Courses
Courses for Kids
Free study material
Free LIVE classes
More
LIVE
Join Vedantu’s FREE Mastercalss

The total surface area of hollow cylinder, which is open from both sides, is 3575$c{{m}^{2}}$; area of the base ring is 357.5$c{{m}^{2}}$ and height is 14 cm. Find the thickness of the cylinder.

Answer
VerifiedVerified
363k+ views
Hint: Use the formula for Total Surface Area of a hollow cylinder, $TSA=2\pi h\left( {{r}_{1}}+{{r}_{2}} \right)+2\pi \left( r_{1}^{2}-r_{2}^{2} \right)$ where TSA is the total surface area of the cylinder, h is the height of the cylinder and ${{r}_{1}}$ and ${{r}_{2}}$ are the outer and inner radius of the base ring. The expression ${{r}_{1}}-{{r}_{2}}$ denotes the thickness of the cylinder. Substitute the given values to find the thickness.

“Complete step-by-step answer:”

It is given that the total surface area of the hollow cylinder is 3575$c{{m}^{2}}$. Now, we know that the formula for the total surface area of hollow cylinder is $TSA=2\pi h\left( {{r}_{1}}+{{r}_{2}} \right)+2\pi \left( r_{1}^{2}-r_{2}^{2} \right)$.
In this formula, h denotes the height of the cylinder and the expression $\pi \left( r_{1}^{2}-r_{2}^{2} \right)$ denotes the area of the base ring. Substituting these values in the formula we get,
\[\begin{align}
  & 2\pi h\left( {{r}_{1}}+{{r}_{2}} \right)+2\left( 357.5 \right)=3575c{{m}^{2}} \\
 & \Rightarrow 2\pi \left( 14cm \right)\left( {{r}_{1}}+{{r}_{2}} \right)+715c{{m}^{2}}=3575c{{m}^{2}} \\
 & \Rightarrow 2\pi \left( 14cm \right)\left( {{r}_{1}}+{{r}_{2}} \right)=\left( 3575-715 \right)c{{m}^{2}} \\
 & \Rightarrow 2\left( \dfrac{22}{7} \right)\left( 14cm \right)\left( {{r}_{1}}+{{r}_{2}} \right)=\left( 2860 \right)c{{m}^{2}} \\
 & \Rightarrow 2\left( 2cm \right)\left( {{r}_{1}}+{{r}_{2}} \right)=2860c{{m}^{2}} \\
 & \Rightarrow {{r}_{1}}+{{r}_{2}}=\dfrac{2860c{{m}^{2}}}{4cm} \\
 & \Rightarrow {{r}_{1}}+{{r}_{2}}=715cm\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) \\
\end{align}\]
The second piece of information given to us, which is the area of the base ring can be substituted in the formula for area of base ring separately to give
\[\pi \left( r_{1}^{2}-r_{2}^{2} \right)=357.5c{{m}^{2}}\]
Further using the formula for ${{a}^{2}}-{{b}^{2}}=\left( a-b \right)\left( a+b \right)$ we get
\[\pi \left( {{r}_{1}}-{{r}_{2}} \right)\left( {{r}_{1}}+{{r}_{2}} \right)=357.5c{{m}^{2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right)\]
Using the value of \[{{r}_{1}}+{{r}_{2}}=715cm\] in the above equation (2)
\[\pi \left( {{r}_{1}}-{{r}_{2}} \right)\left( 715cm \right)=357.5c{{m}^{2}}\]
\[\begin{align}
  & \Rightarrow \left( \dfrac{22}{7} \right)\left( {{r}_{1}}-{{r}_{2}} \right)\left( 715cm \right)=357.5c{{m}^{2}} \\
 & \Rightarrow {{r}_{1}}-{{r}_{2}}=\dfrac{357.5c{{m}^{2}}}{715cm}\times \dfrac{7}{22} \\
 & \Rightarrow {{r}_{1}}-{{r}_{2}}=\dfrac{7}{44}cm \\
 & \Rightarrow {{r}_{1}}-{{r}_{2}}=0.16cm \\
\end{align}\]
Thus the thickness comes out to be 0.16cm.

Note: The hollow cylinder mentioned in the question is a thick hollow cylinder. It should not be confused with a thin hollow cylinder which does not have a base ring area. One more thing to keep in mind while solving the question is that the area of the base ring is only \[\pi \left( r_{1}^{2}-r_{2}^{2} \right)\] and not \[2\pi \left( r_{1}^{2}-r_{2}^{2} \right)\]. The additional 2 is only to indicate that two rings are being considered (one at the top and the other at the bottom).

Last updated date: 02nd Oct 2023
Total views: 363k
Views today: 10.63k