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Surface Areas and Volumes: 4 marks Questions (CBSE Class 10)

27 different 4 marks questions on Surface Areas and Volumes from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (54)2 marks (22)3 marks (17)4 marks (27)5 marks (41)

Q384 marksCase StudySurface Areas and Volumes CBSE 2026 · Basic 430/4/1

'Gilli Danda' is a very popular traditional game of India which is played with two wooden sticks – the larger one is called 'Danda' and smaller one 'Gilli'.
'Danda' – It is cylindrical in shape with diameter 4 cm and length 42 cm.
Gilli – It is cylindrical in middle with identical conical ends of same radius 1.5 cm and length 2.8 cm. The length of cylindrical part is 7 cm.
Based on the above, answer the following questions :
(i) Find the volume of wood used in making both the conical parts of Gilli. (1)
(ii) Find the volume of wood used in making cylindrical part of Gilli. (1)
(iii) (a) A cylindrical log of wood of radius 1.5 cm and length 14 cm is used to make Gilli. Find the volume of the wood scrapped. (2)
OR
(b) Find the total surface area of 'Danda'. (2)

Diagram for CBSE 2026 Class 10 Maths question 38

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Answer: (i) 13.2

c m^{3}

(ii) 49.5

c m^{3}

(iii) (a) 36.3

c m^{3}

(b)

\frac{3872}{7} \approx 553.14

c m^{2}

(i) Each cone: $\frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 1.5 \times 1.5 \times 2.8 = 6.6$ $c m^{3}$ ; both cones = 13.2 $c m^{3}$ .
(ii) Cylinder: $π r^{2} h = \frac{22}{7} \times 1.5 \times 1.5 \times 7 = 49.5$ $c m^{3}$ .
(iii) (a) Volume of log = $\frac{22}{7} \times 1.5 \times 1.5 \times 14 = 99$ $c m^{3}$ .
Volume of Gilli = 13.2 + 49.5 = 62.7 $c m^{3}$ .
Wood scrapped = 99 – 62.7 = 36.3 $c m^{3}$ .
(iii) (b) Danda: r = 2 cm, h = 42 cm. TSA = $2 π r (h + r) = 2 \times \frac{22}{7} \times 2 \times 44 = \frac{3872}{7} \approx 553.14$ $c m^{2}$ .

Also asked in: 2026 Basic 430/4/2, 2026 Basic 430/4/3

Q374 marksCase StudySurface Areas and Volumes CBSE 2026 · Basic 430/5/1

There are many varieties of mushrooms available in the world. One such mushroom ‘Amanita muscaria’ has a upper part which is like red cap (hemispherical) and lower part is like white stem (cylinderical).
The hemispherical cap’s radius = 3 cm and cylindrical stem is 2 cm high with diameter 1.4 cm. Considering mushroom a solid object, answer the following questions :
(i) What is the total height of a mushroom ?
(ii) Find the volume of the stem.
(iii) (a) Determine the volume of 7 such mushrooms.
OR (b) Find the total surface area of 7 such mushrooms.

Diagram for CBSE 2026 Class 10 Maths question 37

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Answer: (i) 5 cm (ii) 3.08 cm

^{3}

(iii) (a) 417.56 cm

^{3}

OR (b) 655.6 cm

^{2}

(i) Total height = radius of cap + height of stem = 3 + 2 = 5 cm.
(ii) Stem radius = 0.7 cm. Volume $= π r^{2} h = \frac{22}{7} \times 0.7 \times 0.7 \times 2 = 3.08$ cm $^{3}$
(iii)(a) Volume of cap $= \frac{2}{3} π R^{3} = \frac{2}{3} \times \frac{22}{7} \times 27 = \frac{396}{7}$ cm $^{3}$
Volume of 7 mushrooms $= 7 \times (\frac{396}{7} + 3.08) = 396 + 21.56 = 417.56$ cm $^{3}$
(iii)(b) Surface of one mushroom = curved surface of cap $(2 π R^{2})$ + flat underside of cap not covered by stem $(π R^{2} - π r^{2})$ + curved surface of stem $(2 π r h)$ + base of stem $(π r^{2})$ $= 3 π R^{2} + 2 π r h$
$= 3 \times \frac{22}{7} \times 9 + 2 \times \frac{22}{7} \times 0.7 \times 2 = \frac{594}{7} + 8.8$ cm $^{2}$
For 7 mushrooms: $594 + 61.6 = 655.6$ cm $^{2}$

Also asked in: 2026 Basic 430/5/2, 2026 Basic 430/5/3

Q384 marksCase StudySurface Areas and Volumes CBSE 2026 · Standard 30/2/1

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 a strawberry ice-cream for you.
Based on the information given above, answer the following questions :
(i) Find the length of the diagonal of Rubik's cube if each edge measures 6 cm. (1)
(ii) Find the volume of Rubik's cube if the length of the edge is 7 cm. (1)
(iii) (a) What is the curved surface area of hemisphere (ice-cream) if the base radius is 7 cm ? (2)
OR
(iii) (b) If two cubes of edges 4 cm are joined end-to-end, then find the surface area of the resulting cuboid. (2)

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Answer: (i)

