Surface Areas And Volumes Class 10 Mathematics NCERT Solutions

Q1EXERCISE 12.1

2 cubes each of volume $64 \text{ cm}^{3}$ are joined end to end. Find the surface area of the resulting cuboid.

Solution

Given: Volume of each cube =

64 \text{ cm}^3

.

To Find: The surface area of the resulting cuboid.

Solution: Let the edge of each cube be 'a'. The volume of a cube is given by the formula

V = a^3

.

a^3 = 64 \text{ cm}^3

a = \sqrt[3]{64} = 4 \text{ cm}

When two cubes are joined end to end, they form a cuboid. The dimensions of the resulting cuboid are: Length (

l

) =

4 \text{ cm} + 4 \text{ cm} = 8 \text{ cm}

Breadth (

b

) =

4 \text{ cm}

Height (

h

) =

4 \text{ cm}

Formula: The surface area of a cuboid is given by

2(lb + bh + hl)

.

Calculation: Surface Area =

2 \times (8 \times 4 + 4 \times 4 + 4 \times 8)

= 2 \times (32 + 16 + 32)

= 2 \times (80)

= 160 \text{ cm}^2

Final Answer: The surface area of the resulting cuboid is

160 \text{ cm}^2

.

Q2EXERCISE 12.1

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 of the vessel.

Solution

Given: A vessel in the shape of a hollow hemisphere mounted by a hollow cylinder. Diameter of the hemisphere =

14 \text{ cm}

. Total height of the vessel =

13 \text{ cm}

.

To Find: The inner surface area of the vessel.

Solution: Radius of the hemisphere (

r

) = Diameter / 2 =

14 / 2 = 7 \text{ cm}

. Since the cylinder is mounted on the hemisphere, the radius of the cylinder is also

r = 7 \text{ cm}

.

The height of the hemispherical part is equal to its radius, which is

7 \text{ cm}

. Height of the cylindrical part (

h

) = Total height of the vessel - Height of the hemisphere

h = 13 \text{ cm} - 7 \text{ cm} = 6 \text{ cm}

The inner surface area of the vessel is the sum of the curved surface area (CSA) of the cylinder and the CSA of the hemisphere.

Formula: CSA of cylinder =

2\pi rh

CSA of hemisphere =

2\pi r^2

Inner surface area =

2\pi rh + 2\pi r^2 = 2\pi r(h + r)

Calculation: Using

\pi = \frac{22}{7}

: Inner surface area =

2 \times \frac{22}{7} \times 7 \times (6 + 7)

= 2 \times 22 \times 13

= 44 \times 13

= 572 \text{ cm}^2

Final Answer: The inner surface area of the vessel is

572 \text{ cm}^2

.

Q3EXERCISE 12.1

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution

Given: A toy in the shape of a cone mounted on a hemisphere. Radius of the cone and hemisphere (

r

) =

3.5 \text{ cm}

. Total height of the toy =

15.5 \text{ cm}

.

To Find: The total surface area of the toy.

Solution: The height of the hemispherical part is equal to its radius,

r = 3.5 \text{ cm}

. Height of the conical part (

h

) = Total height of the toy - Height of the hemisphere

h = 15.5 \text{ cm} - 3.5 \text{ cm} = 12 \text{ cm}

First, we need to find the slant height (

l

) of the cone. Formula:

l = \sqrt{r^2 + h^2}

l = \sqrt{(3.5)^2 + (12)^2}

l = \sqrt{12.25 + 144}

l = \sqrt{156.25} = 12.5 \text{ cm}

The total surface area of the toy is the sum of the curved surface area (CSA) of the cone and the CSA of the hemisphere.

Formula: CSA of cone =

\pi rl

CSA of hemisphere =

2\pi r^2

Total surface area =

\pi rl + 2\pi r^2 = \pi r(l + 2r)

Calculation: Using

\pi = \frac{22}{7}

: Total surface area =

\frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5)

= \frac{22}{7} \times \frac{7}{2} \times (12.5 + 7)

= 11 \times (19.5)

= 214.5 \text{ cm}^2

Final Answer: The total surface area of the toy is

214.5 \text{ cm}^2

.

