Mathematics — CBSE Class 10 board question

Given: Diameter of golf ball = 4.2 cm → Radius R = 2.1 cm; Radius of each dimple r = 2 mm = 0.2 cm; Number of dimples = 315

(i) Surface area of one dimple (hemispherical):

$$= 2\pi r^2 = 2 \times \frac{22}{7} \times (0.2)^2 = 2 \times \frac{22}{7} \times 0.04 = \frac{1.76}{7} \approx 0.2514 \text{ cm}^2$$

(ii) Volume of material dug out for one dimple:

$$= \frac{2}{3}\pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (0.2)^3 = \frac{2}{3} \times \frac{22}{7} \times 0.008 \approx 0.01676 \text{ cm}^3$$

(iii) Total surface area exposed to surroundings:

Surface area of ball = $4\pi R^2 = 4 \times \frac{22}{7} \times (2.1)^2 = 4 \times \frac{22}{7} \times 4.41 = 55.44 \text{ cm}^2$

Area of 315 circular holes removed = $315 \times \pi r^2 = 315 \times \frac{22}{7} \times 0.04 = 39.6 \text{ cm}^2$

Curved area of 315 dimples added = $315 \times 2\pi r^2 = 315 \times 0.2514 = 79.2 \text{ cm}^2$

Total surface area $= 55.44 - 39.6 + 79.2 = \mathbf{95.04 \text{ cm}^2}$

OR

Volume of golf ball:

$$V = \frac{4}{3}\pi R^3 = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3 = \frac{4}{3} \times \frac{22}{7} \times 9.261 = 38.808 \text{ cm}^3$$

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• Part (i): Curved surface area of hemisphere = $2\pi r^2$ (not $3\pi r^2$, as the base circle is not counted — the dimple opens onto the ball's surface).
• Part (ii): Volume of hemisphere = $\frac{2}{3}\pi r^3$.
• Part (iii): Total exposed area = (sphere's surface) − (315 flat circles cut away) + (315 hemispherical curved surfaces added). Examiners specifically look for all three components.
• For the OR, simply apply $\frac{4}{3}\pi R^3$ with R = 2.1 cm. Both options are 2-mark questions; show substitution clearly for full marks.