63

cm (ii) 343

c m^{3}

(iii) (a) 308

c m^{2}

OR (b) 160

c m^{2}

(i) Diagonal of a cube $= 3 a = 63$ cm
(ii) Volume $= a^{3} = 7^{3} = 343$ $c m^{3}$
(iii)(a) CSA of hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times 7 \times 7 = 308$ $c m^{2}$
(iii)(b) Resulting cuboid is 8 cm × 4 cm × 4 cm.
Surface area $= 2 (8 \times 4 + 4 \times 4 + 4 \times 8) = 2 (32 + 16 + 32) = 160$ $c m^{2}$

Also asked in: 2026 Standard 30/2/2, 2026 Standard 30/2/3

Q384 marksCase StudySurface Areas and Volumes CBSE 2026 · Standard 30/3/1

A model of Leafy Ball Fountain is made to be kept on the tabletop. Water gently cascades down the ball into a decorative cylindrical pool where it is recycled.
The diameter of spherical ball is 21 cm.
Cylindrical pool – Outer diameter is 50 cm and inner diameter is 40 cm.
Height of solid base is 14 cm.
Height of water filled is 7 cm.
Observe the figure and answer the following questions :
(i) Determine the total height of the fountain. (1)
(ii) Find the volume of the ball. (1)
(iii) (a) If one-third of the ball is submerged in the water, find the volume of the water filled in the pool. (2)
OR (iii) (b) Find the sum of the outer curved surface area of the cylindrical part and surface area of the ball. (2)

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Answer: (i) 35 cm (ii) 4851 cm

^{3}

(iii) (a) 7183 cm

^{3}

; OR (b) 4686 cm

^{2}

(i) The ball rests on the solid base: total height $= 14 + 21 = 35$ cm
(ii) $r = 10.5$ cm: $V = \frac{4}{3} \times \frac{22}{7} \times 10. 5^{3} = 4851$ cm $^{3}$
(iii) (a) Water region: inner radius 20 cm, height 7 cm: $\frac{22}{7} \times 2 0^{2} \times 7 = 8800$ cm $^{3}$
Submerged part of ball $= \frac{1}{3} \times 4851 = 1617$ cm $^{3}$
Water $= 8800 - 1617 = 7183$ cm $^{3}$
(iii) (b) Outer radius 25 cm, height of cylindrical part $= 14 + 7 = 21$ cm
Outer CSA $= 2 \times \frac{22}{7} \times 25 \times 21 = 3300$ cm $^{2}$
Surface area of ball $= 4 \times \frac{22}{7} \times 10. 5^{2} = 1386$ cm $^{2}$
Sum $= 3300 + 1386 = 4686$ cm $^{2}$

Also asked in: 2026 Standard 30/3/2, 2026 Standard 30/3/3

Q374 marksCase StudySurface Areas and Volumes CBSE 2026 · Standard 30/5/1

A wall mounted lamp, made of fabric, is shown below. Lamp has cuboidal shape, open from top and bottom. A spherical bulb of diameter 7 cm is latched with a very thin rod. (Ignore the rod while making calculations.)
Dimensions of the cuboid are 24 cm × 12 cm × 17 cm.
(i) Find the surface area of the bulb. (1)
(ii) What could be the maximum diameter of the bulb if at least 1 cm space is left from each side ? (1)
(iii) (a) Find the area of the fabric used if there is a fold of 2 cm on top and bottom edges. (2)
OR (iii) (b) Find the space available inside the lamp. (2)

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Answer: (i) 154

c m^{2}

(ii) 10 cm (iii) (a) 1512

c m^{2}

OR (iii) (b) 4896

c m^{3}

(lamp volume); 4716.33

c m^{3}

after excluding the bulb

(i) r = 3.5 cm. Surface area $= 4 π r^{2} = 4 \times \frac{22}{7} \times 3.5 \times 3.5 = 154$ $c m^{2}$
(ii) The smallest dimension of the lamp is 12 cm. Leaving 1 cm on each side, maximum diameter = 12 − 2 = 10 cm.
(iii) (a) The fabric covers the four vertical faces (open top and bottom). Height with folds = 17 + 2 + 2 = 21 cm.
Area $= 2 (24 + 12) \times 21 = 72 \times 21 = 1512$ $c m^{2}$
(iii) (b) Volume of the lamp $= 24 \times 12 \times 17 = 4896$ $c m^{3}$
Volume of bulb $= \frac{4}{3} \times \frac{22}{7} \times (3.5)^{3} = \frac{539}{3} \approx 179.67$ $c m^{3}$
Space available (excluding the bulb) $= 4896 - 179.67 = 4716.33$ $c m^{3}$

Also asked in: 2026 Standard 30/5/2, 2026 Standard 30/5/3

Q374 marksCase StudySurface Areas and Volumes CBSE 2025 · Basic 430/4/1

Playing in a ball pool is good entertainment for kids. Suhana bought 600 new balls of diameter 7 cm to fill in the pool for her kids. The cuboidal box containing 600 balls has dimensions 42 cm × 91 cm × 50 cm ( $l \times b \times h$ ).
Based on above information, answer the following questions :
(i) Find the volume of one ball. (1)
(ii) 10 balls are painted with neon colours. Determine the area of painted surface. (1)
(iii) (a) Find the volume of empty space in the box. (2)
OR
(b) The lowermost layer of the balls covers the base of the box edge to edge when balls are placed evenly adjacent to each other. (A) How much area is covered by one ball? (B) How many balls are there in lowermost layer? (2)

