CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set G

Practice CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set G provided below. The MCQ Questions for Class 10 Chapter 12 Surface Areas and Volumes Mathematics with answers and follow the latest CBSE/ NCERT and KVS patterns. Refer to more Chapter-wise MCQs for CBSE Class 10 Mathematics and also download more latest study material for all subjects

MCQ for Class 10 Mathematics Chapter 12 Surface Areas and Volumes

Class 10 Mathematics students should review the 50 questions and answers to strengthen understanding of core concepts in Chapter 12 Surface Areas and Volumes

Chapter 12 Surface Areas and Volumes MCQ Questions Class 10 Mathematics with Answers

Question. If the radii of circular ends of a frustum of height 6 cm are 15 cm and 7 cm, respectively. Then, the volume of the frustum is
(a) \(1380.12 \text{ cm}^3\)
(b) \(2380.12 \text{ cm}^3\)
(c) \(3380.12 \text{ cm}^3\)
(d) \(4380.12 \text{ cm}^3\)
Answer: B
Given, radii of both circular ends are, \(r_1 = 15 \text{ cm}\), \(r_2 = 7 \text{ cm}\) and height, \(h = 6 \text{ cm}\).
Volume of the frustum, \(V = \frac{1}{3}\pi h (r_1^2 + r_2^2 + r_1 r_2)\)
\(= \frac{1}{3} \times \pi \times 6 (225 + 49 + 105)\)
\(= 2\pi (379)\) [Take \(\pi = 3.14\)]
\(= 3.14 \times 758 = 2380.12 \text{ cm}^3\)

Question. Volume of a spherical shell is given by
(a) \(4\pi (R^2 - r^2)\)
(b) \(\pi (R^3 - r^3)\)
(c) \(4\pi (R^3 - r^3)\)
(d) \(\frac{4}{3}\pi (R^3 - r^3)\)
Answer: D
Volume of spherical shell = \(\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (R^3 - r^3)\)

Question. The volume of a largest sphere that can be cut from cylindrical log of wood of base radius 1 m and height 4 m, is
(a) \(\frac{16}{3}\pi \text{ m}^3\)
(b) \(\frac{8}{3}\pi \text{ m}^3\)
(c) \(\frac{4}{3}\pi \text{ m}^3\)
(d) \(\frac{10}{3}\pi \text{ m}^3\)
Answer: C
Volume of sphere = \(\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (1)^3 = \frac{4}{3}\pi \text{ m}^3\)

Question. A cubical ice-cream brick of edge 22 cm is to be distributed among some children and filling ice-cream cones of radius 2 cm and height 7 cm upto its brim. How many children will get ice-cream cones?
(a) 163
(b) 263
(c) 363
(d) 463
Answer: C
Given, Volume of brick = \((22)^3 \text{ cm}^3\)
Volume of 1 cone = \(\frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times 7 = \frac{22 \times 4}{3}\)
Let number of cones = \(n\)
Then, \(n \times 22 \times \frac{4}{3} = 22 \times 22 \times 22\)
\(n = \frac{22 \times 22 \times 3}{4}\)
\(n = 121 \times 3 = 363\)

Question. Three identical cones with base radius \(r\) are placed on their bases so that each is touching the other two. The radius of the circle drawn through their vertices is
(a) Smaller than \(r\)
(b) equal to \(r\)
(c) larger than \(r\)
(d) depends on the height of the cones
Answer: C
The centres of the bases of the cones form a triangle of side \(2r\). The circumcircle of this triangle will be identical to a circle drawn through the vertices of cones and thus, it will have a radius of \(2/\sqrt{3}\) times \(r\), which is greater than \(r\).

