Surface Areas and Volumes Class 10 Mathematics

20 Questions | 74 Marks

Two identical solid cubes, each of volume 125 cm³, are joined end to end. Find the total surface area of the resulting cuboid. 3marks

A solid hemisphere of radius 7 cm is mounted on a hollow cylinder of the same radius. If the total height of the vessel formed is 26 cm, find the inner surface area of the vessel. 3marks

A toy is in the form of a cone of radius 3 cm mounted on a hemisphere of the same radius. The total height of the toy is 18 cm. Find the total surface area of the toy. 5marks

A cubical block of side 10 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. 3marks

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. 5marks

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 20 mm and the diameter of the capsule is 6 mm. Find its surface area. 3marks

A tent is in the shape of a cylinder surmounted by a conical top. The height and diameter of the cylindrical part are 4.2 m and 4 m respectively, and the slant height of the conical top is 2.8 m. Find the area of the canvas used for making the tent. 5marks

From a solid cylinder whose height is 2.8 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 to the nearest cm². 3marks

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 12 cm, and its base is of radius 4.2 cm, find the total surface area of the article. 5marks

Two metallic right circular cones, one at the top and one at the bottom, are joined together along their bases. The radius of the common base is 8.4 cm and the heights of the cones are 7 cm and 10.5 cm respectively. Find the volume of the solid formed. (Take π = 22/7) 5marks

A solid toy is in the form of a right circular cone surmounted on a hemisphere. The radius of the hemisphere and that of the cone are both 3.5 cm and the total height of the toy is 15.5 cm. Find the volume of the toy. 5marks

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 10 cm × 8 cm × 4 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand. 5marks

A conical vessel, with base radius 5 cm and height 12 cm, is filled with water. When lead shots, each of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped into the vessel. 5marks

A solid iron pole consists of a cylinder of height 150 cm and base diameter 14 cm, which is surmounted by another cylinder of height 60 cm and radius 7 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 7.5 g mass. (Use π = 3.14) 5marks

A solid consisting of a right circular cone of height 144 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm. 5marks

A spherical glass vessel has a cylindrical neck 7 cm long, 1.5 cm in diameter; the diameter of the spherical part is 8 cm. By measuring the amount of water it holds, a student finds its volume to be 300 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14. 4marks

Two solid cones are joined along their bases. The radius of the common base is 10 cm and the heights of the cones are 12 cm and 15 cm. Find the volume of the solid. 3marks

A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed into some identical cones of height 12 cm and radius 4.5 cm and capped by a hemisphere. Find the number of cones needed to fill the ice-cream. 5marks

A hemispherical bowl is made of iron. The thickness of the bowl is 1 cm. If the inner radius is 5 cm, find the volume of the iron used to make the bowl. 3marks

A solid cube of side 9 cm is melted and recast into a cuboid with base dimensions 18 cm × 12 cm. Find the height of the cuboid. 2marks

Answers

The volume of each cube is 125 cm³. Let the side of each cube be 'a'. Then,

a^{3} = 125 \Rightarrow a = 5

cm.When two identical cubes are joined end to end, the resulting cuboid has dimensions:Length (

l

) = 5 + 5 = 10 cmBreadth (

b

) = 5 cmHeight (

h

) = 5 cmTotal Surface Area of the cuboid =

2 (l b + b h + h l) = 2 (10 \times 5 + 5 \times 5 + 5 \times 10) = 2 (50 + 25 + 50) = 2 (125) = 250

cm². 3marks

Radius of the hemisphere and cylinder,

r = 7

cm.Total height of the vessel = 26 cm.Height of the cylindrical part,

h = 26 - r = 26 - 7 = 19

cm.Inner surface area of the vessel = CSA of hemisphere + CSA of cylinderCSA of hemisphere

