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Surface Areas and Volumes: 5 marks Questions (CBSE Class 10)

41 different 5 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)

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/1/1

A toy is in the form of a cone surmounted on a hemisphere. The cone and hemisphere have the same radii. The height of the conical part of the toy is equal to the diameter of its base. If the radius of the conical part is 5 cm, find the volume of the toy.

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Answer:

\frac{11000}{21}

cm³ ≈ 523.81 cm³

$r = 5$ cm, height of cone $h = 2 r = 10$ cm.
Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} π \times 25 \times 10 = \frac{250}{3} π$ .
Volume of hemisphere $= \frac{2}{3} π r^{3} = \frac{250}{3} π$ .
Total $= \frac{500}{3} π = \frac{500}{3} \times \frac{22}{7} = \frac{11000}{21} \approx 523.81$ cm³.

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q34 (OR) (OR)5 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/1/1

A cubical block is surmounted by a hemisphere of radius 3.5 cm. What is the smallest possible length of the edge of the cube so that the hemisphere can totally lie on the cube ? Find the total surface area of the solid so formed.

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Answer: Edge = 7 cm; total surface area = 332.5 cm²

The base of the hemisphere (diameter 7 cm) must fit on a face, so the smallest edge is 7 cm.
TSA = surface of cube $-$ base of hemisphere $+$ CSA of hemisphere $= 6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$ .
$= 6 \times 49 + \frac{22}{7} \times 3.5 \times 3.5 = 294 + 38.5 = 332.5$ cm².

Also asked in: 2025 Basic 430/1/2, 2025 Basic 430/1/3

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/2/1

A wooden article was made by scooping out a hemisphere (of same diameter) from one end of a solid cylinder as shown in the given figure. If the height of the cylinder is 10 cm and the diameter of the cylinder is 14 cm, find the total surface area of the remaining wooden article.
(Use $π = \frac{22}{7}$ )

Diagram for CBSE 2025 Class 10 Maths question 34

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Answer: 902 cm

^{2}

Radius r = 7 cm, height h = 10 cm.
TSA = curved surface of cylinder + area of the bottom base + inner curved surface of hemisphere.
$= 2 π r h + π r^{2} + 2 π r^{2} = 2 π r h + 3 π r^{2}$ .
$2 π r h = 2 \times \frac{22}{7} \times 7 \times 10 = 440$ cm $^{2}$ .
$3 π r^{2} = 3 \times \frac{22}{7} \times 49 = 462$ cm $^{2}$ .
TSA $= 440 + 462 = 902$ cm $^{2}$ .

Q335 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/2/2

A hemispherical depression is scooped out from the top face of a wooden cubical block of side 14 cm. If the diameter of the hemisphere is equal to the side of the cube, find the total surface area of the remaining solid block. (Use $π = \frac{22}{7}$ )

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Answer: 1330 cm

^{2}

Side of cube a = 14 cm; radius of hemisphere r = 7 cm.
TSA = surface area of cube – area of circular top removed + inner curved surface of hemisphere.
$= 6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$ .
$6 \times 1 4^{2} = 1176$ cm $^{2}$ and $π r^{2} = \frac{22}{7} \times 49 = 154$ cm $^{2}$ .
TSA $= 1176 + 154 = 1330$ cm $^{2}$ .

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/2/3

A spherical glass vessel has a cylindrical neck which is 7 cm long and 2 cm in diameter. The diameter of the spherical part is 14 cm. Find the capacity of the entire glass vessel. (Use $π = \frac{22}{7}$ )

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Answer:

\frac{4378}{3}

^{3}

(about 1459.33 cm

^{3}

)

Sphere: radius R = 7 cm. Volume $= \frac{4}{3} \times \frac{22}{7} \times 7^{3} = \frac{4}{3} \times 22 \times 49 = \frac{4312}{3}$ cm $^{3}$ .
Neck (cylinder): radius r = 1 cm, h = 7 cm. Volume $= \frac{22}{7} \times 1^{2} \times 7 = 22$ cm $^{3}$ .
Capacity $= \frac{4312}{3} + 22 = \frac{4378}{3}$ cm $^{3} \approx 1459.33$ cm $^{3}$ .

