HOTS Questions on Chapter 12 Class 10 Surface Areas and Volumes

HOTS Questions on Class 10 Maths Chapter 12 Surface Areas and Volumes help students understand 3D shapes, formulas, and real life applications in a simple way. They improve knowledge of cubes, cuboids, cylinders, cones, spheres, and mixed solids. This chapter builds a strong base for solving volume and surface area problems and improving math skills. A downloadable PDF is also available for easy revision and practice.

HOTS Questions Based on Cubes and Cuboids

Question 1: A cube of side 5 cm is cut into smaller cubes of side 1 cm. Find: a) The number of smaller cubes formed b) Total surface area of all smaller cubes c) The increase in total surface area compared to the original cube

Answer:

Step 1: Number of smaller cubes

= (Side of big cube / Side of small cube)³

= (5/1)³ = 125 smaller cubes

Step 2: Surface area of each small cube

= 6 × 1² = 6 cm²

Total surface area of 125 cubes

= 125 × 6 = 750 cm²

Step 3: Surface area of original cube

= 6 × 5² = 150 cm²

Step 4: Increase in surface area

= 750 - 150 = 600 cm²

Increase factor = 750/150 = 5 times

(When a cube is cut into n³ smaller cubes,

surface area increases by n times)

Answer: 125 cubes, Total SA = 750 cm², Increase = 600 cm²

Question 2: Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Answer:

Step 1: Find side of each cube

Volume = a³ = 64

a = 4 cm

Step 2: Dimensions of resulting cuboid

When joined end to end:

l = 4 + 4 = 8 cm

b = 4 cm

h = 4 cm

Step 3: Surface area of cuboid

= 2(lb + bh + hl)

= 2(8×4 + 4×4 + 4×8)

= 2(32 + 16 + 32)

= 2 × 80

= 160 cm²

Note: Original total SA = 2 × 6 × 16 = 192 cm²

Joined faces (2 squares) lost = 2 × 16 = 32 cm²

Net SA = 192 - 32 = 160 cm²

Answer: Surface area = 160 cm²

Question 3: A rectangular box has dimensions 8 cm × 6 cm × 4 cm. A cube has the same volume as this box. Which has the greater surface area and by how much?

Answer:

Step 1: Volume of cuboid

= 8 × 6 × 4 = 192 cm³

Step 2: Side of cube with same volume

a³ = 192

a = ∛192 ≈ 5.77 cm

Step 3: Surface area of cuboid

= 2(8×6 + 6×4 + 4×8)

= 2(48 + 24 + 32)

= 2 × 104 = 208 cm²

Step 4: Surface area of cube

= 6 × (5.77)²

= 6 × 33.29

= 199.74 cm²

Step 5: Comparison

Cuboid SA = 208 cm²

Cube SA = 199.74 cm²

Cuboid has greater surface area by:

208 - 199.74 = 8.26 cm²

Answer: Cuboid has greater SA by 8.26 cm²

HOTS Questions Based on Cylinders, Cones and Spheres

Question 4: A solid right circular cone of height 14 cm and base radius 7 cm is placed upright inside a cylinder of same base radius and same height. Find the volume of water that can be poured into the cylinder around the cone.

Answer:

Step 1: Volume of cylinder

= πr²h

= (22/7) × 7² × 14

= (22/7) × 49 × 14

= 22 × 7 × 14

= 2156 cm³

Step 2: Volume of cone

= (1/3)πr²h

= (1/3) × (22/7) × 49 × 14

= (1/3) × 2156

= 718.67 cm³

Step 3: Volume of water

= Volume of cylinder - Volume of cone

= 2156 - 718.67

= 1437.33 cm³

Answer: Water volume = 1437.33 cm³

Question 5: A hemispherical bowl of radius 9 cm is filled completely with pudding. The pudding is served into cones of radius 3 cm and height 3 cm. How many cones can be served?