= 2 π r^{2} = 2 \times \frac{22}{7} \times 7^{2} = 2 \times 22 \times 7 = 308

cm².CSA of cylinder

= 2 π r h = 2 \times \frac{22}{7} \times 7 \times 19 = 2 \times 22 \times 19 = 836

cm².Total inner surface area

= 308 + 836 = 1144

cm². 3marks

Radius of the hemisphere and cone,

r = 3.5

cm.Total height of the toy = 15.5 cm.Height of the conical part,

h = 15.5 - 3.5 = 12

cm.Slant height of the cone,

l = \sqrt{r^{2} + h^{2}} = \sqrt{(3.5)^{2} + (12)^{2}} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5

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

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

cm².CSA of cone

= π r l = \frac{22}{7} \times 3.5 \times 12.5 = 22 \times 0.5 \times 12.5 = 11 \times 12.5 = 137.5

cm².Total surface area

= 77 + 137.5 = 214.5

cm². 5marks

Side of the cubical block = 10 cm.The greatest diameter the hemisphere can have is equal to the side of the cube, so diameter = 10 cm.Radius of the hemisphere,

r = \frac{10}{2} = 5

cm.Surface area of the solid = Surface area of cube - Area of base of hemisphere + CSA of hemisphereSurface area of cube

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

cm².Area of base of hemisphere

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

cm².CSA of hemisphere

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

cm².Total surface area

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

cm². 3marks

Let the edge of the cube be 'a'. The diameter of the hemisphere is 'a', so its radius is

r = a / 2

.Surface area of the remaining solid = Surface area of cube - Area of base of hemisphere + CSA of hemisphereSurface area of cube

= 6 a^{2}

.Area of base of hemisphere

= π r^{2} = π (a / 2)^{2} = \frac{π a^{2}}{4}

.CSA of hemisphere

= 2 π r^{2} = 2 π (a / 2)^{2} = 2 π \frac{a^{2}}{4} = \frac{π a^{2}}{2}

.Total surface area

= 6 a^{2} - \frac{π a^{2}}{4} + \frac{π a^{2}}{2} = 6 a^{2} + \frac{π a^{2}}{4} = a^{2} (6 + \frac{π}{4})

. 5marks

Length of the entire capsule = 20 mm.Diameter of the capsule = 5 mm, so radius

r = 2.5

mm.Height of the cylindrical part,

h = 20 - 2 r = 20 - 2 (2.5) = 20 - 5 = 15

mm.Surface area of the capsule = CSA of cylinder + CSA of two hemispheresCSA of cylinder

= 2 π r h = 2 \times \frac{22}{7} \times 2.5 \times 15 = \frac{1100}{7} \times \frac{3}{1} = \frac{3300}{7}

mm².CSA of two hemispheres = CSA of a sphere

= 4 π r^{2} = 4 \times \frac{22}{7} \times (2.5)^{2} = 4 \times \frac{22}{7} \times 6.25 = \frac{550}{7}

mm².Total surface area

= \frac{3300}{7} + \frac{550}{7} = \frac{3850}{7} = 550

mm². 3marks

Diameter of the cylindrical part = 4 m, so radius

r = 2

m.Height of the cylindrical part = 4.2 m.Slant height of the conical top = 2.8 m.Area of the canvas used for the tent = CSA of cylinder + CSA of coneCSA of cylinder

= 2 π r h = 2 \times \frac{22}{7} \times 2 \times 4.2 = 2 \times 22 \times 2 \times 0.6 = 52.8

m².CSA of cone

= π r l = \frac{22}{7} \times 2 \times 2.8 = 22 \times 2 \times 0.4 = 17.6

m².Total area of canvas

= 52.8 + 17.6 = 70.4

m². 5marks

Height of the cylinder,

h = 2.8

cm.Diameter of the cylinder = 1.4 cm, so radius

r = 0.7

cm.The conical cavity has the same height and diameter.Total surface area of the remaining solid = CSA of cylinder - Area of base of cone + CSA of coneTotal surface area

= 2 π r h - π r^{2} + π r l

, where

l

is the slant height of the cone.

l = \sqrt{r^{2} + h^{2}} = \sqrt{(0.7)^{2} + (2.8)^{2}} = \sqrt{0.49 + 7.84} = \sqrt{8.33} \approx 2.886

cm.Total surface area

= 2 π (0.7) (2.8) - π (0.7)^{2} + π (0.7) (2.886)

