Question

The volume of a hemisphere is $19404$ cubic cm. The total surface area is:
A) $2772$ sq.cm
B) $4158$ sq.cm
C) $5544$ sq.cm
D) $1386$ sq.cm

Answer 1

Here, the question deals with the concept of hemispheres. A set of three-dimensional points whose center is at equidistant with all the points on the surface is known as a sphere. Here, in the question we are given the volume of the hemisphere. By using the formula of volume of hemisphere and equating it with the given volume, we will get the value of radius. Then, using that value of radius in the formula of surface area, we will get our required solution.

Formula used:
Volume of hemisphere: $\dfrac{2}{3}\pi {r^3}$
Total surface area of hemisphere: $3\pi {r^2}$

Complete step-by-step solution:
Given the volume of the hemisphere is $19404$ cubic cm.
We know the formula of volume of hemisphere i.e., $\dfrac{2}{3}\pi {r^3}$ . Equate the formula of volume of hemisphere to the given volume and we get,
Volume = $19404$ cubic cm.
$\Rightarrow \dfrac{2}{3}\pi {r^3} = 19404$
To simplify the calculation, we will take the value of $\pi$ as $\dfrac{{22}}{7}$ . Solving it gives,
$\Rightarrow \dfrac{2}{3} \times \dfrac{{22}}{7} \times {r^3} = 19404 \\\ \Rightarrow {r^3} = \dfrac{{19404 \times 3 \times 7}}{{2 \times 22}} \\\ \Rightarrow {r^3} = 9261 \\\$
Now, we will apply cubic root on both sides of the equation and get,
$\Rightarrow \sqrt[3]{{{r^3}}} = \sqrt[3]{{9261}} \\\ \Rightarrow r = \sqrt[3]{{21 \times 21 \times 21}} \\\ \Rightarrow r = 21\,cm \\\$

As now we have the value of radius, we will use it in the formula of total surface area i.e., $3\pi {r^2}$
$\Rightarrow 3 \times \dfrac{{22}}{7} \times {\left( {21} \right)^2} \\\ \Rightarrow 3 \times \dfrac{{22}}{7} \times 21 \times 21 \\\ \Rightarrow 4158\,sq.cm \\\$
Therefore, the value of the total surface area of the hemisphere is $4158$ sq.cm.

Hence the correct answer is option ‘B’.

Note: Here, in this question although we weren’t given the value of $\pi$ , we assumed it to be $\dfrac{{22}}{7}$ for easy calculations. This question was easy to solve as we knew the formulas for the volume and the total surface area of the hemisphere. Students should avoid making any calculation mistakes.