Question. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is
(a) 12 m
(b) 18 m
(c) 36 m
(d) 66 m
Answer: C
We have, diameter of metallic sphere = 6 cm
Radius of metallic sphere = 3 cm
Also, diameter of cross-section of cylindrical wire = 0.2 cm
Radius of cross-sections of cylindrical wire = 0.1 cm
Let the length of the wire be \(h\) cm.
Since, metallic sphere is converted into a cylindrical shaped wire of length \(h\) cm.
Volume of the metal used in wire = Volume of the sphere,
\(\pi \times (\frac{1}{10})^2 \times h = \frac{4}{3} \times \pi \times 27\)
\(\pi \times \frac{1}{100} \times h = 36\pi\)
\(h = \frac{36\pi \times 100}{\pi} = 3600 \text{ cm} = 36 \text{ m}\)

Question. A 20 m deep well, with diameter 7 m is dug and the earth from digging is evently spread out to form a platform 22 m by 14 m. The height of the platform is
(a) 2.5 m
(b) 3.5 m
(c) 3 m
(d) 2 m
Answer: A
Radius of the well = \(\frac{7}{2}\text{ m} = 3.5 \text{ m}\)
Volume of the earth dug out = \(\frac{22}{7} \times (3.5)^2 \times 20 = \frac{22}{7} \times 3.5 \times 3.5 \times 20 = 770 \text{ m}^3\)
Area of platform = \((22 \times 14) \text{ m}^2 = 308 \text{ m}^2\)
Height = \(\frac{770}{308} = 2.5 \text{ m}\)

Question. From a solid circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and same base is removed, then the volume of remaining solid is
(a) \(280\pi \text{ cm}^3\)
(b) \(330\pi \text{ cm}^3\)
(c) \(240\pi \text{ cm}^3\)
(d) \(440\pi \text{ cm}^3\)
Answer: C
Volume of the remaining solid = Volume of the cylinder - Volume of the cone
\(= \pi \times 6^2 \times 10 - \frac{1}{3} \times \pi \times 6^2 \times 10\)
\(= (360\pi - 120\pi) = 240\pi \text{ cm}^3\)

Question. A shuttle cock, used for playing badminton, has the shape of a frustum of a cone mounted on a hemisphere. If the external diameters of the frustum are 5 cm and 2 cm and height of the entire shuttle cock is 7 cm, then its external surface area is
(a) \(67.98 \text{ cm}^2\)
(b) \(74.26 \text{ cm}^2\)
(c) \(76.89 \text{ cm}^2\)
(d) \(47.62 \text{ cm}^3\)
Answer: B
Given, radius of the lower end of the frustum, \(r_1 = 1 \text{ cm}\)
Radius of the upper end of the frustum, \(r_2 = 2.5 \text{ cm}\)
Height of the frustum, \(h = 6 \text{ cm}\)
Now, slant height of the frustum, \(l = \sqrt{h^2 + (r_2 - r_1)^2} = \sqrt{36 + (2.5 - 1)^2} = \sqrt{38.25} = 6.18 \text{ cm}\)
External surface area of shuttlecock = Curved surface area of the frustum + \(2\pi r_1^2\)
\(= \pi (r_1 + r_2)l + 2\pi r_1^2 = \pi(1 + 2.5) \times 6.18 + 2 \times \pi \times 1^2\)
\(= \frac{22}{7} \times 3.5 \times 6.18 + 2 \times \frac{22}{7} = 67.98 + 6.28 = 74.26 \text{ cm}^2\)

Question. An hexagonal pyramid is 20 m high. Side of the base is 5 m. The volume of the pyramid is
(a) \(250\sqrt{3} \text{ m}^3\)
(b) \(250 \text{ m}^3\)
(c) \(25\sqrt{3} \text{ m}^3\)
(d) \(25 \text{ m}^3\)
Answer: A

Question. A golf ball has diameter equal to 4.1 cm. Its surface has 150 dimples each of radius 2 mm. Calculate total surface area which is exposed to the surroundings, assuming that the dimples are hemispherical.
(a) \(22.81 \text{ cm}^2\)
(b) \(68.71 \text{ cm}^2\)
(c) \(71.68 \text{ cm}^2\)
(d) None of the above
Answer: C

FILL IN THE BLANK

DIRECTION : Complete the following statements with an appropriate word/term to be filled in the blank space(s).