= π [2 (0.7) (2.8) - (0.7)^{2} + (0.7) (2.886)]

= π [3.92 - 0.49 + 2.0202]

= π [5.4502] \approx 3.14 \times 5.4502 \approx 17.11

cm².Alternatively, the question states the conical cavity has the same height and diameter as the cylinder. This implies the cone fits exactly within the cylinder's height.Surface area = CSA of cylinder + CSA of coneCSA of cylinder

= 2 π r h = 2 \times \frac{22}{7} \times 0.7 \times 2.8 = 2 \times 22 \times 0.1 \times 2.8 = 12.32

cm².Slant height of cone,

l = \sqrt{r^{2} + h^{2}} = \sqrt{(0.7)^{2} + (2.8)^{2}} = \sqrt{0.49 + 7.84} = \sqrt{8.33} \approx 2.886

cm.CSA of cone

= π r l = \frac{22}{7} \times 0.7 \times 2.886 = 22 \times 0.1 \times 2.886 = 6.35

cm².Total surface area =

12.32 + 6.35 = 18.67

cm².Let's re-read the problem: "conical cavity ... hollowed out". This means the base of the cone is removed from the cylinder's top surface.Total surface area = (CSA of cylinder - Area of base of cylinder) + Area of base of cone + CSA of cone.Since the cone base is the same as cylinder base:Total surface area = CSA of cylinder + CSA of cone.CSA of cylinder

= 2 π r h = 2 \times \frac{22}{7} \times 0.7 \times 2.8 = 12.32

cm².Slant height of cone

l = \sqrt{(0.7)^{2} + (2.8)^{2}} = \sqrt{0.49 + 7.84} = \sqrt{8.33} \approx 2.886

cm.CSA of cone

= π r l = \frac{22}{7} \times 0.7 \times 2.886 = 6.35

cm².Total Surface Area =

12.32 + 6.35 = 18.67

cm².Rounding to the nearest cm²: 19 cm². 3marks

Height of the cylinder = 12 cm.Radius of the base = 4.2 cm.The article consists of a cylinder with a hemisphere scooped out from each end.Total surface area = CSA of cylinder + 2 * CSA of hemisphereCSA of cylinder

= 2 π r h = 2 \times \frac{22}{7} \times 4.2 \times 12 = 2 \times 22 \times 0.6 \times 12 = 316.8

cm².CSA of hemisphere

= 2 π r^{2} = 2 \times \frac{22}{7} \times (4.2)^{2} = 2 \times \frac{22}{7} \times 17.64 = 2 \times 22 \times 2.52 = 110.88

cm².Total surface area

= 316.8 + 2 \times 110.88 = 316.8 + 221.76 = 538.56

cm². 5marks

Radius of the common base,

r = 8.4

cm.Height of the top cone,

h_{1} = 7

cm.Height of the bottom cone,

h_{2} = 10.5

cm.Volume of the solid = Volume of top cone + Volume of bottom coneVolume of cone

= \frac{1}{3} π r^{2} h

.Volume of top cone

= \frac{1}{3} \times \frac{22}{7} \times (8.4)^{2} \times 7 = \frac{1}{3} \times 22 \times 8.4 \times 8.4 = 22 \times 2.8 \times 8.4 = 517.44

cm³.Volume of bottom cone

= \frac{1}{3} \times \frac{22}{7} \times (8.4)^{2} \times 10.5 = \frac{1}{3} \times 22 \times 8.4 \times 10.5 = 22 \times 2.8 \times 10.5 = 646.8

cm³.Total volume

= 517.44 + 646.8 = 1164.24

cm³. 5marks

Radius of the hemisphere and cone,

r = 3.5

cm.Total height of the toy = 15.5 cm.Height of the conical part,

h = 15.5 - 3.5 = 12

cm.Volume of the toy = Volume of hemisphere + Volume of coneVolume 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 \approx 89.83

cm³.Volume of cone

= \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (3.5)^{2} \times 12 = \frac{1}{3} \times 22 \times 12.25 \times 12 = 22 \times 12.25 \times 4 = 1078

cm³.Total volume

= 89.83 + 1078 = 1167.83

cm³. 5marks

Dimensions of the cuboid:

l = 10

cm,

b = 8

cm,

h = 4

cm.Radius of each conical depression,

r = 0.5

cm.Depth of each conical depression,

h_{c} = 1.4

cm.Volume of wood = Volume of cuboid - 4 * Volume of one conical depressionVolume of cuboid

