Volume & Surface Area of Solids — Exercise 21(B)

The height of a right circular cone is 24 cm and the radius of its base is 7 cm. Calculate :

(i) the slant height of the cone

(ii) the lateral surface area of the cone

(iii) the total surface area of the cone

(iv) the volume of the cone

Answer

Given, h = 24 cm and r = 7 cm

(i) Slant height, l = $\sqrt{\text{h}^2 + \text{r}^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm.}$

Hence, slant height of the cone is 25 cm.

(ii) Lateral surface area = πrl

$= \dfrac{22}{7} \times 7 \times 25 \\[1em] = \dfrac{3850}{7} \\[1em] = 550 \text{ cm}^2$

Hence, lateral surface area of the cone is 550 cm².

(iii) Total surface area = πr(l + r)

$= \dfrac{22}{7} \times 7(25 + 7) \\[1em] = \dfrac{154}{7} \times (32) \\[1em] = \dfrac{4928}{7} \\[1em] = 704 \text{ cm}^2$

Hence, total surface area of the cone is 704 cm².

(iv) Volume of cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 24 \\[1em] = \dfrac{22}{21} \times 49 \times 24 \\[1em] = \dfrac{25872}{21} \\[1em] = 1232 \text{ cm}^3$

Hence, volume of the cone is 1232 cm³.

The height of a right circular cone is 8 cm and the diameter of its base is 12 cm. Calculate :

(i) the slant height of the cone

(ii) the total surface area of the cone

(iii) the volume of the cone

Answer

Given, h = 8 cm and r = $\dfrac{\text{diameter}}{2} = \dfrac{12}{2} = 6$ cm

(i) Slant height, l = $\sqrt{\text{h}^2 + \text{r}^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ cm.}$

Hence, slant height of the cone is 10 cm.

(ii) Total surface area = πr(l + r)

$= \dfrac{22}{7} \times 6(10 + 6) \\[1em] = \dfrac{132}{7} \times (16) \\[1em] = \dfrac{2112}{7} \\[1em] = 301.7 \text{ cm}^2$

Hence, total surface area of the cone is 301.7 cm².

(iii) Volume of cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 6^2 \times 8 \\[1em] = \dfrac{22}{21} \times 36 \times 8 \\[1em] = \dfrac{6336}{21} \\[1em] = 301.7 \text{ cm}^3$

Hence, volume of the cone is 301.7 cm³.

The slant height of a cone is 17 cm and the radius of its base is 15 cm. Find:

(i) the height of the cone

(ii) the volume of the cone

(iii) the total surface area of the cone

Answer

Given, slant height, l = 17 cm and radius, r = 15 cm

(i) l² = r² + h²

⇒ h² = l² - r²

⇒ h² = 17² - 15²

⇒ h² = 289 - 225

⇒ h² = 64

⇒ h = $\sqrt{64} = 8 \text{ cm.}$

Hence, height of the cone is 8 cm.

(ii) Volume of cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 15^2 \times 8 \\[1em] = \dfrac{22}{21} \times 225 \times 8 \\[1em] = \dfrac{39600}{21} \\[1em] = 1885.7 \text{ cm}^3$

Hence, volume of the cone is 1885.7 cm³.

(iii) Total surface area = πr(l + r)

$= \dfrac{22}{7} \times 15(17 + 15) \\[1em] = \dfrac{330}{7} \times (32) \\[1em] = \dfrac{10560}{7} \\[1em] = 1508.6 \text{ cm}^2$

Hence, total surface area of the cone is 1508.6 cm².

The volume of a right circular cone is 660 cm³ and diameter of its base is 12 cm. Calculate:

(i) the height of the cone

(ii) the slant height of the cone

(ii) the total surface area of the cone

Answer

Given, radius, r = $\dfrac{\text{diameter}}{2} = \dfrac{12}{2} = 6 \text{ cm.}$

Volume of cone = 660 cm³

(i) By formula,

Volume of cone = $\dfrac{1}{3}$ πr²h

$\Rightarrow 660 = \dfrac{1}{3} \times \dfrac{22}{7} \times 6^2 \times \text{h} \\[1em] \Rightarrow 660 = \dfrac{22}{21} \times 36 \times \text{h} \\[1em] \Rightarrow \text{h} = \dfrac{660 \times 21}{22 \times 36} \\[1em] \Rightarrow \text{h} = \dfrac{13860}{792} \\[1em] \Rightarrow \text{h} = 17.5 \text{ cm}.$

Hence, the height of the cone is 17.5 cm.

