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 as that of cylinder is hollowed out. Find the total surface area of the remaining solid. $(Use\pi =\frac{22}{7})$

• A. $19.60{\text{cm}}^2$
• B. $18.60{\text{cm}}^2$
• C. $17.60{\text{cm}}^2$
• D. $15.60{\text{cm}}^2$

Answer 1

The correct option is C

$17.60{\text{cm}}^2$

Height of Cylinder $=h=AO=2.4\text{cm}$

Diameter of Cylinder $=1.4\text{cm}$

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

Lets find Slant Height $l$ of cone, using pythagoras theorem on $△AOB$ , we get

$AB=l=\sqrt{h^2+r^2}=\sqrt{(2.4{)}^2+(0.7{)}^2}$

$l=\sqrt{5.76+0.49}=\sqrt{6.25}=2.5\text{cm}$

Surface Area of remaining Solid
= Surface Area of the Cylinder $+$ Inner surface Area of the hollow Cone
$=(2\pi rh+\pi r^2)+\pi rl=\pi r(2h+r+l)=\frac{22}{7}\times 0.7(2\times 2.4+0.7+2.5)=2.2(4.8+0.7+2.5)=2.2\left(8\right)=17.60{\text{cm}}^2$

$\therefore$ Total surface area of the remaining solid is $17.60{\text{cm}}^2$