Surface Areas and Volumes Class 10 — Competency Based Questions

20 Questions | 100 Marks

A farmer has a cylindrical container of radius 1.5 m and height 3.5 m. He wants to convert it into a cone of the same radius and height to store grain. What is the difference in the volume of grain the container can hold in its cylindrical and conical forms? 3marks

A solid is formed by a cone of height 6 cm and a hemisphere of radius 3.5 cm attached to its base. A cylindrical hole of radius 2 cm is drilled through the center of the solid. Find the volume of the remaining solid. (Take

π = \frac{22}{7}

) 5marks

A tent is in the shape of a cylinder surmounted by a conical top. The height and diameter of the cylindrical part are 4 m and 6 m respectively. The slant height of the conical part is 5 m.(a) Find the area of the canvas used for making the tent.(b) Find the cost of the canvas if the rate is ₹ 400 per m². 5marks

Assertion: The total surface area of a solid hemisphere is

3 π r^{2}

.Reason: A solid hemisphere consists of a curved surface area of

2 π r^{2}

and a circular base of area

π r^{2}

. 1mark

A hemispherical bowl of internal radius 12 cm is filled with ice-cream. All the ice-cream from the bowl is to be transferred into a cylinder of radius 6 cm and height 10 cm. Find the number of cylinders that can be filled with ice-cream. (Take

π = 3.14

) 3marks

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder. (Take

π = \frac{22}{7}

) 3marks

A toy is in the form of a cone mounted on a hemisphere of the same radius. The diameter of the base of the cone is 10 cm and the total height of the toy is 12 cm. Find the total surface area of the toy. 3marks

The figure shows a vessel in the form of an inverted cone. Its height is 9 cm and its radius is 4.5 cm. Water is poured into it from a cylindrical pipe of radius 1.5 cm at the rate of 4 cm³/s. How much time will it take to fill the cone? 5marks

A solid consists of a cylinder with hemispherical ends. If the total length of the solid is 90 cm and the radius of each hemisphere is 21 cm, find the volume of the solid. (Take

π = \frac{22}{7}

) 5marks

A cubical block of side 10 cm is surmounted by a hemisphere. Find the greatest diameter the hemisphere can have. What is the surface area of the remaining solid? (Take

π = \frac{22}{7}

) 5marks

Assertion: If a cone is carved out of a hemisphere of the same radius, the volume of the cone is

\frac{1}{3}

of the volume of the hemisphere.Reason: The formula for the volume of a cone is

\frac{1}{3} π r^{2} h

and the formula for the volume of a hemisphere is

\frac{2}{3} π r^{3}

. 1mark

A cylindrical tank of radius 2 m and height 7 m is overflowing with milk. The milk is to be emptied into conical jugs of radius 50 cm and height 1 m. How many jugs are needed to empty the tank? (Take

π = \frac{22}{7}

) 5marks

A decorative wooden block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 10 cm, and the hemisphere fixed on the top has a diameter of 8.4 cm. Find the total surface area of the block. (Take

π = \frac{22}{7}

) 3marks

A solid right circular cone of radius 3 cm and height 4 cm is placed over a hemisphere of radius 3 cm. Find the total surface area of the combined solid. 3marks

A canal is 5 m wide and 3 m deep. Water in the canal flows with a speed of 10 km/hour. How many hectares of land can it irrigate in 30 minutes if 9 cm of standing water is required for irrigation? 5marks

Assertion: The surface area of a solid sphere is

4 π r^{2}

.Reason: A sphere can be thought of as a limiting case of a cylinder where the height equals the diameter and the radius of the base also equals the radius of the sphere. 1mark

A bird-bath is in the form of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.4 m and its radius is 35 cm. Find the total surface area of the bird-bath. (Take

π = \frac{22}{7}

) 3marks

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 5 mm. Find the surface area of the capsule. (Take

π = 3.14

) 3marks

Two identical solid cones, each of height 10 cm and radius 6 cm, are joined at their bases. Find the total surface area of the new solid. 3marks

From a solid cylinder of height 24 cm and diameter 14 cm, a cone of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. (Take

π = \frac{22}{7}

) 5marks

Answers

Volume of cylinder =

π r^{2} h = \frac{22}{7} \times (1.5)^{2} \times 3.5 = \frac{22}{7} \times 2.25 \times 3.5 = 22 \times 2.25 \times 0.5 = 24.75

m³.Volume of cone =

\frac{1}{3} π r^{2} h = \frac{1}{3} \times 24.75 = 8.25

m³.Difference in volume =

24.75 - 8.25 = 16.5

m³. 3marks

Volume of cone =

\frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (3.5)^{2} \times 6 = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 6 = 22 \times 12.25 \times 0.666... = 50.25 \times 2 = 100.5

cm³.Volume of hemisphere =

\frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times (3.5)^{3} = \frac{2}{3} \times \frac{22}{7} \times 42.875 = \frac{2}{3} \times 134.75 = 89.833...