= l \times b \times h = 10 \times 8 \times 4 = 320

cm³.Volume of one conical depression

= \frac{1}{3} π r^{2} h_{c} = \frac{1}{3} \times \frac{22}{7} \times (0.5)^{2} \times 1.4 = \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4 = \frac{1}{3} \times \frac{22}{7} \times 0.35 = \frac{1}{3} \times 22 \times 0.05 = \frac{1.1}{3} \approx 0.367

cm³.Volume of 4 conical depressions

= 4 \times \frac{1.1}{3} = \frac{4.4}{3} \approx 1.467

cm³.Volume of wood

= 320 - 1.467 = 318.533

cm³. 5marks

Radius of the conical vessel,

R = 5

cm.Height of the conical vessel,

H = 12

cm.Volume of water in the vessel

= \frac{1}{3} π R^{2} H = \frac{1}{3} \times \frac{22}{7} \times 5^{2} \times 12 = \frac{1}{3} \times 22 \times 25 \times 12 = 22 \times 25 \times 4 = 2200

cm³.Volume of water that flows out

= \frac{1}{4} \times 2200 = 550

cm³.Radius of each lead shot,

r = 0.5

cm.Volume of one lead shot

= \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times (0.5)^{3} = \frac{4}{3} \times \frac{22}{7} \times 0.125 = \frac{1}{3} \times \frac{22}{7} \times 0.5 = \frac{11}{21}

cm³.Number of lead shots =

\frac{Volume of water that flows out}{Volume of one lead shot} = \frac{550}{\frac{11}{21}} = 550 \times \frac{21}{11} = 50 \times 21 = 1050

. 5marks

Height of the first cylinder,

h_{1} = 150

cm.Diameter of the first cylinder = 14 cm, so radius

r_{1} = 7

cm.Height of the second cylinder,

h_{2} = 60

cm.Radius of the second cylinder,

r_{2} = 8

cm.Volume of the first cylinder

= π r_{1}^{2} h_{1} = 3.14 \times (7)^{2} \times 150 = 3.14 \times 49 \times 150 = 23079

cm³.Volume of the second cylinder

= π r_{2}^{2} h_{2} = 3.14 \times (8)^{2} \times 60 = 3.14 \times 64 \times 60 = 12057.6

cm³.Total volume of the pole

= 23079 + 12057.6 = 35136.6

cm³.Mass of 1 cm³ of iron = 7.5 g.Mass of the pole

= 35136.6 \times 7.5 = 263524.5

= 263.5245

kg. 5marks

Radius of the hemisphere and cone,

r = 60

cm.Height of the cone,

h_{c o n e} = 120

cm.Height of the cylinder,

h_{c y l} = 180

cm.Radius of the cylinder,

R = 60

cm.Volume of the solid = Volume of hemisphere + Volume of coneVolume of hemisphere

= \frac{2}{3} π r^{3} = \frac{2}{3} π (60)^{3} = \frac{2}{3} π (216000) = 144000 π

cm³.Volume of cone

= \frac{1}{3} π r^{2} h_{c o n e} = \frac{1}{3} π (60)^{2} \times 120 = \frac{1}{3} π (3600) \times 120 = 144000 π

cm³.Total volume of the solid

= 144000 π + 144000 π = 288000 π

cm³.Volume of the cylinder

= π R^{2} h_{c y l} = π (60)^{2} \times 180 = π (3600) \times 180 = 648000 π

cm³.Volume of water left = Volume of cylinder - Volume of the solidVolume of water left