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/3/1

A perfume bottle is in the form of a cylinder but the bottom of the bottle has a hemispherical raised portion to reduce the capacity of the bottle. The inner diameter of the bottle is 5 cm and the height of the bottle is 10 cm. Find the capacity of the perfume bottle in mL. (Use $π = 3.14$ and 1 $cm^{3}$ = 1 mL)

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Answer:

\approx 163.54

Radius $r = 2.5$ cm, height $h = 10$ cm.
Volume of cylinder $= π r^{2} h = 3.14 \times 6.25 \times 10 = 196.25$ $cm^{3}$ .
Volume of hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times 3.14 \times 15.625 \approx 32.71$ $cm^{3}$ .
Capacity $= 196.25 - 32.71 = 163.54$ $cm^{3}$ (approx.) $= 163.54$ mL.

Q335 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/3/2

From a solid wooden cylinder of height 10 cm and radius 14 cm, a cylinder of radius 7 cm and height 5 cm is scooped out to form a cavity inside the solid cylinder. Find the total surface area of the remaining solid.

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Answer: 2332

cm^{2}

The cavity is cut from one circular face, so that face loses a circle of radius 7 cm but the cavity floor adds an equal circle back.
TSA = curved surface of big cylinder + 2 circular faces of big cylinder + curved surface of cavity.
Curved surface of big cylinder $= 2 π R h = 2 \times \frac{22}{7} \times 14 \times 10 = 880$ $cm^{2}$ .
Two circular faces $= 2 π R^{2} = 2 \times \frac{22}{7} \times 196 = 1232$ $cm^{2}$ .
Curved surface of cavity $= 2 π r h = 2 \times \frac{22}{7} \times 7 \times 5 = 220$ $cm^{2}$ .
TSA $= 880 + 1232 + 220 = 2332$ $cm^{2}$ .

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Basic 430/3/3

A necklace is made up of wooden beads. Each bead is in the form of a sphere of diameter 4.2 mm. A cylinder is hollowed out from each bead. If the radius of the cylinder is 1 mm, find the volume of wood left in each bead.

Diagram for CBSE 2025 Class 10 Maths question 35

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Answer: 25.608

mm^{3}

Radius of sphere $R = 2.1$ mm; the cylinder has radius $r = 1$ mm and, as in the figure, height $h = 4.2$ mm.
Volume of sphere $= \frac{4}{3} π R^{3} = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 = 38.808$ $mm^{3}$ .
Volume of cylinder $= π r^{2} h = \frac{22}{7} \times 1 \times 4.2 = 13.2$ $mm^{3}$ .
Wood left $= 38.808 - 13.2 = 25.608$ $mm^{3}$ .

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/3/1

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

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Answer: 100

Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} π (5)^{2} (8) = \frac{200 π}{3}$ cm $^{3}$ .
Water that flows out $= \frac{1}{4} \times \frac{200 π}{3} = \frac{50 π}{3}$ cm $^{3}$ = total volume of lead shots.
Volume of one shot $= \frac{4}{3} π (0.5)^{3} = \frac{π}{6}$ cm $^{3}$ .
Number of shots $= \frac{50 π}{3} \div \frac{π}{6} = 100$ .

Also asked in: 2025 Standard 30/3/2

Q325 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/3/3

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. Also, find the surface area of the toy. (Take $π = 3.14$ )

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Answer: Volume

= 8 π = 25.12

^{3}

; surface area

= 4 π (2 + 2) \approx 42.88

^{2}

Radius of hemisphere = radius of cone $= r = 2$ cm; height of cone $h = 2$ cm.
Volume $= \frac{2}{3} π r^{3} + \frac{1}{3} π r^{2} h = \frac{2}{3} π (8) + \frac{1}{3} π (4) (2) = \frac{16 π}{3} + \frac{8 π}{3} = 8 π$ .
Volume $= 8 \times 3.14 = 25.12$ cm $^{3}$ .
Slant height $l = 2^{2} + 2^{2} = 22$ cm.
Surface area = CSA of hemisphere + CSA of cone $= 2 π r^{2} + π r l = 8 π + 42 π = 4 π (2 + 2)$ .
$= 12.56 \times 3.414 \approx 42.88$ cm $^{2}$ (taking $2 = 1.414$ ).

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/4/1

A wooden cubical die is formed by forming hemispherical depressions on each face of the cube such that face 1 has one depression, face 2 has two depressions and so on. The sum of number of hemispherical depressions on opposite faces is always 7. If the edge of the cubical die measures 5 cm and each hemispherical depression is of diameter 1.4 cm, find the total surface area of the die so formed.

