Cylinder, Cone & Sphere (Surface Area & Volume) — Exercise 20(B)

The radius and height of a solid metallic cone are r cm each. The volume of the cone is :

1. $\dfrac{1}{3}πr^2$

2. $\dfrac{4}{3}πr^2$

3. $\dfrac{1}{3}πr^3$

4. $\dfrac{4}{3}πr^3$

Answer

By formula,

Volume of cone = $\dfrac{1}{3}π$ × (radius)² × height

Given,

Radius and Height of a solid metallic cone are r cm each.

Volume of cone = $\dfrac{1}{3}π \times r^2 \times r = \dfrac{1}{3}πr^3$.

Hence, Option 3 is the correct option.

The radius of the base of a solid cone is r cm and its height is h cm; its curved surface area is :

1. π × r × $\sqrt{h^2 - r^2}$

2. π × r × $\sqrt{h^2 + r^2}$

3. π × r × (h + r)

4. πr²(h + r)

Answer

By formula,

Curved surface area = πrl

= $πr\sqrt{h^2 + r^2}$.

Hence, Option 2 is the correct option.

A conical toy tent-house, 28 cm in radius and 21 cm in slant height, is made from a rectangular sheet of paper 22 cm wide. The smallest length of the paper sheet required is :

1. 84 cm

2. 42 cm

3. 72 cm

4. 140 cm

Answer

Since, conical toy is made from the rectangular sheet.

∴ Curved surface area of cone = Area of rectangular sheet.

Let smallest length of the paper sheet required be a cm.

∴ πrl = length × breadth

$\Rightarrow \dfrac{22}{7} \times 28 \times 21 = a \times 22 \\[1em] \Rightarrow 22 \times 4 \times 21 = a \times 22 \\[1em] \Rightarrow 88 \times 21 = a \times 22 \\[1em] \Rightarrow 1848 = a \times 22 \\[1em] \Rightarrow a = \dfrac{1848}{22} \\[1em] \Rightarrow a = 84 \text{ cm}.$

Hence, Option 1 is the correct option.

For a cone, the ratio between the volume and area of its base is 11 : 6. The height of the cone is :

1. $1\dfrac{5}{6}$ units

2. 5.5 units

3. 11 units

4. 5 units

Answer

Let radius of cone be r cm and height be h cm.

Given,

For a cone, the ratio between the volume and area of its base is 11 : 6.

$\Rightarrow \dfrac{\text{Volume of cone}}{\text{Area of base}} = \dfrac{11}{6} \\[1em] \Rightarrow \dfrac{\dfrac{1}{3}πr^2h}{πr^2} = \dfrac{11}{6} \\[1em] \Rightarrow \dfrac{h}{3} = \dfrac{11}{6} \\[1em] \Rightarrow h = \dfrac{11}{6} \times 3 \\[1em] \Rightarrow h = \dfrac{11}{2} = 5.5 \text{ units}.$

Hence, Option 2 is the correct option.

The radii of two solid cones are equal and their slant heights are in the ratio 7 : 4. The ratio between their curved surface areas is :

1. 4 : 7

2. 7 : 4

3. 16 : 49

4. 49 : 16

Answer

Let radii of both the cones be r units each.

Let slant height of two cones be l₁ and l₂.

Curved surface area of first cone = πrl₁

Curved surface area of second cone = πrl₂

$\dfrac{\text{CSA of 1st cone}}{\text{CSA of 2nd cone}} = \dfrac{πrl_1}{πrl_2} = \dfrac{l_1}{l_2} = \dfrac{7}{4}$ = 7 : 4.

Hence, Option 2 is the correct option.

Find the volume of a cone whose slant height is 17 cm and radius of base is 8 cm.

Answer

Given,

Slant height (l) = 17 cm

Radius (r) = 8 cm

Let height of cone be h cm.

We know that,

⇒ l² = h² + r²

⇒ 17² = h² + 8²

⇒ 289 = h² + 64

⇒ h² = 225

⇒ h = $\sqrt{225}$ = 15 cm.

By formula,

Volume of cone = $\dfrac{1}{3}πr^2h$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 8^2 \times 15 \\[1em] = \dfrac{22 \times 64 \times 5}{7} \\[1em] = \dfrac{7040}{7} \\[1em] = 1005.71 \text{ cm}^3.$

Hence, volume of cone = 1005.71 cm³.

The curved surface area of a cone is 12320 cm². If the radius of its base is 56 cm, find its height.

Answer

By formula,

Curved surface area = πrl

$\therefore 12320 = \dfrac{22}{7} \times 56 \times l \\[1em] \Rightarrow l = \dfrac{12320 \times 7}{22 \times 56} \\[1em] \Rightarrow l = \dfrac{86240}{1232} = 70.$

We know that,

⇒ l² = h² + r²

⇒ 70² = h² + 56²

⇒ 4900 = h² + 3136

⇒ h² = 4900 - 3136

⇒ h² = 1764

⇒ h = $\sqrt{1764}$ = 42 cm.

