A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same radius. The radius of hemisphere is 3.5 cm and the total wood used in making the toy is 1665/6\ cm³. Find the height of the conical part. Also, find the cost of painting the hemispherical part of the toy at the rate of ₹\ 15 per cm².

MathsCBSE Class 10 · Standard · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 10 Mathematics Standard sample paper 2026-27: question 34(B)

From the official CBSE sample question paper for 2026-27, section D, question 34(B), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(B) · 5 marks · Long answerCompetency-basedSurface Areas and Volumes

A solid wooden toy is in the form of a hemisphere surmounted by a cone of the same radius. The radius of hemisphere is 3.5 cm and the total wood used in making the toy is $166 \frac{5}{6} cm^{3}$ . Find the height of the conical part. Also, find the cost of painting the hemispherical part of the toy at the rate of $₹ 15$ per $cm^{2}$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

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

Hint 1

Total volume = volume of cone + volume of hemisphere.

Hint 2

Convert $166 \frac{5}{6}$ to $\frac{1001}{6}$ and substitute $r = 3.5$ to find $h$ .

Hint 3

The painted curved part of a hemisphere has area $2 π r^{2}$ .

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.

Figure (in words): a hemisphere of radius $r = 3.5$ cm with a cone of the same radius and height $h$ on its flat face.
Volume of toy = volume of cone + volume of hemisphere:
$166 \frac{5}{6} = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3}$
$\Rightarrow \frac{1001}{6} = \frac{1}{3} \times \frac{22}{7} \times (3.5)^{2} \times h + \frac{2}{3} \times \frac{22}{7} \times (3.5)^{3}$
2 marks
Here $π r^{2} = \frac{22}{7} \times 12.25 = 38.5$ . So:
$\frac{38.5 h}{3} + \frac{539}{6} = \frac{1001}{6}$
$\Rightarrow \frac{38.5 h}{3} = \frac{462}{6} = 77$
$\Rightarrow h = \frac{77 \times 3}{38.5} = 6$ cm
1 mark
CSA of the hemispherical part: $2 π r^{2} = 2 \times \frac{22}{7} \times (3.5)^{2} = 77 cm^{2}$
1 mark
Cost of painting $= 77 \times 15 = ₹ 1155$
1 mark

Answer: Height of cone = 6 cm; cost of painting = ₹ 1155

Where the marks are

Writes the volume equation (cone + hemisphere = 1001/6) with correct substitution (2 marks)
Solves to get h = 6 cm (1 mark)
Finds CSA of hemisphere = 77 cm² (1 mark)
Finds cost ₹ 1155 (1 mark)

Revise the chapter

Chapter: Surface Areas and Volumes (notes, formulas, where marks are lost and every past question)
Topic: Cone on a Hemisphere and Similar Shapes

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 paperSample paper 2026-27 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2026-27: question 34(B)

Hints

Step-by-step solution

Revise the chapter

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