= 648000 π - 288000 π = 360000 π

cm³. 5marks

Radius of the spherical part = 8 cm / 2 = 4 cm.Length of the cylindrical neck = 7 cm.Diameter of the cylindrical neck = 1.5 cm, so radius

r_{n e c k} = 0.75

cm.Volume of the spherical part

= \frac{4}{3} π r^{3} = \frac{4}{3} \times 3.14 \times (4)^{3} = \frac{4}{3} \times 3.14 \times 64 = 267.95

cm³ (approx).Volume of the cylindrical neck

= π r_{n e c k}^{2} h_{n e c k} = 3.14 \times (0.75)^{2} \times 7 = 3.14 \times 0.5625 \times 7 = 12.36

cm³ (approx).Total volume of the vessel

= 267.95 + 12.36 = 280.31

cm³.The student's measured volume is 345 cm³. The calculated volume is approximately 280.31 cm³. Therefore, she is not correct. 4marks

Radius of the common base,

r = 10

cm.Height of the first cone,

h_{1} = 12

cm.Height of the second cone,

h_{2} = 15

cm.Volume of the solid = Volume of first cone + Volume of second coneVolume of first cone

= \frac{1}{3} π r^{2} h_{1} = \frac{1}{3} π (10)^{2} \times 12 = \frac{1}{3} π (100) \times 12 = 400 π

cm³.Volume of second cone

= \frac{1}{3} π r^{2} h_{2} = \frac{1}{3} π (10)^{2} \times 15 = \frac{1}{3} π (100) \times 15 = 500 π

cm³.Total volume

= 400 π + 500 π = 900 π

cm³. 3marks

Radius of the cylinder,

R_{c y l} = 6

cm.Height of the cylinder,

H_{c y l} = 15

cm.Radius of the cone,

r_{c o n e} = 4.5

cm.Height of the cone,

h_{c o n e} = 12

cm.The cones are capped by a hemisphere, so the radius of the hemisphere is also

r_{h e m i} = 4.5

cm.Volume of ice-cream in the cylinder

= π R_{c y l}^{2} H_{c y l} = π \times 6^{2} \times 15 = π \times 36 \times 15 = 540 π

cm³.Volume of one cone with hemisphere cap = Volume of cone + Volume of hemisphereVolume of cone

= \frac{1}{3} π r_{c o n e}^{2} h_{c o n e} = \frac{1}{3} π (4.5)^{2} \times 12 = \frac{1}{3} π (20.25) \times 12 = 81 π

cm³.Volume of hemisphere

= \frac{2}{3} π r_{h e m i}^{3} = \frac{2}{3} π (4.5)^{3} = \frac{2}{3} π (91.125) = 60.75 π

cm³.Volume of one filled cone

= 81 π + 60.75 π = 141.75 π

cm³.Number of cones needed

= \frac{Volume of ice-cream}{Volume of one filled cone} = \frac{540 π}{141.75 π} = \frac{540}{141.75} = 3.809

.Since the number of cones must be an integer and the ice-cream needs to be distributed, we round up. However, the problem implies the given volume of ice-cream exactly fills a certain number of cones. Let's recheck calculations.Volume of ice-cream =

540 π

.Volume of one filled cone =

141.75 π

.Number of cones =

540 / 141.75 = 3.809...

This implies the question might have a typo or the answer is not an integer. If we consider the constraints strictly, it should be an integer. Let's assume the question meant that the ice-cream from the cylinder completely fills the cones.Let's check if rounding to fewer decimal places for

π

or intermediate values makes a difference.Let's assume

π

is a common factor.Number of cones =

540 / 141.75 = 3.809...

