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

- r² = (1570 * 3) / (3.14 * 15)

**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 cm****3****. 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:

- 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:

- 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:

- 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 cm****3****?**

**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 mm****3****) 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.**