CHAPTER 12 — SURFACE AREAS AND VOLUMES

Complete Question Bank

A. MCQs

1. A composite solid is best solved by first:
   A) Finding volume B) Splitting it into basic solids C) Adding all TSAs D) Finding diameter
2. The common surface of two joined solids is generally:
   A) Doubled B) Added C) Not included in external surface area D) Painted
3. CSA of a cylinder is:
   A) $\pi r^2h$ B) $2\pi rh$ C) $2\pi r^2$ D) $\pi rh$
4. CSA of a hemisphere is:
   A) $3\pi r^2$ B) $4\pi r^2$ C) $2\pi r^2$ D) $\pi r^2$
5. TSA of a hemisphere is:
   A) $2\pi r^2$ B) $3\pi r^2$ C) $4\pi r^2$ D) $\pi r^2$
6. Volume of a hemisphere is:
   A) $\frac13\pi r^3$ B) $\frac23\pi r^3$ C) $\frac43\pi r^3$ D) $2\pi r^3$
7. The slant height of a cone is:
   A) $r+h$ B) $r^2+h^2$ C) $\sqrt{r^2+h^2}$ D) $\sqrt{r+h}$
8. The volume of a composite solid formed by joining two solids is generally:
   A) Difference B) Sum C) Product D) Average
9. A cylinder with two hemispherical ends has external surface area equal to:
   A) CSA of cylinder + CSA of two hemispheres
   B) TSA of cylinder + TSA of two hemispheres
   C) CSA of cylinder only
   D) CSA of hemispheres only
10. A hemispherical projection inside a cylindrical glass makes its actual capacity:
    A) Greater B) Same C) Smaller D) Zero
11. If a hemisphere is fixed on a cube, the circular area covered by it is:
    A) Added B) Excluded from external area C) Doubled D) Painted
12. A cone has radius 3 cm and height 4 cm. Its slant height is:
    A) 4 cm B) 5 cm C) 6 cm D) 7 cm
13. The volume of a cuboid with a half-cylinder on top is:
    A) $V_c-V_h$ B) $V_c+V_h$ C) $V_cV_h$ D) $V_c/V_h$
14. If $1\,cm^3$ of iron has mass 8 g, the mass of $100\,cm^3$ is:
    A) 80 g B) 800 g C) 8 g D) 108 g
15. In a painting problem, we calculate:
    A) TSA of all solids B) Only the surfaces to be painted C) Volume D) Only bases

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B. Fill in the Blanks

16. CSA of a cylinder = ______.
17. TSA of a cylinder = ______.
18. Volume of a cylinder = ______.
19. CSA of a cone = ______.
20. TSA of a cone = ______.
21. Volume of a cone = ______.
22. CSA of a hemisphere = ______.
23. TSA of a hemisphere = ______.
24. Volume of a hemisphere = ______.
25. Surface area of a sphere = ______.
26. Volume of a sphere = ______.
27. Slant height of a cone = ______.
28. A hidden joining surface is ______ from external surface area.
29. Volume of a composite solid is generally found by ______ the component volumes.
30. Actual capacity = apparent capacity ______ volume of an internal projection.

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C. True or False

31. The TSA of two joined solids is always the sum of their individual TSAs.
32. A common surface between two joined solids is generally not exposed.
33. Volumes of joined solids can generally be added.
34. CSA of a hemisphere is $3\pi r^2$.
35. TSA of a hemisphere is $3\pi r^2$.
36. The slant height of a cone is $\sqrt{r^2+h^2}$.
37. A hemispherical projection inside a glass increases its capacity.
38. In a painting problem, every surface of every component must be painted.
39. All measurements should preferably be converted to compatible units before calculation.
40. The volume of a cavity is subtracted from the original volume to obtain the remaining volume.

