# 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.

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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).

“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

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