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Answer:

182.34

^{2}

Total depressions $= 1 + 2 + 3 + 4 + 5 + 6 = 21$ ; radius $r = 0.7$ cm
Each depression removes a circle of area $π r^{2}$ and adds a hemispherical surface $2 π r^{2}$ : net gain $π r^{2}$
Surface area of cube $= 6 \times 5^{2} = 150$ cm $^{2}$
Net gain $= 21 \times \frac{22}{7} \times 0.7 \times 0.7 = 32.34$ cm $^{2}$
Total surface area $= 150 + 32.34 = 182.34$ cm $^{2}$

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/4/2

In order to provide shelter to flood victims, a shed was constructed using tin sheets which is in the form of cuboid surmounted by a half cylinder as shown below :
The length, breadth and height of cuboidal portion are 10 m, 7 m and 3 m respectively. The diameter of the cylindrical portion is 7 m. Find the cost of tin sheets required to make the shed at the rate of ₹ 70 per square metre, given that the shed is open from the front side and closed from the back side.

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Answer: ₹ 14,717.50 (tin sheet area

210.25

^{2}

)

Radius of half cylinder $r = 3.5$ m, its length $= 10$ m
Two side walls of cuboid $= 2 \times 10 \times 3 = 60$ m $^{2}$
Back wall (rectangular part) $= 7 \times 3 = 21$ m $^{2}$
Curved roof $= π r l = \frac{22}{7} \times 3.5 \times 10 = 110$ m $^{2}$
Back semicircular part $= \frac{1}{2} π r^{2} = \frac{1}{2} \times \frac{22}{7} \times 3.5 \times 3.5 = 19.25$ m $^{2}$
Total $= 60 + 21 + 110 + 19.25 = 210.25$ m $^{2}$ (front open, floor not covered)
Cost $= 210.25 \times 70 =$ ₹ 14,717.50

Q335 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/4/3

A bat manufacturing company made a huge bat for charity and got it signed by world cup winning team.
The dimensions of the bat which is in the form of a cuboid with a cylindrical handle at the top are as follows :
length = 2 m, width = 0.5 m, thickness = 0.1 m
diameter of cylindrical part = 0.1 m
height of cylindrical part = 0.7 m
Find the volume of wood used in the bat. Also, find the total surface area of the wooden bat.

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Answer: Volume

= 0.1055

^{3}

; total surface area

= 2.72

^{2}

Cuboid volume $= 2 \times 0.5 \times 0.1 = 0.1$ m $^{3}$
Cylinder: $r = 0.05$ m, $h = 0.7$ m; volume $= \frac{22}{7} \times 0.05 \times 0.05 \times 0.7 = 0.0055$ m $^{3}$
Volume of wood $= 0.1 + 0.0055 = 0.1055$ m $^{3}$
TSA of cuboid $= 2 (2 \times 0.5 + 0.5 \times 0.1 + 2 \times 0.1) = 2 (1.25) = 2.5$ m $^{2}$
The handle's base covers a circle on the cuboid but its top circle adds the same area, so only the curved surface is extra.
Curved surface of cylinder $= 2 \times \frac{22}{7} \times 0.05 \times 0.7 = 0.22$ m $^{2}$
Total surface area $= 2.5 + 0.22 = 2.72$ m $^{2}$

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/5/1

From one of the faces of a solid wooden cube of side 14 cm, maximum number of hemispheres of diameter 1.4 cm are scooped out. Find the total number of hemispheres that can be scooped out. Also, find the total surface area of the remaining solid.

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Answer: 100 hemispheres; total surface area

= 1330

^{2}

Along one edge: $\frac{14}{1.4} = 10$ hemispheres, so on one face $10 \times 10 = 100$ hemispheres.
Radius $r = 0.7$ cm.
Surface area of cube $= 6 \times 1 4^{2} = 1176$ cm $^{2}$
Each hemisphere removes a circle $π r^{2}$ and adds a curved surface $2 π r^{2}$ : net gain $π r^{2} = \frac{22}{7} \times 0.49 = 1.54$ cm $^{2}$
Total surface area $= 1176 + 100 \times 1.54 = 1176 + 154 = 1330$ cm $^{2}$

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q34 (OR) (OR)5 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/5/1

From a solid cylinder of height 24 cm and radius 5 cm, two cones of height 12 cm and radius 5 cm are hollowed out. Find the volume and surface area of the remaining solid.