cm³.Total volume of solid =

100.5 + 89.833... = 190.333...

cm³.Volume of cylindrical hole =

π r^{2} h = \frac{22}{7} \times 2^{2} \times 6 = \frac{22}{7} \times 4 \times 6 = \frac{528}{7} = 75.428...

cm³.Volume of remaining solid =

190.333... - 75.428... = 114.905...

cm³.Let's re-calculate carefully:Volume of cone =

\frac{1}{3} π (3.5)^{2} (6) = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 6 = 22 \times 12.25 \times \frac{2}{7} = 22 \times 1.75 \times 2 = 77

cm³.Volume of hemisphere =

\frac{2}{3} π (3.5)^{3} = \frac{2}{3} \times \frac{22}{7} \times 42.875 = \frac{2}{3} \times 134.75 = 89.83

cm³.Total volume =

77 + 89.83 = 166.83

cm³.Volume of cylindrical hole =

π (2)^{2} (6) = π \times 4 \times 6 = 24 π = 24 \times \frac{22}{7} = \frac{528}{7} = 75.43

cm³.Volume of remaining solid =

166.83 - 75.43 = 91.4

cm³. 5marks

(a) Radius of cylinder (r) = Diameter/2 = 6/2 = 3 m. Height of cylinder (h) = 4 m. Slant height of cone (l) = 5 m.Area of canvas = CSA of cylinder + CSA of coneCSA of cylinder =

2 π r h = 2 \times \frac{22}{7} \times 3 \times 4 = \frac{528}{7}

m².CSA of cone =

π r l = \frac{22}{7} \times 3 \times 5 = \frac{330}{7}

m².Total area =

\frac{528}{7} + \frac{330}{7} = \frac{858}{7} = 122.57

m² (approx). [3 marks](b) Cost of canvas = Area × Rate =

\frac{858}{7} \times 400 = \frac{343200}{7} = 49028.57

(approx). 2marks

(a) Assertion is True. A solid hemisphere has a curved surface area

2 π r^{2}

and a circular base of area

π r^{2}

, so its total surface area is

2 π r^{2} + π r^{2} = 3 π r^{2}

. 1mark

Volume of hemispherical bowl =

\frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times (12)^{3} = \frac{2}{3} \times \frac{22}{7} \times 1728 = 2 \times 22 \times 12 \times 8 = 7234.56

cm³ (approx).Volume of cylinder =

π r^{2} h = \frac{22}{7} \times (6)^{2} \times 10 = \frac{22}{7} \times 36 \times 10 = \frac{7920}{7} = 1131.43

cm³ (approx).Number of cylinders = Volume of bowl / Volume of cylinder =

7234.56 / 1131.43 = 6.4

.Since we can only fill complete cylinders, the number of cylinders is 6. 3marks

Volume of sphere =

\frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times (4.2)^{3} = \frac{4}{3} \times \frac{22}{7} \times 74.088 = 4 \times 22 \times 1.4 \times 7.008 = 36.959

cm³.Let's re-calculate: Volume of sphere =

\frac{4}{3} \times \frac{22}{7} \times (4.2)^{3} = \frac{4}{3} \times \frac{22}{7} \times 74.088 = \frac{4 \times 22 \times 74.088}{3 \times 7} = \frac{6521.744}{21} = 310.56

cm³.Volume of cylinder =

π r^{2} h = \frac{22}{7} \times (6)^{2} \times h = \frac{22}{7} \times 36 \times h = \frac{792}{7} h

.Equating volumes:

310.56 = \frac{792}{7} h \Rightarrow h = \frac{310.56 \times 7}{792} = \frac{2173.92}{792} = 2.74

cm (approx). 3marks

Radius of hemisphere (r) = Radius of cone (r) = Diameter/2 = 10/2 = 5 cm.Height of cone (h) = Total height - Height of hemisphere =

12 - 5 = 7

cm.Slant height of cone (l) =

\sqrt{r^{2} + h^{2}} = \sqrt{5^{2} + 7^{2}} = \sqrt{25 + 49} = \sqrt{74}

cm.Total surface area = CSA of hemisphere + CSA of coneCSA of hemisphere =

2 π r^{2} = 2 \times \frac{22}{7} \times 5^{2} = 2 \times \frac{22}{7} \times 25 = \frac{1100}{7}

cm².CSA of cone =

π r l = \frac{22}{7} \times 5 \times \sqrt{74} = \frac{110 \sqrt{74}}{7}

cm².Total surface area =

\frac{1100}{7} + \frac{110 \sqrt{74}}{7} = \frac{110}{7} (10 + \sqrt{74}) \approx \frac{110}{7} (10 + 8.6) = \frac{110 \times 18.6}{7} = \frac{2046}{7} = 292.29

cm² (approx). 3marks

Volume of cone =

\frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (4.5)^{2} \times 9 = \frac{1}{3} \times \frac{22}{7} \times 20.25 \times 9 = 22 \times 20.25 \times 3 = 1336.5

cm³.Rate of water flow = 4 cm³/s.Time to fill the cone = Volume of cone / Rate of flow =

