MathsCBSE Class 10 · Standard · Board exam 2023, set 30/5/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Standard 2023: question 36

Set in the CBSE Class 10 board exam of 2023, paper 30/5/1, section E, question 36, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q36 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes

Case Study – 1

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 (hemi-spherical) of radius 2 mm.

Based on the above, answer the following questions :

(i)
Find the surface area of one such dimple.
1 mark
(ii)
Find the volume of the material dug out to make one dimple.
1 mark
(iii)
(a) Find the total surface area exposed to the surroundings.
2 marks
OR
(iii)
(b) Find the volume of the golf ball.
2 marks

Figure in the paper: Left: a spherical golf ball with a dimpled surface. Right: a circle outline with hemispherical dimples (small semicircular notches) around its edge. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 30/5/1.

Hints

Open one hint at a time and try again before the next.

Hint 1

Convert everything to mm first: the ball's radius is 21 mm and each dimple has radius 2 mm.

Hint 2

A dimple is a hemispherical hollow: its surface is a hemisphere's curved surface and its volume is a hemisphere's volume.

Hint 3

For the exposed area, each dimple removes a flat circle πr² from the sphere but adds a curved surface 2πr²; for the volume, subtract all the dimples from the full sphere.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Each dimple is a hemispherical hollow of radius $r = 2$ mm. Its surface is the curved surface of a hemisphere:
$CSA = 2 π r^{2} = 2 \times \frac{22}{7} \times 4 = \frac{176}{7} mm^{2} \approx 25.14 mm^{2}$
1 mark
(ii) Material dug out = volume of a hemisphere of radius 2 mm:
$V = \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 8 = \frac{352}{21} mm^{3} \approx 16.76 mm^{3}$
1 mark
(iii)(a) Radius of ball $R = \frac{4.2}{2} = 2.1$ cm $= 21$ mm. For each dimple, a circular patch of area $π r^{2}$ is removed from the sphere's surface and a hemispherical surface $2 π r^{2}$ is added.
Exposed area $= 4 π R^{2} - 315 π r^{2} + 315 \times 2 π r^{2} = 4 π (21)^{2} + 315 π (2)^{2}$
1 mark
(iii)(a) $= \frac{22}{7} (1764 + 1260) = \frac{22}{7} \times 3024 = 9504 mm^{2}$
1 mark
(iii)(b) Volume of ball = volume of sphere − volume of 315 hemispherical dimples:
$V = \frac{4}{3} π (21)^{3} - 315 \times \frac{2}{3} π (2)^{3} = π (12348 - 1680)$
1 mark
(iii)(b) $= \frac{22}{7} \times 10668 = 22 \times 1524 = 33528 mm^{3}$
1 mark

Answer: (i) 176/7 mm² ≈ 25.1 mm²; (ii) 352/21 mm³ ≈ 16.76 mm³; (iii)(a) 9504 mm²; (iii)(b) 33528 mm³

Where the marks are

(i) CSA of hemisphere $2 π r^{2} = 176/7$ mm² (1 mark)
(ii) Volume $\frac{2}{3} π r^{3} = 352/21$ mm³ (1 mark)
(iii)(a) Correct expression $4 π R^{2} - 315 π r^{2} + 315 \cdot 2 π r^{2}$ with R = 21 mm (1 mark)
(iii)(a) Exposed area = 9504 mm² (1 mark)
(iii)(b) Correct expression $\frac{4}{3} π R^{3} - 315 \cdot \frac{2}{3} π r^{3}$ (1 mark)
(iii)(b) Volume = 33528 mm³ (1 mark)

Revise the chapter

Chapter: Surface Areas and Volumes (notes, formulas, where marks are lost and every past question)
Topic: Surface Area and Volume Formulae Revision

Practise this chapterBoard-style questions from Surface Areas and Volumes, one at a time, marked against the answer key and marking points.Back to the full paper2023 board paper, set 30/5/1 (Standard): every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.

CBSE Class 10 Mathematics Standard 2023: question 36

Hints

Step-by-step solution

Revise the chapter

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