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Answer: Volume

= 400 π = \frac{8800}{7} \approx 1257.14

^{3}

; surface area

= 370 π = \frac{8140}{7} \approx 1162.86

^{2}

Volume of cylinder $= π r^{2} h = π (25) (24) = 600 π$ cm $^{3}$
Volume of two cones $= 2 \times \frac{1}{3} π (25) (12) = 200 π$ cm $^{3}$
Remaining volume $= 400 π = 400 \times \frac{22}{7} = \frac{8800}{7} \approx 1257.14$ cm $^{3}$
The cones are cut from the two ends, so the remaining surface = curved surface of cylinder + curved surfaces of two cones.
Slant height $l = 1 2^{2} + 5^{2} = 13$ cm
Surface area $= 2 π (5) (24) + 2 π (5) (13) = 240 π + 130 π = 370 π = \frac{8140}{7} \approx 1162.86$ cm $^{2}$

Also asked in: 2025 Standard 30/5/2, 2025 Standard 30/5/3

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/6/1

From one face of a solid cube of side 14 cm, the largest possible cone is carved out. Find the volume and surface area of the remaining solid. $(Use π = \frac{22}{7}, 5 = 2.2)$

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Answer: Volume

= \frac{6076}{3} \approx 2025.33

^{3}

; surface area

= 1360.8

^{2}

Largest cone: radius $r = 7$ cm, height $h = 14$ cm
Volume of cube $= 1 4^{3} = 2744$ cm $^{3}$
Volume of cone $= \frac{1}{3} \times \frac{22}{7} \times 49 \times 14 = \frac{2156}{3} \approx 718.67$ cm $^{3}$
Remaining volume $= 2744 - \frac{2156}{3} = \frac{6076}{3} \approx 2025.33$ cm $^{3}$
Slant height $l = 7^{2} + 1 4^{2} = 245 = 75 = 15.4$ cm
Surface area $= 6 \times 1 4^{2} - π r^{2} + π r l = 1176 - 154 + \frac{22}{7} \times 7 \times 15.4$
$= 1176 - 154 + 338.8 = 1360.8$ cm $^{2}$

Also asked in: 2025 Standard 30/6/2

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/6/2

On the day of her examination, Riya sharpened her pencil from both ends as shown below :
The diameter of the cylindrical and conical part of the pencil is 4.2 mm. If the height of each conical part is 2.8 mm and length of entire pencil is 105.6 mm, find the total surface area of the pencil.

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Answer: 1366.2 mm

^{2}

Radius $r = 2.1$ mm; cone height $h = 2.8$ mm
Slant height $l = 2. 1^{2} + 2. 8^{2} = 12.25 = 3.5$ mm
Length of cylindrical part $= 105.6 - 2 \times 2.8 = 100$ mm
TSA $= 2 π r H + 2 π r l = 2 π r (H + l) = 2 \times \frac{22}{7} \times 2.1 \times (100 + 3.5)$
$= 13.2 \times 103.5 = 1366.2$ mm $^{2}$

Q335 marksLong AnswerSurface Areas and Volumes CBSE 2025 · Standard 30/6/3

Fermentation tanks are designed in the form of cylinder mounted on a cone as shown below :
The total height of the tank is 3.3 m and height of conical part is 1.2 m. The diameter of the cylindrical as well as conical part is 1 m. Find the capacity of the tank. If the level of liquid in the tank is 0.7 m from the top, find the surface area of the tank in contact with liquid.

Diagram for CBSE 2025 Class 10 Maths question 33

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Answer: Capacity

= \frac{55}{28}

^{3} \approx 1.96

^{3}

; surface area in contact

= \frac{451}{70}

^{2} \approx 6.44

^{2}

$r = 0.5$ m; cone height $h = 1.2$ m; cylinder height $H = 3.3 - 1.2 = 2.1$ m
Capacity $= π r^{2} H + \frac{1}{3} π r^{2} h = π r^{2} (H + \frac{h}{3}) = \frac{22}{7} \times 0.25 \times (2.1 + 0.4)$
$= \frac{22}{7} \times 0.625 = \frac{55}{28} \approx 1.96$ m $^{3}$
Liquid height in cylinder $= 2.1 - 0.7 = 1.4$ m; it also fills the cone.
Slant height of cone $l = 0. 5^{2} + 1. 2^{2} = 1.69 = 1.3$ m
Area in contact $= 2 π r (1.4) + π r l = π (0.5) (2.8 + 1.3) = 2.05 π$
$= 2.05 \times \frac{22}{7} = \frac{451}{70} \approx 6.44$ m $^{2}$

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/2/1

A textile industry runs in a shed. This shed is in the shape of a cuboid surmounted by a half cylinder. If the base of the industry is of dimensions 14 m $\times$ 20 m and the height of the cuboidal portion is 7 m, find the volume of air that the industry can hold. Further, suppose the machinery in the industry occupies a total space of 400 m $^{3}$ . Then, how much space is left in the industry ?

