CBSE Class 10 Surface Areas and Volumes Sure Shot Questions Set 15

To make the teaching, learning process easier, creative, and innovative, A teacher brings clay in the classroom to teach the topic mensuration.She thought this method of teaching is more interesting, leave a long-lasting impact. She forms a cylinder of radius 6 cm and height 8 cm with the clay, then she moulds the cylinder into a sphere and asks some question to students. [use p = 3.14]

Question. The radius of the sphere so formed :
(a) 6 cm
(b) 7 cm
(c) 4 cm
(d) 8 cm
Answer: (a) 6 cm
Explanation :
\( \frac{4}{3} \pi r^3 = \pi R^2H \)
\( \frac{4}{3} \pi r^3 = 6 \times 6 \times 8 \)
\( r = 6 \text{ cm} \)

Question. The volume of the sphere so formed :
(a) 902.32 \( \text{cm}^3 \)
(b) 899.34 \( \text{cm}^3 \)
(c) 904.32 \( \text{cm}^3 \)
(d) 999.33 \( \text{cm}^3 \)
Answer: (c) 904.32 \( \text{cm}^3 \)
Explanation :
\( \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6 \)
\( r = 904.32 \text{ cm}^3 \)

Question. What is the ratio of the volume of a sphere to the volume of a cylinder?
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 3 : 1
Answer: (c) 1 : 1
Explanation :
\( \frac{V_s}{V_c} = \frac{\frac{4}{3} \pi r^3}{\pi R^2 H} \)
\( = \frac{\frac{4}{3} \times 6 \times 6 \times 6}{6 \times 6 \times 8} \)
\( = 1 : 1 \)

Question. The total surface area of the cylinder is :
(a) 528 \( \text{cm}^2 \)
(b) 570 \( \text{cm}^2 \)
(c) 540 \( \text{cm}^2 \)
(d) 520 \( \text{cm}^2 \)
Answer: (a) 528 \( \text{cm}^2 \)
Explanation :
\( 2\pi RH + 2\pi R^2 = 2\pi R(H + R) \)
\( = 2 \times \frac{22}{7} \times 6 \times (8 + 6) \)
\( = 2 \times \frac{22}{7} \times 6 \times 14 \)
\( = 528 \text{ cm}^2 \)

Question. During the conversion of a solid from one shape to another the volume of the new shape will:
(a) Increase
(b) Decrease
(c) Remain unaltered
(d) Be doubled
Answer: (c) Remain unaltered
Explanation :
The conversion of a solid from one sphere to another the volume of the new shape will remain unattered, or constant.

Ajay is a Class X student. His class teacher Mrs Kiran arranged a historical trip to great Stupa of Sanchi. She explained that Stupa of Sanchi is great example of architecture in India. Its base part is cylindrical in shape. The dome of this stupa is hemispherical in shape, known as Anda. lt also contains a cubical shape part called Hermika at the top. Path around Anda is known as Pradakshina Path.

Question. Find the lateral surface area of the Hermika, if the side of cubical part is 8 m :
(a) 128 \( \text{m}^2 \)
(b) 256 \( \text{m}^2 \)
(c) 512 \( \text{m}^2 \)
(d) 1024 \( \text{m}^2 \)
Answer: (b) 256 \( \text{m}^2 \)
Explanation :
Lateral surface area of Hermika which is cubical in shape \( = 4a^2 = 4 \times (8)^2 = 256 \text{ m}^2 \)

Question. The diameter and height of the cylindrical base part are respectively 42 m and 12 m. If the volume of each brick used is 0.01m³, then find the number of bricks used to make the cylindrical base :
(a) 1663200
(b) 1580500
(c) 1765000
(d) 8065000
Answer: (a) 1663200
Explanation :
Diameter of cylindrical base \( = 42 \text{ m} \)
Radius of cylindrical base \( (r) = 21 \text{ m} \)
Height of cylindrical base \( (h) = 12 \text{ m} \)
Number of bricks used \( = \frac{22}{7} \times \frac{21 \times 21 \times 12}{0.01} = 1663200 \)

Question. lf the diameter of the Anda is 42 m, then the volume of the Anda is :
(a) 17475 \( \text{m}^3 \)
(b) 18605 \( \text{m}^3 \)
(c) 19404 \( \text{m}^2 \)
(d) 18650 \( \text{m}^3 \)
Answer: (c) 19404 \( \text{m}^2 \)
Explanation :
Given, diameter of Anda which is hemispherical in shape \( = 42 \text{ m} \)
\( \Rightarrow \) Radius of Anda \( (r) = 21\text{m} \)
Volume of Anda \( = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 44 \times 21 \times 21 = 19404 \text{ m}^3 \)

