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SURFACE AREAS AND VOLUMES - Q&A

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.

Given:
Diameter (d) = 10.5 cm
Radius (r) = d/2 = 10.5/2 = 5.25 cm
Slant height (l) = 10 cm

Curved Surface Area (CSA) of a cone = πrl
CSA = (22/7) × 5.25 × 10
CSA = (22/7) × 52.5
CSA = 22 × 7.5
CSA = 165 cm².

Answer: The curved surface area of the cone is 165 cm².

[Image of curved surface area of a cone]

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

Given:
Slant height (l) = 21 m
Diameter (d) = 24 m ⇒ Radius (r) = 12 m

Total Surface Area (TSA) of a cone = πr(l + r)
TSA = (22/7) × 12 × (21 + 12)
TSA = (22/7) × 12 × 33
TSA = (264 × 33) / 7
TSA = 8712 / 7
TSA = 1244.57 m² (approx).

Answer: The total surface area of the cone is 1244.57 m².

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.

Given:
Curved Surface Area (CSA) = 308 cm²
Slant height (l) = 14 cm

(i) Find radius (r):
CSA = πrl
308 = (22/7) × r × 14
308 = 22 × r × 2
308 = 44r
r = 308 / 44
r = 7 cm.

(ii) Find Total Surface Area (TSA):
TSA = πr(l + r)
TSA = (22/7) × 7 × (14 + 7)
TSA = 22 × 21
TSA = 462 cm².

Answer: (i) Radius = 7 cm, (ii) Total Surface Area = 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.

Given:
Height (h) = 10 m
Radius (r) = 24 m

(i) Slant height (l):
l = √(r² + h²)
l = √(24² + 10²)
l = √(576 + 100)
l = √676
l = 26 m.

(ii) Cost of canvas:
Canvas required = Curved Surface Area (CSA) = πrl
CSA = (22/7) × 24 × 26
CSA = 13728 / 7 m²

Cost = Rate × Area
Cost = 70 × (13728 / 7)
Cost = 10 × 13728
Cost = ₹ 137280.

Answer: (i) Slant height = 26 m, (ii) Cost = ₹ 137,280.

5. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? 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).

Given:
Height (h) = 8 m
Radius (r) = 6 m
Width of tarpaulin = 3 m

Step 1: Find slant height (l).
l = √(r² + h²)
l = √(6² + 8²)
l = √(36 + 64) = √100
l = 10 m.

Step 2: Find Curved Surface Area (CSA).
CSA = πrl
CSA = 3.14 × 6 × 10
CSA = 188.4 m².

Step 3: Find length of tarpaulin.
Area of tarpaulin = Length × Width
188.4 = Length × 3
Length = 188.4 / 3 = 62.8 m.

Step 4: Add extra length for wastage.
Extra length = 20 cm = 0.2 m.
Total length = 62.8 + 0.2 = 63 m.

Answer: The required length of tarpaulin is 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.

Given:
Slant height (l) = 25 m
Diameter = 14 m ⇒ Radius (r) = 7 m

Step 1: Find Curved Surface Area (CSA).
CSA = πrl
CSA = (22/7) × 7 × 25
CSA = 22 × 25 = 550 m².

Step 2: Calculate cost.
Rate = ₹ 210 per 100 m².
Cost = (CSA / 100) × 210
Cost = (550 / 100) × 210
Cost = 5.5 × 210
Cost = ₹ 1155.

Answer: The cost of white-washing 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.

Given:
Radius (r) = 7 cm
Height (h) = 24 cm

Step 1: Find slant height (l).
l = √(r² + h²)
l = √(7² + 24²)
l = √(49 + 576) = √625
l = 25 cm.

Step 2: Find CSA of 1 cap.
CSA = πrl
CSA = (22/7) × 7 × 25
CSA = 550 cm².

Step 3: Find area for 10 caps.
Total Area = 10 × 550 = 5500 cm².