Diagram for CBSE 2024 Class 10 Maths question 35

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Answer: Volume of air = 3500 m

^{3}

; space left = 3100 m

^{3}

Volume of cuboid $= 14 \times 20 \times 7 = 1960$ m $^{3}$
Half cylinder: radius $= 7$ m, length $= 20$ m. Volume $= \frac{1}{2} \times \frac{22}{7} \times 7 \times 7 \times 20 = 1540$ m $^{3}$
Volume of air $= 1960 + 1540 = 3500$ m $^{3}$
Space left $= 3500 - 400 = 3100$ m $^{3}$

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q35 (OR) (OR)5 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/2/1

From a solid cylinder of height 8 cm and radius 6 cm, a conical cavity of the same height and same radius is carved out. Find the total surface area of the remaining solid. (Take $π = 3.14$ )

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Answer: 602.88 cm

^{2}

Slant height of cone $l = 8^{2} + 6^{2} = 10$ cm
TSA = curved surface of cylinder + top circular face + curved surface of cone
$= 2 π r h + π r^{2} + π r l = π r (2 h + r + l)$
$= 3.14 \times 6 \times (16 + 6 + 10) = 3.14 \times 6 \times 32$
$= 602.88$ cm $^{2}$

Also asked in: 2024 Basic 430/2/2, 2024 Basic 430/2/3

Q325 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/4/1

Two cubes each of volume 125 cm $^{3}$ are joined end to end. Find the volume and the surface area of the resulting cuboid.

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Answer: Volume

= 250

^{3}

; surface area

= 250

^{2}

Edge of each cube $= 3125 = 5$ cm.
Resulting cuboid: $l = 10$ cm, $b = 5$ cm, $h = 5$ cm.
Volume $= 10 \times 5 \times 5 = 250$ cm $^{3}$ .
Surface area $= 2 (l b + bh + h l) = 2 (50 + 25 + 50) = 250$ cm $^{2}$ .

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q32 (OR) (OR)5 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/4/1

A solid is in the shape of a cone surmounted on a hemisphere with both their diameters being equal to 7 cm and the height of the cone is equal to its radius. Find the volume of the solid.

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Answer:

134.75

^{3}

$r = \frac{7}{2} = 3.5$ cm and height of cone $h = r = 3.5$ cm.
Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{3}$ .
Volume of hemisphere $= \frac{2}{3} π r^{3}$ .
Total $= π r^{3} = \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 = 11 \times 12.25 = 134.75$ cm $^{3}$ .

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q325 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/5/1

A toy is in the form of a cone of radius 7 cm mounted on a hemisphere of same radius. The total height of the toy is 31 cm. Find the surface area of the toy.

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Answer: 858 cm

^{2}

Height of cone $= 31 - 7 = 24$ cm.
Slant height $l = 2 4^{2} + 7^{2} = 625 = 25$ cm.
CSA of cone $= π r l = \frac{22}{7} \times 7 \times 25 = 550$ cm $^{2}$ .
CSA of hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times 49 = 308$ cm $^{2}$ .
Surface area of toy $= 550 + 308 = 858$ cm $^{2}$ .

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q32 (OR) (OR)5 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Basic 430/5/1

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 15 cm and its base is of radius 4.2 cm, then find the total surface area of the article.

Diagram for CBSE 2024 Class 10 Maths question 32 (OR)

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Answer: 617.76 cm

^{2}

TSA = CSA of cylinder + 2 $\times$ CSA of hemisphere.
CSA of cylinder $= 2 π r h = 2 \times \frac{22}{7} \times 4.2 \times 15 = 396$ cm $^{2}$ .
2 $\times$ CSA of hemisphere $= 2 \times 2 π r^{2} = 4 \times \frac{22}{7} \times 4.2 \times 4.2 = 221.76$ cm $^{2}$ .
TSA $= 396 + 221.76 = 617.76$ cm $^{2}$ .

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Standard 30/1/2

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area and the volume of the vessel.

