Question

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.

• A. 20
• B. 16
• C. 18
• D. 12

Answer 1

Answer: (B), (C)

The correct options are
B

16

C

18

Height of Cylinder = h = AO =2.4 cm

Diameter of Cylinder = 1.4 cm

$\Rightarrow$ Radius of Cylinder = Radius of Cone = OB = r =$\frac{1.4}{2}$=0.7 cm

Using Pythagoras Theorem on $△$AOB to find Slant Height (I) of cone, we get

AB = l = $\sqrt{h^2+r^2}=\sqrt{(2.4{)}^2+(0.7{)}^2}=\sqrt{5.76+0.49}=\sqrt{6.25}=2.5cm$

Surface Area of remaining Solid = Surface Area of Cylinder Part + Inner surface Area of Hollow Cone
=(2.$\pi .r.h+\pi .r^2)+\pi .r.l=\pi .r(2h+r+l)$
= $\frac{22}{7}\times 0.7\left(\right(2\times 2.4+0.7+2.5)$

= 2.2(4.8 + 0.7 + 2.5) = 2.2(8) = 17.60 $c{m}^2$

Total surface area of the remaining solid =18 $c{m}^2$ = 0.0018 $m^2$