Question. The length of the diagonal of a cube that can be inscribed in a sphere of radius 7.5 cm is ..........
Answer: 15 cm

Question. The volume of a hemisphere is .......... the volume of a cylinder if its height and radius is same as that of the cylinder.
Answer: two-third

Question. If the volume and the surface area of a solid sphere are numerically equal, then its radius is ..........
Answer: 3 Units

Question. A sharpened pencil is a combination of .......... and .......... shapes.
Answer: cylinder, cone

Question. If we cut a cone by a plane parallel to its base, we obtain a .......... and ..........
Answer: cone, frustum of a cone

Question. If the volume of a cube is \(64 \text{ cm}^3\), then its surface area is ..........
Answer: \(96 \text{ cm}^2\)

Question. Solid figures are .......... while plane figures are ..........
Answer: 3-dimensional. 2-dimensional or cube, cuboid, etc. circle, square etc.

Question. If the radius of a sphere is halved, its volume becomes .......... time the volume of original sphere.
Answer: one-eighth

Question. Surahi is the combination of .......... and ..........
Answer: sphere, cylinder

Question. The volume of a solid is the measurement of the portion of the .......... occupied by it.
Answer: Space

Question. If the heights of two cylinders are equal and their radii are in the ratio of 7 : 5, then the ratio of their volumes ..........
Answer: 49 : 25

Question. The ratio of the surface areas of two spheres is 49 : 25, then the ratio of their radii is ..........
Answer: 49 : 25

TRUE/FALSE

DIRECTION : Read the following statements and write your answer as true or false.

Question. Volume of the solid is measured in cubic units.
Answer: True

Question. A cube has eight faces.
Answer: False

Question. Area is the length of the boundary of a closed figure.
Answer: False

Question. Volume of a frustum of cone \( = \frac{1}{2}\pi h(r_1^2 + r_2^2 + r_1 r_2) \)
Answer: False

Question. Area is the total surface covered by a closed figure.
Answer: True

Question. A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone. The total surface area of the combined solid is \( \pi r[\sqrt{r^2 + h^2} + 3r + 2h] \)
Answer: False

Question. The volume of sphere of diameter d is \( \frac{\pi d^3}{6} \).
Answer: True

Question. A solid ball is exactly fitted inside the cubical box of side a . The volume of the ball is \( \frac{4}{3}\pi a^3 \).
Answer: False

Question. The total surface area of a solid cylinder of radius r and height h is \( 2\pi r(h + r) \).
Answer: True

Question. A cone having thrice the height of a cylinder and equal base radius have the same volume as that of the cylinder.
Answer: True

Question. An open metallic bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The surface area of the metallic sheet used is equal to curved surface area of frustum of a cone + area of circular base + curved surface area of cylinder.
Answer: True

MATCHING QUESTIONS

DIRECTION : Each question contains statements given in two columns which have to be matched. Statements (A, B, C, D) in Column-I have to be matched with statements (p, q, r, s) in Column-II.

Question. From a solid cylinder of height 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and some diameter is hollowed out then match the column.
Column-I

(A) Area of bottom of cylinder
(B) Outer curved surface area
(C) Curved area of conical cavity
(D) Total surface area
Column-II
(p) 10.56
(q) 1.54
(r) 5.5
(s) 17.6
Answer: (A) - q, (B) - p, (C) - r, (D) - s

Question. Match the Following:
Column I

(A) Solid
(B) Area
(C) Volume
(D) Cube
(E) Cuboid
(F) Cylinder
(G) Cone
(H) Sphere
(I) Frustum of a cone
Column II
(p) a set of points in the space which are at equal distances from a fixed point.
(q) a mathematical term used for a rigid three-dimensional shape.
(r) quantitative measure of a plane or curved surface.
(s) a solid whose faces are rectangles.
(t) a solid whose faces are all congruent squares.
(u) a solid with a circular base tapering to a point.
(v) a solid whose cross-sections are all circles of the same radii.
(w) a solid which is obtained by removing the upper portion of the cone by a plane parallel to its base.
(x) amount of space occupied by a solid.
Answer: (A) - q, (B) - r, (C) - x, (D) - t, (E) - s, (F) - v, (G) - u, (H) - p, (I) - w

Question. Match the following:
Column-I

(A) Solids
(B) Road rollers
(C) Ice-cream cone
(D) Volleyball
Column-II
(p) Right circular cone
(q) Sphere
(r) Cylinder
(s) Cuboid
(t) Cube
Answer: (A) - (p, q, r, s, t), (B) - r, (C) - p, (D) - q

ASSERTION AND REASON

DIRECTION : In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.