Answer: The area of sheet required 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 √1.04 = 1.02)

Given:
Diameter = 40 cm = 0.4 m ⇒ Radius (r) = 0.2 m
Height (h) = 1 m

Step 1: Find slant height (l).
l = √(r² + h²)
l = √(0.2² + 1²)
l = √(0.04 + 1) = √1.04
Given √1.04 = 1.02, so l = 1.02 m.

Step 2: Find CSA of 1 cone.
CSA = πrl
CSA = 3.14 × 0.2 × 1.02
CSA = 0.64056 m².

Step 3: Find CSA of 50 cones.
Total Area = 50 × 0.64056
Total Area = 32.028 m².

Step 4: Calculate cost.
Cost = Rate × Area
Cost = 12 × 32.028
Cost = ₹ 384.336.

Answer: The approximate cost of painting is ₹ 384.34.

Exercise 11.2

1. Find the surface area of a sphere of radius:

(i) 10.5 cm

Surface Area = 4πr²
= 4 × (22/7) × 10.5 × 10.5
= 4 × 22 × 1.5 × 10.5
= 1386 cm².

(ii) 5.6 m

Surface Area = 4 × (22/7) × 5.6 × 5.6
= 4 × 22 × 0.8 × 5.6
= 394.24 m².

(iii) 14 cm

Surface Area = 4 × (22/7) × 14 × 14
= 4 × 22 × 2 × 14
= 2464 cm².

2. Find the surface area of a sphere of diameter:

(i) 14 cm

Radius = 7 cm.
Area = 4πr² = 4 × (22/7) × 7 × 7 = 616 cm².

(ii) 21 cm

Radius = 10.5 cm.
Area = 4 × (22/7) × 10.5 × 10.5 = 1386 cm².

(iii) 3.5 m

Radius = 1.75 m.
Area = 4 × (22/7) × 1.75 × 1.75 = 38.5 m².

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

Total Surface Area (TSA) of hemisphere = 3πr²
TSA = 3 × 3.14 × 10 × 10
TSA = 3 × 314
TSA = 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.

Let r₁ = 7 cm and r₂ = 14 cm.
Ratio = 4πr₁² : 4πr₂²
Ratio = r₁² : r₂²
Ratio = 7² : 14²
Ratio = 49 : 196
Ratio = 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.

Radius (r) = 10.5 / 2 = 5.25 cm.
Inner CSA = 2πr²
= 2 × (22/7) × 5.25 × 5.25
= 173.25 cm².

Cost = (Area / 100) × Rate
Cost = (173.25 / 100) × 16
Cost = 1.7325 × 16
Cost = ₹ 27.72.

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

4πr² = 154
4 × (22/7) × r² = 154
88/7 × r² = 154
r² = (154 × 7) / 88
r² = 12.25
r = 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.

Let diameter of Earth = D. Radius of Earth (R) = D/2.
Diameter of Moon = D/4. Radius of Moon (r) = D/8.

Ratio = Surface Area of Moon / Surface Area of Earth
Ratio = 4πr² / 4πR²
Ratio = (D/8)² / (D/2)²
Ratio = (D²/64) / (D²/4)
Ratio = 4 / 64 = 1/16.
Ratio is 1:16.

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.

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

Outer CSA = 2πR²
= 2 × (22/7) × 5.25 × 5.25
= 173.25 cm².

9. A right circular cylinder just encloses a sphere of radius r (see Fig. 11.10). Find
(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

(i) Surface area of sphere = 4πr².

(ii) For the cylinder enclosing the sphere:
Radius of cylinder = r.
Height of cylinder (h) = diameter of sphere = 2r.
CSA of cylinder = 2πrh
= 2πr(2r) = 4πr².

(iii) Ratio = (4πr²) / (4πr²) = 1:1.

Exercise 11.3

1. Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

Volume = (1/3)πr²h
= (1/3) × (22/7) × 6 × 6 × 7
= 22 × 2 × 6
= 264 cm³.

(ii) radius 3.5 cm, height 12 cm

Volume = (1/3) × (22/7) × 3.5 × 3.5 × 12
= (1/3) × 22 × 0.5 × 3.5 × 12
= 154 cm³.