1336.5 / 4 = 334.125

seconds.Convert to minutes:

334.125 / 60 = 5.56875

minutes. 5marks

Radius of hemisphere (r) = 21 cm. Height of cylinder (h) = Total length - 2 * radius of hemisphere =

90 - 2 \times 21 = 90 - 42 = 48

cm.Volume of solid = Volume of cylinder + 2 * Volume of hemisphereVolume of cylinder =

π r^{2} h = \frac{22}{7} \times (21)^{2} \times 48 = \frac{22}{7} \times 441 \times 48 = 22 \times 63 \times 48 = 66528

cm³.Volume of hemisphere =

\frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times (21)^{3} = \frac{2}{3} \times \frac{22}{7} \times 9261 = 2 \times 22 \times 3 \times 3087 = 132 \times 3087 = 407484

cm³. Mistake here,

\frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 2 \times 22 \times 3 \times 21 \times 21 = 132 \times 441 = 58212

cm³.Volume of 2 hemispheres =

2 \times 58212 = 116424

cm³.Total volume =

66528 + 116424 = 182952

cm³. 5marks

Greatest diameter of hemisphere = Edge of the cube = 10 cm. So, radius of hemisphere (r) = 5 cm.Surface area of the solid = Surface area of cube - Area of base of hemisphere + CSA of hemisphereSurface area of cube =

6 \times (s i d e)^{2} = 6 \times (10)^{2} = 600

cm².Area of base of hemisphere =

π r^{2} = \frac{22}{7} \times 5^{2} = \frac{22}{7} \times 25 = \frac{550}{7}

cm².CSA of hemisphere =

2 π r^{2} = 2 \times \frac{22}{7} \times 5^{2} = 2 \times \frac{550}{7} = \frac{1100}{7}

cm².Total surface area =

600 - \frac{550}{7} + \frac{1100}{7} = 600 + \frac{550}{7} = 600 + 78.57 = 678.57

cm² (approx). 5marks

Assertion is True. The volume of a cone is

\frac{1}{3} π r^{2} h

. If the cone is carved out of a hemisphere of the same radius, its height is also r. So, volume of cone =

\frac{1}{3} π r^{2} (r) = \frac{1}{3} π r^{3}

. The volume of hemisphere is

\frac{2}{3} π r^{3}

. Therefore, volume of cone is half the volume of hemisphere, not one-third. The statement is incorrect in saying "one-third". Re-reading the assertion: "If a cone is carved out of a hemisphere of the same radius, the volume of the cone is

\frac{1}{3}

of the volume of the hemisphere." This is incorrect. For a cone carved out of a hemisphere of same radius, the height of the cone is also equal to the radius of the hemisphere. Volume of cone =

\frac{1}{3} π r^{2} (r) = \frac{1}{3} π r^{3}

. Volume of hemisphere =

\frac{2}{3} π r^{3}

. So volume of cone is

\frac{1}{2}

of the volume of hemisphere. Assertion is False. 1mark

Volume of cylinder =

π r^{2} h = \frac{22}{7} \times (2)^{2} \times 7 = \frac{22}{7} \times 4 \times 7 = 88

m³.Volume of conical jug =

\frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (0.5)^{2} \times 1 = \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1 = \frac{5.5}{21} = 0.2619

m³ (approx).Number of jugs = Volume of cylinder / Volume of conical jug =

88 / 0.2619 \approx 336

jugs. 5marks

Radius of hemisphere (r) = Diameter/2 = 8.4/2 = 4.2 cm. Edge of cube (a) = 10 cm.Surface area of the block = Surface area of cube - Area of base of hemisphere + CSA of hemisphereSurface area of cube =

6 \times a^{2} = 6 \times (10)^{2} = 600

cm².Area of base of hemisphere =

π r^{2} = \frac{22}{7} \times (4.2)^{2} = \frac{22}{7} \times 17.64 = 22 \times 2.52 = 55.44

cm².CSA of hemisphere =

2 π r^{2} = 2 \times 55.44 = 110.88

cm².Total surface area =

600 - 55.44 + 110.88 = 600 + 55.44 = 655.44

cm². 3marks

Radius of cone (r) = 3 cm. Height of cone (h) = 4 cm. Radius of hemisphere (r) = 3 cm.Slant height of cone (l) =

