CBSE Class 10 Mathematics Surface Areas And Volume VBQs Set E

OBJECTIVE TYPE QUESTIONS

Question. A cylindrical pencil sharpened at one edge is the 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)
Explanation: The sharpened part of the pencil is cone and unsharpened part is cylinder.

Question. A surahi is the 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)
Explanation: A surahi is the combination of a sphere and a cylinder.

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)
Explanation: Let the spherical marble has radius \( r \).
Diameter of the marble = 0.5 cm
\( \Rightarrow r = \frac{0.5}{2} \) cm = 0.25 cm
Length of side of cube \( l \) = 22 cm
Let \( n \) marbles can fill the cube.
\( \therefore \) Volume of \( n \) marbles = \( (1 - \frac{1}{8}) \) part of volume of cube
\( \Rightarrow n \cdot \frac{4}{3} \pi r^3 = \frac{7}{8} \times l^3 \)
\( n = \frac{7l^3}{8} \times \frac{3}{4\pi r^3} \)
\( \Rightarrow n = \frac{7 \times 3 \times 22 \times 22 \times 22 \times 7}{8 \times 4 \times 22 \times 0.25 \times 0.25 \times 0.25} \)
\( \Rightarrow n = 7 \times 3 \times 22 \times 22 \times 2 \times 7 = 42 \times 484 \times 7 = 142296 \)
So, cube can accommodate 142296 marbles.

Question. A medicine-capsule is in the shape of a cylinder of diameter 0.5 cm with two hemispheres stuck to each to its ends. The length of entire capsule is 2 cm. The capacity of the capsule is :
(a) 0.36 cm³
(b) 0.35 cm³
(c) 0.34 cm³
(d) 0.33 cm³
Answer: (a)
Explanation: Capsule consists of 2 hemispheres and a cylinder.
\( r = \frac{0.5}{2} \) cm = 0.25 cm
Total length of capsule = \( r + h + r \)
\( \Rightarrow 2 = 2r + h \)
\( \Rightarrow 2 = 2 \times 0.25 + h \)
\( \Rightarrow h = 2 - 0.5 = 1.5 \) cm
Volume of capsule = Volume of two hemispheres + Volume of cylinder
\( = 2 \times \left( \frac{4}{3} \pi r^3 \right) + \pi r^2 h \)
\( = \frac{4}{3} \pi r^3 + \pi r^2 h = \pi r^2 \left( \frac{4}{3}r + h \right) \)
\( = \frac{22}{7} \times 0.25 \times 0.25 \left( \frac{4}{3} \times 0.25 + 1.5 \right) \)
\( = \frac{22}{7} \times 0.25 \times 0.25 \left( \frac{1}{3} + \frac{3}{2} \right) = \frac{22}{7} \times \frac{25}{100} \times \frac{25}{100} \times \frac{11}{6} = \frac{121}{336} = 0.3601 \approx 0.36 \text{ cm}^3 \).
\( \therefore \) Volume of capsule = \( 0.3601 \text{ cm}^3 \approx 0.36 \text{ cm}^3 \).

Question. If two solid hemispheres of same base radius \( r \) are joined together along their bases, then curved surface area of this new solid is :
(a) \( 4\pi r^2 \)
(b) \( 6\pi r^2 \)
(c) \( 3\pi r^2 \)
(d) \( 8\pi r^2 \)
Answer: (a)
Explanation: When two hemispheres of equal radii are joined base to base, new solid becomes sphere and curved surface area of sphere is \( 4\pi r^2 \).

Question. A right circular cylinder of radius \( r \) cm and height \( h \) cm (where \( h > 2r \)) just encloses of sphere of diameter :
(a) \( r \) cm
(b) \( 2r \) cm
(c) \( h \) cm
(d) \( 2h \) cm
Answer: (b)
Explanation: As the cylinder just encloses the sphere so the cylinder and diameter of sphere are equal, i.e., \( 2r \) and height \( h > 2r \).