Hence, height of cone = 42 cm.

The circumference of the base of a 12 m high conical tent is 66 m. Find the volume of the air contained in it.

Answer

Given,

Circumference of base = 66 m

⇒ 2πr = 66

⇒ $2 \times \dfrac{22}{7} \times r = 66$

⇒ r = $\dfrac{66 \times 7}{2 \times 22} = \dfrac{21}{2}$ = 10.5 cm

By formula,

Volume of cone = $\dfrac{1}{3}πr^2h$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times (10.5)^2 \times 12 \\[1em] = \dfrac{22 \times 110.25 \times 4}{7} \\[1em] = \dfrac{9702}{7} \\[1em] = 1386 \text{ m}^3.$

Hence, volume of air contained in cone = 1386 m³.

The radius and the height of a right circular cone are in the ratio 5 : 12 and its volume is 2512 cubic cm. Find the radius and slant height of the cone. (Take π = 3.14)

Answer

Given,

Radius : Height = 5 : 12

Let radius (r) = 5x and height (h) = 12x.

By formula,

Volume of cone = $\dfrac{1}{3}πr^2h$

$\Rightarrow 2512 = \dfrac{1}{3} \times 3.14 \times (5x)^2 \times (12x) \\[1em] \Rightarrow 2512 = \dfrac{1}{3} \times 3.14 \times 300x^3 \\[1em] \Rightarrow x^3 = \dfrac{2512 \times 3}{3.14 \times 300} \\[1em] \Rightarrow x^3 = \dfrac{7536}{942} \\[1em] \Rightarrow x^3 = 8 \\[1em] \Rightarrow x^3 = 2^3 \\[1em] \Rightarrow x = 2 \text{ cm}.$

Radius (r) = 5x = 5(2) = 10 cm and Height (h) = 12x = 12(2) = 24 cm.

By formula,

⇒ l² = h² + r²

⇒ l² = (24)² + (10)²

⇒ l² = 576 + 100

⇒ l² = 676

⇒ l = $\sqrt{676}$ = 26 cm.

Hence, radius = 10 cm and slant height = 26 cm.

The diameters of two cones are equal. If their slant heights are in the ratio 5 : 4, find the ratio of their curved surface areas.

Answer

Given, ratio of slant height = 5 : 4

Let slant height of 1st cone be 5x cm and 2nd cone be 4x cm.

For 1st cone,

⇒ Diameter = d₁

⇒ Radius = r₁

⇒ Slant height (l₁) = 5x

For 2nd cone,

⇒ Diameter = d₂

⇒ Radius = r₂

⇒ Slant height (l₂) = 4x

Given,

⇒ d₁ = d₂

∴ r₁ = r₂.

$\Rightarrow \dfrac{\text{CSA of 1st cone}}{\text{CSA of 2nd cone}} = \dfrac{πr_1l_1}{πr_2l_2} \\[1em] = \dfrac{l_1}{l_2} = \dfrac{5x}{4x} \\[1em] = \dfrac{5}{4} = 5 : 4.$

Hence, ratio of curved surface areas = 5 : 4.

There are two cones. The curved surface area of one is twice that of the other. The slant height of the latter is twice that of the former. Find the ratio of their radii.

Answer

Let curved surface area of 1st cone be twice than that of 2nd cone.

For 1st cone,

⇒ Slant height (l₁) = l

⇒ Radius = r₁

⇒ Curved surface area (C₁) = 2C .........(1)

For 2nd cone,

⇒ Slant height (l₂) = 2l

⇒ Radius = r₂

⇒ Curved surface area (C₂) = C .........(2)

Dividing equation (1) by (2), we get :

$\Rightarrow \dfrac{\text{CSA of 1st cone}}{\text{CSA of 2nd cone}} = \dfrac{2C}{C} \\[1em] \Rightarrow \dfrac{πr_1l_1}{πr_2l_2} = \dfrac{2C}{C} \\[1em] \Rightarrow \dfrac{πr_1.l}{πr_2.2l} = 2 \\[1em] \Rightarrow \dfrac{r_1}{2r_2} = 2 \\[1em] \Rightarrow \dfrac{r_1}{r_2} = \dfrac{4}{1}.$

⇒ r₁ : r₂ = 4 : 1.

Hence, ratio of radii = 4 : 1.

A heap of wheat is in the form of a cone of diameter 16.8 m and height 3.5 m. Find its volume. How much cloth is required to just cover the heap?