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Answer: (i)

\frac{539}{3} \approx 179.67

^{3}

(ii) 1540 cm

^{2}

(iii) (a) 83300 cm

^{3}

OR (b) (A) 49 cm

^{2}

(B) 78

Radius of a ball $r = 3.5$ cm.
(i) Volume $= \frac{4}{3} \times \frac{22}{7} \times 3. 5^{3} = \frac{539}{3} \approx 179.67$ cm $^{3}$ .
(ii) Surface area of one ball $= 4 \times \frac{22}{7} \times 3. 5^{2} = 154$ cm $^{2}$ ; for 10 balls $= 1540$ cm $^{2}$ .
(iii) (a) Volume of box $= 42 \times 91 \times 50 = 191100$ cm $^{3}$ . Volume of 600 balls $= 600 \times \frac{539}{3} = 107800$ cm $^{3}$ . Empty space $= 191100 - 107800 = 83300$ cm $^{3}$ .
(iii) (b) (A) Each ball occupies a square of side equal to its diameter: $7 \times 7 = 49$ cm $^{2}$ . (B) Number of balls $= \frac{42}{7} \times \frac{91}{7} = 6 \times 13 = 78$ .

Also asked in: 2025 Basic 430/4/2, 2025 Basic 430/4/3

Q374 marksCase StudySurface Areas and Volumes CBSE 2025 · Basic 430/6/1

A hemispherical bowl is packed in a cuboidal box. The bowl just fits in the box. Inner radius of the bowl is 10 cm. Outer radius of the bowl is 10.5 cm.
Based on the above, answer the following questions :
(i) Find the dimensions of the cuboidal box. (1)
(ii) Find the total outer surface area of the box. (1)
(iii) (a) Find the difference between the capacity of the bowl and the volume of the box. (use $π = 3.14$ ) (2)
OR
(iii) (b) The inner surface of the bowl and the thickness is to be painted. Find the area to be painted. (2)

Diagram for CBSE 2025 Class 10 Maths question 37

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Answer: (i) 21 cm

\times

21 cm

\times

10.5 cm (ii) 1764 cm

^{2}

(iii) (a) about 2537.17 cm

^{3}

OR (b) about 660.79 cm

^{2}

(i) The bowl just fits, so length = breadth = outer diameter = 21 cm and height = outer radius = 10.5 cm.
(ii) TSA $= 2 (l b + bh + h l) = 2 (21 \times 21 + 21 \times 10.5 + 10.5 \times 21) = 2 (441 + 220.5 + 220.5) = 1764$ cm $^{2}$ .
(iii) (a) Capacity of bowl $= \frac{2}{3} π r^{3} = \frac{2}{3} \times 3.14 \times 1000 \approx 2093.33$ cm $^{3}$ .
Volume of box $= 21 \times 21 \times 10.5 = 4630.5$ cm $^{3}$ .
Difference $= 4630.5 - 2093.33 \approx 2537.17$ cm $^{3}$ .
(iii) (b) Inner curved surface $= 2 π r^{2} = 2 \times \frac{22}{7} \times 100 = \frac{4400}{7}$ cm $^{2}$ .
Rim (thickness) $= π (R^{2} - r^{2}) = \frac{22}{7} (110.25 - 100) = \frac{225.5}{7}$ cm $^{2}$ .
Area to be painted $= \frac{4625.5}{7} \approx 660.79$ cm $^{2}$ .

Also asked in: 2025 Basic 430/6/2, 2025 Basic 430/6/3

Q384 marksCase StudySurface Areas and Volumes CBSE 2025 · Standard 30/2/1

A skilled carpenter decided to craft a special rolling pin for the local baker. He carefully joined three cylindrical pieces of wood – two small ones on the ends and one larger in the centre to create a perfect tool. The baker loved the rolling pin, as it rolled out the smoothest dough for breads and pastries.
The length of the bigger cylindrical part is 12 cm and diameter is 7 cm and the length of each smaller cylindrical part is 5 cm and diameter is 2.1 cm.
Based on the above information, answer the following questions :
(i) Find the volume of the bigger cylindrical part. (1)
(ii) Find the curved surface area of the bigger cylindrical part. (1)
(iii) (a) Find the ratio of the volume of the bigger cylindrical part to the total volume of the two smaller (identical) cylindrical parts. (2)
OR (iii) (b) Find the sum of the curved surface areas of the two identical smaller cylindrical parts. (2)

Diagram for CBSE 2025 Class 10 Maths question 38

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Answer: (i) 462 cm

^{3}

(ii) 264 cm

^{2}

(iii) (a)