Q4EXERCISE 12.1

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution

Given: A cubical block of side

a = 7 \text{ cm}

. The block is surmounted by a hemisphere.

To Find:

The greatest diameter the hemisphere can have.
The surface area of the solid.

Solution:

The greatest diameter the hemisphere can have is equal to the side of the cube. Therefore, the greatest diameter is $7 \text{ cm}$ .
To find the surface area of the solid: Radius of the hemisphere ( $r$ ) = Diameter / 2 = $7 / 2 = 3.5 \text{ cm}$ . The total surface area of the solid is the sum of the total surface area (TSA) of the cube, the curved surface area (CSA) of the hemisphere, minus the base area of the hemisphere (since it is covered).

Formula: Surface area of solid = TSA of cube + CSA of hemisphere - Base area of hemisphere

= 6a^2 + 2\pi r^2 - \pi r^2

= 6a^2 + \pi r^2

Calculation: Using

a = 7 \text{ cm}

,

r = 3.5 \text{ cm}

, and

\pi = \frac{22}{7}

: Surface area =

6 \times (7)^2 + \frac{22}{7} \times (3.5)^2

= 6 \times 49 + \frac{22}{7} \times (\frac{7}{2}) \times (\frac{7}{2})

= 294 + \frac{22 \times 7}{2 \times 2}

= 294 + \frac{154}{4}

= 294 + 38.5

= 332.5 \text{ cm}^2

Final Answer: The greatest diameter the hemisphere can have is

7 \text{ cm}

. The surface area of the solid is

332.5 \text{ cm}^2

.

Q5EXERCISE 12.1

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

Solution

Given: A hemispherical depression is cut out from one face of a cubical wooden block. The diameter of the hemisphere is

l

. The edge of the cube is also

l

.

To Find: The surface area of the remaining solid.

Solution: Edge of the cube (

a

) =

l

. Diameter of the hemisphere =

l

, so its radius (

r

) =

l/2

.

The surface area of the remaining solid is the sum of the total surface area (TSA) of the cube, the curved surface area (CSA) of the hemispherical depression, minus the area of the circular opening on the face of the cube.

Formula: Surface area of remaining solid = TSA of cube + CSA of hemisphere - Area of circle

= 6a^2 + 2\pi r^2 - \pi r^2

= 6a^2 + \pi r^2

Calculation: Substitute

a = l

and

r = l/2

into the formula: Surface area =

6(l)^2 + \pi (\frac{l}{2})^2

= 6l^2 + \pi \frac{l^2}{4}

We can factor out

l^2

:

= l^2 (6 + \frac{\pi}{4})

Or, we can take a common denominator:

= \frac{24l^2 + \pi l^2}{4} = \frac{l^2(24 + \pi)}{4}

Final Answer: The surface area of the remaining solid is

\frac{l^2}{4}(24 + \pi)

.

Q6EXERCISE 12.1

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 5 mm. Find its surface area.

Solution

Given: A medicine capsule in the shape of a cylinder with two hemispherical ends. Total length of the capsule =

14 \text{ mm}

. Diameter of the capsule =

5 \text{ mm}

.

To Find: The surface area of the capsule.

Solution: Diameter =

5 \text{ mm}

, so radius (

r

) =

5/2 = 2.5 \text{ mm}

. The radius is the same for the cylindrical part and the hemispherical ends.

The length of the cylindrical part (

h

) is the total length minus the lengths of the two hemispherical ends (which is

2 \times

radius).

h = 14 \text{ mm} - 2 \times 2.5 \text{ mm}

h = 14 - 5 = 9 \text{ mm}

The surface area of the capsule is the sum of the curved surface area (CSA) of the cylinder and the surface area of the two hemispheres.