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D. Match the Following

41. Match Column I with Column II.

Column I	Column II
1. Cylinder CSA	A. $\frac23\pi r^3$
2. Cone volume	B. $2\pi rh$
3. Hemisphere volume	C. $\pi rl$
4. Cone CSA	D. $\frac13\pi r^2h$
5. Sphere volume	E. $\frac43\pi r^3$

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E. Very Short Answer Questions

42. What is the first step in solving a composite-solid problem?
43. Why is the common surface of two joined solids excluded from external surface area?
44. Write the formula for the slant height of a cone.
45. Write the formula for the volume of a hemisphere.
46. What happens to the capacity of a glass when a hemispherical projection is present inside it?
47. What is the difference between apparent capacity and actual capacity?
48. Which surfaces of a cylinder with two hemispherical ends are externally visible?
49. Why must units be made consistent before calculation?
50. What operation is generally used to find the volume of a composite solid?

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F. Short Answer Questions

51. Explain why the TSA of two joined solids cannot normally be obtained by simply adding their individual TSAs.
52. A cone is placed on a hemisphere of the same radius. Which surfaces are included in its external surface area?
53. Explain how the external surface area of a cube with a hemisphere fixed on top can be calculated.
54. Explain how to calculate the actual capacity of a cylindrical glass having a hemispherical projection at its bottom.
55. A rocket consists of a cone mounted on a cylinder. Why might a part of the cone’s base need to be included in the painted area?
56. Explain how to calculate the volume of a shed made from a cuboid and a half-cylinder.

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G. Numerical Questions

57. Two cubes, each having volume $64\,cm^3$, are joined face-to-face. Find the surface area of the resulting cuboid.
58. A cone of radius $3.5$ cm is mounted on a hemisphere of the same radius. The total height is $15.5$ cm. Find the external surface area.
59. A cubical block has side 7 cm and a hemisphere is placed on its top. Find the greatest possible diameter of the hemisphere.
60. A hemispherical depression is cut from one face of a cube. The diameter of the depression equals the edge of the cube. Find the surface area of the remaining solid.
61. A capsule consists of a cylinder with two hemispherical ends. Its total length is 14 mm and diameter is 5 mm. Find its surface area.
62. A tent consists of a cylindrical part of height 2.1 m and diameter 4 m, with a conical top of slant height 2.8 m. Find the area of canvas required.
63. A cylinder of height 2.4 cm and diameter 1.4 cm has a conical cavity of the same height and diameter removed. Find the surface area of the remaining solid.
64. A cylinder of height 10 cm and radius 3.5 cm has hemispherical portions scooped out from both ends. Find its total surface area.
65. A shed consists of a cuboid $15m\times7m\times8m$ and a half-cylinder of diameter 7 m and length 15 m. Find its volume.
66. A cylindrical glass has radius 2.5 cm and height 10 cm. A hemisphere of radius 2.5 cm is raised at its base. Find its apparent and actual capacities.
67. A toy consists of a cone of height 2 cm mounted on a hemisphere of radius 2 cm. Find its volume.
68. A solid iron pole consists of two cylinders: the first has height 220 cm and diameter 24 cm, while the second has height 60 cm and radius 8 cm. If $1\,cm^3$ of iron has mass 8 g, find the mass of the pole.

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H. Assertion–Reason

Choose:
A) Both A and R are true, and R explains A.
B) Both are true, but R does not explain A.
C) A is true, R is false.
D) A is false, R is true.

69. A: The common surface of two joined solids is excluded from external TSA.
    R: It is no longer exposed.
70. A: Volumes of joined solids are generally added.
    R: Joining does not remove the volume occupied by the components.
71. A: Actual capacity of a glass with an internal projection is less than its apparent capacity.
    R: The projection occupies some space inside the glass.
72. A: TSA of a hemisphere is $2\pi r^2$.
    R: $2\pi r^2$ represents its curved surface area.
73. A: Slant height is required to calculate the CSA of a cone.
    R: CSA of a cone is $\pi rl$.

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I. Case-Based Questions

Case 1

A toy is formed by placing a cone on a hemisphere. Both have the same radius.

74. Which surface is hidden at the junction?
75. Which surfaces contribute to the external surface area?
76. Write the expression for the external surface area.
77. Which additional measurement is required for the cone’s CSA?

Case 2

A cylindrical glass has a hemispherical projection at its bottom.

78. What gives the apparent capacity?
79. Why is the actual capacity smaller?
80. Which volume must be subtracted?
81. Write the formula for the volume of the projection.

Case 3

A shed consists of a cuboid with a half-cylinder roof.