Let's review the question statement and NCERT examples. Usually these questions result in integer answers.If we assume the radii were different, e.g., if cone radius was such that it leads to integer answer.Let's stick to the calculated value. The phrasing "Find the number of cones needed" suggests an integer. This indicates a potential issue with the problem's numbers.However, if we must provide an answer, and assuming the problem intends for the ice cream to be distributed, perhaps it is asking for the minimum number of cones required to hold all the ice cream. In that case, 4 cones would be needed.Let's re-read the NCERT text. Example 5: volume of air = 1128.75 m3. Example 6: actual capacity = 163.54 cm3. Example 7: volume of toy = 25.12 cm3. These are not always integers.Let's re-calculate carefully:Volume of ice-cream in cylinder =

540 π

cm³.Volume of cone =

\frac{1}{3} π (4.5)^{2} \times 12 = \frac{1}{3} π (20.25) \times 12 = 81 π

cm³.Volume of hemisphere =

\frac{2}{3} π (4.5)^{3} = \frac{2}{3} π (91.125) = 60.75 π

cm³.Volume of one filled cone =

81 π + 60.75 π = 141.75 π

cm³.Number of cones =

\frac{540 π}{141.75 π} = \frac{540}{141.75} = 3.8095...

Given the context of a school problem, it's highly probable there's a rounding issue or intended integer result. If we must choose an integer, and assuming no ice cream is wasted, this problem setup is problematic.However, if we consider that the question implies the ice cream is fully distributed, and the number of cones must be an integer, then 4 cones would be the logical answer to hold all the ice-cream, with some left over.Let's assume there's a typo and proceed as if it were an integer. Suppose the radius of the cone was 4 cm.Volume of cone =

\frac{1}{3} π (4)^{2} \times 12 = \frac{1}{3} π (16) \times 12 = 64 π

.Volume of hemisphere =

\frac{2}{3} π (4)^{3} = \frac{2}{3} π (64) = 42.67 π

.Total =

106.67 π

540 / 106.67 \approx 5

.Let's re-examine the possibility of a calculation error or a common factor mistake.

141.75 = 141 \frac{3}{4} = \frac{564 + 3}{4} = \frac{567}{4}

.Number of cones =

\frac{540 π}{\frac{567}{4} π} = \frac{540 \times 4}{567} = \frac{2160}{567}

.Both are divisible by 9:

2160 / 9 = 240

567 / 9 = 63

. So,

\frac{240}{63}

.Both are divisible by 3:

240 / 3 = 80

63 / 3 = 21

. So,

\frac{80}{21}

\frac{80}{21} \approx 3.8095

.This non-integer result is consistently obtained. It is highly likely that the question has been designed such that the result is not an integer, or there's a typo. In a real exam, this might be a 4 or 5 marker, testing calculation ability and interpretation. If the student must fill cones, 4 would be the minimum number. If it implies an exact fill, then the numbers are flawed for an integer answer.Let's assume the question implies exactly filling the cones.If we are forced to give an integer, and it is "needed", it implies capability to hold. So 4 cones.However, if the question implies an exact amount of ice cream distributed into an integer number of cones, there is an error.Let's assume there are 4 cones. 5marks

Inner radius of the hemispherical bowl,

r = 5

cm.Thickness of the bowl = 1 cm.Outer radius of the hemispherical bowl,

R = r + thickness = 5 + 1 = 6

cm.Volume of iron used = Volume of outer hemisphere - Volume of inner hemisphereVolume of iron

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

Volume of iron

= \frac{2}{3} π (6^{3} - 5^{3}) = \frac{2}{3} π (216 - 125) = \frac{2}{3} π (91) = \frac{182 π}{3}

cm³. 3marks

Side of the solid cube = 9 cm.Volume of the cube

= (side)^{3} = 9^{3} = 729

cm³.The cube is melted and recast into a cuboid with base dimensions 18 cm × 12 cm.Let the height of the cuboid be 'h'.Volume of the cuboid

= length \times breadth \times height = 18 \times 12 \times h = 216 h

cm³.Since the volume of the material remains the same:Volume of cuboid = Volume of cube

216 h = 729

$h = \frac{729}{216}$ .
Dividing numerator and denominator by 9: $h = \frac{81}{24}$ .
Dividing numerator and denominator by 3: $h = \frac{27}{8}$ .
$h = 3.375$ cm. [2 marks]

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