CBSE Class 10 Mathematics Surface Areas And Volume VBQs Set B

Multiple Choice Questions

Question. The radius of a sphere (in cm) whose volume is \(12\pi \text{ cm}^3\), is:
(a) 3
(b) \(3\sqrt{3}\)
(c) \(3^{2/3}\)
(d) \(3^{1/3}\) 
Answer: (c)

Question. A cylindrical pencil sharpened at one edge is a combination of:
(a) a cone and a cylinder
(b) frustum of a cone and a cylinder
(c) a hemisphere and a cylinder
(d) two cylinders. 
Answer: (a)

Question. A surahi is a combination of:
(a) a sphere and a cylinder
(b) a hemisphere and a cylinder
(c) two hemispheres
(d) a cylinder and a cone. 
Answer: (a)

Question. A shuttle cock used for playing badminton has the shape of a combination of:
(a) a cylinder and a sphere
(b) a cylinder and a hemisphere
(c) a sphere and a cone
(d) frustum of a cone and a hemisphere 
Answer: (d)

Question. A metallic sphere of diameter 20 cm is recast into a right circular cone of base radius 10 cm. What is the height of the cone?
(a) 4 cm
(b) 40 cm
(c) 60 cm
(d) 120 cm
Answer: (b)

Question. Three cubes each of side 15 cm are joined end to end. The total surface area of the cuboid is:
(a) \(3150 \text{ cm}^2\)
(b) \(1575 \text{ cm}^2\)
(c) \(1012.5 \text{ cm}^2\)
(d) \(576.4 \text{ cm}^2\)
Answer: (a)

Question. A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and it is assumed that \(\frac{1}{8}\) space of the cube remains unfilled. Then the number of marbles that the cube can accomodate is:
(a) 142296
(b) 142396
(c) 142496
(d) 142596 
Answer: (a)

Trick Applied

• If we divide the total volume filled by marbles in a cube by volume of a marble, we get required number of marbles.

Question. A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively is melted and recast into the form of a cone of base diameter 8cm. The height of the cone is:
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm 
Answer: (b)

Trick Applied

• When a solid shape is melted and recast into other solid shape, then volume of both shapes are equal.

Question. 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:
(a) \(0.36 \text{ cm}^3\)
(b) \(0.35 \text{ cm}^3\)
(c) \(0.34 \text{ cm}^3\)
(d) \(0.33 \text{ cm}^3\) 
Answer: (a)

Question. Volumes of two spheres are in the ratio 64:27. The ratio of their surface areas is:
(a) 3:4
(b) 4:3
(c) 9:16
(d) 16:9 
Answer: (d)

Question. If the radius of the sphere is increased by 100%, the volume of the corresponding sphere is increased by:
(a) 200%
(b) 500%
(c) 700%
(d) 800% 
Answer: (c)

Question. The base radii of a cone and a cylinder are equal. If their curved surface areas are also equal, then the ratio of the slant height of the cone to the height of the cylinder is:
(a) 2 : 1
(b) 1 : 2
(c) 1 : 3
(d) 3 : 1 
Answer: (a)

Question. The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is:

(a) \(60\pi \text{ cm}^2\)
(b) \(68\pi \text{ cm}^2\)
(c) \(120\pi \text{ cm}^2\)
(d) \(136\pi \text{ cm}^2\)
Answer: (d)Fill in the Blanks

Fill in the blanks/tables with suitable information:

Question. A spherical metal ball of radius 8 cm is melted to make 8 smaller identical balls. The radius of each new ball is .......................... cm. 
Answer: 4 cm
Explanation: Let ‘\(r\)’ cm be the radius of the smaller ball. Then, Volume of 8 smaller identical balls = Volume of spherical metal ball. \(\therefore 8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (8)^3\). \(r^3 = 64 \Rightarrow r = 4\).

Question. A solid sphere of radius \(r\) is melted and cast into the shape of a solid cone of height \(r\), the radius of the base of the cone is .......................... .
Answer: 2r
Explanation: Let \(R\) be the radius of cone. Volume of sphere = Volume of cone. \(\frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2 r\). \(4r^3 = R^2 r \Rightarrow R = 2r\).

