Surface Areas And Volumes

This chapter introduces you to the world of 3D shapes and their measurements. You’ll learn about how to calculate the surface area and volume of various solid figures.

Key Concepts:

• Solid Shapes: Understanding different types of solid shapes like cubes, cuboids, cylinders, cones, and spheres.
• Surface Area: The total area of all the faces of a solid figure.
  • Lateral Surface Area: The area of all the sides except the top and bottom faces.
  • Total Surface Area: The sum of the lateral surface area and the area of the bases.
• Volume: The amount of space occupied by a solid figure.
• Formulas: Learning and applying formulas to calculate surface area and volume for different shapes.
• Problem-solving: Using the learned concepts to solve real-world problems involving surface area and volume.

Important Shapes and Formulas:

• Cube:
  • All sides are equal.
  • Lateral Surface Area = 4a²
  • Total Surface Area = 6a²
  • Volume = a³
• Cuboid:
  • Length, breadth, and height are different.
  • Lateral Surface Area = 2h(l + b)
  • Total Surface Area = 2(lb + bh + hl)
  • Volume = l * b * h
• Cylinder:
  • Circular base.
  • Lateral Surface Area = 2πrh
  • Total Surface Area = 2πr(h + r)
  • Volume = πr²h
• Cone:
  • Circular base with a pointed top.
  • Slant height (l) is the distance from the vertex to any point on the base’s circumference.
  • Lateral Surface Area = πrl
  • Total Surface Area = πr(l + r)
  • Volume = (1/3)πr²h
• Sphere:
  • A round solid figure with all points equidistant from the center.
  • Surface Area = 4πr²
  • Volume = (4/3)πr³

Applications: This chapter has practical applications in various fields like architecture, engineering, and daily life. For example, calculating the amount of paint needed for a wall, finding the volume of a water tank, or determining the capacity of a cylindrical container.

Exercise 11.1 

1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Ans : 

1. Find the radius (r):

• Radius = Diameter / 2 = 10.5 / 2 = 5.25 cm

2. Use the formula for curved surface area:

• Curved Surface Area = πrl
  • π = 22/7 (approximately)
  • r = 5.25 cm
  • l = 10 cm
• Curved Surface Area = (22/7) * 5.25 * 10 = 165 cm²

Therefore, the curved surface area of the cone is 165 cm².

2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Ans : 

1. Find the radius (r):

• Radius = Diameter / 2
•  = 24 / 2 
• = 12 m

2. Total surface area:

• Total Surface Area = πr(l + r)
  • π = 22/7 (approximately)
  • r = 12 m
  • l = 21 m
• Total Surface Area = (22/7) * 12 * (21 + 12) = (22/7) * 12 * 33 = 1244.57 m²

Therefore, the total surface area of the cone is 1244.57 square meters.

3. Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find

(i) radius of the base and

(ii) total surface area of the cone

Ans : 

(i) Finding the Radius of the Base

• Curved Surface Area (CSA) of a cone = πrl where r is the radius and l is the slant height.
• Given: CSA = 308 cm², l = 14 cm
• So, 308 = (22/7) * r * 14
• Simplifying, we get: r = (308 * 7) / (22 * 14) = 7 cm

Therefore, the radius of the base is 7 cm.

(ii) Finding the Total Surface Area of the Cone

• Total Surface Area (TSA) of a cone = πr(l + r) where r is the radius and l is the slant height.
• We already know r = 7 cm and l = 14 cm
• So, TSA = (22/7) * 7 * (14 + 7) = 22 * 21 = 462 cm²

4. A conical tent is 10 m high and the radius of its base is 24 m. Find

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is ₹70.

Ans : 

(i) Finding the Slant Height

• l² = h² + r² where h is the height and r is the radius.
• Substituting the given values: l² = 10² + 24² l² 
• = 100 + 576 l² = 676 l = √676 l = 26 m

(ii) Finding the Cost of Canvas

We need to find the curved surface area of the cone to determine the amount of canvas required.

• Curved Surface Area (CSA) = πrl where r is the radius and l is the slant height.
• Substituting the values: CSA = (22/7) * 24 * 26 = 19728/7 m²

Now, we can calculate the cost of the canvas.

• Cost of 1 m² canvas = ₹70
• Cost of 19728/7 m² canvas = (19728/7) * 70 = ₹197280

Therefore, the cost of the canvas required to make the tent is ₹197280.

