Question

What is the surface area of a cylindrical ring where the diameter of the cross section is 6.3 in and the centerline has a length of 48in

Answer 1

The correct answer is 950.02 in^2

Answer 2

Answer: The surface area of a cylindrical ring is 950.4 square in.

Explanation:

We know that the surface area of an open cylinder is given by :-

`S.A.=2\pi rh` , where r= radius and h= height

As per given , diameter of cylindrical ring = 6.3 in

then ,
`r=\frac{6.3\text{ in}}{2}=3.15\text{ in}`

h= 48 in

Now , the surface area of the cylindrical ring =
`2((22)/(7))(3.15)(48)`

`=950.4\text{ square in.}`

Hence, the surface area of a cylindrical ring is 950.4 square in.