MathsCBSE Class 10 · Basic · Board exam 2023, set 430/5/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2023: question 37

Set in the CBSE Class 10 board exam of 2023, paper 430/5/1, section E, question 37, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q37 · 4 marks · Case-basedCompetency-basedSurface Areas and Volumes

Singing bowls (hemispherical in shape) are commonly used in sound healing practices. Mallet (cylindrical in shape) is used to strike the bowl in a sequence to produce sound and vibration.

One such bowl is shown here whose dimensions are: Hemispherical bowl has outer radius 6 cm and inner radius 5 cm. Mallet has height of 10 cm and radius 2 cm.

Based on the above, answer the following questions:

(i)
What is the volume of the material used in making the mallet?
1 mark
(ii)
The bowl is to be polished from inside. Find the inner surface area of the bowl.
1 mark
(iii)(a)
Find the volume of metal used to make the bowl.
2 marks
OR
(iii)(b)
Find total surface area of the mallet. (Use $π = 3.14$ )
2 marks

Figure in the paper: Photograph of a hemispherical singing bowl resting on a wooden surface, with a cylindrical mallet lying across the bowl, with its handle end and striking end visible. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2023 board paper, set 430/5/1.

Hints

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

Hint 1

The mallet is a cylinder and the bowl is a hemispherical shell.

Hint 2

Inner surface of the bowl uses only the inner radius; curved surface of a hemisphere is $2 π r^{2}$ .

Hint 3

Metal in the bowl = outer hemisphere volume − inner hemisphere volume; TSA of a cylinder is $2 π r (r + h)$ .

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) Mallet is a cylinder with $r = 2$ cm, $h = 10$ cm. Volume $= π r^{2} h = 3.14 \times 2 \times 2 \times 10 = 125.6 cm^{3}$ .
1 mark
(ii) Inner curved surface of a hemisphere with $r = 5$ cm: $2 π r^{2} = 2 \times 3.14 \times 25 = 157 cm^{2}$ .
1 mark
(iii)(a) Volume of metal $= \frac{2}{3} π (R^{3} - r^{3}) = \frac{2}{3} \times 3.14 \times (6^{3} - 5^{3}) = \frac{2}{3} \times 3.14 \times (216 - 125) = \frac{2}{3} \times 3.14 \times 91$ .
1 mark
(iii)(a) $= \frac{571.48}{3} \approx 190.5 cm^{3}$ .
1 mark
(iii)(b) TSA of cylinder $= 2 π r (r + h) = 2 \times 3.14 \times 2 \times (2 + 10)$ .
1 mark
(iii)(b) $= 12.56 \times 12 = 150.72 \approx 150.7 cm^{2}$ .
1 mark

Answer: (i) 125.6 cm³; (ii) 157 cm²; (iii)(a) ≈ 190.5 cm³; (iii)(b) ≈ 150.7 cm²

Where the marks are

(i) Uses $π r^{2} h$ to get 125.6 cm³ (1 mark)
(ii) Uses $2 π r^{2}$ with r = 5 to get 157 cm² (1 mark)
(iii)(a) Writes $\frac{2}{3} π (6^{3} - 5^{3})$ (1 mark)
(iii)(a) Gets 190.5 cm³ (1 mark)
(iii)(b) Writes $2 π r (r + h)$ with values (1 mark)
(iii)(b) Gets 150.7 cm² (1 mark)

Revise the chapter

Chapter: Surface Areas and Volumes (notes, formulas, where marks are lost and every past question)

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 430/5/1 (Basic): 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 Basic 2023: question 37

Hints

Step-by-step solution

Revise the chapter

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