5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π = 3.14)

Ans : 

• We need to find the length of tarpaulin required to make a conical tent.
• Height of the tent (h) = 8 m
• Base radius of the tent (r) = 6 m
• Width of the tarpaulin = 3 m
• Extra length for stitching and wastage = 20 cm = 0.2 m
• We will use π = 3.14

Solution:

Step 1: Find the slant height (l) of the cone:

• Using the Pythagorean theorem: l² = h² + r²
  • l² = 8² + 6² = 64 + 36 = 100
  • l = √100 = 10 m

Step 2: Calculate the curved surface area (CSA) of the cone:

• CSA of a cone = πrl
  • CSA = 3.14 * 6 * 10 = 188.4 m²

Step 3: Find the length of the tarpaulin:

• Length = Area / width = CSA / width
• Length = 188.4 / 3 = 62.8 m

Step 4: Add extra length for stitching and wastage:

• Total length of tarpaulin = 62.8 + 0.2 = 63 m

6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹ 210 per 100 m2 .

Ans : 

• Slant height (l) of the conical tomb = 25 m
• Base diameter = 14 m, so radius (r) = 7 m
• Cost of white-washing = ₹210 per 100 m²

Solution

Step 1: Find the curved surface area (CSA) of the tomb:

• CSA of a cone = πrl
  • CSA = (22/7) * 7 * 25 = 550 m²

Step 2: Find the cost of white-washing:

• Cost of white-washing 100 m² = ₹210
• Cost of white-washing 1 m² = ₹210/100
• Cost of white-washing 550 m² = (210/100) * 550 = ₹1155

Therefore, the cost of white-washing the curved surface of the conical tomb is ₹1155

7. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Ans : 

• The joker’s cap is shaped like a right circular cone.
• Base radius (r) = 7 cm
• Height (h) = 24 cm
• We need to find the area of the sheet required for 10 caps.

Solution

Step 1: Find the slant height (l) of the cone:

• Using the Pythagorean theorem: l² = r² + h²
  • l² = 7² + 24² = 49 + 576 = 625
  • l = √625 = 25 cm

Step 2: Find the curved surface area (CSA) of one cap:

• CSA of a cone = πrl
  • CSA = (22/7) * 7 * 25 
  • = 550 cm²

Step 3: Find the total area for 10 caps:

• Total area = CSA of one cap * number of caps
  • Total area = 550 cm² * 10 = 5500 cm²

Therefore, the area of the sheet required to make 10 joker’s caps is 5500 cm².

8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m², what will be the cost of painting all these cones? (Use π = 3.14 and take 104= 1.02)

Ans : 

• Number of cones = 50
• Base diameter of each cone = 40 cm = 0.4 m
• Height of each cone = 1 m
• Cost of painting per m² = ₹12
• We need to find the total cost of painting all cones.

Solution:

Step 1: Calculate the radius of the cone:

• Radius (r) = Diameter / 2 = 0.4 / 2 = 0.2 m

Step 2: Calculate the slant height (l) of the cone:

• Using the Pythagorean theorem: l² = h² + r²
  • l² = 1² + 0.2² = 1 + 0.04 = 1.04
  • l = √1.04 = 1.02 m (given)

Step 3: Calculate the curved surface area (CSA) of one cone:

• CSA of a cone = πrl
  • CSA = 3.14 * 0.2 * 1.02 = 0.64046 m²

Step 4: Calculate the total curved surface area of all cones:

• Total CSA = CSA of one cone * number of cones
  • Total CSA = 0.64046 * 50 = 32.023 m²

Step 5: Calculate the total cost of painting:

• Cost of painting 1 m² = ₹12
• Cost of painting 32.023 m² = 12 * 32.023 = ₹384.276 ≈ ₹384.34

Therefore, the total cost of painting all the cones is approximately ₹384.34.

Exercise 11.2

1. Find the surface area of a sphere of radius

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Ans : 

Formula:

• Surface Area = 4πr² where r is the radius of the sphere.