Formula: Surface Area = CSA of cylinder + 2

\times

CSA of hemisphere

= 2\pi rh + 2 \times (2\pi r^2)

= 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r)

Calculation: Using

\pi = \frac{22}{7}

: Surface area =

2 \times \frac{22}{7} \times 2.5 \times (9 + 2 \times 2.5)

= 2 \times \frac{22}{7} \times 2.5 \times (9 + 5)

= 2 \times \frac{22}{7} \times 2.5 \times 14

= 2 \times 22 \times 2.5 \times 2

= 44 \times 5

= 220 \text{ mm}^2

Final Answer: The surface area of the medicine capsule is

220 \text{ mm}^2

.

Q7EXERCISE 12.1

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹500 per $\mathrm{m}^{2}$ . (Note that the base of the tent will not be covered with canvas.)

Solution

Given: A tent in the shape of a cylinder surmounted by a cone. For the cylindrical part: Height (

h

) =

2.1 \text{ m}

. Diameter =

4 \text{ m}

, so radius (

r

) =

2 \text{ m}

. For the conical part: Slant height (

l

) =

2.8 \text{ m}

. Radius (

r

) =

2 \text{ m}

(same as the cylinder). Rate of canvas = ₹500 per

\text{m}^2

.

To Find:

The area of the canvas used.
The cost of the canvas.

Solution:

The area of the canvas used is the sum of the curved surface area (CSA) of the cylindrical part and the CSA of the conical part. The base is not covered.

Formula: Area of canvas = CSA of cylinder + CSA of cone

= 2\pi rh + \pi rl = \pi r(2h + l)

Calculation: Using

\pi = \frac{22}{7}

: Area =

\frac{22}{7} \times 2 \times (2 \times 2.1 + 2.8)

= \frac{44}{7} \times (4.2 + 2.8)

= \frac{44}{7} \times 7

= 44 \text{ m}^2

The cost of the canvas is the area of the canvas multiplied by the rate.

Calculation: Cost = Area

\times

Rate

= 44 \text{ m}^2 \times ₹500/\text{m}^2

= ₹22000

Final Answer: The area of the canvas used is

44 \text{ m}^2

, and the cost of the canvas is ₹22,000.

Q8EXERCISE 12.1

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 to the nearest $\mathrm{cm}^{2}$ .

Solution

Given: A solid cylinder with a conical cavity hollowed out. Height of cylinder and cone (

h

) =

2.4 \text{ cm}

. Diameter of cylinder and cone =

1.4 \text{ cm}

.

To Find: The total surface area of the remaining solid to the nearest

\text{cm}^2

.

Solution: Radius of the cylinder and cone (

r

) = Diameter / 2 =

1.4 / 2 = 0.7 \text{ cm}

.

First, we need to find the slant height (

l

) of the conical cavity. Formula:

l = \sqrt{r^2 + h^2}

l = \sqrt{(0.7)^2 + (2.4)^2}

l = \sqrt{0.49 + 5.76}

l = \sqrt{6.25} = 2.5 \text{ cm}

The total surface area of the remaining solid is the sum of the curved surface area (CSA) of the cylinder, the area of the base of the cylinder, and the CSA of the conical cavity.

Formula: Total Surface Area = CSA of cylinder + Area of base + CSA of cone

= 2\pi rh + \pi r^2 + \pi rl

= \pi r(2h + r + l)

Calculation: Using

\pi = \frac{22}{7}

: Area =

\frac{22}{7} \times 0.7 \times (2 \times 2.4 + 0.7 + 2.5)

= 22 \times 0.1 \times (4.8 + 0.7 + 2.5)

= 2.2 \times (8.0)

= 17.6 \text{ cm}^2

The question asks for the answer to the nearest

\text{cm}^2

. Rounding

17.6 \text{ cm}^2

to the nearest integer gives

18 \text{ cm}^2

.