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Answer: Inner surface area = 572

cm^{2}

; volume

= \frac{4928}{3} \approx 1642.67

cm^{3}

Radius $r = 7$ cm; height of cylinder $h = 13 - 7 = 6$ cm.
Inner surface area $= 2 π r^{2} + 2 π r h = 2 π r (r + h) = 2 \times \frac{22}{7} \times 7 \times 13 = 572$ $cm^{2}$ .
Volume $= \frac{2}{3} π r^{3} + π r^{2} h = π r^{2} (\frac{2 r}{3} + h) = \frac{22}{7} \times 49 \times (\frac{14}{3} + 6)$
$= 154 \times \frac{32}{3} = \frac{4928}{3} \approx 1642.67$ $cm^{3}$ .

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Standard 30/3/1

A solid iron pole consists of a solid cylinder of height 200 cm and base diameter 28 cm, which is surmounted by another cylinder of height 50 cm and radius 7 cm. Find the mass of the pole, given that 1 cm $^{3}$ of iron has approximately 8 g mass.

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Answer: 1047200 g = 1047.2 kg

Lower cylinder: $r = 14$ cm, $h = 200$ cm; volume = $\frac{22}{7} \times 1 4^{2} \times 200 = 123200$ cm $^{3}$ .
Upper cylinder: $r = 7$ cm, $h = 50$ cm; volume = $\frac{22}{7} \times 7^{2} \times 50 = 7700$ cm $^{3}$ .
Total volume = 123200 + 7700 = 130900 cm $^{3}$ .
Mass = $130900 \times 8 = 1047200$ g = 1047.2 kg.

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

Q35 (OR) (OR)5 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Standard 30/3/1

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 4 mm, find its surface area. Also, find its volume.

Diagram for CBSE 2024 Class 10 Maths question 35 (OR)

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Answer: Surface area = 176 mm

^{2}

; volume =

\frac{3344}{21} \approx 159.24

^{3}

Radius $r = 2$ mm; length of the cylindrical part = $14 - 2 - 2 = 10$ mm.
Surface area = $2 π r h + 2 \times 2 π r^{2} = 2 π (2) (10) + 4 π (2)^{2} = 40 π + 16 π = 56 π$ .
$= 56 \times \frac{22}{7} = 176$ mm $^{2}$ .
Volume = $π r^{2} h + 2 \times \frac{2}{3} π r^{3} = π (4) (10) + \frac{4}{3} π (8) = 40 π + \frac{32}{3} π = \frac{152}{3} π$ .
$= \frac{152}{3} \times \frac{22}{7} = \frac{3344}{21} \approx 159.24$ mm $^{3}$ .

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

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Standard 30/5/1

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 5.8 cm and its base is of radius 2.1 cm, find the total surface area of the article.

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Answer: 132 cm²

TSA = CSA of cylinder + 2 × CSA of hemisphere $= 2 π r h + 2 (2 π r^{2}) = 2 π r (h + 2 r)$ .
$= 2 \times \frac{22}{7} \times 2.1 \times (5.8 + 4.2) = 13.2 \times 10 = 132$ cm².

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Standard 30/5/2

A tent is in the shape of a cylinder, surmounted by a conical top. If the height and diameter of the cylindrical part are 3.5 m and 6 m, and slant height of the top is 4.2 m, find the area of canvas used for making the tent. Also, find the cost of canvas of the tent at the rate of ₹ 500 per $m^{2}$ .

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Answer: 105.6 m²; ₹ 52,800

Radius $r = 3$ m, cylinder height $h = 3.5$ m, slant height $l = 4.2$ m.
Canvas = CSA of cylinder + CSA of cone $= 2 π r h + π r l = π r (2 h + l)$ .
$= \frac{22}{7} \times 3 \times (7 + 4.2) = \frac{22}{7} \times 3 \times 11.2 = 105.6$ m².
Cost $= 105.6 \times 500 = ₹ 52, 800$ .

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2024 · Standard 30/5/3

A juice seller was serving his customers using glasses as shown in the figure. The inner diameter of the cylindrical glass was 5.6 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of the glass was 10 cm, find the apparent capacity and the actual capacity of the glass.

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Answer: Apparent capacity = 246.4 cm³; actual capacity ≈ 200.41 cm³

Radius $r = 2.8$ cm, height $h = 10$ cm.
Apparent capacity $= π r^{2} h = \frac{22}{7} \times 2.8 \times 2.8 \times 10 = 246.4$ cm³.
Volume of the hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 2. 8^{3} = \frac{2}{3} \times \frac{22}{7} \times 21.952 \approx 45.99$ cm³.
Actual capacity $= 246.4 - 45.99 \approx 200.41$ cm³.

Q335 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Basic 430/1/1

A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder of same diameter. The diameter of the hemispherical bowl is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. Also, find the volume of the vessel.