(ii) By formula,

Slant height (l) = $\sqrt{\text{h}^2 + \text{r}^2}$

$= \sqrt{(17.5)^2 + 6^2} = \sqrt{306.25 + 36} = \sqrt{342.25} = 18.5 \text{ cm.}$

Hence, slant height of the cone is 18.5 cm.

(iii) Total surface area = πr(l + r)

$= \dfrac{22}{7} \times 6 \times (18.5 + 6) \\[1em] = \dfrac{132}{7} \times (24.5) \\[1em] = \dfrac{3234}{7} \\[1em] = 462 \text{ cm}^2$

Hence, total surface area of the cone is 462 cm².

The total surface of a right circular cone of slant height 20 cm is 384π cm². Calculate:

(i) its radius in cm

(ii) its volume in cm³, in terms of π

Answer

Given, slant height, l = 20 cm and total surface area of cone = 384π cm²

(i) By formula,

Total surface area = πr(l + r)

⇒ 384π = πr(20 + r)

⇒ 384 = 20r + r²

⇒ r² + 20r - 384 = 0

⇒ r² + 32r - 12r - 384 = 0

⇒ r(r + 32) - 12(r + 32) = 0

⇒ (r + 32) = 0 or (r - 12) = 0

⇒ r = - 32 or r = 12

Since, radius cannot be negative.

∴ r = 12 cm.

Hence, radius of the cone is 12 cm.

(ii) l² = r² + h²

⇒ h² = l² - r²

⇒ h² = 20² - 12²

⇒ h² = 400 - 144

⇒ h² = 256

⇒ h = $\sqrt{256} = 16 \text{ cm.}$

Volume of cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times π \times 12^2 \times 16 \\[1em] = \dfrac{1}{3} \times π \times 144 \times 16 \\[1em] = \dfrac{2304}{3} π \\[1em] = 768 π \text{ cm}^3$

Hence, volume of the cone is 768 π cm³.

The radius and the height of a right circular cone are in the ratio of 5 : 12 and its volume is 2512 cm³. Find:

(i) the radius and height of the cone

(ii) the curved surface area of the cone

(iii) the total surface area of the cone

(Take π = 3.14)

Answer

Given, radius(r) : height(h) = 5 : 12

(i) Let r = 5x and h = 12x

Volume of cone = $\dfrac{1}{3}$ πr²h

$\Rightarrow 2512 = \dfrac{1}{3} \times 3.14 \times (5\text{x})^2 \times 12\text{x} \\[1em] \Rightarrow 2512 = 3.14 \times 25\text{x}^2 \times 4\text{x} \\[1em] \Rightarrow \text{x}^3 = \dfrac{2512}{3.14 \times 25 \times 4} \\[1em] \Rightarrow \text{x}^3 = \dfrac{2512}{314} \\[1em] \Rightarrow \text{x}^3 = 8 \\[1em] \Rightarrow \text{x} = \sqrt[3]{8} \\[1em] \Rightarrow \text{x} = 2$

⇒ r = 5x = 5 × 2 = 10 cm

⇒ h = 12x = 12 × 2 = 24 cm

Hence, radius of the cone is 10 cm and height of the cone is 24 cm.

(ii) Curved surface area = πrl

l² = r² + h²

⇒ l² = 10² + 24²

⇒ l² = 100 + 576

⇒ l² = 676

⇒ l = $\sqrt{676}$ = 26 cm

$= 3.14 \times 10 \times 26 \\[1em] = 816.4 \text{ cm}^2$

Hence, curved surface area of the cone is 816.4 cm².

(iii) Total surface area = πr(l + r)

$= 3.14 \times 10(26 + 10) \\[1em] = 3.14 \times 10 \times 36 \\[1em] = 1130.4 \text{ cm}^2$

Hence, total surface area of the cone is 1130.4 cm².

A right circular cone of radius 20 cm has its volume 8800 cm³. Find its:

(i) height

(ii) curved surface area.

Give your answer to the nearest whole number.

Answer

Let height of cone be h cm.

