Question

The internal and external diameter of a hollow hemispherical shell are 6 cm and 10 cm respectively. What is the total surface area of the shell?

• A. $44\pi c{m}^2$
• B. $84\pi c{m}^2$
• C. $124\pi c{m}^2$
• D. $294\pi c{m}^2$

Answer 1

The correct option is B $84\pi c{m}^2$


Observe from the figure that the total surface area of a hemispherical shell is the sum of the inner curved surface area, outer curved surface area and the area of the ring.

Given, inner diameter of the shell = 6 cm
So, inner radius(r) = 3 cm
Therefore, inner curved surface area of the shell
= $2\pi r^2$
= $2\pi \times 3^2$
= $18\pi c{m}^2$

Given, outer diameter of the shell = 10 cm
So, outer radius of the shell(R) = 5 cm
Therefore, outer curved surface area of the shell
= $2\pi R^2$
= $2\pi \times 5^2$
= $50\pi c{m}^2$

Now, inner radius of the ring is 3 cm and outer radius of the ring is 5 cm.
Therefore, area of the ring
= $\pi (R^2-r^2)$
= $\pi (5^2-3^2)$
= $16\pi c{m}^2$

$\therefore$ Total surface area = Inner curved surface area + Outer curved surface area + Area of the ring
= $50\pi c{m}^2$ + $18\pi c{m}^2$ + $16\pi c{m}^2$
= $84\pi c{m}^2$