Question. Assertion : Total surface area of the cylinder having radius of the base 14 cm and height 30 cm is \( 3872 \text{ cm}^2 \).
Reason : If r be the radius and h be the height of the cylinder, then total surface area \( = 2\pi rh + 2\pi r^2 \).

(a) Choice (a)
(b) Choice (b)
(c) Choice (c)
(d) Choice (d)
Answer: A
Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). Total surface area \( = 2\pi rh + 2\pi r^2 = 2\pi r(h + r) = 2 \times \frac{22}{7} \times 14(30 + 14) = 88(44) = 3872 \text{ cm}^2 \)

Question. Assertion : The slant height of the frustum of a cone is 5 cm and the difference between the radii of its two circular ends is 4 cm. Than the height of the frustum is 3 cm.
Reason : Slant height of the frustum of the cone is given by \( l = \sqrt{(R - r)^2 + h^2} \).

(a) Choice (a)
(b) Choice (b)
(c) Choice (c)
(d) Choice (d)
Answer: A
Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A). We have, \( l = 5 \text{ cm} \), \( R - r = 4 \text{ cm} \). \( 5 = \sqrt{(4)^2 + h^2} \implies 16 + h^2 = 25 \implies h^2 = 25 - 16 = 9 \implies h = 3 \text{ cm} \)

Question. Assertion : If the height of a cone is 24 cm and diameter of the base is 14 cm, then the slant height of the cone is 15 cm.
Reason : If r be the radius and h the slant height of the cone, then slant height \( = \sqrt{h^2 + r^2} \).

(a) Choice (a)
(b) Choice (b)
(c) Choice (c)
(d) Choice (d)
Answer: D
Assertion (A) is false but reason (R) is true. Slant height \( = \sqrt{(14/2)^2 + (24)^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \)

Question. Assertion : Two identical solid cube of side 5 cm are joined end to end. Then total surface area of the resulting cuboid is \( 300 \text{ cm}^2 \).
Reason : Total surface area of a cuboid is \( 2(lb + bh + lh) \)

(a) Choice (a)
(b) Choice (b)
(c) Choice (c)
(d) Choice (d)
Answer: D
A is false but R is true. When cubes are joined end to end, it will form a cuboid. \( l = 2 \times 5 = 10 \text{ cm} \), \( b = 5 \text{ cm} \), \( h = 5 \text{ cm} \). Total surface area \( = 2(lb + bh + lh) = 2(10 \times 5 + 5 \times 5 + 10 \times 5) = 2(125) = 250 \text{ cm}^2 \)

Question. Assertion : If the radius of a cone is halved and volume is not changed, then height remains same.
Reason : If the radius of a cone is halved and volume is not changed then height must become four times of the original height.

(a) Choice (a)
(b) Choice (b)
(c) Choice (c)
(d) Choice (d)
Answer: D
Assertion (A) is false but reason (R) is true. \( \frac{V_1}{V_2} = \frac{(1/3)\pi r^2 h_1}{(1/3)\pi (r/2)^2 h_2} = \frac{4h_1}{h_2} \). As \( V_1 = V_2 \), \( h_2 = 4h_1 \).

MCQs for Chapter 12 Surface Areas and Volumes Mathematics Class 10

Students can use these MCQs for Chapter 12 Surface Areas and Volumes to quickly test their knowledge of the chapter. These multiple-choice questions have been designed as per the latest syllabus for Class 10 Mathematics released by CBSE. Our expert teachers suggest that you should practice daily and solving these objective questions of Chapter 12 Surface Areas and Volumes to understand the important concepts and better marks in your school tests.

Chapter 12 Surface Areas and Volumes NCERT Based Objective Questions

Our expert teachers have designed these Mathematics MCQs based on the official NCERT book for Class 10. We have identified all questions from the most important topics that are always asked in exams. After solving these, please compare your choices with our provided answers. For better understanding of Chapter 12 Surface Areas and Volumes, you should also refer to our NCERT solutions for Class 10 Mathematics created by our team.

Online Practice and Revision for Chapter 12 Surface Areas and Volumes Mathematics

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Where can I access latest CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set G?

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Are Assertion-Reasoning and Case-Study MCQs included in the Mathematics Class 10 material?

Yes, our CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set G include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

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