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MERITROOT FREE LEARNING · CLASS 10 / MATHEMATICS

Surface Areas and Volumes

Your chapter workspace for understanding concepts, revising key ideas and practising questions. Choose an available resource below.

PYQs6Formulae9Derivations2
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PYQs

Practise questions grouped by chapter and source year.

Practice questions are organised using the years and labels in our source collection. Work through the solutions with your teacher; these are not a complete official examination paper.

Practice question 1

Source year: 20253 marks

Cone, hemisphere, cylinder on equal bases. Ratio of volumes. [CBSE 2025]

Practice question 2

Source year: 20243 marks

Sphere r=4.2cm melted into cylinder r=6cm. Find height. [CBSE 2024]

Practice question 3

Source year: 20253 marks

A cone, hemisphere and cylinder stand on equal bases and have same height. Find ratio of their volumes. [CBSE 2025]

Practice question 4

Source year: 20243 marks

A metallic sphere of radius 4.2cm is melted and recast into shape of cylinder of radius 6cm. Find height of cylinder. [CBSE 2024]

Practice question 5

Source year: 20233 marks

Find volume and surface area of sphere of diameter 14cm. [CBSE 2023]

Practice question 6

Source year: 20223 marks

A toy is in form of cone mounted on hemisphere. Radius is 3.5cm and total height is 15.5cm. Find volume. [CBSE 2022]

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Formulae

Revisit important relationships and quick revision formulae.

Cylinder Volume

V = πr²h

Cone Volume

V = (1/3)πr²h

Sphere Volume

V = (4/3)πr³

Cuboid

Volume = l×b×h\nTotal Surface Area = 2(lb+bh+hl)\nLateral Surface Area = 2h(l+b)

Cube

Volume = a³\nTotal Surface Area = 6a²\nLateral Surface Area = 4a²

Cylinder

Volume = πr²h\nCurved Surface Area = 2πrh\nTotal Surface Area = 2πr(r+h)

Cone

Volume = (1/3)πr²h\nCurved Surface Area = πrl (l=slant height)\nTotal Surface Area = πr(r+l)\nSlant height: l = √(r²+h²)

Sphere

Volume = (4/3)πr³\nSurface Area = 4πr²

Hemisphere

Volume = (2/3)πr³\nCurved Surface Area = 2πr²\nTotal Surface Area = 3πr²

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Derivations

Work through the reasoning behind key concepts.

Volume Conservation

When a solid is melted and recast into different shape, volume remains constant. Volume before melting = Volume after recasting. Use this principle: Find volume of original shape, equate to volume of new shape, solve for unknown dimension.

Volume Conservation

When object is melted and recast: Volume before = Volume after\nUse this to find unknown dimension

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