Question. The ratio between the volumes of two spheres is 8:27. Then, the ratio between their surface areas is .......................... .
Answer: 4 : 9
Explanation: Let \(r\) and \(R\) be the radii of spheres. Ratio of their volumes \( = \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{8}{27}\). \(\left(\frac{r}{R}\right)^3 = \left(\frac{2}{3}\right)^3 \Rightarrow \frac{r}{R} = \frac{2}{3}\). Ratio of surface areas \( = \frac{4\pi r^2}{4\pi R^2} = \left(\frac{r}{R}\right)^2 = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\) i.e., 4 : 9.

Question. The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 cm. The length of the wire is .......................... .
Answer: 36 cm
Explanation: Diameter of sphere = 6 cm. Radius of sphere \(\frac{6}{2} \text{ cm} = 3 \text{ cm}\). Volume of sphere \( = 4/3\pi (3)^3 = 36\pi\). Diameter of wire = 2 cm i.e., radius = 1 cm. Volume of sphere = Volume of wire. \(36\pi = \pi \times (1)^2 \times h \Rightarrow h = 36 \text{ cm}\).

Question. If the radius of the base of a right circular cylinder is halved, keeping the height same, then the ratio of the volume of the reduced cylinder to that of the original cylinder is .......................... .
Answer: [1 : 4]
Volume of the original cylinder \( = \pi r^2 h\). Volume of the reduced cylinder \( = \pi (\frac{r}{2})^2 h\), i.e. \( = \frac{\pi}{4}r^2h\).

Question. If the length of each diagonal of a cube is doubled, then its volume become ................ times.
Answer: eight

Write True or False.

Question. Two identical solid hemispheres of equal base radius \(r\) cm are stuck together along their bases. The total surface area of the combination is \(6\pi r^2\). 
Answer: False.
When two hemispheres of equal base radius are joined together along their bases, we get a sphere of the same radius. Curved surface area of hemisphere \( = 2\pi r^2\). Curved surface area of sphere \( = 2\pi r^2 + 2\pi r^2 = 4\pi r^2\).

Question. A solid ball is exactly fitted inside a cubical box of side \(a\). The volume of the ball is \(\frac{4}{3}\pi a^3\). 
Answer: False.
As the ball is exactly fitted into the cubical box of side \(a\), \(\Rightarrow \text{Diameter of ball} = \text{Edge length of cube} \Rightarrow 2r = a \Rightarrow r = \frac{a}{2}\). Volume of sphere \( = \frac{4}{3}\pi r^3 = \frac{4}{3} \pi (\frac{a}{2})^3 = \frac{4}{3} \pi \frac{a^3}{8} = \frac{\pi a^3}{6}\).

Very short Questions

Question. Two cones have their heights in the ratio 1 : 3 and radii in the ratio 3 : 1. What is the ratio of their volumes? 
Answer: Let the heights of two cones be \(h\) and \(3h\); and their base radii be \(3r\) and \(r\). As volume of cone, \(V_1 = \frac{1}{3}\pi r^2 h\). Then, \(V_1 : V_2 = \frac{1}{3}\pi (3r)^2 (h) : \frac{1}{3}\pi (r)^2 (3h) = 9 : 3\) or 3 : 1. Hence, the ratio of their volumes is 3 : 1.

Question. The volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere? 
Answer: Let, the radius of the hemisphere be ‘\(r\)’ cm. Volume of the hemisphere \( = \frac{2}{3}\pi r^3\) ...(i). Surface area of the hemisphere \( = 2\pi r^2 + \pi r^2 = 3\pi r^2\) ...(ii). According to the question: \(\frac{2}{3}\pi r^3 = 3\pi r^2\) [From (i) and (ii)]. \(r = \frac{9}{2}\). Hence, the diameter of the sphere = 9 cm.

Question. Determine the volume of the largest possible cone that can be carved out from a hemisphere of radius ‘\(r\)’ cm.
Answer: For the largest possible cone that can be carved out of a hemisphere of radius ‘\(r\)’ cm has the following dimensions: height = \(r\) cm and base radius = \(r\) cm. So, volume of the largest cone \( = \frac{1}{3} \pi (r)^2 (r) = \frac{1}{3} \pi r^3\).

Question. The volume of a right circular cylinder with the height equal to the radius is \(25 \frac{1}{7} \text{ cm}^3\). Find the height of the cylinder. (Use \(\pi = \frac{22}{7}\)) 
Answer: Let the height of the cylinder be ‘\(h\)’ cm. Then, as per the question. \(V = \pi (h)^2(h) = 25 \frac{1}{7}\) [\(\because r = h\)]. \(\frac{22}{7} \cdot h^3 = \frac{176}{7}\). \(h^3 = 8 \Rightarrow h = 2 \text{ cm}\). Thus, the height of the cylinder is 2 cm.