Answer

Given,

Diameter of heap of cone = 16.8 m

Radius of heap of cone (r) = $\dfrac{16.8}{2}$ = 8.4 m

Height (h) = 3.5 m

Volume of cone = $\dfrac{1}{3}πr^2h$

$= \dfrac{1}{3} \times \dfrac{22}{7} \times (8.4)^2 \times 3.5 \\[1em] = \dfrac{22}{21} \times 246.96 \\[1em] = \dfrac{5433.12}{21} \\[1em] = 258.72 \text{ m}^3.$

We know that,

⇒ l² = r² + h²

⇒ l² = (8.4)² + (3.5)²

⇒ l² = 70.56 + 12.25

⇒ l² = 82.81

⇒ l = $\sqrt{82.81}$ = 9.1 cm.

Cloth required to cover the heap = Curved surface area of heap = πrl

= $\dfrac{22}{7} \times 8.4 \times 9.1$

= 240.24 m².

Hence, volume = 258.72 m³ and cloth required to cover the heap = 240.24 m².

If you are given a rectangular canvas of 1.5 m in width, what length of this canvas would you require to make a conical tent that is 48 m in diameter and 7 m in height? Note that 10% of the canvas is used (wasted) in folds and stitching.

Also, find the cost of the canvas at the rate of ₹ 24 per meter.

Answer

Given,

Diameter of conical tent = 48 m

Radius of conical tent (r) = $\dfrac{48}{2}$ = 24 m

Height of conical tent (h) = 7 m

Let l be the slant height of the conical tent.

By formula,

$\Rightarrow l = \sqrt{r^2 + h^2} \\[1em] \Rightarrow l = \sqrt{(24)^2 + 7^2} \\[1em] \Rightarrow l = \sqrt{576 + 49} \\[1em] \Rightarrow l = \sqrt{625} \\[1em] \Rightarrow l = 25 \text{ m}.$

By formula,

Surface area of conical tent (S) = πrl

$S = \dfrac{22}{7} \times 24 \times 25 \\[1em] = \dfrac{13200}{7} \text{ m}^2.$

Given,

10% of the canvas is used in folds and stitching.

Thus, 90% of the total canvas area is used for making the tent.

$\Rightarrow \dfrac{90}{100} \times \text{ Total canvas area} = \text{Surface area of conical tent} \\[1em] \Rightarrow \dfrac{90}{100} \times \text{ Total canvas area} = \dfrac{13200}{7} \\[1em] \Rightarrow \text{ Total canvas area} = \dfrac{13200}{7} \times \dfrac{100}{90} \\[1em] \Rightarrow \text{ Total canvas area} = \dfrac{132000}{63} \text{ m}^2.$

Thus,

⇒ Total canvas area = Length of canvas × Width of canvas

⇒ $\dfrac{132000}{63}$ = Length of canvas × 1.5

⇒ Length of canvas = $\dfrac{132000}{63 \times 1.5} = \dfrac{132000}{94.5}$ = 1396.83 m

Cost of canvas = Rate per m × Length

= 24 × 1396.83

= ₹ 33,523.92

Hence, length of canvas required = 1396.83 m and cost of canvas = ₹ 33,523.92

A solid cone of height 8 cm and base radius 6 cm is melted and recast into identical cones, each of height 2 cm and diameter 1 cm. Find the number of cones formed.

Answer

For larger cone,

Height (h₁) = 8 cm

Radius (r₁) = 6 cm

For smaller cones,

Height (h₂) = 2 cm

Radius (r₂) = $\dfrac{1}{2}$ = 0.5 cm

Let no. of smaller cones be n.

Volume of larger cone = n × Volume of smaller cones

$\Rightarrow \dfrac{1}{3}πr_1^2h_1 = n \times \dfrac{1}{3}πr_2^2h_2 \\[1em] \Rightarrow r_1^2h_1 = n \times r_2^2h_2 \\[1em] \Rightarrow n = \dfrac{r_1^2h_1}{r_2^2h_2} \\[1em] \Rightarrow n = \dfrac{6^2 \times 8}{(0.5)^2 \times 2} \\[1em] \Rightarrow n = \dfrac{288}{0.5} \\[1em] \Rightarrow n = 576.$

Hence, no. of cones formed = 576.

The total surface area of a right circular cone of slant height 13 cm is 90π cm². Calculate :

(i) its radius in cm

(ii) its volume in cm³.

[Take π = 3.14]

Answer

(i) Given,

Total surface area = 90π

∴ πrl + πr² = 90π

⇒ πr(l + r) = 90π

⇒ r(l + r) = 90

⇒ r(13 + r) = 90

⇒ r² + 13r - 90 = 0

⇒ r² + 18r - 5r - 90 = 0

⇒ r(r + 18) - 5(r + 18) = 0

⇒ (r - 5)(r + 18) = 0

⇒ (r - 5) = 0 or (r + 18) = 0

⇒ r = 5 or r = -18.