82. How is its total volume calculated?
83. What fraction of the cylinder’s volume represents the roof?
84. If machinery occupies some space, what should be done?
85. If workers also occupy space, how is the available air volume affected?

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J. HOTS / Application Questions

86. A student adds the TSA of a cone and hemisphere after joining them. Identify and explain the mistake.
87. A student includes the circular joining face while calculating the external surface area of a composite solid. Is this correct? Give a reason.
88. Two solids can have equal volumes but different surface areas. Is this possible? Explain.
89. A container has a cavity cut from it. Should the cavity’s volume be added or subtracted while finding the remaining volume? Explain.
90. Why is breaking a complicated solid into familiar solids useful in surface-area and volume problems?
91. A cylindrical glass contains a raised hemispherical portion at its bottom. Why is its actual capacity different from the volume calculated using only the cylinder formula?

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ANSWER KEY

A. MCQs

1-B, 2-C, 3-B, 4-C, 5-B, 6-B, 7-C, 8-B, 9-A, 10-C, 11-B, 12-B, 13-B, 14-B, 15-B.

B. Fill in the Blanks

16. $2\pi rh$
17. $2\pi r(h+r)$
18. $\pi r^2h$
19. $\pi rl$
20. $\pi r(l+r)$
21. $\frac13\pi r^2h$
22. $2\pi r^2$
23. $3\pi r^2$
24. $\frac23\pi r^3$
25. $4\pi r^2$
26. $\frac43\pi r^3$
27. $\sqrt{r^2+h^2}$
28. excluded
29. adding
30. minus

C. True/False

31-F, 32-T, 33-T, 34-F, 35-T, 36-T, 37-F, 38-F, 39-T, 40-T.

D. Match

41. 1-B, 2-D, 3-A, 4-C, 5-E

E. Very Short Answers

42. Split/identify the composite solid into basic solids.
43. Because it is no longer exposed.
44. $l=\sqrt{r^2+h^2}$
45. $V=\frac23\pi r^3$
46. It decreases.
47. Apparent capacity is the capacity calculated without considering the projection; actual capacity accounts for the space occupied by it.
48. The curved surface of the cylinder and the curved surfaces of the two hemispheres.
49. To obtain a correct numerical result and unit.
50. Addition.

F. Short Answers

51. Joining hides the common surfaces, so those areas must not be counted.
52. CSA of the cone + CSA of the hemisphere.
53. TSA of cube − covered circular area + CSA of hemisphere.
54. Cylinder volume − hemisphere volume.
55. If the cone is wider than the cylinder, the exposed ring of the cone’s base is also painted.
56. Volume of cuboid + half the volume of the corresponding cylinder.

G. Numerical Answers

57. 128 cm²
58. Approximately 171.6 cm²
59. 7 cm
60. 6a2−π(a/2)2+2π(a/2)26a^2-\pi(a/2)^2+2\pi(a/2)^2, where aa is the cube’s edge
61. Approximately 219.9 mm²
62. Approximately 30.2 m²
63. Approximately 18 cm²
64. Approximately 308 cm²
65. 1128.75 m³
66. Apparent = 196.25 cm³; Actual ≈ 163.54 cm³
67. 25.12 cm³
68. Approximately 112.6 kg

H. Assertion–Reason

69-A, 70-A, 71-A, 72-D, 73-A.

I. Case-Based

74. Common circular face.
75. Cone’s curved surface + hemisphere’s curved surface.
76. $\pi rl+2\pi r^2$
77. Slant height.
78. Volume of the cylinder.
79. The projection occupies internal space.
80. Volume of hemisphere.
81. $\frac23\pi r^3$
82. Cuboid volume + half-cylinder volume.
83. $\frac12$
84. Subtract machinery volume.
85. Subtract the space occupied by workers.

J. HOTS

86. The common circular surface is hidden and must not be included.
87. No; it is not externally exposed.
88. Yes. Different shapes can occupy the same volume while having different exposed areas.
89. Subtract it because material has been removed.
90. It converts a difficult composite problem into familiar surface-area/volume calculations.
91. The hemisphere occupies some of the space that would otherwise hold liquid.