40 : 3

; OR (b) 66 cm

^{2}

(i) Bigger part: $r = 3.5$ cm, $h = 12$ cm; $V = \frac{22}{7} \times 3.5 \times 3.5 \times 12 = 462$ cm $^{3}$
(ii) CSA $= 2 \times \frac{22}{7} \times 3.5 \times 12 = 264$ cm $^{2}$
(iii) (a) Each smaller part: $r = 1.05$ cm, $h = 5$ cm; $V = \frac{22}{7} \times 1.05 \times 1.05 \times 5 = 17.325$ cm $^{3}$
Two smaller parts: $34.65$ cm $^{3}$
Ratio $= 462 : 34.65 = 40 : 3$
(iii) (b) CSA of one smaller part $= 2 \times \frac{22}{7} \times 1.05 \times 5 = 33$ cm $^{2}$
Sum for two parts $= 66$ cm $^{2}$

Also asked in: 2025 Standard 30/2/2, 2025 Standard 30/2/3

Q374 marksCase StudySurface Areas and Volumes CBSE 2024 · Standard 30/2/1

The word ‘circus’ has the same root as ‘circle’. In a closed circular area, various entertainment acts including human skill and animal training are presented before the crowd.
A circus tent is cylindrical upto a height of 8 m and conical above it. The diameter of the base is 28 m and total height of tent is 18.5 m.
Based on the above, answer the following questions :
(i) Find slant height of the conical part. (1)
(ii) Determine the floor area of the tent. (1)
(iii) (a) Find area of the cloth used for making tent. (2)
OR
(b) Find total volume of air inside an empty tent. (2)

Show answer & solution

Answer: (i) 17.5 m (ii) 616 m

^{2}

(iii) (a) 1474 m

^{2}

OR (b) 7084 m

^{3}

Radius $r = 14$ m, cylinder height $h = 8$ m, cone height $H = 18.5 - 8 = 10.5$ m.
(i) $l = r^{2} + H^{2} = 196 + 110.25 = 306.25 = 17.5$ m
(ii) Floor area $= π r^{2} = \frac{22}{7} \times 14 \times 14 = 616$ m $^{2}$
(iii) (a) Cloth = CSA of cylinder + CSA of cone $= 2 π r h + π r l = π r (2 h + l) = \frac{22}{7} \times 14 \times (16 + 17.5) = 44 \times 33.5 = 1474$ m $^{2}$
OR
(b) Volume $= π r^{2} h + \frac{1}{3} π r^{2} H = 616 \times 8 + \frac{1}{3} \times 616 \times 10.5 = 4928 + 2156 = 7084$ m $^{3}$

Also asked in: 2024 Standard 30/2/2, 2024 Standard 30/2/3

Q384 marksCase StudySurface Areas and Volumes CBSE 2024 · Standard 30/4/1

Tamper-proof tetra-packed milk guarantees both freshness and security. This milk ensures uncompromised quality, preserving the nutritional values within and making it a reliable choice for health-conscious individuals.
500 mL milk is packed in a cuboidal container of dimensions 15 cm $\times$ 8 cm $\times$ 5 cm. These milk packets are then packed in cuboidal cartons of dimensions 30 cm $\times$ 32 cm $\times$ 15 cm.
Based on the above given information, answer the following questions :
(i) Find the volume of the cuboidal carton. (1)
(ii) Find the total surface area of a milk packet. (2)
OR How many milk packets can be filled in a carton ? (2)
(iii) How much milk can the cup (as shown in the figure) hold ? (1)

Diagram for CBSE 2024 Class 10 Maths question 38

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Answer: (i) 14400 cm

^{3}

(ii) 470 cm

^{2}

; OR 24 packets (iii) 550 cm

^{3}

, i.e. 550 mL

(i) Volume of carton = $30 \times 32 \times 15 = 14400$ cm $^{3}$ .
(ii) TSA of a packet = $2 (l b + bh + h l) = 2 (15 \times 8 + 8 \times 5 + 5 \times 15) = 2 (120 + 40 + 75) = 470$ cm $^{2}$ .
OR: Volume of a packet = $15 \times 8 \times 5 = 600$ cm $^{3}$ ; number of packets = $\frac{14400}{600} = 24$ (they fit exactly: $\frac{30}{15} \times \frac{32}{8} \times \frac{15}{5} = 2 \times 4 \times 3 = 24$ ).
(iii) The cup is a cylinder of radius 5 cm (marked from the centre of the base) and height 7 cm.
Volume = $π r^{2} h = \frac{22}{7} \times 5 \times 5 \times 7 = 550$ cm $^{3}$ = 550 mL.

Also asked in: 2024 Standard 30/4/2, 2024 Standard 30/4/3

Q384 marksCase StudySurface Areas and Volumes CBSE 2023 · Basic 430/2/1

A wooden toy is shown in the picture. This is a cuboidal wooden block of dimensions 14 cm $\times$ 17 cm $\times$ 4 cm. On its top there are seven cylindrical hollows for bees to fit in. Each cylindrical hollow is of height 3 cm and radius 2 cm.
Based on the above, answer the following questions :
(i) Find the volume of wood carved out to make one cylindrical hollow. (1)
(ii) Find the lateral surface area of the cuboid to paint it with green colour. (1)
(iii) (a) Find the volume of wood in the remaining cuboid after carving out seven cylindrical hollows. (2)
OR (iii) (b) Find the surface area of the top surface of the cuboid to be painted yellow. (2)

Show answer & solution

Answer: (i)