Final Answer: The total surface area of the remaining solid is approximately

18 \text{ cm}^2

.

Q9EXERCISE 12.1

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

Solution

Given: A wooden article made from a cylinder with a hemisphere scooped out from each end. Height of the cylinder (

h

) =

10 \text{ cm}

. Radius of the base (

r

) =

3.5 \text{ cm}

.

To Find: The total surface area of the article.

Solution: The total surface area of the article is the sum of the curved surface area (CSA) of the cylinder and the CSA of the two hemispherical scoops.

Formula: Total Surface Area = CSA of cylinder + 2

\times

CSA of hemisphere

= 2\pi rh + 2 \times (2\pi r^2)

= 2\pi rh + 4\pi r^2 = 2\pi r(h + 2r)

Calculation: Using

\pi = \frac{22}{7}

: Area =

2 \times \frac{22}{7} \times 3.5 \times (10 + 2 \times 3.5)

= 2 \times \frac{22}{7} \times \frac{7}{2} \times (10 + 7)

= 22 \times 17

= 374 \text{ cm}^2

Final Answer: The total surface area of the article is

374 \text{ cm}^2

.

Q1EXERCISE 12.2

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$ .

Solution

Given: A solid in the shape of a cone standing on a hemisphere. Radius of cone and hemisphere (

r

) =

1 \text{ cm}

. Height of the cone (

h

) = radius =

1 \text{ cm}

.

To Find: The volume of the solid in terms of

\pi

.

Solution: The volume of the solid is the sum of the volume of the cone and the volume of the hemisphere.

Formula: Volume of solid = Volume of cone + Volume of hemisphere

V = \frac{1}{3}\pi r^2 h + \frac{2}{3}\pi r^3

Calculation: Substitute

r = 1 \text{ cm}

and

h = 1 \text{ cm}

into the formula:

V = \frac{1}{3}\pi (1)^2 (1) + \frac{2}{3}\pi (1)^3

V = \frac{1}{3}\pi + \frac{2}{3}\pi

V = \frac{3}{3}\pi

V = \pi \text{ cm}^3

Final Answer: The volume of the solid is

\pi \text{ cm}^3

.

Q2EXERCISE 12.2

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution

Given: A model shaped like a cylinder with two cones at its ends. Total length of the model =

12 \text{ cm}

. Diameter of the model =

3 \text{ cm}

. Height of each cone (

h_{\text{cone}}

) =

2 \text{ cm}

.

To Find: The volume of air contained in the model.

Solution: Radius of the model (

r

) = Diameter / 2 =

3 / 2 = 1.5 \text{ cm}

. The radius is the same for the cylinder and the cones.

Height of the cylindrical part (

h_{\text{cyl}}

) = Total length - 2

\times

Height of one cone

h_{\text{cyl}} = 12 \text{ cm} - 2 \times 2 \text{ cm} = 12 - 4 = 8 \text{ cm}

The volume of air in the model is the sum of the volume of the cylinder and the volumes of the two cones.

Formula: Volume of air = Volume of cylinder + 2

\times

Volume of cone

V = \pi r^2 h_{\text{cyl}} + 2 \times (\frac{1}{3}\pi r^2 h_{\text{cone}})

V = \pi r^2 (h_{\text{cyl}} + \frac{2}{3}h_{\text{cone}})

Calculation: Using

\pi = \frac{22}{7}

:

V = \frac{22}{7} \times (1.5)^2 \times (8 + \frac{2}{3} \times 2)

V = \frac{22}{7} \times 2.25 \times (8 + \frac{4}{3})

V = \frac{22}{7} \times 2.25 \times (\frac{24+4}{3})

V = \frac{22}{7} \times 2.25 \times \frac{28}{3}

V = 22 \times 2.25 \times \frac{4}{3}

V = 22 \times (0.75 \times 3) \times \frac{4}{3}

V = 22 \times 0.75 \times 4

V = 22 \times 3 = 66 \text{ cm}^3

Final Answer: The volume of air contained in the model is

66 \text{ cm}^3

.