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

(i) radius 7 cm, slant height 25 cm

First find height (h):
h = √(l² - r²) = √(25² - 7²) = √(625 - 49) = √576 = 24 cm.
Volume = (1/3)πr²h
= (1/3) × (22/7) × 7 × 7 × 24
= 1232 cm³.
Capacity in litres = 1232 / 1000 = 1.232 litres.

(ii) height 12 cm, slant height 13 cm

First find radius (r):
r = √(l² - h²) = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.
Volume = (1/3)πr²h
= (1/3) × (22/7) × 5 × 5 × 12
= 2200 / 7 cm³.
Capacity in litres = (2200/7) / 1000 = 11/35 litres (approx 0.314 litres).

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)

Volume = 1570 cm³, h = 15 cm.
(1/3)πr²h = 1570
(1/3) × 3.14 × r² × 15 = 1570
3.14 × 5 × r² = 1570
15.7 × r² = 1570
r² = 100
r = 10 cm.

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

Volume = 48π, h = 9 cm.
(1/3)πr²h = 48π
(1/3) × r² × 9 = 48
3r² = 48
r² = 16 ⇒ r = 4 cm.
Diameter = 2r = 8 cm.

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

Radius (r) = 3.5 / 2 = 1.75 m. Height (h) = 12 m.
Volume = (1/3)πr²h
= (1/3) × (22/7) × 1.75 × 1.75 × 12
= 22 × 0.25 × 1.75 × 4
= 38.5 m³.
Since 1 m³ = 1 kilolitre.
Capacity = 38.5 kilolitres.

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

Given: Diameter = 28 cm ⇒ r = 14 cm. Volume = 9856 cm³.

(i) Height (h):
(1/3)πr²h = 9856
(1/3) × (22/7) × 14 × 14 × h = 9856
(1/3) × 22 × 2 × 14 × h = 9856
616h / 3 = 9856
h = (9856 × 3) / 616
h = 16 × 3 = 48 cm.

(ii) Slant height (l):
l = √(r² + h²) = √(14² + 48²) = √(196 + 2304) = √2500 = 50 cm.

(iii) Curved Surface Area (CSA):
CSA = πrl = (22/7) × 14 × 50 = 22 × 2 × 50 = 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.

When revolved about side 12 cm, it forms a cone with:
Height (h) = 12 cm
Radius (r) = 5 cm
Volume = (1/3)πr²h
= (1/3) × π × 5 × 5 × 12
= 100π cm³.

8. If the triangle ABC in the Question 7 above is revolved about 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.

When revolved about side 5 cm:
Height (h) = 5 cm
Radius (r) = 12 cm
Volume = (1/3)πr²h
= (1/3) × π × 12 × 12 × 5
= 240π cm³.

Ratio of volumes (Q7 : Q8) = 100π : 240π = 10 : 24 = 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.

Given: Diameter = 10.5 m ⇒ r = 5.25 m. Height (h) = 3 m.

Volume = (1/3)πr²h
= (1/3) × (22/7) × 5.25 × 5.25 × 3
= 22 × 0.75 × 5.25
= 86.625 m³.

Area of canvas = CSA = πrl
First find l = √(r² + h²) = √(5.25² + 3²) = √(27.5625 + 9) = √36.5625 = 6.05 m (approx).
CSA = (22/7) × 5.25 × 6.05
= 22 × 0.75 × 6.05
= 99.825 m².

Exercise 11.4

1. Find the volume of a sphere whose radius is

(i) 7 cm

Volume = (4/3)πr³
= (4/3) × (22/7) × 7 × 7 × 7
= (4/3) × 22 × 49
= 4312 / 3 = 1437.33 cm³.

(ii) 0.63 m

Volume = (4/3) × (22/7) × 0.63 × 0.63 × 0.63
= 4 × 22 × 0.03 × 0.63 × 0.63
= 1.047816 m³ (approx).

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

(i) 28 cm

Radius (r) = 14 cm.
Volume = (4/3)πr³
= (4/3) × (22/7) × 14 × 14 × 14
= (4/3) × 22 × 2 × 196
= 34496 / 3 = 11498.67 cm³.