\frac{264}{7}

^{3}

≈ 37.71 cm

^{3}

(ii) 248 cm

^{2}

(iii) (a) 688 cm

^{3}

OR (iii) (b) 150 cm

^{2}

(i) Volume of one hollow $= π r^{2} h = \frac{22}{7} \times 2^{2} \times 3 = \frac{264}{7} \approx 37.71$ cm $^{3}$
(ii) Lateral surface area $= 2 h (l + b) = 2 \times 4 \times (14 + 17) = 248$ cm $^{2}$
(iii) (a) Volume of cuboid $= 14 \times 17 \times 4 = 952$ cm $^{3}$
Wood removed $= 7 \times \frac{264}{7} = 264$ cm $^{3}$ , so remaining wood $= 952 - 264 = 688$ cm $^{3}$
(iii) (b) Area of top face $= 14 \times 17 = 238$ cm $^{2}$
Area of the 7 circular openings $= 7 \times \frac{22}{7} \times 2^{2} = 88$ cm $^{2}$
Area to be painted yellow $= 238 - 88 = 150$ cm $^{2}$

Also asked in: 2023 Basic 430/2/2, 2023 Basic 430/2/3

Q374 marksCase StudySurface Areas and Volumes CBSE 2023 · Basic 430/5/1

Singing bowls (hemispherical in shape) are commonly used in sound healing practices. Mallet (cylindrical in shape) is used to strike the bowl in a sequence to produce sound and vibration.
One such bowl is shown here whose dimensions are :
Hemispherical bowl has outer radius 6 cm and inner radius 5 cm.
Mallet has height of 10 cm and radius 2 cm.
Based on the above, answer the following questions :
(i) What is the volume of the material used in making the mallet ? (1)
(ii) The bowl is to be polished from inside. Find the inner surface area of the bowl. (1)
(iii) (a) Find the volume of metal used to make the bowl. (2)
OR
(iii) (b) Find total surface area of the mallet. (Use $π = 3.14$ ) (2)

Show answer & solution

Answer: (i)

\frac{880}{7} \approx 125.71

^{3}

(ii)

\frac{1100}{7} \approx 157.14

^{2}

(iii) (a)

\frac{572}{3} \approx 190.67

^{3}

OR (b) 150.72 cm

^{2}

(i) Volume of mallet $= π r^{2} h = \frac{22}{7} \times 2^{2} \times 10 = \frac{880}{7} \approx 125.71$ cm $^{3}$ .
(ii) Inner surface area $= 2 π r^{2} = 2 \times \frac{22}{7} \times 5^{2} = \frac{1100}{7} \approx 157.14$ cm $^{2}$ .
(iii) (a) Volume of metal $= \frac{2}{3} π (R^{3} - r^{3}) = \frac{2}{3} \times \frac{22}{7} \times (216 - 125) = \frac{2}{3} \times \frac{22}{7} \times 91 = \frac{572}{3} \approx 190.67$ cm $^{3}$ .
(iii) (b) TSA of mallet $= 2 π r (h + r) = 2 \times 3.14 \times 2 \times (10 + 2) = 150.72$ cm $^{2}$ .

Also asked in: 2023 Basic 430/5/2, 2023 Basic 430/5/3

Q374 marksCase StudySurface Areas and Volumes CBSE 2023 · Standard 30/2/1

In a coffee shop, coffee is served in two types of cups. One is cylindrical in shape with diameter 7 cm and height 14 cm and the other is hemispherical with diameter 21 cm.
Based on the above, answer the following questions :
(i) Find the area of the base of the cylindrical cup. (1)
(ii) (a) What is the capacity of the hemispherical cup ? (2)
OR (ii) (b) Find the capacity of the cylindrical cup. (2)
(iii) What is the curved surface area of the cylindrical cup ? (1)

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Answer: (i) 38.5 cm² (ii) (a) 2425.5 cm³ OR (b) 539 cm³ (iii) 308 cm²

(i) $r = 3.5$ cm; base area $= \frac{22}{7} \times 3.5 \times 3.5 = 38.5$ cm²
(ii)(a) $R = 10.5$ cm; capacity $= \frac{2}{3} \times \frac{22}{7} \times 10. 5^{3} = 2425.5$ cm³
(ii)(b) Capacity $= 38.5 \times 14 = 539$ cm³
(iii) CSA $= 2 \times \frac{22}{7} \times 3.5 \times 14 = 308$ cm²

Also asked in: 2023 Standard 30/2/2, 2023 Standard 30/2/3

Q364 marksCase StudySurface Areas and Volumes CBSE 2023 · Standard 30/5/1

A golf ball is spherical with about 300 – 500 dimples that help increase its velocity while in play. Golf balls are traditionally white but available in colours also. In the given figure, a golf ball has diameter 4.2 cm and the surface has 315 dimples (hemi-spherical) of radius 2 mm.
Based on the above, answer the following questions :
(i) Find the surface area of one such dimple. (1)
(ii) Find the volume of the material dug out to make one dimple. (1)
(iii) Find the total surface area exposed to the surroundings. (2)
OR (iii) Find the volume of the golf ball. (2)