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Answer: Inner surface area = 572 cm²; volume =

\frac{4928}{3}

cm³ ≈ 1642.67 cm³

Radius $r = 7$ cm. Height of cylindrical part $h = 13 - 7 = 6$ cm.
Inner surface area $= 2 π r^{2} + 2 π r h = 2 π r (r + h)$
$= 2 \times \frac{22}{7} \times 7 \times 13 = 572$ cm²
Volume $= \frac{2}{3} π r^{3} + π r^{2} h = π r^{2} (\frac{2}{3} r + h)$
$= \frac{22}{7} \times 49 \times (\frac{14}{3} + 6) = 154 \times \frac{32}{3}$
$= \frac{4928}{3} \approx 1642.67$ cm³

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Basic 430/1/2

The sum of the radius of the base and height of a solid right-circular cylinder is 37 cm. If the total surface area of the solid cylinder is 1628 cm², find the volume of the cylinder.

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Answer: 4620 cm³

Let radius = $r$ and height = $h$ ; $r + h = 37$ .
Total surface area $= 2 π r (r + h) = 1628$
$2 \times \frac{22}{7} \times r \times 37 = 1628$ , so $r = \frac{1628 \times 7}{44 \times 37} = 7$ cm
$h = 37 - 7 = 30$ cm
Volume $= π r^{2} h = \frac{22}{7} \times 49 \times 30 = 4620$ cm³

Q335 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Basic 430/1/3

A heap of rice is in the form of a cone of base diameter 24 m and height $\frac{7}{2}$ m. Find the volume of rice. How much canvas cloth is required to just cover the heap ?

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Answer: Volume = 528 m³; canvas required =

\frac{3300}{7}

m² ≈ 471.43 m²

Radius $r = 12$ m, height $h = \frac{7}{2}$ m.
Volume $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 144 \times \frac{7}{2} = 528$ m³
Slant height $l = r^{2} + h^{2} = 144 + \frac{49}{4} = \frac{625}{4} = \frac{25}{2}$ m
Canvas needed = curved surface area $= π r l = \frac{22}{7} \times 12 \times \frac{25}{2} = \frac{3300}{7} \approx 471.43$ m²

Q355 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Basic 430/4/1

The boilers are used in thermal power plants to store water and then used to produce steam. One such boiler consists of a cylindrical part in middle and two hemispherical parts at its both ends.
Length of the cylindrical part is 7m and radius of cylindrical part is $\frac{7}{2}$ m.
Find the total surface area and the volume of the boiler. Also, find the ratio of the volume of cylindrical part to the volume of one hemispherical part.

Diagram for CBSE 2023 Class 10 Maths question 35

Show answer & solution

Answer: TSA

= 308

^{2}

; volume

= \frac{2695}{6} \approx 449.17

^{3}

; ratio

= 3 : 1

$r = \frac{7}{2}$ m, $h = 7$ m.
TSA = CSA of cylinder + 2 $\times$ CSA of hemisphere $= 2 π r h + 4 π r^{2}$ .
$= 2 \times \frac{22}{7} \times \frac{7}{2} \times 7 + 4 \times \frac{22}{7} \times \frac{49}{4} = 154 + 154 = 308$ m $^{2}$ .
Volume of cylinder $= π r^{2} h = \frac{22}{7} \times \frac{49}{4} \times 7 = \frac{539}{2}$ m $^{3}$ .
Volume of two hemispheres $= \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{343}{8} = \frac{539}{3}$ m $^{3}$ .
Total volume $= \frac{539}{2} + \frac{539}{3} = \frac{2695}{6} \approx 449.17$ m $^{3}$ .
Ratio $= π r^{2} h : \frac{2}{3} π r^{3} = h : \frac{2}{3} r = 7 : \frac{7}{3} = 3 : 1$ .

Also asked in: 2023 Basic 430/4/2, 2023 Basic 430/4/3

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Basic 430/6/1

An ice-cream filled cone having radius 5 cm and height 10 cm is as shown in the figure. Find the volume of the ice-cream in 7 such cones.

Diagram for CBSE 2023 Class 10 Maths question 34

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Answer:

\frac{11000}{3}

^{3}

\approx 3666.67

^{3}

The ice-cream fills the cone (r = 5 cm, h = 10 cm) and a hemisphere of radius 5 cm on top.
Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 25 \times 10 = \frac{5500}{21}$ cm $^{3}$ .
Volume of hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 125 = \frac{5500}{21}$ cm $^{3}$ .
Ice-cream in one cone $= \frac{11000}{21}$ cm $^{3}$ .
In 7 cones $= 7 \times \frac{11000}{21} = \frac{11000}{3} \approx 3666.67$ cm $^{3}$ .