Calculations:

i) Radius = 10.5 cm

• Surface Area = 4 * (22/7) * (10.5)² = 4 * (22/7) * 110.25 = 1386 cm²

ii) Radius = 5.6 cm

• Surface Area = 4 * (22/7) * (5.6)² = 4 * (22/7) * 31.36 = 394.24 cm²

iii) Radius = 14 cm

• Surface Area = 4 * (22/7) * (14)² = 4 * (22/7) * 196 = 2464 cm²

2. Find the surface area of a sphere of diameter

(i) 14 cm

(ii) 21 cm

(iii) 3.5 m

Ans : 

Formula:

• Surface Area = 4πr² where r is the radius of the sphere.

Calculations:

i) Diameter = 14 cm

• Radius (r) = 14/2 = 7 cm
• Surface Area = 4 * (22/7) * 7² = 4 * 22 * 7 = 616 cm²

ii) Diameter = 21 cm

• Radius (r) = 21/2 = 10.5 cm
• Surface Area = 4 * (22/7) * (10.5)² = 4 * (22/7) * 110.25 = 1386 cm²

iii) Diameter = 3.5 m

• Radius (r) = 3.5/2 = 1.75 m
• Surface Area = 4 * (22/7) * (1.75)² = 4 * (22/7) * 3.0625 = 38.5 m²

3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Ans : 

Total Surface Area = 3πr²

Substituting the given values:

Total Surface Area = 3 * 3.14 * (10)² = 3 * 3.14 * 100 = 942 cm²

Therefore, the total surface area of the hemisphere is 942 cm².

4. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans :

• Initial radius (r1) = 7 cm
• Final radius (r2) = 14 cm
• We need to find the ratio of the surface areas of the balloon in the two cases.

Solution

• Surface area of the balloon initially (S1) = 4πr1² = 4π(7)² = 196π cm²
• Surface area of the balloon finally (S2) = 4πr2² = 4π(14)² = 784π cm²
• Ratio of surface areas = S1 : S2 = 196π : 784π = 1 : 4

5. A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².

Ans : 

• Inner diameter of the hemispherical bowl = 10.5 cm
• Radius (r) = 10.5 / 2 = 5.25 cm
• Cost of tin-plating = ₹16 per 100 cm²

Solution:

Step 1: Find the curved surface area (CSA) of the hemisphere:

• CSA of a hemisphere = 2πr²
  • CSA = 2 * (22/7) * (5.25)² = 173.25 cm²

Step 2: Find the cost of tin-plating:

• Cost of tin-plating 100 cm² = ₹16
• Cost of tin-plating 173.25 cm² = (16/100) * 173.25 = ₹27.72

Therefore, the cost of tin-plating the inside of the hemispherical bowl is ₹27.72.

6. Find the radius of a sphere whose surface area is 154 cm².

Ans : 

Surface Area = 4πr² where r is the radius of the sphere.

Substituting the given values:

154 = 4 * (22/7) * r²

Simplifying the equation:

r² = (154 * 7) / (4 * 22) = 49/4

r = √(49/4) 

= 7/2 

= 3.5 cm

Therefore, the radius of the sphere is 3.5 cm.

7. The diameter of the Moon is approximately one-fourth of the diameter of the Earth. Find the ratio of their surface areas.

Ans : 

Let the diameter of Earth be D. So, the diameter of the Moon is D/4.

• Radius of Earth (R) = D/2
• Radius of Moon (r) = (D/4)/2 = D/8

Surface area of a sphere = 4πr²

• Surface area of Earth (S1) = 4πR² = 4π(D/2)² = πD²
• Surface area of Moon (S2) = 4πr² = 4π(D/8)² = πD²/16

Ratio of surface areas (S1 : S2) = (πD²) : (πD²/16) = 16 : 1

Therefore, the ratio of the surface areas of the Earth to the Moon is 16:1.

8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Ans : 

Step 1: Find the outer radius (R):

• Outer radius (R) = Inner radius + Thickness = 5 cm + 0.25 cm = 5.25 cm

Step 2: Find outer hemisphere:

• Outer curved surface area = 2 * (22/7) * (5.25)²
•  = 173.25 cm²

9. A right circular cylinder just encloses a sphere of radius r (see figure). Find

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in (i) and (ii).