Diagram for CBSE 2023 Class 10 Maths question 36

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Answer: (i) 0.2514 cm

^{2}

(≈ 25.14 mm

^{2}

) (ii) 0.01676 cm

^{3}

(≈ 16.76 mm

^{3}

) (iii) 95.04 cm

^{2}

; OR: 33.528 cm

^{3}

Dimple radius r = 2 mm = 0.2 cm; ball radius R = 2.1 cm.
(i) Curved surface of a hemispherical dimple $= 2 π r^{2} = 2 \times \frac{22}{7} \times 0.04 = 0.2514$ cm $^{2}$ (approx.).
(ii) Volume dug out $= \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 0.008 = 0.01676$ cm $^{3}$ (approx.).
(iii) Surface of sphere $= 4 π R^{2} = 4 \times \frac{22}{7} \times 4.41 = 55.44$ cm $^{2}$ .
Each dimple removes a circle $π r^{2}$ and adds $2 π r^{2}$ , a net gain of $π r^{2}$ .
Total exposed area $= 55.44 + 315 \times \frac{22}{7} \times 0.04 = 55.44 + 39.6 = 95.04$ cm $^{2}$ .
OR: Volume of sphere $= \frac{4}{3} \times \frac{22}{7} \times 2. 1^{3} = 38.808$ cm $^{3}$ .
Volume removed by dimples $= 315 \times \frac{2}{3} \times \frac{22}{7} \times 0.008 = 5.28$ cm $^{3}$ .
Volume of golf ball = 38.808 – 5.28 = 33.528 cm $^{3}$ .

Also asked in: 2023 Standard 30/5/2, 2023 Standard 30/5/3

Q114 marksLong AnswerSurface Areas and Volumes CBSE 2022 · Basic 430/1/1

A spherical glass vessel has a cylindrical neck 8 cm long and 1 cm in radius. The radius of the spherical part is 9 cm. Find the amount of water (in litres) it can hold, when filled completely.

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Answer: 3.08 litres

Volume of spherical part $= \frac{4}{3} π (9)^{3} = 972 π$ cm $^{3}$
Volume of neck $= π (1)^{2} (8) = 8 π$ cm $^{3}$
Total volume $= 980 π = 980 \times \frac{22}{7} = 3080$ cm $^{3}$
$= \frac{3080}{1000} = 3.08$ litres

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q11 (OR) (OR)4 marksLong AnswerSurface Areas and Volumes CBSE 2022 · Basic 430/1/1

From a solid cylinder, whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.

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Answer: 17.6 cm

^{2}

$r = 0.7$ cm, $h = 2.4$ cm
Slant height $l = 2. 4^{2} + 0. 7^{2} = 6.25 = 2.5$ cm
TSA = curved surface of cylinder + top base + curved surface of cone
$= 2 π r h + π r^{2} + π r l = π r (2 h + r + l)$
$= \frac{22}{7} \times 0.7 \times (4.8 + 0.7 + 2.5) = 2.2 \times 8 = 17.6$ cm $^{2}$

Also asked in: 2022 Basic 430/1/2, 2022 Basic 430/1/3

Q144 marksCase StudySurface Areas and Volumes CBSE 2022 · Basic 430/2/1

A solid cuboidal toy is made of wood. It has five cone shaped cavities to hold toy carrots.
The dimensions of the toy are cuboid – 10 cm × 10 cm × 8 cm.
Each cone carved out – Radius = 2.1 cm and Height = 6 cm.
(1) Find the volume of wood carved out to make five conical cavities. (2)
(2) Find the volume of the wood in the final product. (2)

Diagram for CBSE 2022 Class 10 Maths question 14

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Answer: (1) 138.6 cm³ (2) 661.4 cm³

(1) Volume of one cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 6 = 27.72$ cm³.
Volume of five cones $= 5 \times 27.72 = 138.6$ cm³.
(2) Volume of cuboid $= 10 \times 10 \times 8 = 800$ cm³.
Volume of wood left $= 800 - 138.6 = 661.4$ cm³.

Also asked in: 2022 Basic 430/2/2, 2022 Basic 430/2/3

Q144 marksCase StudySurface Areas and Volumes CBSE 2022 · Basic 430/3/1

The technique of Rainwater harvesting through Recharge pit is very useful. Rainwater is collected on the roof and then flowing through the Recharge pit it goes to the ground. Observe the picture given below :
B (BREADTH) = 3 m
D (DEPTH = 2 m
L (LENGTH) = 3 m
The surface area of the roof floor is 100 m $^{2}$ . The cuboidal pit measures 3 m $\times$ 3 m $\times$ 2 m.
(a) Water standing on the roof is released into the cuboidal pit. If the cuboidal pit is filled completely by the roof water, then find the height of standing water on the roof. (2)
(b) Instead of a cuboidal pit, if a cylindrical pit with diameter 3 m and height 2 m had been built, then which tank would hold more water ? (2)

Show answer & solution

Answer: (a) 0.18 m (18 cm) (b) The cuboidal pit (18 m

^{3}

) holds more water than the cylindrical pit (

\approx 14.14

^{3}

)

(a) Volume of cuboidal pit $= 3 \times 3 \times 2 = 18$ m $^{3}$
Volume of water on roof $= 100 \times h$ ; $100 h = 18 \Rightarrow h = 0.18$ m $= 18$ cm
(b) Cylinder: $r = 1.5$ m, $h = 2$ m; volume $= π r^{2} h = \frac{22}{7} \times 2.25 \times 2 \approx 14.14$ m $^{3}$
Since $18 > 14.14$ , the cuboidal pit would hold more water.

