A hemispherical bowl is packed in a cuboidal box. The bowl just fits in the box. Inner radius of the bowl is 10 cm. Outer radius of the bowl is 10.5 cm. Based on the above, answer the following questions: (i) Find the dimensions of the cuboidal box. (ii) Find the total outer surface area of the box. (iii)(a) Find the difference between the capacity of the bowl and the volume of the box. (use π = 3.14) (iii)(b) The inner surface of the bowl and the thickness is to be painted. Find the area to be painted.

MathsCBSE Class 10 · Basic · Board exam 2025, set 430/6/1 · 4 marks · Case-based

CBSE Class 10 Mathematics Basic 2025: question 37

Set in the CBSE Class 10 board exam of 2025, paper 430/6/1, 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

A hemispherical bowl is packed in a cuboidal box. The bowl just fits in the box. Inner radius of the bowl is 10 cm. Outer radius of the bowl is 10.5 cm.

Based on the above, answer the following questions:

(i)
Find the dimensions of the cuboidal box.
1 mark
(ii)
Find the total outer surface area of the box.
1 mark
(iii)(a)
Find the difference between the capacity of the bowl and the volume of the box. (use $π = 3.14$ )
2 marks
OR
(iii)(b)
The inner surface of the bowl and the thickness is to be painted. Find the area to be painted.
2 marks

Figure in the paper: Left: an open cuboidal box with a hemispherical bowl inside. Right: a hemispherical bowl sitting inside a cuboid, with the cuboid's top face at the bowl's rim level, showing a dashed rim ellipse and the bowl's curved bottom touching the base of the cuboid. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 430/6/1.

Hints

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

Hint 1

The box must hold the outside of the bowl: use the outer radius for its size.

Hint 2

Capacity of the bowl uses the inner radius in $\frac{2}{3} π r^{3}$ .

Hint 3

The 'thickness' painted is the flat ring at the rim, area $π (R^{2} - r^{2})$ .

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) The bowl just fits, so the box's length and breadth equal the outer diameter $= 2 \times 10.5 = 21$ cm and its height equals the outer radius $= 10.5$ cm. Dimensions: $21 cm \times 21 cm \times 10.5 cm$ .
1 mark
(ii) TSA $= 2 (l b + bh + h l) = 2 (21 \times 21 + 21 \times 10.5 + 10.5 \times 21) = 2 (441 + 220.5 + 220.5)$
½ mark
(ii) $= 2 \times 882 = 1764 cm^{2}$
½ mark
(iii)(a) Capacity of bowl (inner radius 10 cm) $= \frac{2}{3} π r^{3} = \frac{2}{3} \times 3.14 \times 1 0^{3}$
½ mark
(iii)(a) $= \frac{6280}{3} \approx 2093.33 cm^{3}$
½ mark
(iii)(a) Volume of box $= 21 \times 21 \times 10.5 = 4630.5 cm^{3}$
½ mark
(iii)(a) Difference $= 4630.5 - 2093.33 = \frac{15223}{6} \approx 2537.17 cm^{3}$
½ mark
(iii)(b) Area to be painted = inner curved surface + ring at the rim $= 2 π r^{2} + π (R^{2} - r^{2}) = 2 \times \frac{22}{7} \times 1 0^{2} + \frac{22}{7} \times (10. 5^{2} - 1 0^{2})$
1 mark
(iii)(b) $= \frac{4400}{7} + \frac{22}{7} \times 10.25 = \frac{4400}{7} + \frac{451}{14} = \frac{9251}{14} \approx 660.79 cm^{2}$
1 mark

Answer: (i) 21 cm × 21 cm × 10.5 cm; (ii) 1764 cm²; (iii)(a) ≈ 2537.17 cm³; (iii)(b) ≈ 660.79 cm²

Where the marks are

(i) Box dimensions 21 cm × 21 cm × 10.5 cm (1 mark)
(ii) Correct TSA formula/substitution (½ mark) and 1764 cm² (½ mark)
(iii)(a) Capacity of bowl $\frac{2}{3} \times 3.14 \times 1 0^{3}$ (½ mark) $= 2093.33$ cm³ (½ mark)
(iii)(a) Volume of box 4630.5 cm³ (½ mark); difference ≈ 2537.17 cm³ (½ mark)
(iii)(b) Expression $2 π (10)^{2} + π (10. 5^{2} - 1 0^{2})$ (1 mark); area ≈ 660.79 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 paper2025 board paper, set 430/6/1 (Basic): every question, by section.

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CBSE Class 10 Mathematics Basic 2025: question 37

Hints

Step-by-step solution

Revise the chapter

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