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 8 cm. The height of the cone is:
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm
Answer: (b)
Explanation: During recasting a shape into another shape it's volume does not change.
For spherical shell
\( r_1 = \frac{4}{2} = 2 \) cm
\( r_2 = \frac{8}{2} = 4 \) cm
For cone
\( r = \frac{8}{2} = 4 \) cm
\( h = ? \)
During recasting volume remains same so,
Volume of cone = Volume of hollow spherical shell
\( \frac{1}{3} \pi r^2 h = \frac{4}{3} \pi r_2^3 - \frac{4}{3} \pi r_1^3 \)
\( \frac{1}{3} \pi r^2 h = \frac{4}{3} \pi (r_2^3 - r_1^3) \)
\( r^2 h = 4(r_2^3 - r_1^3) \)
\( (4 \times 4)h = 4[(4)^3 - (2)^3] \)
\( 4h = 64 - 8 \)
\( h = \frac{56}{4} = 14 \) cm.

Question. A solid piece of iron in the form of a cuboid of dimensions 49 cm × 33 cm × 24 cm, is moulded to form a solid sphere. The radius of the sphere is:
(a) 21 cm
(b) 23 cm
(c) 25 cm
(d) 19 cm
Answer: (a)
Explanation: Solid cuboid of iron is moulded into solid sphere.
Hence, volume of cuboid and sphere are equal.
For sphere \( r = ? \)
For cuboid \( l = 49 \) cm, \( b = 33 \) cm, \( h = 24 \) cm.
\( \therefore \) Volume of sphere (solid) = Volume of cuboid
\( \frac{4}{3} \pi r^3 = l \times b \times h \)
\( r^3 = \frac{(l \times b \times h \times 3)}{(4 \times \pi)} \)
\( = \frac{(49 \times 33 \times 24 \times 3 \times 7)}{(4 \times 22)} \)
\( r^3 = 7 \times 7 \times 7 \times 3 \times 3 \times 3 \)
\( r = 21 \) cm.

Question. A mason constructs a wall of dimensions 270 cm × 300 cm × 350 cm with the bricks each of size 22.5 cm × 11.25 cm × 8.75 cm and it is assumed that \( \frac{1}{8} \) space is covered by the mortar. Then the number of bricks used to construct the wall is:
(a) 11100
(b) 11200
(c) 11000
(d) 11300
Answer: (b)
Explanation: The volume of the wall covered by mortar = \( \frac{1}{8} \) part.
So, the volume of wall covered by bricks = \( (1 - \frac{1}{8}) \) volume of wall = \( \frac{7}{8} \) volume of wall.
Bricks (Cuboid): \( l_1 = 22.5 \) cm, \( b_1 = 11.25 \) cm, \( h_1 = 8.75 \) cm.
Wall (Cuboid): \( l = 270 \) cm, \( b = 300 \) cm, \( h = 350 \) cm.
Let \( n \) be the number of bricks.
According to the question, we have
Volume of \( n \) bricks = \( \frac{7}{8} \) Volume of wall
\( \Rightarrow n \times l_1 \times b_1 \times h_1 = \frac{7}{8} \times l \times b \times h \)
\( \Rightarrow n = \frac{(7 \times l \times b \times h)}{(8 \times l_1 \times b_1 \times h_1)} \)
\( = \frac{(7 \times 270 \times 300 \times 350)}{(8 \times 22.5 \times 11.25 \times 8.75)} \)
\( \Rightarrow n = \frac{(7 \times 270 \times 300 \times 350 \times 100 \times 10 \times 100)}{(8 \times 225 \times 1125 \times 875)} \)
\( \Rightarrow n = 2 \times 4 \times 350 \times 4 = 32 \times 350 = 11,200 \) bricks.

Question. Twelve solid sphere of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is:
(a) 4 cm
(b) 3 cm
(c) 2 cm
(d) 6 cm
Answer: (c)
Explanation: Solid cylinder is recasted into 12 spheres.
So, the volume of 12 spheres will be equal to the volume of the cylinder.
For spheres \( R = ? \)
For cylinder \( r = \frac{2}{2} = 1 \) cm, \( h = 16 \) cm.
\( \therefore \) Volume of 12 spheres = Volume of cylinder
\( \Rightarrow 12 \times \frac{4}{3} \pi R^3 = \pi r^2 h \)
\( \Rightarrow R^3 = \frac{(3r^2 h)}{(4 \times 12)} = \frac{(3 \times 1 \times 1 \times 16)}{(4 \times 12)} = 1 \)
\( \Rightarrow R = 1 \) cm. Hence, diameter = \( 2R = 2 \times 1 = 2 \) cm.

