Question

The radius of a solid copper cylinder is 4.2 cm and its height is 1.6 cm. How many discs of diameter 1.4 cm and thickness 0.2 cm can be made from this cylinder melting it?

Answer 1

Hint: Assume the number of the discs that can be formed from the cylinder be x. The radius and height of the cylinder are equal to 4.2 cm and 1.6 cm respectively. Now use the formula, \[Volume=\pi {{\left( radius \right)}^{2}}\left( height \right)\] and get the volume of the cylinder. The diameter and thickness of the disc are 1.4 cm and 0.2 cm respectively. We know that radius is half of the diameter. Now, get the radius of the disc. Now, use the formula, \[Volume=\pi {{\left( radius \right)}^{2}}\left( thickness \right)\] and get the volume of the disc. Now, calculate the volume of x discs. Since the discs are formed from the cylinder after melting it so, the total volume of discs is equal to the volume of the cylinder. Therefore, The volume of the cylinder = The total volume of all the discs. Solve it further and get the value of x.

Complete step by step solution:

According to the question, we have a solid copper cylinder that has a radius equal to 4.2 cm and its height is 1.6 cm. After melting the cylinder, we have to form discs of diameter 1.4 cm and thickness 0.2 cm. We have to find out the number of discs that can be formed after melting the cylinder.
The radius of the cylinder = 4.2 cm ………………..(1)
The height of the cylinder = 1.6 cm ………………….(2)
We know the formula of volume of the cylinder, \[\text{Volume}=\pi {{\left( \text{radius} \right)}^{2}}\left( \text{height} \right)\] …………………(3)
Now, putting the values of the radius from equation (1) and the height from equation (2), in the formula of equation (3), we get
\[\begin{align}
  & \Rightarrow \text{Volume}=\pi {{\left( 4.2 \right)}^{2}}\left( 1.6 \right)c{{m}^{3}} \\
 & \Rightarrow \text{Volume}=\dfrac{22}{7}\times \left( \dfrac{42}{10} \right)\left( \dfrac{42}{10} \right)\left( 1.6 \right)c{{m}^{3}} \\
\end{align}\]
\[\Rightarrow \text{Volume}=22\times \left( \dfrac{6}{10} \right)\left( \dfrac{42}{10} \right)\left( 1.6 \right)c{{m}^{3}}\] ……………………………..(4)
Let the number of discs that can be formed after melting the cylinder be x.
The thickness of the disc = 0.2 cm …………………………..(5)
The diameter of the disc = 1.4 cm
We know that the radius is half of the diameter.
The radius of the disc = \[\dfrac{1.4}{2}\] cm = 0.7 cm ……………………………(6)
We know the formula of volume of the disc, \[Volume=\pi {{\left( radius \right)}^{2}}\left( thickness \right)\] ………………………..(7)
Now, putting the values of the radius from equation (6) and the thickness from equation (5), in the formula of equation (7), we get
\[\begin{align}
  & \Rightarrow Volume=\pi {{\left( 0.7 \right)}^{2}}\left( 0.2 \right)c{{m}^{3}} \\
 & \Rightarrow Volume=\dfrac{22}{7}\times \left( \dfrac{7}{10} \right)\left( \dfrac{7}{10} \right)\left( 0.2 \right)c{{m}^{3}} \\
\end{align}\]
\[\Rightarrow Volume=22\times \left( \dfrac{1}{10} \right)\left( \dfrac{7}{10} \right)\left( 0.2 \right)c{{m}^{3}}\]
The volume of one disc is \[\text{22}\times \left( \dfrac{1}{10} \right)\left( \dfrac{7}{10} \right)\left( 0.2 \right)c{{m}^{3}}\] .
So, the total volume of x discs = \[\text{22}\times \left( \dfrac{1}{10} \right)\left( \dfrac{7}{10} \right)\left( 0.2 \right)x\,c{{m}^{3}}\] ………………………….(8)
Since the discs are formed from the cylinder after melting it so, the total volume of discs is equal to the volume of the cylinder.
From equation (4), we have the volume of the cylinder.
The volume of the cylinder = The total volume of all the discs
\[\begin{align}
  & \Rightarrow 22\times \left( \dfrac{6}{10} \right)\left( \dfrac{42}{10} \right)\left( 1.6 \right)c{{m}^{3}}=\text{22}\times \left( \dfrac{1}{10} \right)\left( \dfrac{7}{10} \right)\left( 0.2 \right)x\,c{{m}^{3}} \\
 & \Rightarrow \dfrac{22\times 6\times 42\times 1.6}{22\times 7\times 0.2}=x \\
 & \Rightarrow \dfrac{6\times 6\times 1.6}{0.2}=x \\
 & \Rightarrow 6\times 6\times 8=x \\
 & \Rightarrow 288=x \\
\end{align}\]
Therefore, the total number of discs that can be formed from the cylinder is 288.

Note: In this question, one can take the volume of the disc as \[\text{ }\!\!\pi\!\!\text{ }{{\left( 1.4 \right)}^{\text{2}}}\left( 0.2 \right)c{{m}^{3}}\] . This is wrong because 1.4 cm is the measure of the diameter and we need the measure of the radius to find the volume of the disc. We know that radius is half of the diameter. So, the measure of the radius is \[\dfrac{1.4}{2}=0.7\] cm. Therefore, the volume of the disc is \[\text{ }\!\!\pi\!\!\text{ }{{\left( 0.7 \right)}^{\text{2}}}\left( 0.2 \right)c{{m}^{3}}\] .