Since, radius cannot be negative.

∴ radius = 5 cm.

Hence, radius = 5 cm.

(ii) By formula,

⇒ l² = r² + h²

⇒ 13² = 5² + h²

⇒ h² = 169 - 25

⇒ h² = 144

⇒ h = $\sqrt{144}$

⇒ h = 12 cm.

Volume = $\dfrac{1}{3}πr^2h$

= $\dfrac{1}{3} \times 3.14 \times (5)^2 \times 12$

= $\dfrac{942}{3}$

= 314 cm³.

Hence, volume of circular cone = 314 cm³.

The area of the base of a conical solid is 38.5 cm² and its volume is 154 cm³. Find the curved surface area of the solid.

Answer

Given,

Area of the base of conical solid = 38.5 cm²

⇒ πr² = 38.5 ............(1)

⇒ $\dfrac{22}{7}r^2$ = 38.5

⇒ r² = $\dfrac{38.5 \times 7}{22}$ = 12.25

⇒ r = $\sqrt{12.25}$ = 3.5 cm

Given,

$\Rightarrow \text{Volume } = 154 \\[1em] \Rightarrow \dfrac{1}{3}πr^2h = 154 \\[1em] \Rightarrow \dfrac{1}{3} \times 38.5 \times h = 154 \quad [\text{From (1)}] \\[1em] \Rightarrow h = \dfrac{154 \times 3}{38.5} \\[1em] \Rightarrow h = 12 \text{ cm}.$

By formula,

⇒ l² = r² + h²

⇒ l² = (3.5)² + (12)²

⇒ l² = 12.25 + 144

⇒ l² = 156.25

⇒ l = $\sqrt{156.25}$ = 12.5 cm.

Curved surface area = πrl

= $\dfrac{22}{7} \times 3.5 \times 12.5$

= 137.5 cm².

Hence, curved surface area = 137.5 cm².

A vessel, in the form of an inverted cone, is filled with water to the brim. Its height is 32 cm and diameter of the base is 25.2 cm. Six equal solid cones are dropped in it, so that they are fully submerged. As a result one-fourth of water in the original cone overflows. What is the volume of each of the solid cones submerged?

Answer

Radius of vessel (R) = $\dfrac{25.2}{2}$ = 12.6 cm.

Total volume of water in vessel = $\dfrac{1}{3}πR^2H$

On submerging six equal cones in vessel, one-fourth of water in the original cone overflows.

Let radius of small cones be r and height be h,

$\therefore 6 \times \text{Vol. of each cone} = \dfrac{1}{4} \times \dfrac{1}{3}πR^2H \\[1em] \Rightarrow \text{Vol. of each cone} = \dfrac{1}{72}πR^2H \\[1em] \Rightarrow \text{Vol. of each cone} = \dfrac{1}{72} \times \dfrac{22}{7} \times (12.6)^2 \times 32 \\[1em] \Rightarrow \text{Vol. of each cone} = 221.76 \text{ cm}^3.$

Hence, volume of each cone = 221.76 cm³.

The volume of a conical tent is 1232 m³ and the area of the base floor is 154 m². Calculate the :

(i) radius of the floor,

(ii) height of the tent,

(iii) length of the canvas required to cover this conical tent if its width is 2 m.

Answer

(i) Given,

Area of base floor = 154 m²

$\Rightarrow πr^2 = 154 \\[1em] \Rightarrow \dfrac{22}{7} \times r^2 = 154 \\[1em] \Rightarrow r^2 = \dfrac{154 \times 7}{22} \\[1em] \Rightarrow r^2 = 7 \times 7 \\[1em] \Rightarrow r = 7 \text{ m}.$

Hence, radius of the floor = 7 m.

(ii) Given,

Volume of tent = 1232 m³

$\Rightarrow \dfrac{1}{3}πr^2h = 1232 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times (7)^2 \times h = 1232 \\[1em] \Rightarrow \dfrac{154}{3} \times h = 1232 \\[1em] \Rightarrow h = \dfrac{1232 \times 3}{154} \\[1em] \Rightarrow h = 24 \text{ m}.$

Hence, height of tent = 24 m.

(iii) By formula,

⇒ l² = r² + h²

⇒ l² = (7)² + (24)²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$

⇒ l = 25 m.

Curved surface area of cone = πrl

= $\dfrac{22}{7} \times 7 \times 25$

= 550 m².

Let length of canvas be l.

Area of canvas = Curved surface area of cone

⇒ l × b = 550

⇒ l × 2 = 550

⇒ l = $\dfrac{550}{2}$

⇒ l = 275 m.

Hence, length of canvas required = 275 m.