Q3EXERCISE 12.2

A gulab jamun, contains sugar syrup up to about $30\%$ of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution

Given: Number of gulab jamuns = 45. Shape of each gulab jamun: a cylinder with two hemispherical ends. Total length =

5 \text{ cm}

. Diameter =

2.8 \text{ cm}

. Syrup content =

30\%

of volume.

To Find: Approximate volume of syrup in 45 gulab jamuns.

Solution: Radius of the gulab jamun (

r

) = Diameter / 2 =

2.8 / 2 = 1.4 \text{ cm}

. Length of the cylindrical part (

h

) = Total length - 2

\times

Radius of hemisphere

h = 5 \text{ cm} - 2 \times 1.4 \text{ cm} = 5 - 2.8 = 2.2 \text{ cm}

Volume of one gulab jamun = Volume of cylinder + Volume of two hemispheres (a sphere).

Formula: Volume of one gulab jamun =

\pi r^2 h + \frac{4}{3}\pi r^3 = \pi r^2 (h + \frac{4}{3}r)

Calculation: Using

\pi = \frac{22}{7}

: Volume =

\frac{22}{7} \times (1.4)^2 \times (2.2 + \frac{4}{3} \times 1.4)

= \frac{22}{7} \times 1.96 \times (2.2 + \frac{5.6}{3})

= 22 \times 0.28 \times (\frac{6.6 + 5.6}{3})

= 6.16 \times (\frac{12.2}{3}) = \frac{75.152}{3} \approx 25.05 \text{ cm}^3

Volume of 45 gulab jamuns =

45 \times \frac{75.152}{3}

= 15 \times 75.152 = 1127.28 \text{ cm}^3

Volume of syrup =

30\%

of the total volume. Volume of syrup =

0.30 \times 1127.28

= 338.184 \text{ cm}^3

Approximately, the volume of syrup is

338 \text{ cm}^3

.

Final Answer: Approximately

338 \text{ cm}^3

of syrup would be found in 45 gulab jamuns.

Q4EXERCISE 12.2

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 15 cm by 10 cm by 3.5 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.

Solution

Given: A pen stand in the shape of a cuboid with 4 conical depressions. Cuboid dimensions:

l = 15 \text{ cm}

,

b = 10 \text{ cm}

,

h = 3.5 \text{ cm}

. Conical depression: radius (

r

) =

0.5 \text{ cm}

, depth (

h_{\text{cone}}

) =

1.4 \text{ cm}

.

To Find: The volume of wood in the entire stand.

Solution: Volume of the cuboid =

l \times b \times h

= 15 \times 10 \times 3.5 = 525 \text{ cm}^3

Volume of one conical depression =

\frac{1}{3}\pi r^2 h_{\text{cone}}

Using

\pi = \frac{22}{7}

:

= \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 22 \times 0.25 \times 0.2

= \frac{1}{3} \times 22 \times 0.05 = \frac{1.1}{3} \text{ cm}^3

Volume of four conical depressions =

4 \times \frac{1.1}{3} = \frac{4.4}{3} \text{ cm}^3

.

Volume of wood in the stand = Volume of cuboid - Volume of four conical depressions

= 525 - \frac{4.4}{3}

= \frac{1575 - 4.4}{3} = \frac{1570.6}{3}

\approx 523.53 \text{ cm}^3

Final Answer: The volume of wood in the entire stand is approximately

523.53 \text{ cm}^3

.

Q5EXERCISE 12.2

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.

Solution

Given: An inverted conical vessel filled with water. Cone dimensions: height (

H

) =

8 \text{ cm}

, radius (

R

) =

5 \text{ cm}

. Lead shots are spheres with radius (

r

) =

0.5 \text{ cm}

. Volume of water that flows out =

\frac{1}{4}

of the total volume of water.