Given,

Radius of cone (r) = 20 cm

(i) Given,

Volume = 8800 cm³

By formula,

Volume of cone = $\dfrac{1}{3}\pi r^2 h$

∴ $\dfrac{1}{3}\pi r^2 h = 8800$

$\Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times 20^2 \times h = 8800 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times 400 \times h = 8800 \\[1em] \Rightarrow h = \dfrac{8800 \times 7 \times 3}{400 \times 22} \\[1em] \Rightarrow h = 7 \times 3 \\[1em] \Rightarrow h = 21 \text{ cm}.$

Hence, height of cone = 21 cm.

(ii) By formula,

Slant Height of cone (l) = $\sqrt{(r^2 + h^2)}$

$l = \sqrt{(20)^2 + (21)^2} \\[1em] = \sqrt{400 + 441} \\[1em] = \sqrt{841} \\[1em] = 29 \text{ cm}$

By formula,

Curved Surface Area of cone = πrl

Curved Surface Area of cone = $\dfrac{22}{7} \times 20 \times 29$

⇒ $\dfrac{12760}{7}$

⇒ 1822.857 ≈ 1823 cm².

Hence, curved surface area of cone = 1823 cm².

How many metres of canvas 1.25 m will be needed to make a conical tent whose base radius is 17.5 m and height 6 m?

Answer

Given, r = 17.5 m and h = 6 m

l² = r² + h²

⇒ l² = 17.5² + 6²

⇒ l² = 306.25 + 36

⇒ l² = 342.25

⇒ l = $\sqrt{342.25}$ = 18.5 m

So, the total curved surface area of the tent = πrl

$= \dfrac{22}{7} \times 17.5 \times 18.5 \\[1em] = \dfrac{7122.5}{7} \\[1em] = 1017.5 \text{ m}^2$

Width of the canvas used = 1.25 m

Length of canvas = $\dfrac{\text{area of canvas}}{\text{width of canvas}} = \dfrac{1017.5}{1.25}$ = 814 m.

Hence, 814 metres of canvas will be needed to make a conical tent.

A circus tent is cylindrical to a height of 3 meters and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 2.5 m wide to make the required tent.

Answer

From figure,

Radius of conical part = radius of cylindrical part = r.

Radius of the cylindrical part of the tent (r) = $\dfrac{\text{diameter}}{2} = \dfrac{105}{2} \text{ m}$

Radius of the conical part (r) = $\dfrac{105}{2} \text{ m}$

Slant height (l) = 53 m

So, the total curved surface area of the tent = 2πrh + πrl

$= 2 \times \dfrac{22}{7} \times \dfrac{105}{2} \times 3 + \dfrac{22}{7} \times \dfrac{105}{2} \times 53 \\[1em] = \dfrac{13860}{14} + \dfrac{122430}{14} \\[1em] = 990 + 8745 \\[1em] = 9735 \text{ m}^2.$

Width of the canvas used = 2.5 m

Length of canvas = $\dfrac{\text{area of canvas}}{\text{width of canvas}} = \dfrac{9735}{2.5}$ = 3894 m.

Hence, length of the canvas required to make tent is 3894 m.

An iron pillar consists of a cylindrical portion, 2.8 m high and 20 cm in diameter and a cone 42 cm high is surrounding it. Find the weight of the pillar, given that 1 cm³ of iron weighs 7.5 g.

Answer

Radius of cylindrical portion, r = $\dfrac{\text{diameter}}{2} = \dfrac{20}{2}$ = 10 cm

Height of the cylindrical portion, h = 2.8 m = 2.8 × 100 = 280 cm

Height of the conical portion, H = 42 cm

From figure,

Radius of the conical part = radius of the cylindrical portion = r = 10 cm

Volume of iron pole = Volume of cylindrical portion + Volume of conical portion

= πr²h + $\dfrac{1}{3}$ πr²H

$= \dfrac{22}{7} \times 10^2 \times 280 + \dfrac{1}{3} \times \dfrac{22}{7} \times 10^2 \times 42 \\[1em] = \dfrac{22}{7} \times 100 \times 280 + \dfrac{1}{3} \times \dfrac{22}{7} \times 100 \times 42 \\[1em] = \dfrac{22}{7} \times 100 \times 280 + \dfrac{22}{7} \times 100 \times 14 \\[1em] = \dfrac{616000}{7} + \dfrac{30800}{7} \\[1em] = \dfrac{616000 + 30800}{7} \\[1em] = \dfrac{646800}{7} \\[1em] = 92400 \text{ cm}^3.$

Given,

Weight of 1 cm³ of iron = 7.5 gm.