Answer:

Step 1: Volume of hemispherical bowl

= (2/3)πr³

= (2/3) × (22/7) × 9³

= (2/3) × (22/7) × 729

= (2 × 22 × 729)/(3 × 7)

= 32076/21

= 1527.43 cm³

Step 2: Volume of one cone

= (1/3)πr²h

= (1/3) × (22/7) × 3² × 3

= (1/3) × (22/7) × 27

= (22 × 27)/(3 × 7)

= 594/21

= 28.29 cm³

Step 3: Number of cones

= Volume of bowl / Volume of one cone

= 1527.43 / 28.29

= 54 cones (approximately)

Exact calculation:

= [(2/3) × (22/7) × 729] / [(1/3) × (22/7) × 9]

= [2 × 729] / [1 × 9]

= 1458/9

= 162 cones

Answer: 162 cones can be served

Question 6: A cylinder, a cone, and a hemisphere all have the same base radius r = 7 cm and the same height h = 7 cm. Arrange them in ascending order of their volumes. Find the ratio of their volumes.

Answer:

Given: r = 7, h = 7

Volume of cone = (1/3)πr²h

= (1/3) × (22/7) × 49 × 7

= (1/3) × 1078

= 359.33 cm³

Volume of hemisphere = (2/3)πr³

(Since h = r = 7 for hemisphere)

= (2/3) × (22/7) × 343

= (2/3) × 1078

= 718.67 cm³

Volume of cylinder = πr²h

= (22/7) × 49 × 7

= 1078 cm³

Ratio:

Cone : Hemisphere : Cylinder

= 359.33 : 718.67 : 1078

= 1 : 2 : 3

This is the famous 1:2:3 ratio.

Ascending order:

Cone < Hemisphere < Cylinder

Answer: Ratio = 1:2:3 (Cone : Hemisphere : Cylinder)

HOTS Questions Based on Combination of Solids

Question 7: A toy is in the form of a cone mounted on a hemisphere. The diameter of the base of the cone is 6 cm and its height is 4 cm. Find the total surface area of the toy.

Answer:

Diagram:

r = 3 cm (common radius)

h = 4 cm (height of cone)

Step 1: Slant height of cone

l = √(r² + h²)

= √(9 + 16)

= √25 = 5 cm

Step 2: Curved surface area of cone

= πrl

= (22/7) × 3 × 5

= 330/7

= 47.14 cm²

Step 3: Curved surface area of hemisphere

= 2πr²

= 2 × (22/7) × 9

= 396/7

= 56.57 cm²

Step 4: Total surface area of toy

= CSA of cone + CSA of hemisphere

(The flat circular base is hidden inside)

= 47.14 + 56.57

= 103.71 cm²

Answer: Total surface area = 103.71 cm²

Question 8: A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 10 cm and its base radius is 3.5 cm, find the total surface area of the article.

Answer:

r = 3.5 cm, h = 10 cm

Step 1: Curved surface area of cylinder

= 2πrh

= 2 × (22/7) × 3.5 × 10

= 2 × 22 × 0.5 × 10

= 220 cm²

Step 2: Curved surface area of each hemisphere

= 2πr²

= 2 × (22/7) × 3.5²

= 2 × (22/7) × 12.25

= 2 × 38.5

= 77 cm²

Step 3: Total surface area

= CSA of cylinder + 2 × CSA of hemisphere

= 220 + 2 × 77

= 220 + 154

= 374 cm²

Answer: Total surface area = 374 cm²

Question 9: A tent is in the shape of a cylinder surmounted by a conical top. The height of the cylindrical part is 3 m and total height of the tent is 8 m. The diameter of the base is 12 m. Find the cost of canvas needed at ₹120 per m².

Answer:

r = 6 m

Height of cylinder = 3 m

Total height = 8 m

Height of cone = 8 - 3 = 5 m

Step 1: Slant height of cone

l = √(r² + h²)

= √(36 + 25)

= √61

≈ 7.81 m

Step 2: CSA of cylinder

= 2πrh

= 2 × (22/7) × 6 × 3

= 792/7

= 113.14 m²

Step 3: CSA of cone

= πrl

= (22/7) × 6 × 7.81

= (22 × 6 × 7.81)/7

= 1030.92/7

= 147.27 m²

Step 4: Total canvas needed

= CSA of cylinder + CSA of cone

= 113.14 + 147.27

= 260.41 m²

Step 5: Cost

= 260.41 × 120

= ₹31,249.20

Answer: Cost of canvas = ₹31,249.20

HOTS Questions Based on Conversion of Solids

Question 10: A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones formed.

