Question Bank — Surface Areas and Volumes

NCERT Class 10 Mathematics · Chapter 12 · CBSE Board Pattern

Class 10 Mathematics Chapter 12 Question Bank CBSE Board Pattern

Section A — MCQs (1 mark each)

  1. The curved surface area of a hemisphere of radius \(r\) is
    (a) \(2\pi r^2\) (b) \(3\pi r^2\) (c) \(4\pi r^2\) (d) \(\pi r^2\)

  2. A solid is formed by joining a cylinder and two hemispheres at its ends. The total surface area of the solid is
    (a) CSA of cylinder + CSA of two hemispheres
    (b) TSA of cylinder + TSA of two hemispheres
    (c) CSA of cylinder only
    (d) Volume of cylinder + volume of hemispheres

  3. In the formula for volume of a cone surmounted on a hemisphere of same radius \(r\), the total volume is
    (a) \(\frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h\) (b) \(\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3\)
    (c) \(\pi r^2 h + \frac{4}{3}\pi r^3\) (d) \(\frac{4}{3}\pi r^3\)

  4. When a hemisphere is cut out from one face of a cube of edge equal to the diameter, the surface area of the remaining solid includes
    (a) 5 faces of cube + curved surface of hemisphere
    (b) 6 faces of cube – base area of hemisphere + curved surface of hemisphere
    (c) only curved surface of hemisphere
    (d) volume of cube – volume of hemisphere

  5. The slant height of the conical part is required when calculating
    (a) volume of the cone (b) curved surface area of the cone
    (c) volume of the hemisphere (d) total surface area of the cylinder

Assertion-Reason Questions

  1. Assertion (A): The total surface area of a toy in the shape of a cone surmounted by a hemisphere is equal to CSA of cone + CSA of hemisphere.
    Reason (R): The flat circular faces where the cone and hemisphere meet are not included in the external surface area.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false. (d) A is false but R is true.

  2. Assertion (A): The volume of a solid formed by joining a cylinder and a hemisphere is the sum of their individual volumes.
    Reason (R): When two solids are joined, no part of their volumes disappears.
    (a) Both A and R are true and R is the correct explanation of A.
    (b) Both A and R are true but R is not the correct explanation of A.
    (c) A is true but R is false. (d) A is false but R is true.

  3. The inner surface area of a vessel formed by a hollow hemisphere mounted on a hollow cylinder is calculated using
    (a) only curved surfaces (b) curved surface of cylinder + curved surface of hemisphere
    (c) total surface areas of both (d) volume of water it can hold

  4. For a medicine capsule (cylinder with two hemispheres at ends), the surface area excludes
    (a) the two hemispherical curved surfaces (b) the two circular bases of the cylinder
    (c) the curved surface of the cylinder (d) none of these

  5. In Example 3 of the chapter (rocket shape), the area painted orange includes
    (a) CSA of cone only (b) CSA of cone + base area of cone – base area of cylinder
    (c) CSA of cylinder (d) volume of cone

Section B — Very Short Answer (2 marks each)

  1. A hemisphere of radius 7 cm is surmounted on a cone of same radius and height 24 cm. Find the total surface area of the solid so formed. (\(\pi = 22/7\))

  2. State the formula for the total surface area of a solid consisting of a cylinder with two hemispheres attached at its ends.

  3. The diameter of a hemispherical depression cut from a cube of edge 14 cm is equal to the edge of the cube. Write the expression for the surface area of the remaining solid.

  4. Find the volume of a solid consisting of a right circular cone of height 8 cm and base radius 6 cm standing on a hemisphere of the same radius. (\(\pi = 22/7\))

  5. A vessel is in the form of an inverted cone of height 8 cm and radius 5 cm. Write the expression for its capacity.

  6. Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Section C — Short Answer (3 marks each)

  1. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. (\(\pi = 22/7\))

  2. From a solid cylinder of height 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid (nearest cm²). (\(\pi = 22/7\))

  3. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid if the edge of the cube is 14 cm. (\(\pi = 22/7\))

  4. A wooden article is made by scooping out a hemisphere from each end of a solid cylinder of height 10 cm and base radius 3.5 cm. Find the total surface area of the article. (\(\pi = 22/7\))

  5. A solid is in the shape of a cone standing on a hemisphere with both radii 1 cm and height of cone equal to its radius. Find the volume of the solid in terms of \(\pi\).

Section D — Long Answer (5 marks each)

  1. A tent is in the shape of a cylinder surmounted by a conical top. The height and diameter of the cylindrical part are 2.1 m and 4 m respectively and the slant height of the conical top is 2.8 m. Find the area of the canvas used for making the tent. Also find the cost of the canvas at ₹500 per m². (Base of the tent is not covered.)

  2. Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (\(\pi = 22/7\))

  3. A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (\(\pi = 3.14\))

Section E — Case/Source-Based (4 marks each)

Case 1

A juice seller uses glasses that are cylindrical with a hemispherical raised portion at the bottom. The inner diameter of the glass is 5 cm and height is 10 cm.

Sub-questions:
(i) Find the apparent capacity of the glass.
(ii) Find the volume of the hemispherical portion.
(iii) Calculate the actual capacity of the glass.
(iv) If 50 such glasses are filled, how much juice is actually served (in litres)?

Case 2

A decorative block is made of a cube of edge 5 cm surmounted by a hemisphere of diameter 4.2 cm.

Sub-questions:
(i) Find the total surface area of the cube.
(ii) Find the base area of the hemisphere.
(iii) Calculate the total surface area of the block.
(iv) If the block is to be painted at ₹2 per cm², find the cost of painting.

▶ Answer Key Attempt all questions first,
then tap to reveal

Section A

  1. (a)
  2. (a)
  3. (b)
  4. (b)
  5. (b)
  6. (a)
  7. (a)
  8. (b)
  9. (b)
  10. (b)

Section B

  1. CSA hemisphere = \(2\pi r^2 = 308\) cm²; CSA cone = \(\pi r l = 275\) cm² (l = 25 cm); Total = 583 cm².
  2. TSA = CSA cylinder + 2 × CSA hemisphere.
  3. 5 × (edge)² – πr² + 2πr².
  4. Volume = \(\frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3 = 264\) cm³.
  5. \(\frac{1}{3}\pi r^2 h\).
  6. 6 × 5 × 5 = 150 cm².

Section C

  1. Height of cone = 12 cm, l = 12.5 cm; TSA = 214.5 cm² (marking: formula 1, substitution 1, answer 1).
  2. Remaining TSA = 17.6 cm² (nearest).
  3. 5 × 196 – πr² + 2πr² = 1052.57 cm² (approx.).
  4. TSA = 374 cm².
  5. \(\frac{5}{3}\pi\) cm³.

Section D

  1. Canvas area = 44 m²; Cost = ₹22 000 (multi-step: CSA cylinder + CSA cone, exclusion of base).
  2. TSA = 3.3 m² (formula 2πr(h + r), substitution, answer with unit).
  3. Volume of toy = 25.12 cm³; Difference = 25.12 cm³ (volume of circumscribing cylinder – toy volume).

Section E

Case 1

(i) 196.25 cm³ (ii) 32.71 cm³ (iii) 163.54 cm³ (iv) 8.177 litres.

Case 2

(i) 150 cm² (ii) 13.86 cm² (iii) 163.86 cm² (iv) ₹327.72.

← Revision Summary All Chapters

All questions are answerable from the NCERT chapter text. Reviewed by GFIS faculty.