Total weight = 92400 × 7.5 = 693000 gm = $\dfrac{693000}{1000}$ kg = 693 kg.

Hence, the weight of the pillar is 693 kg.

Water flows at the rate of 10 m per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm?

Answer

Radius of cylindrical pipe, r = $\dfrac{\text{diameter}}{2} = \dfrac{0.5}{2}$ = 0.25 cm

Given, water flows at the rate of 10 m per minute.

Length of the cylindrical portion, h = 10 m = 10 × 100 = 1000 cm

Height of the conical portion, H = 24 cm

Radius of conical pipe, R = $\dfrac{\text{diameter}}{2} = \dfrac{40}{2}$ = 20 cm

Volume of water that flows in 1 min = πr²h

$= \dfrac{22}{7} \times (0.25)^2 \times 1000 \\[1em] = \dfrac{22}{7} \times 0.0625 \times 1000 \\[1em] = \dfrac{1375}{7}$

Volume of the conical vessel = $\dfrac{1}{3}$ πR²H

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 20^2 \times 24 \\[1em] = \dfrac{22}{7} \times 400 \times 8 \\[1em] = \dfrac{70400}{7}$

Required time = $\dfrac{\text{Volume of conical vessel}}{\text{Volume of water that flows in 1 min}}$

$= \dfrac{\dfrac{70400}{7}}{\dfrac{1375}{7}} \\[1em] = \dfrac{70400}{7} \times \dfrac{7}{1375} \\[1em] = \dfrac{70400}{1375} \\[1em] = 51.2 \text{ min.}$

= 51 min 12 sec.

Hence, time required to fill a conical vessel is 51 min 12 sec.

A conical tent is to accommodate 11 persons. Each person must have 4 m² of the space on the ground and 20m³ of air to breathe. Find the height of the cone.

Answer

Given,

Each person must have 20 m³ of air to breathe.

∴ 11 persons need 11 × 20 m³ = 220 m³

Each person must have 4 m² of the space on the ground.

∴ 11 persons need 11 × 4 m² = 44 m²

Base of the conical tent = area of the circle = πr²

$\Rightarrow 44 = \dfrac{22}{7} \times \text{r}^2 \\[1em] \Rightarrow \text{r}^2 = \dfrac{7 \times 44}{22} \\[1em] \Rightarrow \text{r}^2 = \dfrac{308}{22} \\[1em] \Rightarrow \text{r}^2 = 14 \text{ m.}$

Let height of the conical tent be h meters.

Since, conical tent needs to accomodate 11 persons, so its volume will be equal to volume of air required for 11 persons.

$\Rightarrow \dfrac{1}{3}π \text{r}^2 \text{h} = 220 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times 14 \times \text{h} = 220 \\[1em] \Rightarrow \text{h} = \dfrac{220 \times 7 \times 3}{22 \times 14} \\[1em] \Rightarrow \text{h} = \dfrac{4620}{308} \\[1em] \Rightarrow \text{h} = 15 \text{ m.}$

Hence, height of the tent is 15 m..

A conical tent is to accommodate 77 persons. Each person must have 16 m³ of air to breathe. Given the radius of the tent as 7 m, find the height of the tent and also its curved surface area.

Answer

Given,

Each person must have 16 m³ of air to breathe.

∴ 77 persons need 77 × 16 m³ = 1232 m³

Radius of the tent (r) = 7 m

Let height of the conical tent be h meters.

Since, conical tent needs to accomodate 77 persons, so its volume will be equal to volume of air required for 77 persons.

$\Rightarrow \dfrac{1}{3}π \text{r}^2 \text{h} = 1232 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times \text{h} = 1232 \\[1em] \Rightarrow \dfrac{1}{3} \times \dfrac{22}{7} \times 49 \times \text{h} = 1232 \\[1em] \Rightarrow \text{h} = \dfrac{1232 \times 7 \times 3}{22 \times 49} \\[1em] \Rightarrow \text{h} = \dfrac{25872}{1078} \\[1em] \Rightarrow \text{h} = 24 \text{ m.}$

By formula,

l² = r² + h²

⇒ l² = 7² + 24²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$ = 25 m

Curved surface area of the tent = πrl

$= \dfrac{22}{7} \times 7 \times 25 \\[1em] = 22 \times 25 \\[1em] = 550 \text{ m}^2$

Hence, height of the tent is 24 m and curved surface area of the tent is 550 m².