Question. The radius of the Pradakshina path is 25 m. If Buddhist priest walks 14 rounds on this path, then find the distance covered by the priest :
(a) 1850 m
(b) 3600 m
(c) 2400 m
(d) 2200 m
Answer: (d) 2200 m
Explanation :
Given, radius of Pradakshina Path \( (r) = 25 \text{ m} \)
Distance covered by priest \( = 14 \times 2 \times \frac{22}{7} \times 25 = 2200 \text{ m} \)

Question. The curved surface area of the Anda is :
(a) 2856 \( \text{m}^2 \)
(b) 2772 \( \text{m}^2 \)
(c) 2473 \( \text{m}^2 \)
(d) 2652 \( \text{m}^2 \)
Answer: (b) 2772 \( \text{m}^2 \)
Explanation :
Radius of Anda \( (r) = 21\text{m} \)
Curved surface area of Anda \( = 2\pi r^2 = 2 \times \frac{22}{7} \times 21 \times 21 = 2772 \text{ m}^2 \)

A carpenter used to make and sell different kinds of wooden pen stands like rectangular, cuboidal, cylindrical, conical. Aarav went to his shop and asked him to make a pen stand as explained below. Pen stand must be of the cuboidal shape with three conical depressions, which can hold 3 pens. The dimensions of the cuboidal part must be 20 cm × 15 cm × 5 cm and the radius and depth of each conical depression must be 0.6 cm and 2.1 cm respectively. Based on the above information, answer the following questions.

Question. The volume of the cuboidal part is :
(a) 1250 \( \text{cm}^3 \)
(b) 1500 \( \text{cm}^3 \)
(c) 1625 \( \text{cm}^3 \)
(d) 1500 \( \text{cm}^3 \)
Answer: (b) 1500 \( \text{cm}^3 \)
Explanation :
Volume of cuboidal part \( = l \times b \times h = (20 \times 15 \times 5) \text{ cm}^3 = 1500 \text{ cm}^3 \)

Question. Total volume of conical depressions is:
(a) 2.508 \( \text{cm}^3 \)
(b) 1.5 \( \text{cm}^3 \)
(c) 2.376 \( \text{cm}^3 \)
(d) 3.6 \( \text{cm}^3 \)
Answer: (c) 2.376 \( \text{cm}^3 \)
Explanation :
Radius of conical depression, \( r = 0.6 \text{ cm} \)
Height of conical depression, \( h = 2.1 \text{ cm} \)
Total volume of conical depressions \( = 3 \times \frac{1}{3} \pi r^2 h = \frac{22}{7} \times 0.6 \times 0.6 \times 2.1 = 2.376 \text{ cm}^3 \)

Question. Volume of the wood used in the entire stand is :
(a) 631.31 \( \text{cm}^3 \)
(b) 3564 \( \text{cm}^3 \)
(c) 1502.376 \( \text{cm}^3 \)
(d) 1497.624 \( \text{cm}^3 \)
Answer: (d) 1497.624 \( \text{cm}^3 \)
Explanation :
Volume of wood used in the entire stand \( = \text{Volume of cuboidal part} - \text{Total volume of conical depressions} = 1500 - 2.376 = 1497.624 \text{ cm}^3 \)

Question. Total surface area of cone of radius r is given by :
(a) \( prl + pr^2 \)
(b) \( 2prl + pr^2 \)
(c) \( pr^2l + pr^2 \)
(d) \( prl + 2pr^3 \)
Answer: (a) \( prl + pr^2 \)
Explanation :
Total surface area of cone of radius \( r \) is given by \( prl + pr^2 \).

Question. If the cost of wood used is ` 5 per cm³, then the total cost of making the pen stand is :
(a) ` 8450.50
(b) ` 7480
(c) ` 9962.14
(d) ` 7488.12
Answer: (d) ` 7488.12
Explanation :
Cost of wood per \( \text{cm}^3 = \text{` } 5 \)
Total cost of making the pen stand \( = \text{` } (5 \times 1497.624) = \text{` } 7488.12 \)

To promote cooperation, culture, creativity, sharing, self-confidence, and other social values, a student adventure camp was organized by the school for X-class students and their accommodation was planned in texts. The teacher divides the students into groups, each group of four students was given to prepare a conical tent of radius 7 m and canvas of area 551 m² in which 1 m² is used in stitching and wasting of canvas :

Question. Curved surface of conical tent :
(a) prl
(b) \( pr^2h \)
(c) 13prl
(d) \( 2pr(r + l) \)
Answer: (a) prl
Explanation :
Curved surface of conical tent is \( prl \).

