Question

A golf ball has diameter equal to $4.1{\text{cm}}$. Its surface has $150$ dimples each of radius $2{\text{mm}}$. Calculate total surface area exposed to the surroundings assuming that each dimple is a hemisphere.

Answer 1

We have to know that the normal ball is surrounded by some hemispherical dimples. While calculating the total area of the golf ball, the main thing we have to know is that half of the hemisphere is found inside the ball.

Formula used: The formulae used in this question is,
The area of normal ball is, $4\pi {{\text{R}}^2}$
The area of the hemisphere is, $2\pi {{\text{r}}^2}$
Where,
${\text{R}}$ is the radius of normal ball
${\text{r}}$ is the radius of the dimple

Complete step-by-step answer:
The given values in the question are,
The diameter of the golf ball, ${\text{D}} = \dfrac{{\text{R}}}{2} = 4.1{\text{cm}}$
The number of dimples surrounded by the ball is $150$
The radius of each dimple, ${\text{r}} = 2{\text{mm}}$

The area of the normal ball is ${\text{A = }}$ $4\pi {{\text{R}}^2}$
While substituting the ${\text{R}}$ value we get,
${\text{A}} = 4 \times \pi \times {\left( {\dfrac{{4.1}}{2}} \right)^2}$
While the above equation we get,
${\text{A}} = 16.81\pi {\text{c}}{{\text{m}}^2}--------------(1)$
Area of one dimple $a = 2\pi {{\text{r}}^2}$
By substituting the ${\text{r}}$ in ${\text{cm}}$value we get,
$\Rightarrow a = 2\pi {\left( {\dfrac{2}{{10}}} \right)^2}$
By solving we get,
$\Rightarrow a = \dfrac{{2\pi }}{{25}}{\text{c}}{{\text{m}}^2}----------------(2)$
The area of $150$ dimples $= 150 \times {\text{area of one dimple}}$
By substituting the value, we get,
$\Rightarrow 150 \times \dfrac{{2\pi }}{{25}}$
While solving the above we get,
$\Rightarrow 12\pi {\text{c}}{{\text{m}}^2}--------------(3)$
Finally, the total area of golf ball $\left( {{A_t}} \right)$
Total area $=$ area of normal ball + area of $150$ dimples - $150 \times$ area of half of one dimple
While substituting the values we get,
$\Rightarrow {A_t} = 16.81\pi + 12\pi - 150 \times \dfrac{\pi }{{25}}$
Solving the above equation we get,
$\Rightarrow {A_t} = 16.81\pi + 6\pi$
By adding it we get,
$\Rightarrow {A_t} = 22.8\pi \,c{m^2}$
Therefore, the total area of the golf ball is $22.8\pi \,c{m^2}$.
Hence, the total surface area exposed to the surroundings is $22.8\pi \,c{m^2}$ ${\text{c}}{{\text{m}}^2}$ or $71.68\,c{m^2}$.

So, the correct answer is “Option A”.

Note: From the diagram shown above the half of the dimple is hiding inside the golf ball. This is the reason for subtracting $150 \times$area of half of one dimple in the total area of the golf ball. The main thing we have to keep in mind is that the ball is in the shape of a sphere.