Answer:

Step 1: Volume of sphere

= (4/3)πr³

= (4/3) × (22/7) × 10.5³

= (4/3) × (22/7) × 1157.625

= (4 × 22 × 1157.625)/(3 × 7)

= 101871/21

= 4851 cm³

Step 2: Volume of one cone

= (1/3)πr²h

= (1/3) × (22/7) × 3.5² × 3

= (1/3) × (22/7) × 36.75

= (22 × 36.75)/(7 × 3)

= 808.5/21

= 38.5 cm³

Step 3: Number of cones

= Volume of sphere / Volume of one cone

= 4851 / 38.5

= 126 cones

Answer: 126 cones are formed

Question 11: A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys in the shape of a cone of radius 3 cm and height 9 cm with a hemispherical top of radius 3 cm. Find the number of toys.

Answer:

Step 1: Volume of cylinder

r = 6 cm, h = 15 cm

= π × 36 × 15 = 540π cm³

Step 2: Volume of one toy

Toy = Cone + Hemisphere (same radius r = 3)

Volume of cone part:

= (1/3)π × 9 × 9 = 27π cm³

Volume of hemisphere part:

= (2/3)π × 27 = 18π cm³

Total volume of one toy = 27π + 18π = 45π cm³

Step 3: Number of toys

= 540π / 45π

= 12 toys

(π cancels out clean calculation)

Answer: 12 toys are formed

Question 12: A sphere of diameter 6 cm is dropped into a cylinder of diameter 12 cm partially filled with water. If the sphere is completely submerged, find the rise in water level.

Answer:

Sphere r = 3 cm

Cylinder r = 6 cm

Step 1: Volume of sphere

= (4/3)π × 27 = 36π cm³

Step 2: The sphere displaces water equal

to its own volume.

Rise in water level = Volume of sphere / Base area of cylinder

= 36π / (π × 36)

= 36π / 36π

= 1 cm

Rise in water level = 1 cm

Answer: Water level rises by 1 cm

Question 13: A conical vessel of radius 6 cm and height 8 cm is completely filled with water. A sphere is lowered into the vessel. Find the largest sphere that can be placed in the cone without overflowing, and find the volume of water that overflows when it is placed.

Answer:

Given: Cone r = 6, h = 8

Slant height l = √(36 + 64) = √100 = 1

Step 1: Find radius of largest inscribed sphere

The sphere touches the base and the slant surface.

Using formula for inscribed sphere radius:

r_sphere = (r × h) / (r + l)

= (6 × 8) / (6 + 10)

= 48 / 16

= 3 cm

Step 2: Volume of sphere

= (4/3)π × 27

= 36π

= 113.14 cm³

Step 3: Volume of cone

= (1/3)π × 36 × 8

= 96π

= 301.71 cm³

Step 4: Volume of overflow

Since cone is full and sphere is placed:

Water overflow = Volume of sphere

= 113.14 cm³

Answer: Largest sphere radius = 3 cm, Water overflow = 113.14 cm³

Question 14: A hemispherical tank of radius 1.75 m is full of water. It is connected to a pipe which empties it at 7 liters per second. How much time will it take to empty the tank completely? (1 m³ = 1000 litres)

Answer:

Step 1: Volume of hemispherical tank

= (2/3)πr³

= (2/3) × (22/7) × (1.75)³

= (2/3) × (22/7) × 5.359

= (2 × 22 × 5.359)/(3 × 7)

= 235.79/21

= 11.228 m³

Step 2: Convert to litres

= 11.228 × 1000

= 11228 litres

Step 3: Time to empty

Rate = 7 litres per second

Time = 11228 / 7

= 1604 seconds

= 26 minutes 44 seconds

Answer: Time = 1604 seconds ≈ 26 minutes 44 seconds

Question 15: A manufacturer makes cylindrical cans of radius 3.5 cm and height 10 cm. These cans are packed in a rectangular box of dimensions 14 cm × 14 cm × 10 cm. How many cans fit in one box? What percentage of the box volume is wasted?