Question. Height of the conical tent :
(a) 23 m
(b) 24 m
(c) 25 m
(d) 26 m
Answer: (b) 24 m
Explanation :
\( prl = 550 \)
\( l = \frac{7 \times 550}{22 \times 7} = 125 \)
\( h = \sqrt{l^2 - r^2} \)
\( = \sqrt{25^2 - 7^2} = 24 \text{ m} \)

Question. Volume of tent :
(a) 1234 \( \text{m}^3 \)
(b) 1232 \( \text{m}^3 \)
(c) 1332 \( \text{m}^3 \)
(d) 1343 \( \text{m}^3 \)
Answer: (b) 1232 \( \text{m}^3 \)
Explanation :
\( pr^2h = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 \)
\( = 1232 \text{ m}^3 \)

Question. How much space is occupied by each student in the tent?
(a) 318 \( \text{m}^3 \)
(b) 813 \( \text{m}^3 \)
(c) 308 \( \text{m}^3 \)
(d) 391 \( \text{m}^3 \)
Answer: (c) 308 \( \text{m}^3 \)
Explanation :
\( \frac{1232}{4} = 308 \text{ m}^3 \)

Question. The cost of canvas required for making the tent, if the canvas cost ` 70 per sq. m :
(a) Rs 40,000
(b) Rs 38570
(c) Rs 38575
(d) Rs 48470
Answer: (b) ` 38570
Explanation :
Cost \( = 551 \times 70 \)
\( = \text{` } 38570 \)

Question. To make the learning process more interesting, creative and innovative, Amayras' class teacher brings clay in the classroom, to teach the topic-surface areas and volumes. With clay, she forms a cylinder of radius \( 6 \text{ cm} \) and height \( 8 \text{ cm} \). Then she moulds the cylinder into a sphere and asks some questions to students.

Question. The radius of the sphere so formed is :
(a) \( 4 \text{ cm} \)
(b) \( 6 \text{ cm} \)
(c) \( 7 \text{ cm} \)
(d) \( 8 \text{ cm} \)
Answer: (b) \( 6 \text{ cm} \)
Explanation :
\( \frac{4}{3} \pi R^3 = \pi r^2 h \)
where \( R, r \) are the radii of sphere and cylinder respectively.
\( \Rightarrow R^3 = 6 \times 6 \times 8 \times \frac{3}{4} = (6)^3 \)
\( \Rightarrow R = 6 \text{ cm} \)
\( \therefore \) Radius of sphere = \( 6 \text{ cm} \)

Question. The volume of the sphere so formed is :
(a) \( 905.14 \text{ cm}^3 \)
(b) \( 903.27 \text{ cm}^3 \)
(c) \( 1296.5 \text{ cm}^3 \)
(d) \( 1156.63 \text{ cm}^3 \)
Answer: (a) \( 905.14 \text{ cm}^3 \)
Explanation :
Volume of sphere = \( \frac{4}{3} \pi R^3 \)
\( = \frac{4}{3} \times \frac{22}{7} \times 6 \times 6 \times 6 \)
\( = 905.14 \text{ cm}^3 \)

Question. Find the ratio of the volume of sphere to the volume of cylinder.
(a) \( 2 : 1 \)
(b) \( 1 : 2 \)
(c) \( 1 : 1 \)
(d) \( 3 : 1 \)
Answer: (c) \( 1 : 1 \)
Explanation :
\( \because \) Volume of sphere = Volume of cylinder
\( \therefore \) Required ratio = \( 1 : 1 \)

Question. Total surface area of the cylinder is :
(a) \( 528 \text{ cm}^2 \)
(b) \( 756 \text{ cm}^2 \)
(c) \( 625 \text{ cm}^2 \)
(d) \( 636 \text{ cm}^2 \)
Answer: (a) \( 528 \text{ cm}^2 \)
Explanation :
Total surface area of the cylinder
\( = 2\pi r(r + h) \)
\( = 2 \times \frac{22}{7} \times 6(6 + 8) = 2 \times \frac{22}{7} \times 6 \times 14 \)
\( = 528 \text{ cm}^2 \)

Question. During the conversion of a solid from one shape to another the volume of new shape will :
(a) Be increase
(b) Be decrease
(c) Remain unaltered
(d) Be double
Answer: (c) Remain unaltered
Explanation :
Volume will remain constant.