To Find: The number of lead shots dropped in the vessel.

Solution: Volume of water in the cone = Volume of the cone

V_{\text{cone}} = \frac{1}{3}\pi R^2 H

= \frac{1}{3}\pi (5)^2 (8) = \frac{200\pi}{3} \text{ cm}^3

Volume of water that flows out =

\frac{1}{4} \times V_{\text{cone}}

= \frac{1}{4} \times \frac{200\pi}{3} = \frac{50\pi}{3} \text{ cm}^3

Volume of one spherical lead shot:

V_{\text{shot}} = \frac{4}{3}\pi r^3

= \frac{4}{3}\pi (0.5)^3 = \frac{4}{3}\pi (0.125) = \frac{0.5\pi}{3} \text{ cm}^3

Let 'n' be the number of lead shots dropped. The total volume of 'n' lead shots is equal to the volume of water that flows out.

n \times V_{\text{shot}} = \text{Volume of water flowed out}

n \times \frac{0.5\pi}{3} = \frac{50\pi}{3}

Divide both sides by

\frac{\pi}{3}

:

n \times 0.5 = 50

n = \frac{50}{0.5} = 100

Final Answer: The number of lead shots dropped in the vessel is 100.

Q6EXERCISE 12.2

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that $1 \text{ cm}^{3}$ of iron has approximately 8 g mass. (Use $\pi=3.14$ )

Solution

Given: A solid iron pole made of two cylinders. Large cylinder: height (

H

) =

220 \text{ cm}

, diameter =

24 \text{ cm}

. Small cylinder: height (

h

) =

60 \text{ cm}

, radius (

r

) =

8 \text{ cm}

. Density of iron =

8 \text{ g/cm}^3

.

To Find: The mass of the pole.

Solution: Radius of the large cylinder (

R

) = Diameter / 2 =

24 / 2 = 12 \text{ cm}

.

Total volume of the pole = Volume of large cylinder + Volume of small cylinder.

Formula: Volume =

\pi R^2 H + \pi r^2 h = \pi (R^2 H + r^2 h)

Calculation: Using

\pi = 3.14

: Volume =

3.14 \times ((12)^2 \times 220 + (8)^2 \times 60)

= 3.14 \times (144 \times 220 + 64 \times 60)

= 3.14 \times (31680 + 3840)

= 3.14 \times 35520

= 111532.8 \text{ cm}^3

Mass of the pole = Volume

\times

Density

= 111532.8 \text{ cm}^3 \times 8 \text{ g/cm}^3

= 892262.4 \text{ g}

To convert the mass to kilograms, divide by 1000: Mass =

\frac{892262.4}{1000} = 892.2624 \text{ kg}

.

Final Answer: The mass of the pole is approximately

892.26 \text{ kg}

.

Q7EXERCISE 12.2

A solid consisting of a right circular cone of height 120 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.

Solution

Given: A solid (cone on a hemisphere) is placed in a cylinder filled with water. Solid dimensions: Cone: height (

h_{\text{cone}}

) =

120 \text{ cm}

, radius (

r

) =

60 \text{ cm}

. Hemisphere: radius (

r

) =

60 \text{ cm}

. Cylinder dimensions: Height (

H_{\text{cyl}}

) =

180 \text{ cm}

, radius (

R_{\text{cyl}}

) =

60 \text{ cm}

.

To Find: The volume of water left in the cylinder.

Solution: The volume of water left in the cylinder is the volume of the cylinder minus the volume of the solid immersed in it.

