A circular test is cylinderical to …

According to the question,
diameter of cylindrical portion = 105 m
Therefore, radius of cylindrical portion= 52.5 m and
height of cylindrical portion (h) = 4 m
{tex}{/tex}Now, Curved Surface area of cylindrical portion = 2{tex}\pi r h{/tex}
{tex}= 2 \times \frac { 22 } { 7 } \times 52.5 \times 4{/tex}
{tex}= \frac { 9240 } { 7 }{/tex}
= 1320 m²
Given slant height of conical portion (l) = 40 m and
Radius of conical portion = Radius of cylindrical portion = 52.5 m
{tex}\Rightarrow{/tex} Curved Surface area of conical portion = {tex}\pi r l{/tex}
{tex}= \frac { 22 } { 7 } \times 52.5 \times 40{/tex}
{tex}= \frac { 46200 } { 7 }{/tex}
= 6600 m²
Therefore, Total surface area = Curved Surface area of cylindrical portion + Curved Surface area of conical portion
= 1320 + 6600
= 7920 m²