} \\
 & \Rightarrow V=60-2.095 \\
 & \Rightarrow V=57.905c{{m}^{3}} \\
\end{align}$
Given that the weight of the toy is $W=21.129$ grams. So, the density of the cone is given by the dividing the mass by true volume of the cone, mathematically
$\begin{align}
  & d=\dfrac{W}{V} \\
 & \Rightarrow d=\dfrac{21.129gr}{57.905c{{m}^{3}}} \\
 & \Rightarrow d=0.365gr/c{{m}^{3}} \\
\end{align}$

So, the correct answer is “Option B”.

Note: We need to have a better understanding of the quantities and their units while solving the problems of $3D$ shapes. Here they have asked to calculate the density, in some cases they may give options with different units of the density like $g/c{{m}^{3}}$, $kg/{{m}^{3}}$ and also, they may give options with the units that are not related to density. In those cases we don’t need to solve the whole problem from options we can choose the correct one.