Ans : 

(i) For the sphere, radius = r
∴ Surface area of the sphere = 4πR2

(ii) For the right circular cylinder,
Radius of the cylinder = Radius of the sphere
∴Radius of the cylinder = r
∴ Height of the cylinder (h) 2r

= 2πr(2r) = 4πr2

Exercise 11.3

1. Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

Ans : 

Formula:

• Volume (V) = (1/3) * π * r² * h Where:
  • r = radius of the base
  • h = height of the cone

Calculations:

(i) Radius = 6 cm, Height = 7 cm

• V = (1/3) * (22/7) * 6² * 7 = (1/3) * (22/7) * 36 * 7 = 264 cm³

(ii) Radius = 3.5 cm, Height = 12 cm

• V = (1/3) * (22/7) * (3.5)² * 12 = (1/3) * (22/7) * 12.25 * 12 = 154 cm³

2. Find the capacity in litres of a conical vessel with

(i) radius 7 cm, slant height 25 cm

(ii) height 12 cm, slant height 13 cm

Ans : 

Formula for the volume of a cone:

• Volume (V) = (1/3) * π * r² * h Where:
  • r = radius of the base
  • h = height of the cone

Note: 1 liter = 1000 cm³

(i) First, we need to find the height using the Pythagorean theorem:

• h² = l² – r²
  • h² = 25² – 7² =
  •  625 – 49 = 576
  • h = √576 = 24 cm

Now, we can calculate the volume:

• V = (1/3) * (22/7) * 7² * 24 = 1232 cm³

To convert to liters:

• Capacity = 1232 cm³ * (1 L / 1000 cm³) = 1.232 liters  

(ii) First, we need to find the radius using the Pythagorean theorem:

• r² = l² – h²
  • r² = 13² – 12² = 169 – 144 = 25
  • r = √25 = 5 cm

Now, we can calculate the volume:

• V = (1/3) * (22/7) * 5² * 12 = 2200/7 cm³

To convert to liters:

• Capacity = (2200/7) cm³ * (1 L / 1000 cm³) = 22/70 liters ≈ 0.314 liters

3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Ans : 

• The formula for the volume of a cone is:
  • V = (1/3)πr²h
• Substituting the given values:
  • 1570 
  • = (1/3) * 3.14 * r² * 15
• Simplifying the equation:
  • r² = (1570 * 3) / (3.14 * 15)
    • r² = 100
  • r = √100
    • r = 10 cm

Therefore, the radius of the base of the cone is 10 cm.

4. If the volume of a right circular cone of height 9 cm is 48 cm³, find the diameter of its base.

Ans : 

Diameter = 2 x radius .

∴ Diameter of the base of the cone = (2 x 4)cm = 8 cm

5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Ans : 

Step 1: Find the radius of the pit:

• Radius (r) = Diameter / 2 
• = 3.5 / 2 = 1.75 m

Step 2: Find the volume of the pit:

• The formula for the volume of a cone is: V = (1/3)πr²h
  • V =

 (1/3) * (22/7) * (1.75)² * 12

• V = 38.5 m³

Step 3: Convert the volume to kiloliters:

• 1 m³ = 1 kiloliter
• So, 38.5 m³ = 38.5 kiloliters

6. The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the cone

Ans : 

(i) Find the height of the cone (h)

We can find the height (h) of the cone using the formula for the volume of a cone and the diameter of the base.

h = (3 * V) / (π * r²))

where:

• r is the radius of the base, which is equal to half the diameter (d/2)

Steps to solve:

1. Calculate the radius (r)
   • r = d / 2 = 28 cm / 2 
   • = 14 cm
   • h = (3 * 9856 cm³) / ((22/7) * (14 cm)²)
   • h ≈ 48.02 cm

Therefore, the height of the cone is approximately 48.02 cm.

(ii) Find the slant height of the cone (l)

The slant height (l) of the cone is the hypotenuse of a right triangle where the height (h) is one leg and the radius (r) is the other leg. 

l² = h² + r²

l = √(h² + r²)

Steps to solve:

1. Substitute the known values into the formula and solve for l
   • l² = (48.02 cm)² + (14 cm)²
   • l ≈ 50.02 cm

Therefore, the slant height of the cone is approximately 50.02 cm.

(iii) Find the curved surface area of the cone (CSA)

The curved surface area (CSA) of the cone is the lateral surface area that excludes the base. We can calculate it using the formula for the curved surface area of a cone.

CSA = π * r * l

Steps to solve:

1. Substitute the known values into the formula and solve for CSA
   • CSA = (22/7) * (14 cm) * (50.02 cm)
   • CSA ≈ 2200 cm²

Therefore, the curved surface area of the cone is approximately 2200 cm².

