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 to the nearest cm².

Cylinder: 

Height of the cylinder, $h=2.4 cm$

Diameter of the cylinder, $d=1.4 cm$

Radius of the cylinder, $r=\frac{1.4}{2}=0.7 cm$

Cone: 

Height of the cone, $h=2.4 cm$ [From the figure] 

Diameter of the cone, $d=1.4 cm$

Radius of the cone, $r=\frac{1.4}{2}=0.7 cm$ 

Slant height of the cone, 

$l=\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2}$

$l=\sqrt{{0.7}^2+{2.4}^2}$

$l=\sqrt{0.49+5.76}$

$l=\sqrt{6.25}$

$l=2.5$

So, the total surface area of the remaining solid 

= Curved surface area of the cylinder + Area of the base of the cylinder + Curved surface area of the cone.

$=2\mathit{\pi rh}+\pi r^2+\pi rl$

$=2\times \frac{22}{7}\times 0.7\times 2.4+\frac{22}{7}\times \mathrm{ }0.7\times 0.7+\frac{22}{7}\times 0.7\times 2.5$

$=4.4\times 2.4+2.2\times 0.7+2.2\times 2.5$

$=10.56+1.54+5.5$

$=17.60{\mathrm{ cm}}^2$

Hence, the total surface area of the remaining solid to the nearest cm² = $18 {\mathrm{cm}}^2$