15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
The volume of a hemisphere of radius \(3\) cm is
Why: V = (2/3)πr³ = (2/3)π × 27.
\(\dfrac23 \pi (3)^3 = \dfrac23 \times 27\pi = 18\pi\ \text{cm}^3\).
The curved surface area of a cylinder of radius \(7\) cm and height \(10\) cm is (Take \(\pi = \dfrac{22}{7}\).)
Why: CSA = 2πrh = 2 × (22/7) × 7 × 10.
\(2\pi rh = 2 \times \dfrac{22}{7} \times 7 \times 10 = 440\ \text{cm}^2\).
A solid is made of a cylinder of radius \(r\) and height \(h\) with a hemisphere of radius \(r\) fixed on its top. Its total surface area is
Why: Cylinder's curved surface + its bottom disc + the hemisphere's curved surface.
Exposed parts: curved surface of the cylinder \(2\pi rh\), its base \(\pi r^2\) and the curved surface of the hemisphere \(2\pi r^2\). Total \(= 2\pi rh + 3\pi r^2\). (The joined circle is not on the outside.)
A toy is a cone of radius \(3\) cm and height \(4\) cm mounted on a hemisphere of the same radius. The volume of the toy is
Why: Cone (1/3)π × 9 × 4 = 12π plus hemisphere (2/3)π × 27 = 18π.
\(\dfrac13\pi(3)^2(4) + \dfrac23\pi(3)^3 = 12\pi + 18\pi = 30\pi\ \text{cm}^3\).
A cone has base radius \(7\) cm and slant height \(25\) cm. Its vertical height is
Why: h = √(l² − r²) = √(625 − 49).
\(h = \sqrt{25^2 - 7^2} = \sqrt{576} = 24\) cm.
A solid cuboid of size \(8\) cm \(\times\) \(4\) cm \(\times\) \(4\) cm is cut into two equal cubes. The increase in the total surface area is
Why: The cut creates two new 4 cm × 4 cm faces.
Cuboid: \(2(32 + 16 + 32) = 160\ \text{cm}^2\). Two cubes: \(2 \times 6 \times 16 = 192\ \text{cm}^2\). Increase \(= 32\ \text{cm}^2\) (two new faces of \(16\ \text{cm}^2\)).
The surface area of a sphere of radius \(7\) cm is (Take \(\pi = \dfrac{22}{7}\).)
Why: S = 4πr² = 4 × (22/7) × 49.
\(4 \times \dfrac{22}{7} \times 49 = 616\ \text{cm}^2\).
A solid is a cylinder of radius \(7\) cm with a cone of the same radius and height \(24\) cm on top. The curved surface area of the conical part is (Take \(\pi = \dfrac{22}{7}\).)
Why: Slant height √(49 + 576) = 25; CSA = πrl.
\(l = \sqrt{7^2 + 24^2} = 25\) cm. CSA \(= \pi r l = \dfrac{22}{7} \times 7 \times 25 = 550\ \text{cm}^2\).
A hemisphere and a cone have the same radius \(r\) and the same volume. The height of the cone is
Why: (2/3)πr³ = (1/3)πr²h gives h = 2r.
\(\dfrac23\pi r^3 = \dfrac13\pi r^2 h \Rightarrow h = 2r\).
The largest sphere is carved out of a solid cube of edge \(14\) cm. The volume of the sphere is (Take \(\pi = \dfrac{22}{7}\).)
Why: The sphere's diameter equals the edge, so r = 7.
\(r = 7\) cm. \(V = \dfrac43 \times \dfrac{22}{7} \times 343 = \dfrac{4312}{3}\ \text{cm}^3\).
A solid cylinder of radius \(3\) cm and height \(10\) cm has a hemispherical hollow of radius \(3\) cm scooped out of one end. The volume of the remaining solid is
Why: Cylinder 90π minus hemisphere 18π.
\(\pi(3)^2(10) - \dfrac23\pi(3)^3 = 90\pi - 18\pi = 72\pi\ \text{cm}^3\).
A tent is a cylinder of radius \(7\) m and height \(3\) m with a conical top of the same radius and slant height \(10\) m. The area of canvas needed (no floor) is (Take \(\pi = \dfrac{22}{7}\).)
Why: Canvas = 2πrh + πrl = (22/7) × 7 × (6 + 10).
\(2\pi rh + \pi r l = \dfrac{22}{7} \times 7 \times (2 \times 3 + 10) = 22 \times 16 = 352\ \text{m}^2\).
A solid cylinder of radius \(2\) cm and height \(6\) cm has a cone of the same radius fixed on top. The volume of the whole solid is \(28\pi\ \text{cm}^3\). The height of the cone is
Why: Cylinder 24π, so the cone is 4π = (1/3)π × 4 × h.
Cylinder \(= \pi(2)^2(6) = 24\pi\). Cone \(= 28\pi - 24\pi = 4\pi = \dfrac13\pi(2)^2 h \Rightarrow h = 3\) cm.
If the radius of a sphere is doubled, its surface area becomes
Why: Surface area is proportional to r².
\(\dfrac{4\pi(2r)^2}{4\pi r^2} = 4\): the surface area becomes \(4\) times.
A solid is a cone of radius \(r\) and height \(h\) standing on a cylinder of radius \(r\) and height \(h\). Its volume is
Why: πr²h + (1/3)πr²h.
\(\pi r^2 h + \dfrac13\pi r^2 h = \dfrac43\pi r^2 h\).
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
For the school science fair, Zoya builds a solid model rocket: a cylinder of radius \(3\) cm and height \(12\) cm with a cone of the same radius and height \(4\) cm fixed on top. She will paint the whole outside, including the flat circular base. Leave answers in terms of \(\pi\).
A craftsman in Channapatna turns a wooden spinning top. It is a hemisphere of radius \(7\) cm with a cone of the same radius and height \(24\) cm on its flat face, so that the cone points upwards. (Take \(\pi = \dfrac{22}{7}\).)
This free set is separate from the chapter's question bank. On the Surface Areas and Volumes chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 37 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.