7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Ans : 

• Radius (r) = 5 cm
• Height (h) = 12 cm

Calculation:

• V = (1/3) * π * (5 cm)² * 12 cm
•  = (1/3) * π * 25 cm² * 12 cm 
• = 100π cm³

Therefore, the volume of the solid formed is 100π cubic centimeters.

8. If the triangle ABC in Question 7 above is revolved around the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Ans : 

When the right-angled triangle is revolved around the side of 5 cm, it forms a cone with:

• Height (h) = 5 cm
• Radius (r) = 12 cm

Volume of the cone (V2):

• V2 = (1/3) * π * r² * h = (1/3) * π * (12 cm)² * 5 cm = 240π cm³

• Volume of the first solid (V1) = 100π cm³ (from the previous question)
• Volume of the second solid (V2) = 240π cm³
• Ratio of V1 to V2 = V1 / V2 = (100π cm³) / (240π cm³) = 5/12

Therefore, the volume of the solid obtained when the triangle is revolved around the 5 cm side is 240π cm³, and the ratio of the volumes of the two solids is 5:12.

9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Ans : 

1. Finding the Volume of the Heap

• Radius (r) = Diameter / 2 = 10.5 m / 2 = 5.25 m
• Height (h) = 3 m
• Volume (V) 

= (1/3) * π * r² * h

Substituting the values:

• V = (1/3) * (22/7) * (5.25 m)² * 3 m = 86.625 m³

Therefore, the volume of the wheat heap is 86.625 cubic meters.

2. Finding the Area of Canvas Required

The canvas required will cover the curved surface area of the cone.

• l = √(r² + h²) = √((5.25 m)² + (3 m)²) ≈ 6.046 m
• Curved Surface Area (CSA) 

= π * r * l

Substituting the values:

• CSA = (22/7) * 5.25 m * 6.046 m ≈ 99.76 m²

Therefore, the area of the canvas required is approximately 99.76 square meters.

Exercise 11.4

1. Find the volume of a sphere whose radius is

(i) 7 cm

(ii) 0.63 cm

Ans : 

i) Radius = 7 cm

• Volume = (4/3) * (22/7) * (7)³ = (4/3) * 22 * 7 * 7 = 1386 cm³  

ii) Radius = 0.63 cm

• Volume = (4/3) * (22/7) * (0.63)³ ≈ 1.05 cm³

2. Find the amount of water displaced by a solid spherical ball of diameter

(i) 28 cm

(ii) 0.21 m

Ans : 

i) Diameter = 28 cm

• Radius (r) = Diameter / 2 = 28 cm / 2 = 14 cm  
• Volume (V) = (4/3) * (22/7) * (14 cm)³ = 11494.04 cm³

ii) Diameter = 0.21 m

• Convert diameter to cm: Diameter = 0.21 m * 100 cm/m = 21 cm
• Radius (r) = Diameter / 2 = 21 cm / 2 = 10.5 cm  
• Volume (V) = (4/3) * (22/7) * (10.5 cm)³ = 4851 cm³  

3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?

Ans : 

1. Find the radius:

• Radius (r) = Diameter / 2 = 4.2 cm / 2 = 2.1 cm  

2. Find the volume of the ball:

• The formula for the volume of a sphere is: V = (4/3) * π * r³
  • V = (4/3) * π * (2.1 cm)³  

3. Find the mass of the ball:

• Density = Mass / Volume
• Mass = Density * Volume
  • Mass = 8.9 g/cm³ * 38.808 cm³
  • Mass ≈ 345.39 g

Therefore, the mass of the metallic ball is approximately 345.39 grams.   

4. The diameter of the Moon is approximately one-fourth of the diameter of the Earth. What fraction of the volume of the Earth is the volume of the Moon?

Ans : 

• The diameter of the Earth as D
• The diameter of the Moon as d

We know that:

• d = (1/4)D

Finding the Radii

• Radius of the Earth (R) = D/2
• Radius of the Moon (r) = d/2 = (1/4)D / 2 = D/8

Finding the Volumes

• Volume of the Earth (V_E) = (4/3)πR³ = (4/3)π(D/2)³
• Volume of the Moon (V_M) = (4/3)πr³ = (4/3)π(D/8)³

Finding the Ratio of Volumes

• V_M / V_E = [(4/3)π(D/8)³] / [(4/3)π(D/2)³]
• Simplifying the equation, we get:
• V_M / V_E = (D/8)³ / (D/2)³
• V_M / V_E = (1/8)³ / (1/2)³
• V_M / V_E = (1/512) / (1/8)
• V_M / V_E = 1/64

Therefore, the volume of the Moon is 1/64th the volume of the Earth.

5. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Ans :

Diameter of the bowl = 10.5 cm

We need to find the volume of the bowl in liters.

Step 1: 

Radius (r) = Diameter / 2 = 10.5 cm / 2 = 5.25 cm   

Step 2: Find the volume of the hemispherical bowl:

Volume of a hemisphere = (2/3) * π * r³

V = (2/3) * (22/7) * (5.25 cm)³

V ≈ 303.19 cm³

Step 3: Convert the volume to liters:

1 liter = 1000 cm³

V = 303.19 cm³ * (1 liter / 1000 cm³) ≈ 0.303 liters   

Therefore, the hemispherical bowl can hold approximately 0.303 liters of milk.   

6. A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Ans : 

Inner radius (r) = 1 m

∵ Thickness = 1 cm = 1100m = 0 .01m

∴ Outer radius (R) = 1 m + 0.01 m = 1.01 m

= 0.06348 m3

7. Find the volume of a sphere whose surface area is 154 cm2.

Ans : 

. Find the radius:

• Surface area of a sphere = 4πr²
  • 154 = 4 * (22/7) * r²
  • r² 

= (154 * 7) / (4 * 22)

• r² = 49/4
• r = √(49/4) = 7/2 
• = 3.5 cm

2. Find the volume:

• Volume of a sphere = (4/3)πr³
  • Volume = (4/3) * (22/7) * (3.5)³
  • Volume = (4/3) * (22/7) * 42.875
  • Volume ≈ 179.67 cm³

Therefore, the volume of the sphere is approximately 179.67 cubic centimeters.

8. A dome of a building is in the form of a hemisphere. From inside, it was white washed at the cost of ₹498.96. If the cost of white washing is ₹2.00 per square metre, find the

(i) inside surface area of the dome,

(ii) volume of the air inside the dome.

Ans : 

i) Inside surface area of the dome

• Total cost = Surface area * Cost per square meter
• ₹498.96 = Surface area * ₹2.00
• Surface area = ₹498.96 / ₹2.00 = 249.48 m²

Therefore, the inside surface area of the dome is 249.48 square meters.

ii) 

Surface area of a hemisphere = 2πr²

• 249.48 = 2 * π * r²
• r² = 249.48 / (2 * π)
• r ≈ 6.3 m (approximate value)
• Volume of a hemisphere = (2/3)πr³
  • Volume = (2/3) * π * (6.3 m)³
  • Volume ≈ 524.01 m³

Therefore, the volume of the air inside the dome is approximately 524.01 cubic meters.

9. Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S’. Find the

(i) radius r’ of the new sphere,

(ii) ratio of S and S’.

Ans : 

i) Finding the radius r’ of the new sphere

Since the volume of the material remains constant, the total volume of the 27 smaller spheres is equal to the volume of the larger sphere.

• Volume of one small sphere = (4/3)πr³
• Total volume of 27 spheres = 27 * (4/3)πr³ = 36πr³
• Let the radius of the new sphere be r’.
• Volume of the new sphere = (4/3)πr’³

Equating the volumes:

• 36πr³ = (4/3)πr’³
• r’³ = 27r³
• r’ = 3r

ii) Finding the ratio of S and S’

• Surface area of one small sphere (S) = 4πr²
• Surface area of the new sphere (S’) = 4πr’² = 4π(3r)² = 36πr²
• Ratio of S to S’ = S/S’ = (4πr²) / (36πr²) = 1/9

Therefore, the ratio of S to S’ is 1:9.

10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?

Ans : 

Step 1: Find the radius of the capsule:

• Radius (r) = Diameter / 2 = 3.5 mm / 2 = 1.75 mm

Step 2: Find the volume of the capsule:

• The formula for the volume of a sphere is: V = (4/3) * π * r³
  • V = (4/3) * (22/7) * (1.75 mm)³
  • V ≈ 22.46 mm³

Therefore, approximately 22.46 mm³ of medicine is needed to fill the capsule.