Question. During conversion of a solid from one shape to another, the volume of new shape will:
(a) increase
(b) decrease
(c) remains unaltered
(d) be doubled
Answer: (c)
Explanation: During reshaping a solid, the volume of new solid will be equal to old one or remains unaltered.

Question. A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is:
(a) 3.5
(b) 7
(c) \( \frac{80}{7} \)
(d) 5
Answer: (a)
Explanation: Circumference = 22 cm
\( 2\pi r = 22 \)
\( 2 \times \frac{22}{7} \times r = 22 \)
\( r = 3.5 \) cm.

Question. The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is:
(a) 3
(b) 5
(c) 4
(d) 6
Answer: (b)
Explanation: No. of solid spheres = Volume of cylinder / Volume of sphere
\( = \frac{\pi R^2 h}{\frac{4}{3} \pi r^3} = \frac{\pi (2)^2 \times 45}{\frac{4}{3} \pi (3)^3} = 5 \).

Case-based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark. I. Read the following text and answer the questions that follows, on the basis of the same: The Great Stupa at Sanchi is one of the oldest stone structures in India, and an important monument of Indian Architecture. It was originally commissioned by the emperor Ashoka in the 3rd century BCE. Its nucleus was a simple hemispherical brick structure built over the relics of the Buddha. It is a perfect example of combination of solid figures. A big hemispherical dome with a cuboidal structure mounted on it. Take \( \pi = \frac{22}{7} \).

Question. Calculate the volume of the hemispherical dome if the height of the dome is 21 m:
(a) 19404 cu. m
(b) 2000 cu. m
(c) 15000 cu. m
(d) 19000 cu. m
Answer: (a)
Explanation: Height of hemispherical dome = Radius of hemispherical dome = 21 m.
Volume of dome = \( \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 19,404 \text{ m}^3 \).

Question. The formula to find the volume of sphere is:
(a) \( \frac{2}{3} \pi r^3 \)
(b) \( \frac{4}{3} \pi r^3 \)
(c) \( 4\pi r^2 \)
(d) \( 2\pi r^2 \)
Answer: (b)

Question. The cloth require to cover the hemispherical dome if the radius of its base is 14 m is:
(a) 1222 sq.m
(b) 1232 sq.m
(c) 1200 sq.m
(d) 1400 sq.m
Answer: (b)
Explanation: The cloth required to cover the hemispherical dome it the radius of its base is 14 cm is
\( 2\pi R^2 = 2 \times \frac{22}{7} \times (14)^2 \text{ m}^2 = 1232 \text{ sq. m} \).

Question. The total surface area of the combined figure i.e. hemispherical dome with radius 14 m and cuboidal shaped top with dimensions 8 m × 6 m × 4 m is:
(a) 1200 sq. m
(b) 1232 sq. m
(c) 1392 sq.m
(d) 1932 sq. m
Answer: (c)
Explanation: Total surface Area of Combined figure = \( 2\pi r^2 + 2(lb + bh + hl) - lb \)
\( = 2 \times \frac{22}{7} \times 14 \times 14 + 2(8 \times 6 + 6 \times 4 + 4 \times 8) - 8 \times 6 \text{ m}^2 \)
\( = [1232 + 208 - 48] \text{ m}^2 = 1392 \text{ m}^2 \).

Question. The volume of the cuboidal shaped top is with dimensions mentioned in question 4 is:
(a) 182.45 m³
(b) 282.45 m³
(c) 292 m³
(d) 192 m³
Answer: (d)
Explanation: Volume of the cuboidal shaped top = \( l \times b \times h = 8 \text{ m} \times 6 \text{ m} \times 4 \text{ m} = 192 \text{ m}^3 \).