Also asked in: 2022 Basic 430/3/2, 2022 Basic 430/3/3

Q144 marksCase StudySurface Areas and Volumes CBSE 2022 · Basic 430/4/1

A company deals in casting and moulding of metal on orders received from its clients.
In one such order, company is supposed to make 50 toys in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of hemisphere. If the radius of the base of the cone is 21 cm and height is 28 cm,
(a) find the volume of 50 toys; (2)
(b) find the ratio of the volume of hemisphere to the volume of cone. (2)

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Answer: (a) 1617000 cm

^{3}

(b) 3 : 2

(a) $r = 21$ cm, $h = 28$ cm. Volume of one toy $= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} (2 r + h)$ .
$= \frac{1}{3} \times \frac{22}{7} \times 441 \times 70 = 32340 cm^{3}$ .
Volume of 50 toys $= 50 \times 32340 = 1617000 cm^{3}$ .
(b) $\frac{\frac{2}{3} π r ^{3}}{\frac{1}{3} π r ^{2} h} = \frac{2 r}{h} = \frac{42}{28} = \frac{3}{2}$ , so the ratio is 3 : 2.

Q144 marksCase StudySurface Areas and Volumes CBSE 2022 · Basic 430/4/2

A company deals in casting and moulding of metal on orders received from its clients.
In one such order, company is supposed to make 50 toys in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of hemisphere. If the radius of the base of the cone is 21 cm and height is 28 cm, then
(a) find the volume of 50 toys; (2)
(b) find the ratio of the volume of hemisphere to the volume of cone. (2)

Show answer & solution

Answer: (a) 1617000 cm

^{3}

(b) 3 : 2

(a) $r = 21$ cm, $h = 28$ cm. Volume of one toy $= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} (2 r + h)$ .
$= \frac{1}{3} \times \frac{22}{7} \times 441 \times 70 = 32340 cm^{3}$ .
Volume of 50 toys $= 50 \times 32340 = 1617000 cm^{3}$ .
(b) $\frac{\frac{2}{3} π r ^{3}}{\frac{1}{3} π r ^{2} h} = \frac{2 r}{h} = \frac{42}{28} = \frac{3}{2}$ , so the ratio is 3 : 2.

Also asked in: 2022 Basic 430/4/3

Q114 marksLong AnswerSurface Areas and Volumes CBSE 2022 · Standard 30/1/1

From a solid cylinder of height 30 cm and radius 7 cm, a conical cavity of height 24 cm and same radius is hollowed out. Find the total surface area of the remaining solid.

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Answer: 2024 cm²

Slant height of cone $l = 2 4^{2} + 7^{2} = 625 = 25$ cm.
TSA = curved surface of cylinder + area of base + curved surface of cone.
Curved surface of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 7 \times 30 = 1320$ cm².
Base $= π r^{2} = \frac{22}{7} \times 7 \times 7 = 154$ cm².
Curved surface of cone $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm².
TSA $= 1320 + 154 + 550 = 2024$ cm².

Also asked in: 2022 Standard 30/1/2

Q11 (OR) (OR)4 marksLong AnswerSurface Areas and Volumes CBSE 2022 · Standard 30/1/1Not in current syllabus

Water in a canal, 8 m wide and 6 m deep, is flowing with a speed of 12 km/hour. How much area will it irrigate in one hour, if 0.05 m of standing water is required ?

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Answer: 11520000 m² (1152 hectares)

In one hour the water travels 12 km = 12000 m.
Volume flowing in one hour $= 8 \times 6 \times 12000 = 576000$ m³.
Area irrigated $\times 0.05 = 576000$
Area $= \frac{576000}{0.05} = 11520000$ m² $= 1152$ hectares.

Also asked in: 2022 Standard 30/1/2

Q124 marksLong AnswerSurface Areas and Volumes CBSE 2022 · Standard 30/1/3

In Figure 2, from a solid cube of side 7 cm, a cylinder of radius 2.1 cm and height 7 cm is scooped out. Find the total surface area of the remaining solid.

Diagram for CBSE 2022 Class 10 Maths question 12

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Answer: 358.68 cm²

TSA = surface of cube − 2 circular ends + curved surface of cylinder.
Surface of cube $= 6 \times 7^{2} = 294$ cm².
Two circular ends $= 2 π r^{2} = 2 \times \frac{22}{7} \times 2.1 \times 2.1 = 27.72$ cm².
Curved surface of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 2.1 \times 7 = 92.4$ cm².
TSA $= 294 - 27.72 + 92.4 = 358.68$ cm².

Q12 (OR) (OR)4 marksLong AnswerSurface Areas and Volumes CBSE 2022 · Standard 30/1/3Not in current syllabus

A well of diameter 5 m is dug 24 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 3 m to form an embankment. Find the height of the embankment.

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Answer: 6.25 m

Volume of earth dug out $= π (2.5)^{2} \times 24 = 150 π$ m³.
Embankment: inner radius 2.5 m, outer radius $2.5 + 3 = 5.5$ m.
Area of ring $= π (5. 5^{2} - 2. 5^{2}) = π (30.25 - 6.25) = 24 π$ m².
Height $= \frac{150 π}{24 π} = 6.25$ m.