A right circular cone is 3.6 cm high and the radius of its base is 1.6 cm. It is melted and recast into a right circular cone with radius of its base as 1.2 cm. Find its height.

Answer

Radius of cone, r = 1.6 cm

Height of the cone, h = 3.6 cm

Volume of circular cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times \dfrac{22}{7} \times (1.6)^2 \times 3.6 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 2.56 \times 3.6 \\[1em] = \dfrac{202.752}{21}$

Volume of cone of radius (R) = 1.2 cm and height (H)

Volume of circular cone = $\dfrac{1}{3}$ πR²H

$= \dfrac{1}{3} \times \dfrac{22}{7} \times (1.2)^2 \times \text{H} \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 1.44 \times \text{H} \\[1em] = \dfrac{31.68}{21} \text{H}$

Since, cone is melted and recasted into right circular cone of radius 1.2 cm, the volume remains the same.

$\therefore \dfrac{202.752}{21} = \dfrac{31.68}{21} \text{H} \\[1em] \Rightarrow \text{H} = \dfrac{202.752 \times 21}{31.68 \times 21} \\[1em] \Rightarrow \text{H} = 6.4 \text{ cm.}$

Hence, height of the cone is 6.4 cm.

A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to form cones, each of height 1 cm and base radius 1 mm. Find the number of cones.

Answer

For larger cylinder,

Height (H) = 5 cm

Radius (R) = 3 cm

For smaller cones,

Height (h) = 1 cm

Radius (r) = 1 mm = 0.1 cm

Let no. of smaller cones formed be n.

Volume of larger cylinder = n × Volume of smaller cones

$\Rightarrow π\text{R}^2\text{H} = \text{n} \times \dfrac{1}{3}π\text{r}^2\text{h} \\[1em] \Rightarrow \dfrac{22}{7} \times (3)^2 \times 5 = \text{n} \times \dfrac{1}{3} \times \dfrac{22}{7} \times (0.1)^2 \times 1 \\[1em] \Rightarrow \dfrac{22}{7} \times 9 \times 5 = \text{n} \times \dfrac{1}{3} \times \dfrac{22}{7} \times 0.01 \times 1 \\[1em] \Rightarrow \dfrac{990}{7} = \text{n} \times \dfrac{0.22}{21} \\[1em] \Rightarrow \text{n} = \dfrac{990 \times 21}{0.22 \times 7} \\[1em] \Rightarrow \text{n} = \dfrac{20790}{1.54} \\[1em] \Rightarrow \text{n} = 13500$

Hence, the number of cones formed = 13500.

A conical vessel, whose internal radius is 12 cm and height 50 cm, is full of liquid. The contents are emptied into a cylindrical vessel with internal radius 10 cm. Find the height to which the liquid rises in the cylindrical vessel.

Answer

For cylindrical vessel,

Let height be H cm

Radius (R) = 10 cm

For conical vessel,

Height (h) = 50 cm

Radius (r) = 12 cm

Since, contents in conical vessel are emptied into a cylindrical vessel, hence there volume will be same.

∴ Volume of cylinder = Volume of conical vessel

$\Rightarrow π\text{R}^2\text{H} = \dfrac{1}{3}π\text{r}^2\text{h} \\[1em] \Rightarrow \text{R}^2\text{H} = \dfrac{1}{3}\text{r}^2\text{h} \\[1em] \Rightarrow 10^2 \times \text{H} = \dfrac{1}{3} \times 12^2 \times 50 \\[1em] \Rightarrow 100 \times \text{H} = \dfrac{1}{3} \times 144 \times 50 \\[1em] \Rightarrow \text{H} = \dfrac{144 \times 50}{3 \times 100} \\[1em] \Rightarrow \text{H} = \dfrac{7200}{300} \\[1em] \Rightarrow \text{H} = 24 \text{ cm.}$

Hence, height to which the liquid rises in the cylindrical vessel is 24 cm.

The height of a cone is 40 cm. A small cone is cut off at the top by a plane parallel to its base. If its volume be $\dfrac{1}{64}$ of the volume of the given cone, at what height above the base is the section cut?