Question. Arun a X standard student makes a project on corona virus in science for an exhibition in his school. In this project, he picks a sphere which has volume \( 38808 \text{ cm}^3 \) and \( 11 \) cylindrical shapes, each of volume \( 1540 \text{ cm}^3 \) with length \( 10\text{cm} \).

Question. Diameter of the base of the cylinder is :
(a) \( 7 \text{ cm} \)
(b) \( 14 \text{ cm} \)
(c) \( 12 \text{ cm} \)
(d) \( 16 \text{ cm} \)
Answer: (b) \( 14 \text{ cm} \)
Explanation :
We know that,
Volume of cylinder = \( \pi r^2 h \)
\( \Rightarrow 1540 = \frac{22}{7} \times r^2 \times 10 \)
\( \Rightarrow 154 \times \frac{7}{22} = r^2 = 49 \)
\( \Rightarrow r = 7 \text{ cm} \)
\( \Rightarrow \) Diameter of the base of cylinder \( = 2r = 2 \times 7 = 14 \text{ cm} \)

Question. Diameter of the sphere is :
(a) \( 40 \text{ cm} \)
(b) \( 42 \text{ cm} \)
(c) \( 21 \text{ cm} \)
(d) \( 20 \text{ cm} \)
Answer: (b) \( 42 \text{ cm} \)
Explanation :
We know that,
Volume of sphere = \( \frac{4}{3} \pi r^3 \)
\( \Rightarrow 38808 = \frac{4}{3} \times \frac{22}{7} \times r^3 \)
\( \Rightarrow r^3 = 38808 \times 3 \times \frac{7}{88} \)
\( = 441 \times 21 = (21)^3 \)
\( \Rightarrow r = 21 \text{ cm} \)
\( \therefore \) Diameter of sphere = \( 42 \text{ cm} \)

Question. Total volume of the shape formed is :
(a) \( 85541 \text{ cm}^3 \)
(b) \( 45738 \text{ cm}^3 \)
(c) \( 24625 \text{ cm}^3 \)
(d) \( 55748 \text{ cm}^3 \)
Answer: (d) \( 55748 \text{ cm}^3 \)
Explanation :
Total volume of shape formed
= Volume of cylindrical shape + Volume of sphere
\( = 11 \times 1540 + 38808 \)
\( = 16940 + 38808 \)
\( = 55748 \text{ cm}^3 \)

Question. Curved surface area of the one cylindrical shape is :
(a) \( 850 \text{ cm}^2 \)
(b) \( 221 \text{ cm}^2 \)
(c) \( 440 \text{ cm}^2 \)
(d) \( 540 \text{ cm}^2 \)
Answer: (c) \( 440 \text{ cm}^2 \)
Explanation :
Curved surface area of one cylindrical shape
\( = 2\pi rh \)
\( = 2 \times \frac{22}{7} \times 7 \times 10 \)
\( = 440 \text{ cm}^2 \)

Question. Total area covered by cylindrical shapes on the surface of sphere is :
(a) \( 1694 \text{ cm}^2 \)
(b) \( 1580 \text{ cm}^2 \)
(c) \( 1896 \text{ cm}^2 \)
(d) \( 1740 \text{ cm}^2 \)
Answer: (a) \( 1694 \text{ cm}^2 \)
Explanation :
Area covered by cylindrical shapes on the surface of sphere
\( = 11 \times \pi r^2 \)
\( = 11 \times \frac{22}{7} \times 7 \times 7 \)
\( = 1694 \text{ cm}^2 \)

Question. Pinki's class teacher explained the students about the benefits of drinking fruit juice in the morning. So, Pinki went to a juice stall with her friend Bipin. On the stall, they observed that shopkeeper has three types of glasses of inner diameter \( 4.6 \text{ cm} \) to serve customers. The height of each glass is \( 11 \text{ cm} \). Seeing this, certain questions came into their mind. Help Pinki and Bipin to solve these questions.

Question. Volume of the type (A) glass is :
(a) \( 275.6 \text{ cm}^3 \)
(b) \( 250.6 \text{ cm}^3 \)
(c) \( 182.88 \text{ cm}^3 \)
(d) \( 208 \text{ cm}^3 \)
Answer: (c) \( 182.88 \text{ cm}^3 \)
Explanation :
Volume of type (A) glass
\( = \pi r^2 h \)
\( = \frac{22}{7} \times 2.3 \times 2.3 \times 11 \)
\( = 182.88 \text{ cm}^3 \)