(ii) 0.21 m

Radius (r) = 0.105 m.
Volume = (4/3) × (22/7) × 0.105 × 0.105 × 0.105
= 0.004851 m³.

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?

Radius (r) = 2.1 cm.
Volume = (4/3)πr³
= (4/3) × (22/7) × 2.1 × 2.1 × 2.1
= 4 × 22 × 0.1 × 4.41
= 38.808 cm³.

Mass = Density × Volume
= 8.9 × 38.808
= 345.39 g (approx).

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?

Let Radius of Earth = R. Radius of Moon (r) = R/4.
Volume of Earth = (4/3)πR³.
Volume of Moon = (4/3)π(R/4)³ = (4/3)π(R³/64) = (1/64) × Volume of Earth.
So, the volume of the moon is 1/64 of the volume of the earth.

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

Radius (r) = 5.25 cm.
Volume of hemisphere = (2/3)πr³
= (2/3) × (22/7) × 5.25 × 5.25 × 5.25
= 303.1875 cm³.
Capacity in litres = 303.1875 / 1000 = 0.303 litres (approx).

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.

Inner radius (r) = 1 m.
Outer radius (R) = 1 + 0.01 = 1.01 m (since 1 cm = 0.01 m).
Volume of iron = Volume of outer shell - Volume of inner shell
= (2/3)πR³ - (2/3)πr³
= (2/3)π(R³ - r³)
= (2/3) × (22/7) × (1.01³ - 1³)
= (44/21) × (1.030301 - 1)
= (44/21) × 0.030301
= 0.06348 m³ (approx).

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

Surface area = 4πr² = 154 ⇒ r = 3.5 cm (Calculated in Ex 11.2 Q6).
Volume = (4/3)πr³
= (4/3) × (22/7) × 3.5 × 3.5 × 3.5
= 179.67 cm³.

8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 4989.60. If the cost of white-washing is ₹ 20 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.

(i) Inside surface area = Total Cost / Rate
= 4989.60 / 20
= 249.48 m².

(ii) Find radius first. 2πr² = 249.48
2 × (22/7) × r² = 249.48
r² = (249.48 × 7) / 44
r² = 39.69 ⇒ r = 6.3 m.

Volume of air = Volume of hemisphere = (2/3)πr³
= (2/3) × (22/7) × 6.3 × 6.3 × 6.3
= 523.9 m³ (approx).

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'.

(i) Volume of 27 spheres = Volume of new sphere
27 × (4/3)πr³ = (4/3)π(r')³
27r³ = (r')³
(3r)³ = (r')³
r' = 3r.

(ii) Ratio of S and S'
S = 4πr²
S' = 4π(r')² = 4π(3r)² = 36πr²
S : S' = 4πr² : 36πr² = 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?

Radius (r) = 3.5 / 2 = 1.75 mm.
Volume = (4/3)πr³
= (4/3) × (22/7) × 1.75 × 1.75 × 1.75
= 22.46 mm³ (approx).

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Quick Review Flashcards - Click to flip and test your knowledge!

Question

Which specific type of solid is a cone considered to be, as it is not a prism?

Answer

A cone is considered a pyramid-type solid.

Question

What three-dimensional shape is generated by rotating a right-angled triangle about one of its perpendicular sides?

Answer

A right circular cone is generated by this rotation.

Question

In the geometry of a right circular cone, what does the term 'vertex' refer to?

Answer

The vertex is the topmost point of the cone where the slant sides meet.

Question

What is the name of the line segment joining the vertex of a cone to the centre of its circular base?

Answer

This line segment is called the height ($h$) of the cone.

Question

In a right circular cone, what is the 'slant height' ($l$)?

Answer

The slant height is the distance from the vertex to any point on the perimeter of the circular base.

Question

Under what condition is a cone specifically called a 'right circular' cone?

Answer

It is a right circular cone if the line joining the vertex to the base centre is at a right angle to the base.

Question

What shape is formed by the surface of a paper cone when it is cut along its slant height and flattened?