II. Read the following text and answer the questions that follows, on the basis of the same: On a Sunday, your Parents took you to a fair. You could see lot of toys displayed, and you wanted them to buy a RUBIK’s cube and strawberry ice cream for you.

Question. The length of the diagonal if each edge measures 6 cm is:
(a) \( 3\sqrt{3} \) cm
(b) \( 3\sqrt{6} \) cm
(c) \( \sqrt{12} \) cm
(d) \( 6\sqrt{3} \) cm
Answer: (d)
Explanation: We know that, Length of the diagonal = \( \sqrt{3} \times \text{side} \).
\( \therefore \) Length of the diagonal of cube with edge 6 cm = \( \sqrt{3} \times 6 = 6\sqrt{3} \) cm.

Question. Volume of the solid figure if the length of the edge is 7 cm is:
(a) 256 cm³
(b) 196 cm³
(c) 343 cm³
(d) 434 cm³
Answer: (c)
Explanation: Volume of cube = \( (\text{side})^3 = (7)^3 = 343 \text{ cm}^3 \).

Question. What is the curved surface area of hemisphere (ice cream) if the base radius is 7 cm ?
(a) 309 cm²
(b) 308 cm²
(c) 803 cm²
(d) 903 cm²
Answer: (b)
Explanation: We know that, CSA(curved surface area) of hemisphere = \( 2\pi r^2 \).
Given, \( r = 7 \) cm. So, CSA = \( 2 \times \frac{22}{7} \times 7 \times 7 = 308 \text{ cm}^2 \).

Question. Slant height of a cone if the radius is 7 cm and the height is 24 cm___.
(a) 26 cm
(b) 25 cm
(c) 52 cm
(d) 62 cm
Answer: (b)
Explanation: Slant height of cone, \( l = \sqrt{r^2 + h^2} \).
Hence, \( r = 7 \) cm and \( h = 24 \) cm.
\( \therefore l = \sqrt{(7)^2 + (24)^2} = \sqrt{625} = 25 \) cm.

Question. The total surface area of cone with hemispherical ice cream is:
(a) 858 cm²
(b) 885 cm²
(c) 588 cm²
(d) 855 cm²
Answer: (a)
Explanation: Total surface area of the cone with hemispherical ice-cream = curved surface area of the cone + curved surface area of the hemisphere = \( 550 + 308 = 858 \text{ cm}^2 \).

III. Read the following text and answer the following questions that follows, on the basis of the same. Adventure camps are the perfect place for the children to practice decision making for themselves without parents and teachers guiding their every move. Some students of a school reached for adventure at Sakleshpur. At the camp, the waiters served some students with a welcome drink in a cylindrical glass and some students in a hemispherical cup whose dimensions are shown below. After that they went for a jungle trek. The jungle trek was enjoyable but tiring. As dusk fell, it was time to take shelter. Each group of four students was given a canvas of area 551 m². Each group had to make a conical tent to accommodate all the four students. Assuming that all the stitching and wasting incurred while cutting, would amount to 1 m², the students put the tents. The radius of the tent is 7 m.

Question. The volume of cylindrical cup is:
(a) 295.75 cm³
(b) 7415.5 cm³
(c) 384.88 cm³
(d) 404.25 cm³
Answer: (d)
Explanation: diameter, \( d = 7 \) cm, radius, \( r = \frac{d}{2} = 3.5 \) cm, height, \( h = 10.5 \) cm.
Volume of cylindrical cup = \( \pi r^2 h = \frac{22}{7} \times 3.5 \times 3.5 \times 10.5 = 404.25 \text{ cm}^2 \).

Question. The volume of hemispherical cup is:
(a) 179.67 cm³
(b) 89.83 cm³
(c) 172.25 cm³
(d) 210.60 cm³
Answer: (b)
Explanation: For hemispherical cup, \( d = 7 \Rightarrow r = \frac{d}{2} = \frac{7}{2} \) cm.
Volume of hemisphere = \( \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (\frac{7}{2})^3 \text{ cm}^3 = 89.83 \text{ cm}^3 \).

