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A golf ball is spherical with about 300–500 dimples that help increase its velocity while in play. Golf balls are traditionally white but available in colours also. In the given figure

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Question

A golf ball is spherical with about 300–500 dimples that help increase its velocity while in play. Golf balls are traditionally white but available in colours also. In the given figure, a golf ball has diameter 4.2 cm and the surface has 315 dimples (hemispherical) of radius 2 mm.

Based on the above information, answer the following questions.

  1. Find the surface area of one such dimple.
  2. Find the volume of the material dug out to make one dimple.
  3. Find the total surface area exposed to the surroundings.
  4. Find the volume of the golf ball.
Case Study
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Solution

i. Surface area of one dimple

Given: Radius of hemispherical dimple, \[r = 2\text{ mm} = 0.2\text{ cm}\]

Formula: \[\text{CSA} = 2\pi r^2\]

Calculation: \[\text{CSA} = 2 \times \frac{22}{7} \times 0.2 \times 0.2\]

\[\text{CSA} = \frac{1.76}{7} \approx 0.2514\text{ cm}^2\]

ii. Volume of material dug out to make one dimple

Formula: \[V = \frac{2}{3}\pi r^3\]

Calculation: \[V = \frac{2}{3} \times \frac{22}{7} \times (0.2)^3\]

\[V = \frac{2}{3} \times \frac{22}{7} \times 0.008 \approx 0.01676\text{ cm}^3\]

iii. Total surface area exposed to the surroundings

Given: Radius of ball, \[R = \frac{4.2}{2} = 2.1\text{ cm}\]

Formula: \[\text{Exposed Area} = \text{Surface Area of sphere} - 315 \times (\pi r^2) + 315 \times (2\pi r^2)\]

\[\text{Exposed Area} = 4\pi R^2 + 315\pi r^2 = \pi \left(4R^2 + 315r^2\right)\]

Calculation: \[\text{Exposed Area} = \frac{22}{7} \left[4(2.1)^2 + 315(0.2)^2\right]\]

\[\text{Exposed Area} = \frac{22}{7} [4(4.41) + 315(0.04)]\]

\[\text{Exposed Area} = \frac{22}{7} [17.64 + 12.60]\] 

\[= \frac{22}{7} \times 30.24\]

\[= 95.04\text{ cm}^2\]

iv. Volume of the golf ball

Formula: \[\text{Volume} = \text{Volume of sphere} - 315 \times \text{Volume of one dimple}\]

Calculation: \[\text{Volume of sphere} = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3 = 38.808\text{ cm}^3\] 

\[\text{Volume of 315 dimples} = 315 \times \left(\frac{2}{3} \times \frac{22}{7} \times (0.2)^3\right) = 5.28\text{ cm}^3\] 

\[\text{Volume of golf ball} = 38.808 - 5.28 = 33.528\text{ cm}^3\]

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Chapter 20: Additional Questions - Volume and Surface Areas of Solids [Page 1054]

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R.S. Aggarwal Mathematics [English] Class 10
Chapter 20 Additional Questions
Volume and Surface Areas of Solids | Q 6. | Page 1054
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