Volume of the solid = Volume of cone + Volume of hemisphere

V_{\text{solid}} = \frac{1}{3}\pi r^2 h_{\text{cone}} + \frac{2}{3}\pi r^3

= \frac{1}{3}\pi r^2 (h_{\text{cone}} + 2r)

= \frac{1}{3} \times \pi \times (60)^2 \times (120 + 2 \times 60)

= \frac{1}{3} \pi \times 3600 \times (120 + 120)

= 1200\pi \times 240 = 288000\pi \text{ cm}^3

Volume of the cylinder:

V_{\text{cyl}} = \pi R_{\text{cyl}}^2 H_{\text{cyl}}

Since

R_{\text{cyl}} = r = 60 \text{ cm}

:

= \pi (60)^2 \times 180

= 3600\pi \times 180 = 648000\pi \text{ cm}^3

Volume of water left =

V_{\text{cyl}} - V_{\text{solid}}

= 648000\pi - 288000\pi

= 360000\pi \text{ cm}^3

Using

\pi = \frac{22}{7}

: Volume left =

360000 \times \frac{22}{7} = \frac{7920000}{7} \approx 1131428.57 \text{ cm}^3

.

To convert to

\text{m}^3

, divide by

100^3 = 1,000,000

. Volume left =

\frac{1131428.57}{1000000} \approx 1.131 \text{ m}^3

.

Final Answer: The volume of water left in the cylinder is

1.131 \text{ m}^3

(approximately).

Q8EXERCISE 12.2

A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be $345 \text{ cm}^{3}$ . Check whether she is correct, taking the above as the inside measurements, and $\pi=3.14$ .

Solution

Given: A glass vessel with a spherical body and a cylindrical neck. Cylindrical neck: length (

h

) =

8 \text{ cm}

, diameter =

2 \text{ cm}

. Spherical part: diameter =

8.5 \text{ cm}

. Child's measurement of volume =

345 \text{ cm}^3

. Use

\pi = 3.14

.

To Find: Check if the child's measurement is correct.

Solution: Radius of the cylindrical neck (

r

) = Diameter / 2 =

2 / 2 = 1 \text{ cm}

. Radius of the spherical part (

R

) = Diameter / 2 =

8.5 / 2 = 4.25 \text{ cm}

.

The total volume of the vessel is the sum of the volume of the sphere and the volume of the cylinder.

Formula: Total Volume = Volume of sphere + Volume of cylinder

V = \frac{4}{3}\pi R^3 + \pi r^2 h

Calculation: Volume of sphere =

\frac{4}{3} \times 3.14 \times (4.25)^3

= \frac{4}{3} \times 3.14 \times 76.765625 \approx 321.39 \text{ cm}^3

Volume of cylinder =

3.14 \times (1)^2 \times 8

= 3.14 \times 8 = 25.12 \text{ cm}^3

Total Volume =

321.39 + 25.12 = 346.51 \text{ cm}^3

.

The calculated volume is

346.51 \text{ cm}^3

, while the child's measurement is

345 \text{ cm}^3

. These values are not equal.

Final Answer: The child is incorrect. The correct volume of the vessel is approximately

346.51 \text{ cm}^3

.

Surface Areas And VolumesClass 10 Mathematics NCERT Solutions

2 cubes each of volume 64 cm364 \text{ cm}^{3}64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Solution

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 of the vessel.

Solution

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution

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

Solution

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 its surface area.

Solution

Solution

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 to the nearest cm2\mathrm{cm}^{2}cm2.

Solution

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 is of radius 3.5 cm, find the total surface area of the article.

Solution

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π\piπ.

Solution

Solution

A gulab jamun, contains sugar syrup up to about 30%30\%30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

Solution

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 15 cm by 10 cm by 3.5 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.

Solution

Solution

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm31 \text{ cm}^{3}1 cm3 of iron has approximately 8 g mass. (Use π=3.14\pi=3.14π=3.14)

Solution

Solution

Solution

2 cubes each of volume $64 \text{ cm}^{3}$ are joined end to end. Find the surface area of the resulting cuboid.

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

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 to the nearest $\mathrm{cm}^{2}$ .

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of $\pi$ .

A gulab jamun, contains sugar syrup up to about $30\%$ of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm.

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that $1 \text{ cm}^{3}$ of iron has approximately 8 g mass. (Use $\pi=3.14$ )