Question

A metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm. find its
(i) inner curved surface area.
(ii) outer curved surface area.
(iii) total surface area.

Answer 1

Hint: We will use the formula of the surface area of cylindrical things. A metal pipe is in the shape of a cylinder. The length is given. The surface area will be different for outer surface and inner surface. The formula for surface area is dependent on the radius of the figure. For the inner and outer surface area the radii are different. So, we will get different values. For the total surface area, we need to add up both inner and outer surface area.

Complete step-by-step solution:
The given figure of the metal pipe is of a cylindrical shape. It has both inner and outer diameter which means the figure has a measurable thickness. So, the inner and outer surface area will be different.
Now, the length of the metal pipe is 77 cm. So, $l=77$.
Let, the inner and outer radius of the pipe be $r$ and $R$.
Given that, the inner diameter of a cross-section is 4 cm and the outer diameter being 4.4 cm.
So, the radius is half of the diameter.
$2r=4\Rightarrow r=\dfrac{4}{2}=2$ cm and $2R=4.4\Rightarrow R=\dfrac{4.4}{2}=2.2$ cm.
We know that formula of the surface area of a cylinder is $S=length\times perimeter$.
So, for the inner surface area ${{S}_{r}}=2\pi rl$ and for the outer surface area ${{S}_{R}}=2\pi Rl$.
We put the values in the formula to get the surface area.
The inner surface area ${{S}_{r}}=2\pi rl=2\times \dfrac{22}{7}\times 2\times 77=968c{{m}^{2}}$.
 The outer surface area ${{S}_{R}}=2\pi Rl=2\times \dfrac{22}{7}\times 2.2\times 77=1064.8c{{m}^{2}}$.
The total surface area is the sum of the inner and outer surface areas.
So, total surface area is $S={{S}_{r}}+{{S}_{R}}=968+1064.8=2032.8c{{m}^{2}}$.

Note: We need to always remember the difference between the inner and outer surface area as it’s just because of the thickness of the figure. If the thickness is missing then we would only have one kind of thickness i.e. both side surface area will be of the same value. We tried to process the formula using a radius. But to minimize our problem we can also use the diameter as instead of using $r$ and $R$, we can use $2r$ and $2R$ in the perimeter. Also, we need to care about the lid of the figure if the lid exists. In our figure, it’s not mentioned. So, we didn’t use the lid factor. In that case, we need to add the area of the circle which is $2\times \pi {{R}^{2}}=2\pi {{R}^{2}}$. There are two lids on two sides.