CBSE Class 10 Mathematics Surface Areas and Volumes MCQs Set 11

Question. The radius of a sphere (in cm) whose volume is \( 12\pi \) cm\(^3\), is:
(a) 3
(b) \( 3\sqrt{3} \)
(c) \( 3^{2/3} \)
(d) \( 3^{1/3} \)
Answer: (c) \( 3^{2/3} \)
Answer: Explanation:
Let ‘r’ cm be the radius of the sphere. Then,
\( \frac{4}{3}\pi r^3 = 12\pi \)
\( \implies \) \( r^3 = 9 \)
i.e., \( r^3 = 3^2 \)
\( \implies \) \( r = (3^2)^{1/3} \), i.e., \( 3^{2/3} \)

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) a cone and a cylinder
Answer: Explanation:
The tip of a sharpened pencil is conical in shape and the rest of the part is cylindrical in shape. The shape of a sharpened pencil is a cylinder and a cone.

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) a sphere and a cylinder
Answer: Explanation:
The top part of surahi is cylindrical in shape and bottom part is spherical in shape. Therefore, surahi is a combination of sphere and a cylinder.

Question. The shape of a gilli, in the gilli-danda game (see the given figure), is a combination of
(a) two cylinders
(b) a cone and a cylinder
(c) two cones and a cylinder
(d) two cylinders and a cone
Answer: (c) two cones and a cylinder
Answer: Explanation:
As the left and right part of a gilli are conical and the central part is cylindrical. Therefore, Given figure = Cone + Cylinder + Cone = Two cones and a cylinder.

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) frustum of a cone and a hemisphere
Answer: Explanation:
The upper part of a shuttle cock is hemispherical in shape and the lower part is in the shape of frustum of a cone. Therefore, it is a combination of frustum of a cone and a hemisphere.

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) 40 cm
Answer: Explanation:
Here, \( \frac{1}{3}\pi(10)^2 h = \frac{4}{3}\pi(10)^3 \)
This gives, \( h = 40 \) cm

Question. Three cubes each of side 15 cm are joined end to end. The total surface area of the cuboid is:
(a) 3150 cm\(^2\)
(b) 1575 cm\(^2\)
(c) 1012.5 cm\(^2\)
(d) 576.4 cm\(^2\)
Answer: (a) 3150 cm\(^2\)
Answer: Explanation:
The dimensions of the cuboid are 45 cm \( \times \) 15 cm \( \times \) 15 cm
So, total surface area of the cuboid \( = 2(lb + bh + hl) \) cm\(^2\)
\( = 2(675 + 225 + 675) \) cm\(^2\)
\( = 3150 \) cm\(^2\)

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 accommodate is:
(a) 142296
(b) 142396
(c) 142496
(d) 142596
Answer: (a) 142296
Answer: Explanation:
Given, internal edge of cube = 22 cm
\( \therefore \) Volume of cube = \( (\text{Side})^3 = (22)^3 = 10648 \) cm\(^3\)
Let the spherical marble has radius \( r \).
Diameter of the marble, 0.5 cm
\( \therefore \) radius of marble, \( r = \frac{0.5}{2} = 0.25 \) cm
\( \therefore \) Volume of 1 marble = \( \frac{4}{3}\pi r^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times (0.25)^3 \)
\( = \frac{1.375}{21} \) cm\(^3\)
As \( \frac{1}{8} \) part of the cube remains unfilled, only \( \frac{7}{8} \) part of cube remains filled.
\( \therefore \) Volume of Filled cube \( = \frac{7}{8} \times \text{Volume of cube} \)
\( = \frac{7}{8} \times 10648 \)
\( = 7 \times 1331 = 9317 \) cm\(^3\)
\( \therefore \) Required no. of marbles \( = \frac{\text{Total space filled by marbles}}{\text{Volume of 1 marble}} \)
\( = \frac{9317}{1.375/21} = \frac{9317 \times 21}{1.375} = 142296 \)
Hence, cube can accommodate is 142296 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) 14 cm
Answer: Explanation:
We know that during recasting a shape into another its’ volume does not change.
Spherical shell:
Internal diameter, \( d_1 = 4 \) cm
\( \therefore \) Internal radius, \( r_1 = 2 \) cm
External diameter, \( d_2 = 8 \) cm
\( \therefore \) External radius, \( r_2 = 4 \) cm
Volume of spherical shell \( = \frac{4}{3}\pi [r_2^3 - r_1^3] \)
\( = \frac{4}{3}\pi [4^3 - 2^3] = \frac{4}{3}\pi [64 - 8] \)
\( = \frac{4}{3}\pi \times 56 = \frac{224\pi}{3} \) cm\(^3\)
Let \( h \) be the height of cone.
Diameter of base of cone = 8 cm
\( \therefore \) Radius = \( \frac{8}{2} \) cm = 4 cm
During recasting volume remains same so,
Volume of cone = Volume of spherical shell
\( \implies \) \( \frac{1}{3}\pi r^2 h = \frac{4}{3}\pi [r_2^3 - r_1^3] \)
\( \implies \) \( \frac{1}{3} \times \pi \times (4)^2 \times h = \frac{4}{3} \times \pi \times 56 \)
\( \implies \) \( h = 14 \) cm
Hence, the height of the cone is 14 cm.

