Question

The figure shows the 3D view of a metal nut. The metal nut has two regular hexagonal faces. The side of a hexagonal face is 2 cm, and a thickness of 1 cm. A hole of diameter 2 cm is present in the metal nut. Given that the density of the metal is \[7.9\] grams per cubic cm, calculate the mass of the nut.

Answer 1

Hint: Here, we need to find the mass of the metal nut. The metal nut is in the shape of a hexagonal prism, with a cylindrical hole in the middle. First, we will find the volume of the cylindrical hole, and the volume of the hexagonal prism (without the hole). Then, we will use the two volumes to find the volume of metal used to make the nut. Finally, we will multiply the volume by the density per cubic cm to find the mass of the metal nut.

Formula Used:
 We will use the following formulas:
1.The volume of a prism is the product of the area of its base and the height of the prism.
2.The area of a regular hexagon is given by the formula \[\dfrac{{3\sqrt 3 }}{2}{a^2}\], where \[a\] is the length of the side of the hexagon.
3.The volume of a cylinder is given by the formula \[\pi {r^2}h\], where \[r\] is the radius of the base, and \[h\] is the height of the cylinder.

Complete step-by-step answer:
First, we need to find the volume of the metal nut.
The metal nut is in the shape of a hexagonal prism, with a cylindrical hole in the middle.
Thus, the volume of the metal nut is the difference in the volume of the hexagonal nut (without the hole), and the volume of the cylindrical hole.
We will find the dimensions of the cylindrical hole.
The height of the cylinder is equal to the thickness of the nut.
Therefore, we get
Height of cylindrical hole \[ = 1\] cm
Now, we will find the radius of the hole.
The radius of the base of the cylindrical hole is half of its diameter, that is \[r = \dfrac{d}{2}\].
Substituting \[d = 2\]cm in the formula, we get
Radius of the base of the hole \[ = \dfrac{2}{2} = 1\] cm.
Now, we need to find out the volume of the hexagonal prism (without the hole), and the volume of the cylindrical hole.
The volume of a cylinder is given by the formula \[\pi {r^2}h\], where \[r\] is the radius of the base, and \[h\] is the height of the cylinder.
Substituting \[r = 1\]cm and \[h = 1\]cm in the formula, we get
Volume of the cylindrical hole \[ = \pi {\left( 1 \right)^2}{\rm{1 c}}{{\rm{m}}^3} = \pi \times 1 \times 1{\rm{ c}}{{\rm{m}}^3} = \pi {\rm{ c}}{{\rm{m}}^3}\]
Substituting \[\pi = 3.14\] in the expression, we get
\[ \Rightarrow \] Volume of the cylindrical hole \[ = 3.14{\rm{ c}}{{\rm{m}}^3}\]
Now, we will find the volume of the hexagonal prism (without the hole).
The volume of a prism is the product of the area of its base and the height of the prism.
The area of a regular hexagon is given by the formula \[\dfrac{{3\sqrt 3 }}{2}{a^2}\], where \[a\] is the length of the side of the hexagon.
Therefore, we get
Volume of hexagonal prism (without the hole) \[ = \dfrac{{3\sqrt 3 }}{2}{a^2} \times h\]
Substituting \[a = 2\]cm and \[h = 1\]cm in the formula, we get
\[ \Rightarrow \] Volume of hexagonal prism (without the hole) \[ = \dfrac{{3\sqrt 3 }}{2}{\left( 2 \right)^2} \times \left( 1 \right)\]
Simplifying the expression, we get
\[ \Rightarrow \] Volume of hexagonal prism (without the hole) \[ = \dfrac{{3\sqrt 3 }}{2} \times 4 = 6\sqrt 3 {\rm{ c}}{{\rm{m}}^3}\]
Substituting \[\sqrt 3 = 1.732\] in the expression, we get
\[ \Rightarrow \] Volume of hexagonal prism (without the hole) \[ = 6 \times 1.732{\rm{ c}}{{\rm{m}}^3} = 10.392{\rm{ c}}{{\rm{m}}^3}\]
Finally, we can calculate the volume of the metal nut.
The volume of the metal nut is the difference in the volume of the hexagonal nut (without the hole), and the volume of the cylindrical hole.
Therefore, we get
Volume of the metal nut \[ = \left( {10.392 - 3.14} \right){\rm{ c}}{{\rm{m}}^3} = 7.252{\rm{ c}}{{\rm{m}}^3}\]
Thus, the volume of metal used to make the metal nut is \[7.252{\rm{ c}}{{\rm{m}}^3}\].
It is given that 1 cubic cm of metal has density \[7.9\] grams.
The mass of the metal nut is the product of the volume of the metal nut in cubic cm, and the density of each cubic cm of metal.
Thus, we get
Mass of the metal nut \[ = 7.252 \times 7.9{\rm{ g}}\]
Multiplying the terms, we get
\[ \Rightarrow \] Mass of the metal nut \[ = 57.2908{\rm{ g}}\]
Therefore, we get the mass of the metal nut as \[57.2908\] grams, or approximately 57 grams.

Note: Here we can make a mistake by using 2 cm as the radius of the base of the hole. This will result in the wrong volume of the cylindrical hole, and thus, incorrect volume of the metal nut. Also we should remember to convert the diameter to radius first. Radius is exactly half of the diameter.