Q144 marksCase StudySurface Areas and Volumes CBSE 2022 · Standard 30/2/1

A ‘circus’ is a company of performers who put on shows of acrobats, clowns etc. to entertain people started around 250 years back, in open fields, now generally performed in tents.
One such ‘Circus Tent’ is shown below.
The tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of cylindrical part are 9 m and 30 m respectively and height of conical part is 8 m with same diameter as that of the cylindrical part, then find
(1) the area of the canvas used in making the tent; (3)
(2) the cost of the canvas bought for the tent at the rate ₹ 200 per sq m, if 30 sq m canvas was wasted during stitching. (1)

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Answer: (1) 1650 sq m (2) ₹ 3,36,000

(1) Radius $r = 15$ m, cylinder height $h = 9$ m, cone height 8 m.
Slant height $l = 1 5^{2} + 8^{2} = 17$ m.
Canvas = curved surface of cylinder + curved surface of cone $= 2 π r h + π r l = π r (2 h + l)$ .
$= \frac{22}{7} \times 15 \times (18 + 17) = \frac{22}{7} \times 15 \times 35 = 1650$ sq m.
(2) Canvas bought $= 1650 + 30 = 1680$ sq m.
Cost $= 1680 \times 200 = ₹ 3, 36, 000$ .

Also asked in: 2022 Standard 30/2/2, 2022 Standard 30/2/3

Q144 marksCase StudySurface Areas and Volumes CBSE 2022 · Standard 30/3/1

John planned a birthday party for his younger sister with his friends. They decided to make some birthday caps by themselves and to buy a cake from a bakery shop. For these two items, they decided the following dimensions :
Cake : Cylindrical shape with diameter 24 cm and height 14 cm.
Cap : Conical shape with base circumference 44 cm and height 24 cm.
Based on the above information, answer the following questions :
(a) How many square cm paper would be used to make 4 such caps ? (2)
(b) The bakery shop sells cakes by weight (0.5 kg, 1 kg, 1.5 kg, etc.). To have the required dimensions, how much cake should they order, if 650 cm $^{3}$ equals 100 g of cake ? (2)

Show answer & solution

Answer: (a) 2200 cm

^{2}

(b) 1 kg (cake needed ≈ 975 g)

(a) Base circumference $2 π r = 44 \Rightarrow r = 7$ cm (taking $π = \frac{22}{7}$ ).
Slant height $l = 7^{2} + 2 4^{2} = 25$ cm.
Paper for one cap $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm $^{2}$ ; for 4 caps $= 2200$ cm $^{2}$ .
(b) Cake radius $= 12$ cm; volume $= π r^{2} h = \frac{22}{7} \times 144 \times 14 = 6336$ cm $^{3}$ .
Weight $= \frac{6336}{650} \times 100 \approx 974.8$ g ≈ 0.97 kg.
Since cakes are sold in steps of 0.5 kg, they should order 1 kg of cake.

Also asked in: 2022 Standard 30/3/2, 2022 Standard 30/3/3

Q144 marksCase StudySurface Areas and Volumes CBSE 2022 · Standard 30/4/1Not in current syllabus

Khurja is a city in the Indian state of Uttar Pradesh famous for the pottery. Khurja pottery is traditional Indian pottery work which has attracted Indians as well as foreigners with a variety of tea-sets, crockery and ceramic tile works. A huge portion of the ceramics used in the country is supplied by Khurja and is also referred as ‘The Ceramic Town’.
One of the private schools of Bulandshahr organised an Educational Tour for class 10 students to Khurja. Students were very excited about the trip. Following are the few pottery objects of Khurja.
Students found the shapes of the objects very interesting and they could easily relate them with mathematical shapes viz sphere, hemisphere, cylinder etc. Maths teacher who was accompanying the students asked following questions :
(a) The internal radius of hemispherical bowl (filled completely with water) in I is 9 cm and radius and height of cylindrical jar in II is 1.5 cm and 4 cm respectively. If the hemispherical bowl is to be emptied in cylindrical jars, then how many cylindrical jars are required ? (2)
(b) If in the cylindrical jar full of water, a conical funnel of same height and same diameter is immersed, then how much water will flow out of the jar ? (2)

Show answer & solution

Answer: (a) 54 (b)

3 π

cm^{3} = \frac{66}{7}

cm^{3} \approx 9.43

cm^{3}

(a) Volume of hemispherical bowl $= \frac{2}{3} π (9)^{3} = 486 π$ $cm^{3}$ .
Volume of one jar $= π (1.5)^{2} \times 4 = 9 π$ $cm^{3}$ .
Number of jars $= \frac{486 π}{9 π} = 54$ .
(b) Water that flows out = volume of the cone with $r = 1.5$ cm, $h = 4$ cm.
$= \frac{1}{3} π (1.5)^{2} \times 4 = 3 π = 3 \times \frac{22}{7} = \frac{66}{7} \approx 9.43$ $cm^{3}$ .

Also asked in: 2022 Standard 30/4/2, 2022 Standard 30/4/3

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