Answer

The flattened surface forms a sector of a circle.

Question

The curved boundary of the flattened surface of a cone is equal to the _____ of the cone's base.

Answer

Circumference

Question

Formula: Curved Surface Area (CSA) of a right circular cone.

Answer

The formula is $\pi rl$, where $r$ is the radius and $l$ is the slant height.

Question

How is the slant height ($l$) of a cone calculated if only the radius ($r$) and height ($h$) are known?

Answer

It is calculated using the formula $l = \sqrt{r^2 + h^2}$.

Question

Formula: Total Surface Area (TSA) of a right circular cone.

Answer

The formula is $\pi r(l + r)$.

Question

Calculate the curved surface area of a cone with a slant height of $10$ cm and a base radius of $7$ cm.

Answer

The curved surface area is $220$ $cm^2$.

Question

The curved surface area of a cone is $308$ $cm^2$ and its slant height is $14$ cm. Find the radius of the base.

Answer

The radius of the base is $7$ cm.

Question

What is the relationship between the slant height ($l$), height ($h$), and radius ($r$) in a right circular cone?

Answer

The relationship is defined by the equation $l^2 = r^2 + h^2$.

Question

Definition: Sphere

Answer

A sphere is a three-dimensional solid made of all points in space at a constant distance from a fixed centre.

Question

Formula: Surface Area of a sphere with radius $r$.

Answer

The formula is $4 \pi r^2$.

Question

The surface area of a sphere is equal to _____ times the area of a circle with the same radius.

Answer

Four

Question

What is a hemisphere?

Answer

A hemisphere is the solid formed by dividing a sphere into two equal parts using a plane through its centre.

Question

How many faces does a solid hemisphere have?

Answer

A solid hemisphere has two faces: one curved face and one flat circular base.

Question

Formula: Curved Surface Area (CSA) of a hemisphere.

Answer

The formula is $2 \pi r^2$.

Question

Formula: Total Surface Area (TSA) of a solid hemisphere.

Answer

The formula is $3 \pi r^2$.

Question

What is the surface area of a sphere with a radius of $7$ cm?

Answer

The surface area is $616$ $cm^2$.

Question

Calculate the total surface area of a hemisphere with a radius of $21$ cm.

Answer

The total surface area is $4158$ $cm^2$.

Question

If a sphere's surface area is $154$ $cm^2$, find its radius.

Answer

The radius is $3.5$ cm.

Question

A cylinder just encloses a sphere of radius $r$. What is the ratio of the sphere's surface area to the cylinder's curved surface area?

Answer

The ratio is $1:1$.

Question

What is the ratio of the volume of a right circular cone to the volume of a cylinder with the same base radius and height?

Answer

The ratio is $1:3$.

Question

Formula: Volume of a right circular cone.

Answer

The formula is $\frac{1}{3} \pi r^2 h$.

Question

To calculate the volume of a cone, which specific measurement of height must be used?

Answer

The perpendicular height ($h$) must be used.

Question

If a cone has a height of $15$ cm and its volume is $1570$ $cm^3$, what is the radius of its base? (Use $\pi = 3.14$)

Answer

The radius is $10$ cm.

Question

Formula: Volume of a sphere of radius $r$.

Answer

The formula is $\frac{4}{3} \pi r^3$.

Question

Formula: Volume of a hemisphere of radius $r$.

Answer

The formula is $\frac{2}{3} \pi r^3$.

Question

The volume of a sphere is $\frac{4}{3} \pi$ times the _____ of its radius.

Answer

Cube

Question

How is the radius of a sphere found if only the surface area ($SA$) is provided?

Answer

The radius is found using $r = \sqrt{\frac{SA}{4\pi}}$.

Question

A hemispherical bowl has a radius of $3.5$ cm. Calculate the volume of water it can contain.

Answer

The volume is approximately $89.8$ $cm^3$.

Question

If the diameter of the moon is approximately one-fourth the diameter of the earth, what is the ratio of their surface areas?

Answer

The ratio is $1:16$.