Question. Which container had more juice and by how much?
(a) Hemispherical cup, 195 cm³
(b) Cylindrical glass, 207 cm³
(c) Hemispherical cup, 280.85 cm³
(d) Cylindrical glass, 314.42 cm³
Answer: (d)
Explanation: Volume of cylindrical cup = \( \pi r^2 h \) (Given, \( r = \frac{7}{2} \) cm and \( h = 10.5 \) cm) = \( 404.25 \text{ cm}^3 \).
Also, volume of hemispherical cup = 89.83 cm³ [As calculated in Q.2.].
Thus, cylindrical cup has more volume. So, difference in volumes = \( (404.25 - 89.83) \text{ cm}^3 = 314.42 \text{ cm}^3 \).

Question. The height of the conical tent prepared to accommodate four students is:
(a) 18 m
(b) 10 m
(c) 24 m
(d) 14 m
Answer: (c)
Explanation: Radius = 7 m. Area of conical tent = 551 m² – 1 m² = 550 m².
\( \pi rl = 550 \Rightarrow \frac{22}{7} \times 7 \times \sqrt{r^2 + h^2} = 550 \Rightarrow 22 \times \sqrt{7^2 + h^2} = 550 \)
\( \sqrt{7^2 + h^2} = \frac{550}{22} = 25 \Rightarrow 7^2 + h^2 = (25)^2 \)
\( h^2 = 625 - 49 = 576 \Rightarrow h = \sqrt{576} = 24 \text{ m} \).

Question. How much space on the ground is occupied by each student in the conical tent?
(a) 54 m²
(b) 38.5 m²
(c) 86 m²
(d) 24 m²
Answer: (b)
Explanation: Area of Base of conical tent = \( \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154 \text{ m}^2 \).
Area of occupied by each student = \( \frac{1}{4} \times 154 \text{ m}^2 = 38.5 \text{ m}^2 \).

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm.

Question. What is the volume of cuboid?
(a) \(525 \text{ cm}^3\)
(b) \(225 \text{ cm}^3\)
(c) \(552 \text{ cm}^3\)
(d) \(255 \text{ cm}^3\)
Answer: (a)
Explanation: For cuboid \(l = 15 \text{ cm}, b = 10 \text{ cm}\) and \(h = 3.5 \text{ cm}\).
Volume of the cuboid \( = l \times b \times h\)
\( = 15 \times 10 \times 3.5\)
\( = 525 \text{ cm}^3\)

Question. What is the volume of a conical depression ?
(a) \(\frac{11}{3} \text{ cm}^3\)
(b) \(\frac{11}{30} \text{ cm}^3\)
(c) \(\frac{3}{11} \text{ cm}^3\)
(d) \(\frac{30}{11} \text{ cm}^3\)
Answer: (b)
Explanation: For conical depression: \(r = 0.5 \text{ cm}, h = 1.4 \text{ cm}\)
Volume of conical depression \( = \frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4\)
\( = \frac{11}{30} \text{ cm}^3\)

Question. What is the total volume of conical depressions?
(a) \(1.74 \text{ cm}^3\)
(b) \(1.44 \text{ cm}^3\)
(c) \(1.47 \text{ cm}^3\)
(d) \(1.77 \text{ cm}^3\)
Answer: (c)
Explanation: Volume of four conical depressions \( = 4 \times \frac{11}{30} \approx 1.47 \text{ cm}^3\)

Question. What is the volume of wood in the entire stand?
(a) \(522.35 \text{ cm}^3\)
(b) \(532.53 \text{ cm}^3\)
(c) \(523.35 \text{ cm}^3\)
(d) \(523.53 \text{ cm}^3\)
Answer: (d)
Explanation: Volume of the wood in the entire stand = Volume of cuboid – Volume of 4 conical depressions
\( = 525 - 1.47\)
\( = 523.53 \text{ cm}^3\)

Question. The given problem is based on which mathematical concept?
(a) Triangle
(b) Surface Areas and Volumes
(c) Height and Distances
(d) None of these
Answer: (b)