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) 16:9
Answer: Explanation:
We know that \( r_1 \) and \( r_2 \) be the radii of two spheres respectively
Volume of sphere \( = \frac{4}{3}\pi r^3 \)
Given : Ratio of volume is 64:27.
\( \frac{\text{Volume of } 1^{\text{st}} \text{ sphere}}{\text{Volume of } 2^{\text{nd}} \text{ sphere}} = \frac{64}{27} \)
\( \implies \) \( \frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{64}{27} \)
\( \implies \) \( \frac{r_1^3}{r_2^3} = \frac{64}{27} \implies \frac{r_1}{r_2} = \frac{4}{3} \)
Surface area of sphere \( = 4\pi r^2 \)
\( \frac{\text{Surface area of } 1^{\text{st}} \text{ sphere}}{\text{Surface area of } 2^{\text{nd}} \text{ sphere}} = \frac{4\pi r_1^2}{4\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2 \)
\( = \left(\frac{4}{3}\right)^2 = \frac{16}{9} \)
Hence, the ratio of their surface area is 16:9.

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) 700%
Answer: Explanation: When the radius is increased by 100%, the corresponding volume becomes 800% and thus increase is 700%.

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) 2 : 1
Answer: Explanation: Since, the radius of cone and cylinder are equal i.e., \( r \).
Then, \( \pi r l = 2\pi r h \)
\( \frac{l}{h} = \frac{2}{1} = 2 : 1 \)

Question. The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is:
(a) \( 60\pi \) cm\(^2\)
(b) \( 68\pi \) cm\(^2\)
(c) \( 120\pi \) cm\(^2\)
(d) \( 136\pi \) cm\(^2\)
Answer: (d) \( 136\pi \) cm\(^2\)
Answer: Explanation: Curved surface area of the cone \( = \pi r l \)
Here, \( r = 8 \) cm and \( h = 15 \) cm
\( \therefore l = \sqrt{h^2 + r^2} \)
C.S.A. \( = \pi(8) \sqrt{8^2 + 15^2} \) cm\(^2 = \pi(8)(17) \) cm\(^2 = 136\pi \) cm\(^2 \)

Fill in the Blanks

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
Answer: Explanation: Let ‘r’ cm be the radius of the smaller ball. Then,
Volume of 8 smaller identical balls = Volume of spherical metal ball.
\( \therefore 8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (8)^3 \)
\( \implies \) \( r^3 = 64 \)
\( \implies \) \( r = 4 \)

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: \( 2r \)
Answer: Explanation: Let \( R \) be the radius of cone
Volume of sphere = Volume of cone
\( \frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2 r \)
\( \implies \) \( 4r^3 = R^2 r \)
\( \implies \) \( R = 2r \)

Question. The ratio between the volumes of two spheres is 8:27. Then, the ratio between their surface areas is .......................... .
Answer: 4 : 9
Answer: Explanation: Let \( r \) and \( R \) be the radii of spheres
Ratio of their volumes = \( \frac{\frac{4}{3}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{8}{27} \)
\( \implies \) \( \left(\frac{r}{R}\right)^3 = \left(\frac{2}{3}\right)^3 \)
\( \implies \) \( \frac{r}{R} = \frac{2}{3} \)
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
Answer: Explanation: Diameter of sphere = 6 cm
Radius of sphere \( = \frac{6}{2} \) cm = 3 cm
Volume of sphere \( = \frac{4}{3}\pi (3)^3 = 36\pi \)
Diameter of wire = 2 cm i.e., radius = 1 cm
Volume of sphere = Volume of wire
\( 36\pi = \pi \times (1)^2 \times h \)
\( \implies \) \( h = 36 \) 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]
Answer: Volume of the original cylinder \( = \pi r^2 h \)
Volume of the reduced cylinder \( = \pi \left(\frac{r}{2}\right)^2 h \), i.e. \( \frac{\pi}{4} r^2 h \).

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

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.
Answer: 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.
Answer: As the ball is exactly fitted into the cubical box of side \( a \),
\( \implies \) Diameter of ball = Edge length of cube
\( \implies \) \( 2r = a \)
\( \implies \) \( r = \frac{a}{2} \)
Volume of sphere \( = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \left(\frac{a}{2}\right)^3 = \frac{4}{3} \times \frac{\pi a^3}{8} = \frac{\pi a^3}{6} \)