Internal radius of the cylindrical container = 10 cm = r
Height of water = 7 cm = h
So, the surface area of the wetted surface
![\[\begin{array}{*{35}{l}} =\text{ }2\pi rh\text{ }+\text{ }\pi {{r}^{2}}~=\text{ }\pi r\left( 2h\text{ }+\text{ }r \right)\text{ }=\text{ }22/7\text{ x }10\text{ x }\left( 2\text{ x }7\text{ }+\text{ }10 \right) \\ =\text{ }220/7\text{ x }24 \\ =\text{ }754.29\text{ }c{{m}^{2}} \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b9834a57bfbf315f9088d1a4ba1dc57c_l3.png "Rendered by QuickLaTeX.com")
Internal radius of the cylindrical container = 10 cm = r
Height of water = 7 cm = h
So, the surface area of the wetted surface