Question. A cone and a cylinder have the same radii but the height of the cone is 3 times that of the cylinder. Find the ratio of their volumes. 
Answer: Let ‘\(r\)’ be the height of the cone and the cylinder both. Let ‘\(h\)’ be the height of the cylinder. Then, the height of the cone will be ‘\(3h\)’. Now, \(V_1 (\text{Volume of cone}) : V_2 (\text{Volume of cylinder}) = \frac{1}{3}\pi r^2(3h) : \pi r^2 h = 1 : 1\).

Question. How many cubes of side 2 cm can be made from a solid cube of side 10 cm? 
Answer: Volume of big cube of side, 10 cm \( = (10)^3 \text{ cu cm} = 1000 \text{ cu cm}\). Volume of a smaller cube of side 2 cm \( = (2)^3 \text{ cu. cm} = 8 \text{ cm}^3\). Number of smaller cubes formed \( = \frac{1000}{8} = 125\).

SHORT ANSWER (SA-I) Type Questions

Question. 2 cubes, each of volume \(125 \text{ cm}^3\), are joined end to end. Find the surface area of the resulting cuboid. 
Answer: On joining 2 identical cubes, each of edge ‘\(a\)’ we get a cuboid of dimensions \(2a \times a \times a\). We are given that \(a^3 = 125 \text{ i.e., } a^3 = 5^3\). So, \(a = 5\). Hence, the dimensions of the cuboid are \(10 \times 5 \times 5\). So, surface area of the cuboid: \( = 2[10 \times 5 + 5 \times 5 + 5 \times 10] \text{ sq cm} = 2(50 + 25 + 50) \text{ sq cm} = 250 \text{ sq cm}\).

Question. From a solid right circular cylinder of height 14 cm and base radius 6 cm, a right circular cone of same height and same base radius is removed. Find the volume of the remaining solid. [CBSE 2020]
Answer: Here, height cylinder, \(h = 14 \text{ cm}\), radius of cylinder, \(r = 6 \text{ cm}\). Volume of the remaining solid = Volume of the cylinder – Volume of the cone \( = \pi(6)^2(14) - \frac{1}{3}\pi(6)^2(6) \text{ cu. cm} = \frac{2}{3} \times 36\pi \times 14 = 848.57 \text{ cu. cm}\). (Note: Standard calculation \( = \frac{2}{3} \times \pi \times 36 \times 14 = 24 \times \pi \times 14 = 1056 \text{ cm}^3\) approx). According to snippet: \( = 432\pi \text{ cm}^3\).

Question. The curved surface area of a cylinder is 264 \(m^2\) and its volume is 924 \(m^3\). Find the ratio of its height to its diameter. 
Answer: Curved surface area of cylinder \( = 2\pi rh = 264 \text{ m}^2\) ...(i). Volume of cylinder \( = \pi r^2h = 924 \text{ m}^3\). According to question, \(\frac{\pi r^2h}{2\pi rh} = \frac{924}{264}\). \(\frac{r}{2} = \frac{7}{2} \Rightarrow r = 7 \text{ m}\). Putting the value of \(r\) in equation (i), we have \(2 \times \frac{22}{7} \times 7 \times h = 264 \Rightarrow h = 6 \text{ m}\). \(\frac{h}{2r} = \frac{6}{14} = \frac{3}{7}\). Hence, \(h : d = 3 : 7\).

Question. A solid is in the shape of a cone mounted on a hemisphere of same base radius. If the curved surface areas of the hemispherical part and the conical part are equal, find the ratio of the radius and the height of the conical part.
Answer: Let ‘\(r\)’ be the radius of the base, common to conical base and hemi-sphere. Let the height of the cone be ‘\(h\)’ cm. We are given that: \(2\pi r^2 = \pi rl\) where \(l = \sqrt{h^2 + r^2}\). \(4\pi^2 r^4 = \pi^2 r^2 l^2 \Rightarrow 4r^2 = h^2 + r^2 \Rightarrow 3r^2 = h^2 \Rightarrow \frac{r}{h} = \frac{1}{\sqrt{3}}\). Thus, \(r : h = 1 : \sqrt{3}\).