Answer

Let OAB be the given cone of height 40 cm and base radius R cm. Let this cone be cut by the plane CND (parallel to the base plane AMB) to obtain cone OCD with height h cm and base radius r cm as shown in the figure below :

From figure,

∠NOD = ∠MOB (Common angle)

∠OND = ∠OMB = 90° (Heights are perpendicular to radii)

∠ODN = ∠OBM (Corresponding angles since ND || MB)

∴ △OND ~ △OMB (By AA similarity)

We know that,

Ratio of corresponding sides of similar triangle are proportional.

∴ $\dfrac{\text{r}}{\text{R}} = \dfrac{\text{h}}{40}$ ...(1)

According to given,

Volume of cone OCD = $\dfrac{1}{64}$ Volume of cone OAB

∴ $\dfrac{1}{3}$ πr²h = $\dfrac{1}{64} \times \dfrac{1}{3}$ πR² × 40

Dividing both sides by π and multiplying by 3 we get,

$\Rightarrow \text{r}^2 \text{h} = \dfrac{40}{64} \times \text{R}^2 \\[1em] \Rightarrow \dfrac{\text{r}^2}{\text{R}^2} = \dfrac{5}{8 \text{h}} \\[1em] \Rightarrow \Big(\dfrac{\text{r}}{\text{R}}\Big)^2 = \dfrac{5}{8 \text{h}} \\[1em]$

Using eq.(1),

$\Rightarrow \Big(\dfrac{\text{h}}{40}\Big)^2 = \dfrac{5}{8 \text{h}} \\[1em] \Rightarrow \dfrac{\text{h}^2}{1600} = \dfrac{5}{8 \text{h}} \\[1em] \Rightarrow \text{h}^3 = \dfrac{5 \times 1600}{8} \\[1em] \Rightarrow \text{h}^3 = 5 \times 200 \\[1em] \Rightarrow \text{h}^3 = 1000 \\[1em] \Rightarrow \text{h} = \sqrt[3]{1000} \\[1em] \Rightarrow \text{h} = 10 \text{ cm.}$

The height of the cone OCD = 10 cm

∴ The section is cut at the height of 40 - 10 = 30 cm.

Hence, the section is cut above 30 cm from the base.

A hollow metallic cylindrical tube has an internal radius of 3 cm and height 21 cm. The thickness of the metal is 0.5 cm. The tube is melted and cast into a right circular cone of height 7 cm. Find the radius of the cone, correct to one decimal place.

Answer

Internal radius (r) = 3 cm

Height (h) = 21 cm

Thickness = External radius (R) - Internal radius

⇒ 0.5 = External radius - 3 cm

⇒ External radius = 0.5 + 3 = 3.5 cm

Volume of hollow cylinder = π(R² - r²)h,

Putting values we get,

∴ Volume of metal = π(3.5² - 3²) × 21

$= \dfrac{22}{7} \times (12.25 - 9) \times 21 \\[1em] = 22 \times 3.25 \times 3 \\[1em] = 214.5 \text{ cm}^3$

Given the tube is melted and cast into a right circular cone of height (H) 7 cm.

So, the volume of metal and volume of cone will be same.

∴ 214.5 = $\dfrac{1}{3}$ πr²H

$\Rightarrow 214.5 = \dfrac{1}{3} \times \dfrac{22}{7} \times \text{r}^2 \times 7 \\[1em] \Rightarrow 214.5 = \dfrac{1}{3} \times 22 \times \text{r}^2 \\[1em] \Rightarrow \text{r}^2 = \dfrac{214.5 \times 3}{22} \\[1em] \Rightarrow \text{r}^2 = 29.25 \\[1em] \Rightarrow \text{r} = \sqrt{29.25} \\[1em] \Rightarrow \text{r} = 5.4 \text{ cm.}$

Hence, radius of the cone is 5.4 cm.

