question_answer

From a solid cylinder of height 4 cm and radius 3 cm, a conical cavity of height 4 cm and of base radius 3 cm is hollowed out. What is the total surface area of the remaining solid?

A) \[15\pi \,sq\,cm\]          

B)       \[22\pi \,sq\,cm\]

C) \[33\pi \,sq\,cm\]          

D)       \[48\pi \,sq\,cm\]

E) None of these

Correct Answer: D

Solution :

Explanation Option [d] is correct. The resulting solid look like as follows: Here the unshaded portion is a hollow. Thus, total surface of remaining solid will be the bottom circle of cylindrical portion, curved surface of cylinder and the inner curved surface area of cone. Now, surface area of bottom of circle = \[\pi {{r}^{2}}=9\pi sq\,\,cm\] Curved surface area of cylinder = \[2\pi rh=2\pi \left( 3 \right)\left( 4 \right)=24\pi \,\,sq\,\,cm\] Slant height of cone \[\ell =\sqrt{{{r}^{2}}+{{h}^{2}}}=\sqrt{{{3}^{2}}+{{4}^{2}}}=5\,cm\] Curved surface area of cone = \[\pi r\ell =\pi \left( 3 \right)\left( 5 \right)=15\pi \,sq\,cm\] Total surface area of resulting solid = \[9\pi +24\pi +15\pi =48\pi \,sq\,cm\,\]