CBSE Class 10 · Boards 2027 · NCERT 2026-27 syllabus
NCERT Exemplar Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes
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Boards begin February 2027
Pre-boards are in December
Surface Areas and Volumes is in Mensuration, which carries 10 of the 80 marks in the 2026-27 CBSE Maths paper
13 NCERT Exemplar questions from Chapter 12
Exercise 12.1
Exemplar Ex 12.1 Q141 mark
A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each of its ends. The length of the entire capsule is 2 cm. The capacity of the capsule is
Step 1Radius r = 0.25 cm; cylinder length h = 2 − 2(0.25) = 1.5 cm.
Step-by-step solution
Radius r = 0.25 cm; cylinder length h = 2 − 2(0.25) = 1.5 cm.
State whether true or false and justify: 'Two identical solid hemispheres of equal base radius r cm are stuck together along their bases. The total surface area of the combination is 6πr².'
Step 1Joining two hemispheres along their flat bases gives a sphere of radius r.
Step-by-step solution
Joining two hemispheres along their flat bases gives a sphere of radius r.
Surface area of a sphere = 4πr² (NOT 6πr²).
So the statement is FALSE.
Answer:False — the combined solid is a sphere, whose surface area is 4πr², not 6πr².
State whether true or false and justify: 'A solid cylinder of radius r and height h is placed over another cylinder of the same height and radius. The total surface area of the shape so formed is 4πrh + 4πr².'
Step 1Stacking the two cylinders gives a single cylinder of radius r and total height 2h.
Step-by-step solution
Stacking the two cylinders gives a single cylinder of radius r and total height 2h.
From a solid cube of side 7 cm, a conical cavity of height 7 cm and radius 3 cm is hollowed out. Find the volume of the remaining solid. (Use π = 22/7)
Two cones with the same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed. (Use π = 22/7)
Step 1Joining two cones along their bases hides both circular faces; only the two slant surfaces remain.
Step-by-step solution
Joining two cones along their bases hides both circular faces; only the two slant surfaces remain.
Slant height l = √(r² + h²) = √(8² + 15²) = √(64 + 225) = √289 = 17 cm.
16 glass spheres each of radius 2 cm are packed into a cuboidal box of internal dimensions 16 cm × 8 cm × 8 cm and then the box is filled with water. Find the volume of water filled in the box. (Use π = 22/7)
Step 1[1 mark] Setup: V(water) = V(cuboidal box) − V(16 spheres); sphere volume formula V = (4/3)πr³.
Step-by-step solution
[1 mark] Setup: V(water) = V(cuboidal box) − V(16 spheres); sphere volume formula V = (4/3)πr³.
A rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are 6 cm and 12 cm, respectively. If the slant height of the conical portion is 5 cm, find the total surface area and volume of the rocket. (Use π = 3.14)
Step 1Radius r = 3 cm; cylinder height h₁ = 12 cm; cone slant l = 5 cm → cone height h₂ = √(l² − r²) = √(25 − 9) = 4 cm.
Step-by-step solution
Radius r = 3 cm; cylinder height h₁ = 12 cm; cone slant l = 5 cm → cone height h₂ = √(l² − r²) = √(25 − 9) = 4 cm.
TSA = CSA(cylinder) + area(bottom circle of cylinder) + CSA(cone) = 2πrh₁ + πr² + πrl.
A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains 41 19/21 m³ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building.
Step 1[1 mark] Let the dome radius be r; internal diameter 2r equals the total height H, so H = 2r.
Step-by-step solution
[1 mark] Let the dome radius be r; internal diameter 2r equals the total height H, so H = 2r.
[1 mark] Cylinder height = H − r = 2r − r = r, with radius r.
A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is equal to the radius of the cone.
CBSE board pattern: Assertion-Reason and case-based questions
Written in the CBSE board format. These are not numbered NCERT/Exemplar questions; some case studies are built on textbook problems. Neither NCERT book has Assertion-Reason questions: they, and case-based questions, are a CBSE board-paper format, here on this chapter's 2026-27 syllabus, with solutions by LazyTopper.
CBSE board pattern · Assertion-Reason1 mark
Assertion (A): Two solid hemispheres of equal base radius r joined together along their bases have a total surface area of 4πr².
Reason (R): The combined solid is a sphere of radius r.
Step 1Two hemispheres glued along their flat circular bases form a sphere — Reason is TRUE.
Step-by-step solution
Two hemispheres glued along their flat circular bases form a sphere — Reason is TRUE.
Surface area of a sphere of radius r = 4πr² — Assertion is TRUE.
Reason gives the explanation for Assertion. Hence option (A).
Board examiners mark every step. Photograph your written answer and see the marks for each step, and exactly where they were lost. Your first answer check is free.
Questions from the NCERT Class 10 Mathematics Exemplar Problems (the CBSE board-pattern section is in the board format, not numbered book questions); the solutions are written by LazyTopper. LazyTopper is an independent platform and is not affiliated with CBSE or NCERT.