\sqrt{r^{2} + h^{2}} = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5

cm.Total surface area = CSA of cone + CSA of hemisphereCSA of cone =

π r l = \frac{22}{7} \times 3 \times 5 = \frac{330}{7}

cm².CSA of hemisphere =

2 π r^{2} = 2 \times \frac{22}{7} \times 3^{2} = 2 \times \frac{22}{7} \times 9 = \frac{396}{7}

cm².Total surface area =

\frac{330}{7} + \frac{396}{7} = \frac{726}{7} = 103.71

cm² (approx). 3marks

Volume of water flowing per hour from the canal:Width = 8 m, Depth = 3 m, Speed = 10 km/hr = 10000 m/hr.Volume per hour =

8 \times 3 \times 10000 = 240000

m³.In 30 minutes (0.5 hours), volume of water =

240000 \times 0.5 = 120000

m³.This volume irrigates the land, requiring 9 cm (0.09 m) of standing water.Area of land irrigated = Volume of water / Height of standing water =

120000 / 0.09 = 1333333.33

m².Convert to hectares: 1 hectare = 10000 m².Area in hectares =

1333333.33 / 10000 = 133.33

hectares. 5marks

Assertion is True. The surface area of a solid sphere is indeed

4 π r^{2}

. 1mark

Radius of cylinder (r) = Radius of hemisphere (r) = 35 cm. Height of cylinder (h) = Total height - radius of hemisphere =

1.4 m - 35 c m = 140 c m - 35 c m = 105 c m

.Total surface area = Area of top surface of cylinder + CSA of cylinder + CSA of hemisphereArea of top surface of cylinder =

π r^{2} = \frac{22}{7} \times (35)^{2} = \frac{22}{7} \times 1225 = 22 \times 175 = 3850

cm².CSA of cylinder =

2 π r h = 2 \times \frac{22}{7} \times 35 \times 105 = 2 \times 22 \times 5 \times 105 = 23100

cm².CSA of hemisphere =

2 π r^{2} = 2 \times \frac{22}{7} \times (35)^{2} = 2 \times 3850 = 7700

cm².Total surface area =

3850 + 23100 + 7700 = 34650

cm². 3marks

Length of capsule = 14 mm. Diameter of capsule = 5 mm, so radius (r) = 2.5 mm.Height of cylinder (h) = Total length - 2 * radius of hemisphere =

14 - 2 \times 2.5 = 14 - 5 = 9

mm.Surface area of capsule = CSA of cylinder + 2 * CSA of hemisphereCSA of cylinder =

2 π r h = 2 \times 3.14 \times 2.5 \times 9 = 141.3

mm².CSA of hemisphere =

2 π r^{2} = 2 \times 3.14 \times (2.5)^{2} = 2 \times 3.14 \times 6.25 = 39.25

mm².Total surface area =

141.3 + 2 \times 39.25 = 141.3 + 78.5 = 219.8

mm². 3marks

Height of each cone (h) = 10 cm. Radius of each cone (r) = 6 cm.Slant height of cone (l) =

\sqrt{r^{2} + h^{2}} = \sqrt{6^{2} + 10^{2}} = \sqrt{36 + 100} = \sqrt{136}

cm.When two cones are joined at their bases, the new solid has two conical surfaces.Total surface area = 2 * CSA of coneCSA of cone =

π r l = π \times 6 \times \sqrt{136} = 6 π \sqrt{136}

cm².Total surface area =

2 \times 6 π \sqrt{136} = 12 π \sqrt{136} = 12 π \times 2 \sqrt{34} = 24 π \sqrt{34}

cm².Let's use

π = 3.14

24 \times 3.14 \times \sqrt{136} \approx 24 \times 3.14 \times 11.66 = 877.8

cm² (approx). 3marks

Height of cylinder (H) = 24 cm. Diameter of cylinder = 14 cm, so radius (R) = 7 cm.Height of cone (h) = 24 cm. Diameter of cone = 14 cm, so radius (r) = 7 cm.Total surface area of remaining solid = CSA of cylinder + CSA of cone + Area of base of cylinderCSA of cylinder =

2 π R H = 2 \times \frac{22}{7} \times 7 \times 24 = 2 \times 22 \times 24 = 1056

cm².CSA of cone =

π r h = \frac{22}{7} \times 7 \times 24 = 22 \times 24 = 528

cm².Area of base of cylinder =

π R^{2} = \frac{22}{7} \times 7^{2} = \frac{22}{7} \times 49 = 22 \times 7 = 154

cm².Total surface area =

1056 + 528 + 154 = 1738

cm². 5marks

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Surface Areas and Volumes — Competency Based Questions

Competency based questions

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