Answer:

Step 1: Can dimensions: r = 3.5 cm, h = 10 cm

Diameter of can = 7 cm

Step 2: Number of cans in box

Box base = 14 × 14 cm

Can diameter = 7 cm

Along 14 cm length: 14/7 = 2 cans

Along 14 cm width: 14/7 = 2 cans

Height of box = 10 cm = height of can (1 layer)

Total cans = 2 × 2 × 1 = 4 cans

Step 3: Volume of box

= 14 × 14 × 10 = 1960 cm³

Step 4: Volume of 4 cans

= 4 × πr²h

= 4 × (22/7) × 12.25 × 10

= 4 × 385

= 1540 cm³

Step 5: Wasted volume

= 1960 - 1540 = 420 cm³

Step 6: Percentage wasted

= (420/1960) × 100

= 21.43%

Answer: 4 cans fit, 21.43% of box volume is wasted

Question 16: A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.

Answer:

Internal radius r₁ = 2 cm

External radius r₂ = 4 cm

Cone base radius = 4 cm

Step 1: Volume of hollow sphere

= (4/3)π(r₂³ - r₁³)

= (4/3)π(64 - 8)

= (4/3)π × 56

= 224π/3 cm³

Step 2: Volume of cone = Volume of sphere

(1/3)πr²h = 224π/3

(1/3) × 16 × h = 224/3

16h/3 = 224/3

16h = 224

h = 14 cm

Answer: Height of cone = 14 cm

Mixed HOTS Practice Questions on Surface Areas and Volumes

Question 17: A solid is in the shape of a cone standing on a hemisphere with both having the same radius of 1 cm. If the height of the cone is equal to its radius, find the volume and total surface area of the solid.

Answer:

r = 1 cm, h = 1 cm (cone height = radius)

Volume:

Cone = (1/3)π × 1 × 1 = π/3 cm³

Hemisphere = (2/3)π × 1 = 2π/3 cm³

Total = π/3 + 2π/3 = π = (22/7) = 3.14 cm³

Slant height of cone:

l = √(r² + h²) = √(1 + 1) = √2 cm

Surface area:

CSA of cone = π × 1 × √2 = √2π cm²

CSA of hemisphere = 2π × 1 = 2π cm²

Total = (√2 + 2)π

= (1.414 + 2) × 22/7

= 3.414 × 22/7

= 75.11/7

= 10.73 cm²

Answer: Volume = π ≈ 3.14 cm³, TSA = (√2 + 2)π ≈ 10.73 cm²

Question 18: Twenty-seven solid iron spheres, each of radius r, are melted to form a sphere of radius R. Find the ratio R:r. Also find the ratio of their surface areas.

Answer:

Volume conservation:

27 × (4/3)πr³ = (4/3)πR³

27r³ = R³

R³/r³ = 27

R/r = ∛27 = 3

So R = 3r, Ratio R:r = 3:1

Surface area of 27 small spheres:

= 27 × 4πr²

= 108πr²

Surface area of large sphere:

= 4πR²

= 4π(3r)²

= 36πr²

Ratio of surface areas:

(Large sphere):(27 small spheres)

= 36πr²:108πr²

= 36:108

= 1:3

So 27 small spheres have 3 times

more total surface area than the

single large sphere!

Answer: R:r = 3:1, Surface area ratio = 1:3

Question 19: A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find: a) The capacity of the bucket (volume) b) The metal sheet required to make the bucket (curved surface + base) c) Cost at ₹10 per 100 cm²

Answer:

R = 20 cm (upper), r = 10 cm (lower), h = 30 cm

a) Volume of frustum:

= (πh/3)(R² + r² + Rr)

= (π × 30/3)(400 + 100 + 200)

= 10π × 700

= 7000π

= 7000 × 22/7

= 22000 cm³

= 22 litres

b) Slant height:

l = √(h² + (R-r)²)

= √(900 + 100)

= √1000

= 10√10 cm

CSA of frustum:

= π(R + r)l

= (22/7)(20 + 10)(10√10)

= (22/7) × 30 × 31.62

= (22 × 30 × 31.62)/7

= 20869.2/7

= 2981.31 cm²

Area of bottom circle (base):

= πr² = (22/7) × 100 = 314.28 cm²

Total metal sheet:

= 2981.31 + 314.28

= 3295.59 cm²

c) Cost:

= (3295.59/100) × 10

= 32.96 × 10

= ₹329.56

Answers: a) Capacity = 22000 cm³ = 22 litres b) Metal sheet = 3295.59 cm² c) Cost = ₹329.56

Formula Table:

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