Q335 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Basic 430/6/2

A solid is in the shape of a cone standing on a hemisphere with both their diameters being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid. [Use $π = 3.14$ ]

Show answer & solution

Answer: 0.3925 cm

^{3}

(about 0.39 cm

^{3}

)

Radius $r = \frac{1}{2} = 0.5$ cm; height of cone $h = r = 0.5$ cm.
Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{3}$ ; volume of hemisphere $= \frac{2}{3} π r^{3}$ .
Volume of solid $= \frac{1}{3} π r^{3} + \frac{2}{3} π r^{3} = π r^{3} = 3.14 \times 0.125 = 0.3925$ cm $^{3}$ .

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Standard 30/4/1

A student was asked to make a model shaped like a cylinder with two cones attached to its ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its total length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model.

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Answer: 66 cm³

Radius $r = 1.5$ cm. Each cone has height 2 cm, so cylinder length $= 12 - 2 - 2 = 8$ cm.
Volume of cylinder $= π r^{2} h = π (1.5)^{2} (8) = 18 π$
Volume of two cones $= 2 \times \frac{1}{3} π (1.5)^{2} (2) = 3 π$
Total volume $= 21 π = 21 \times \frac{22}{7} = 66$ cm³

Q325 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Standard 30/4/2

From a solid cylinder of height 20 cm and diameter 12 cm, a conical cavity of height 8 cm and radius 6 cm is hallowed out. Find the total surface area of the remaining solid.

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Answer: 1056 cm²

Cylinder: $r = 6$ cm, $h = 20$ cm. Cone: $r = 6$ cm, $h = 8$ cm, slant height $l = 6^{2} + 8^{2} = 10$ cm.
The cavity is cut from one end, so the remaining solid has: curved surface of cylinder + one base + curved surface of cone.
CSA of cylinder $= 2 π r h = 2 π (6) (20) = 240 π$
Base $= π r^{2} = 36 π$ ; CSA of cone $= π r l = 60 π$
TSA $= 240 π + 36 π + 60 π = 336 π = 336 \times \frac{22}{7} = 1056$ cm²

Q335 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Standard 30/4/3

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm, find the total surface area of the article.

Diagram for CBSE 2023 Class 10 Maths question 33

Show answer & solution

Answer: 374 cm²

$r = 3.5$ cm, $h = 10$ cm
TSA = CSA of cylinder + CSA of two hemispheres $= 2 π r h + 2 (2 π r^{2})$
$2 π r h = 2 \times \frac{22}{7} \times 3.5 \times 10 = 220$ cm²
$4 π r^{2} = 4 \times \frac{22}{7} \times 3.5 \times 3.5 = 154$ cm²
TSA $= 220 + 154 = 374$ cm²

Also asked in: 2023 Basic 430/6/3

Q335 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Standard 30/6/1

A solid is in the shape of a right-circular cone surmounted on a hemisphere, the radius of each of them being 7 cm and the height of the cone is equal to its diameter. Find the volume of the solid.

Show answer & solution

Answer:

1437 \frac{1}{3}

^{3}

(about 1437.33 cm

^{3}

)

r = 7 cm, height of cone h = 14 cm.
Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 49 \times 14 = \frac{2156}{3}$ cm $^{3}$
Volume of hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 343 = \frac{2156}{3}$ cm $^{3}$
Total volume $= \frac{4312}{3} = 1437 \frac{1}{3}$ cm $^{3}$

Also asked in: 2023 Standard 30/6/3

Q345 marksLong AnswerSurface Areas and Volumes CBSE 2023 · Standard 30/6/2

A solid is in the shape of a right-circular cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of the solid is 9.5 cm. Find the volume of the solid.

Show answer & solution

Answer:

166 \frac{5}{6}

^{3}

(about 166.83 cm

^{3}

)

r = 3.5 cm; height of cone h = 9.5 – 3.5 = 6 cm.
Volume of cone $= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 6 = 77$ cm $^{3}$
Volume of hemisphere $= \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 3. 5^{3} = \frac{539}{6} = 89 \frac{5}{6}$ cm $^{3}$
Total volume $= 77 + 89 \frac{5}{6} = 166 \frac{5}{6}$ cm $^{3}$

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