Question

The total surface area of a hollow cylinder, which is open on both sides, is \[3575c{{m}^{2}}\]; area of its base ring is \[357.5c{{m}^{2}}\]and its height is 14cm, then the thickness of the cylinder is

Answer 1

Hint: We solve the given problem by using the simple formulas of total surface area and area of the ring. According to the question the upper surface of cylinder is shown below

We use the total surface area of cylinder when both ends are open is given as
 \[TSA=2\pi Rh+2\pi rh+2\pi \left( {{R}^{2}}-{{r}^{2}} \right)\]
The base area of the cylinder is given as
 \[A=\pi \left( {{R}^{2}}-{{r}^{2}} \right)\]
By using these two formulas we find the thickness of the cylinder as \['R-r'\].

Complete step-by-step answer:
We are given that the height of the cylinder as
 \[h=14cm\]
We are given that the total surface area of cylinder is
 \[TSA=3575c{{m}^{2}}\]
Let us assume that \[R,r\]are the outer and inner radii of the hollow cylinder.
We know that the total surface area of open cylinder is given as
 \[TSA=2\pi Rh+2\pi rh+2\pi \left( {{R}^{2}}-{{r}^{2}} \right)\]
By substituting the area and height values in above equation we get
 \[\begin{align}
  & \Rightarrow 3575=2\pi R\left( 14 \right)+2\pi r\left( 14 \right)+2\pi \left( {{R}^{2}}-{{r}^{2}} \right) \\
 & \Rightarrow 28\pi \left( R+r \right)+2\pi \left( {{R}^{2}}-{{r}^{2}} \right)=3575...........equation(i) \\
\end{align}\]
We are given that the base area of hollow cylinder is given as
 \[A=357.5c{{m}^{2}}\]
We know that the base area of a hollow cylinder is given as
 \[A=\pi \left( {{R}^{2}}-{{r}^{2}} \right)\]
By substituting the value of area in above equation we get
 \[\Rightarrow 357.5=\pi \left( {{R}^{2}}-{{r}^{2}} \right).....equation(ii)\]
Now by combining the equation (i) and equation (ii) we get
 \[\begin{align}
  & \Rightarrow 28\pi \left( R+r \right)+2\left( 357.5 \right)=3575 \\
 & \Rightarrow 28\pi \left( R+r \right)=3575-751 \\
 & \Rightarrow R+r=\dfrac{2824}{28\pi } \\
\end{align}\]
We know that \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\].
Now, by applying the above formula to the equation (ii) we get
 \[\begin{align}
  & \Rightarrow \pi \left( R+r \right)\left( R-r \right)=357.5 \\
 & \Rightarrow \left( R+r \right)\left( R-r \right)=\dfrac{357.5}{\pi } \\
\end{align}\]
Now, by substituting the value of \[\left( R+r \right)\]in above equation we get
 \[\begin{align}
  & \Rightarrow \left( \dfrac{2824}{28\pi } \right)\left( R-r \right)=\dfrac{357.5}{\pi } \\
 & \Rightarrow \left( R-r \right)=\dfrac{357.5\times 28}{2824} \\
 & \Rightarrow \left( R-r \right)=3.54cm \\
\end{align}\]
Therefore the thickness of the hollow cylinder is 3.54 cm.

Note: Students may make mistakes in the formula of total surface area of the hollow cylinder. Here, we are mentioned that the hollow cylinder is open on both sides, so the total surface area is given as
 \[TSA=2\pi Rh+2\pi rh+2\pi \left( {{R}^{2}}-{{r}^{2}} \right)\]
But due to over reading they miss this point of open on both sides and assume the formula of total surface area as
 \[TSA=2\pi Rh+2\pi \left( {{R}^{2}}-{{r}^{2}} \right)\]
This will give us the wrong answer. So, reading the question is important in this problem.