Question

How does mass relate to the volume and density of a metallic sphere?

Answer

Mass is calculated by multiplying the volume of the sphere by the density of the metal.

Question

A conical tent is $10$ m high with a base radius of $24$ m. Calculate the slant height.

Answer

The slant height is $26$ m.

Question

If the volume of a right circular cone of height $9$ cm is $48 \pi$ $cm^3$, what is the diameter of its base?

Answer

The diameter of the base is $8$ cm.

Question

When calculating the amount of canvas required for a conical tent, which surface area formula is typically used?

Answer

The Curved Surface Area (CSA) formula, $\pi rl$, is used.

Question

A capsule of medicine is spherical with a diameter of $3.5$ mm. What volume of medicine is needed to fill it?

Answer

The volume needed is approximately $22.46$ $mm^3$.

Question

Calculate the curved surface area of a hemisphere with a radius of $21$ cm.

Answer

The curved surface area is $2772$ $cm^2$.

Question

The diameter of a metallic ball is $4.2$ cm. What is its mass if the metal's density is $8.9$ g per $cm^3$?

Answer

The mass is approximately $345.39$ g.

Question

If the radius of a spherical balloon increases from $7$ cm to $14$ cm, what is the ratio of their surface areas?

Answer

The ratio is $1:4$.

Question

In the context of the NCERT text, what is the 'usual meaning' of the letter $r$?

Answer

It represents the radius of the circular base or the sphere.

Question

In the context of the NCERT text, what is the 'usual meaning' of the letter $l$?

Answer

It represents the slant height of a cone.

Question

A hemispherical tank has an inner radius of $1$ m and is made of $1$ cm thick iron. How is the volume of iron found?

Answer

It is found by subtracting the inner volume from the outer volume using $\frac{2}{3} \pi (r_{outer}^3 - r_{inner}^3)$.

Question

What fraction of a sphere's volume is the volume of a hemisphere with the same radius?

Answer

The fraction is one-half ($\frac{1}{2}$).

Question

If the circumference of a hemispherical dome is $17.6$ m, what is its radius?

Answer

The radius is $2.8$ m.

Question

A right triangle with sides $5$ cm, $12$ cm, and $13$ cm is revolved about the $12$ cm side. What is the radius of the resulting cone?

Answer

The radius is $5$ cm.

Question

Find the volume of a sphere whose surface area is $154$ $cm^2$.

Answer

The volume is approximately $179.67$ $cm^3$.

Question

A conical pit is $12$ m deep and has a top diameter of $3.5$ m. What is its capacity in kilolitres?

Answer

The capacity is $38.5$ kilolitres.

Question

What constitutes the 'Total Surface Area' of a cone?

Answer

It is the sum of the curved surface area and the area of the circular base.

Question

Which mathematical theorem is used to prove the relationship $l^2 = r^2 + h^2$ in a cone?

Answer

Pythagoras Theorem is used.

Question

In Figure 11.2(b), why is the solid shown NOT a right circular cone?

Answer

It is not a right circular cone because the base is not circular.

Question

The volume of a right circular cone is $9856$ $cm^3$ and the base diameter is $28$ cm. Find its height.

Answer

The height of the cone is $48$ cm.

Question

How is the radius $r'$ of a new sphere calculated when $27$ identical small spheres of radius $r$ are melted together?

Answer

The new radius is found using the relation $r' = 3r$.

Question

A joker's cap is a right circular cone with radius $7$ cm and height $24$ cm. Find the area of sheet required for $10$ caps.

Answer

The area required is $5500$ $cm^2$.

Question

If the height of a cone is $16$ cm and the base radius is $12$ cm, find the slant height.

Answer

The slant height is $20$ cm.

Question

What determines the 'riding space' available for a motorcyclist inside a hollow sphere?

Answer

The riding space is determined by the surface area of the sphere ($4 \pi r^2$).

Question

In a hemisphere, why is the total surface area $3 \pi r^2$?

Answer

It is the sum of the curved surface ($2 \pi r^2$) and the flat circular base ($\pi r^2$).