Question. Volume of type (B) glass is :
(a) \( 208.6 \text{ cm}^3 \)
(b) \( 150.6 \text{ cm}^3 \)
(c) \( 152.4 \text{ cm}^3 \)
(d) \( 157.39 \text{ cm}^3 \)
Answer: (d) \( 157.39 \text{ cm}^3 \)
Explanation :
Volume of type (B) glass
= Volume of type (A) glass - Volume of hemisphere
\( = 182.88 - \frac{2}{3} \pi r^3 \)
\( = 182.88 - \frac{2}{3} \times \frac{22}{7} \times 2.3 \times 2.3 \times 2.3 \)
\( = 182.88 - 25.49 = 157.39 \text{ cm}^3 \)

Question. How much more juice can be filled in type (A) glass than glass of type (C)?
(a) \( 10.48 \text{ mL} \)
(b) \( 9.10 \text{ mL} \)
(c) \( 98.12 \text{ mL} \)
(d) \( 8.86 \text{ mL} \)
Answer: (d) \( 8.86 \text{ mL} \)
Explanation :
Volume of type (C) glass
= Volume of type (A) glass - Volume of cone
\( = 182.88 - \frac{1}{3} \pi r^2 h \)
\( = 182.88 - \frac{1}{3} \times \frac{22}{7} \times 2.3 \times 2.3 \times 1.6 \)
\( = 182.88 - 8.86 = 174.02 \text{ cm}^3 \)
\( \therefore \) Required difference
\( = 182.88 - 174.02 = 8.86 \text{ cm}^3 = 8.86 \text{ mL} \)

Question. Which glass has minimum capacity?
(a) Type (A)
(b) Type (B)
(c) Type (C)
(d) All glasses have same capacity
Answer: (b) Type (B)
Explanation :
Glass of type B has minimum capacity.

Question. Which mathematical concept has been used in above problem?
(a) Curved surface area
(b) Total surface area
(c) Volume
(d) None of the options
Answer: (c) Volume
Explanation :
Concept of volume is used in the question.

Question. A mathematics teacher took her grade X students to the Taj Mahal. It was an educational trip. She was interested in history also. On reaching there she told them about the history and facts about the seven wonder. She also told them that the structure of the monument is a combination of several solid figures. There are \( 4 \) pillars that are cylindrical in shape. A big dome in the center and \( 2 \) more small domes on both sides of the big dome on its side. The domes are hemispherical. The pillars also have domes on them.

Question. How much cloth material will be required to cover a big dome of a diameter of \( 7\text{m} \)?
(a) \( 77 \text{ m}^2 \)
(b) \( 78 \text{ m}^2 \)
(c) \( 79 \text{ m}^2 \)
(d) \( 80 \text{ m}^2 \)
Answer: (a) \( 77 \text{ m}^2 \)
Explanation :
\( 2\pi r^2 = 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
\( = 77 \text{ m}^2 \)

Question. Write the formula to calculate the volume of the pillar.
(a) \( \pi r^2 h + \pi r^3 \)
(b) \( \pi r^2 h + \frac{2}{3} \pi r^2 l \)
(c) \( \pi rl + \frac{2}{3} \pi r^3 \)
(d) \( \pi r^2 h + \frac{2}{3} \pi r^3 \)
Answer: (d) \( \pi r^2 h + \frac{2}{3} \pi r^3 \)
Explanation :
The formula to calculate the volume of pillar is \( \pi r^2 h + \frac{2}{3} \pi r^3 \).

Question. How much is the volume of the hemisphere if the radius of the base is \( 3 \text{ m} \)?
(a) \( 65.57 \text{ m}^3 \)
(b) \( 75.77 \text{ m}^3 \)
(c) \( 56.57 \text{ m}^3 \)
(d) \( 85.57 \text{ m}^3 \)
Answer: (c) \( 56.57 \text{ m}^3 \)
Explanation :
\( V = \frac{2}{3} \pi r^3 \)
\( = \frac{2}{3} \times \pi \times 3 \times 3 \times 3 \)
\( = 56.57 \text{ m}^3 \)

Question. What is the ratio of the sum of volumes of two-cylinder of radius \( 1 \text{ cm} \) and height \( 2 \text{ cm} \) each to the volume of a sphere of radius \( 3 \text{ cm} \)?
(a) \( 2 : 3 \)
(b) \( 3 : 2 \)
(c) \( 1 : 9 \)
(d) \( 1 : 2 \)
Answer: (c) \( 1 : 9 \)
Explanation :
Required ratio = \( \frac{2 \times \pi \times 1^2 \times 2}{\frac{4}{3} \pi \times 3^3} \)
\( = \frac{4 \times 3}{4 \times 27} = \frac{1}{9} \)