From a circular cylinder of diameter 10 cm and height 12 cm, a conical cavity of the same base radius and of the same height is hollowed out. Find the volume and the whole surface of the remaining solid. Leave the answer in π

Answer

Given,

Height of the cylinder (H) = 12 cm

Radius of the base of the cylinder (R) = $\dfrac{\text{diameter}}{2} = \dfrac{10}{2}$ = 5 cm

Height of the cone (h) = 12 cm

Radius of the cone (r) = 5 cm

Volume of the remaining part = Volume of cylinder - Volume of cone

= πR²H - $\dfrac{1}{3}$ πr²h

$= π \times 5^2 \times 12 - π \times \dfrac{1}{3} \times 5^2 \times 12 \\[1em] = π(25 \times 12 - 25 \times 4) \\[1em] = π(300 - 100) \\[1em] = 200 π \text{ cm}^3$

∴ The volume of the remaining solid is 200 π cm³.

By formula,

l² = r² + h²

⇒ l² = 5² + 12²

⇒ l² = 25 + 144

⇒ l² = 169

⇒ l = $\sqrt{169}$ = 13 cm

Total surface area of remaining solid = Curved surface area of cylinder + curved surface area of cone + base area of cylinder

= 2πRH + πrl + πR²

= π(2RH + rl + R²)

= π(2 × 5 × 12 + 5 × 13 + 5²)

= π(120 + 65 + 25)

= 210 π cm².

Hence, the volume of the remaining solid is 200 π cm³ and total surface area of remaining solid is 210 π cm².

From a cube of edge 14 cm, a cone of maximum size is carved out. Find the volume of the cone and of the remaining material, each correct to one place of decimal.

Answer

Edge of a cube = 14 cm

Volume = side³ = 14³ = 2744 cm³.

Cone of maximum size is carved out as shown in figure,

Diameter of the cone cut out from it = 14 cm

Radius, r = $\dfrac{\text{diameter}}{2} = \dfrac{14}{2}$ = 7 cm

Height, h = 14 cm

Volume of cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 14 \\[1em] = \dfrac{1}{3} \times 22 \times 49 \times 2 \\[1em] = \dfrac{2156}{3} \\[1em] = 718.67 \text{ cm}^3.$

Rounding off to one decimal place = 718.8 cm³

Volume of the remaining material = Volume of the cube - Volume of the cone

= 2744 - 718.67

= 2025.33 cm³

Rounding off to one decimal place = 2025.3 cm³

Hence, the volume of the cone is 718.8 cm³ and of the remaining material is 2025.3 cm³.

A cone of maximum volume is carved out of a block of wood of size 20 cm × 10 cm × 10 cm. Find the volume of the cone carved out, correct to one decimal place.

Answer

Volume of block of wood = 20 cm × 10 cm × 10 cm = 2000 cm³

Diameter of the cone for maximum volume = 10 cm

Cone of maximum volume is carved out as shown in figure,

Radius, r = $\dfrac{\text{diameter}}{2} = \dfrac{10}{2}$ = 5 cm.

Height of the cone for maximum volume, h = 20 cm

Volume of cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 5^2 \times 20 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 25 \times 20 \\[1em] = \dfrac{11000}{21} \\[1em] = 523.8 \text{ cm}^3.$

Hence, the volume of the cone carved out is 523.8 cm³.

From a solid wooden cylinder of height 28 cm and diameter 6 cm, two conical cavities are hollowed out. The diameters of the cones are also of 6 cm and height 10.5 cm. Find the volume of the remaining solid.

Answer

Given,

Diameter of solid wooden cylinder (D) = 6 cm

Radius of solid wooden cylinder (R) = $\dfrac{6}{2}$ = 3 cm

Height of solid wooden cylinder (H) = 28 cm

Diameter of cone (d) = 6 cm

Radius of cone (r) = $\dfrac{6}{2}$ = 3 cm

Height of the cone (h) = 10.5 cm

Volume of cylinder = πR²H

$= \dfrac{22}{7} \times 3^2 \times 28 \\[1em] = 22 \times 9 \times 4 \\[1em] = 792 \text{ cm}^3.$

Volume of single cone = $\dfrac{1}{3}$ πr²h

$= \dfrac{1}{3} \times \dfrac{22}{7} \times 3^2 \times 10.5 \\[1em] = \dfrac{22}{7} \times 9 \times 3.5 \\[1em] = \dfrac{693}{7} \\[1em] = 99 \text{ cm}^3.$

Volume of two conical cavities = 2 × 99 = 198 cm³

Volume of remaining solid = Volume of cylinder - Volume of 2 conical cavities

= 792 - 198

= 594 cm³.

Hence, volume of the remaining solid = 594 cm³.