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

An open hollow cylindrical container, made of very thin metal sheet, has radius r cm and height h cm, its total surface area (in cm²) is :

1. πr²h + 2πrh

2. 4πrh + πr²

3. 4πrh + 2πr²

4. 2πrh + 2πr²

Answer

Given,

It is an open hollow cylindrical container.

∴ It will be open at one end and closed at the other end.

Since, hollow cylindrical container is made up of very thin metal sheet.

∴ External radius = Internal radius = r cm.

Given,

Height = h cm.

Total surface area of hollow cylinder = External curved surface area + Internal curved surface area + Area of cross section (lower end)

= 2πrh + 2πrh + πr²

= 4πrh + πr².

Hence, Option 2 is the correct option.

The curved area of a solid cylinder is S cm² and the circumference of its base is C cm; the height of the cylinder is :

1. C × S

2. C + S

3. S ÷ C

4. S - C

Answer

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

Given,

The curved area of a solid cylinder is S cm².

By formula,

Circumference of base = 2πr

Curved surface area = 2πrh

⇒ S = 2πrh

⇒ S = C.h

⇒ h = $\dfrac{S}{C}$ = S ÷ C.

Hence, Option 3 is the correct option.

A cylindrical container has sufficient water to submerge a solid cubical object of its each edge a cm. If the radius of the container is r cm, the rise in level (h cm) of water in the container is :

1. $\sqrt{\dfrac{a^3}{πr}}$

2. $\dfrac{a}{\sqrt{πr}}$

3. $\dfrac{a^3}{πr^2}$

4. $\dfrac{πr^2}{a}$

Answer

Suppose rise in level of water be h cm.

Increase in volume of cylinder = πr²h.

Given,

Edge of cubical solid submerged = a cm.

Volume of cubical solid submerged = a³

We know that,

Volume of cubical solid submerged = Increase in volume of cylinder

⇒ a³ = πr²h

⇒ h = $\dfrac{a^3}{πr^2}$.

Hence, Option 3 is the correct option.

A solid cylinder of radius R cm and height H cm is melted and recast into smaller identical solid cylinders of height h cm and radius r cm. The number of smaller solid cylinders obtained is :

1. $\dfrac{RH}{rh}$

2. $\dfrac{R^2H}{r^2h}$

3. R²H ÷ r²h²

4. $\dfrac{RH^2}{rh}$

Answer

Given,

A solid cylinder of radius R cm and height H cm is melted and recast into smaller identical solid cylinders of height h cm and radius r cm.

Let no. of smaller cylinders formed be n.

∴ Volume of larger cylinder = n × Volume of one smaller cylinder

⇒ πR²H = n × πr²h

⇒ n = $\dfrac{πR^2H}{πr^2h} = \dfrac{R^2H}{r^2h}$

Hence, Option 2 is the correct option.

The total surface area of an open pipe of length H cm, external radius R cm and internal radius r cm, is :

1. πR² - πr² + 2π(R - r)H

2. 2πR² - 2πr² + 2πRH

3. 2πRH - 2πrh + πR² - πr²

4. 2(πR² - πr²) + 2πRH + 2πrH

Answer

Given,

External radius = R cm,

Internal radius = r cm,

Height = H cm.

Total surface area of hollow cylinder = External curved surface area + Internal curved surface area + 2(Area of cross section)

= 2πRH + 2πrH + 2(πR² - πr²).

Hence, Option 4 is the correct option.

The inner radius of a pipe is 2.1 cm. How much water can 12 m of this pipe hold ?

Answer

By formula,

Volume of cylinder = πr²h

Since, pipe is in the form of cylinder.

Given, radius (r) = 2.1 cm and height (h) = 12 m = 1200 cm.

Volume of water inside pipe = $\dfrac{22}{7} \times (2.1)^2 \times 1200$

= 22 × 0.3 × 2.1 × 1200

= 16632 cm³.

Hence, pipe can hold 16632 cm³ of water.

How many cubic meters of earth must be dug out to make a well 28 m deep and 2.8 m in diameter ? Also, find the cost of plastering its inner surface at ₹ 4.50 per sq meter.

Answer

Given,

Diameter of well = 2.8 m

Radius of the well = $\dfrac{2.8}{2}$ = 1.4 m

Depth of the well = 28 m.

Volume of earth dug out = Volume of well = πr²h

= $\dfrac{22}{7}$ x 1.4 x 1.4 x 28

= 172.48 m³.

Area of curved surface = 2πrh

= 2 x $\dfrac{22}{7}$ x 1.4 x 28

= 2 x 22 x 0.2 x 28

= 246.4 m².

Given, the cost of plastering at the rate of ₹ 4.50 per sq meter.

∴ Cost of plastering inner surface = ₹ 246.40 x 4.50

= ₹ 1108.80.

Hence, volume of earth to be dug out = 172.48 m³ and cost of plastering inner surface = ₹ 1108.80.

What length of solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of external diameter 20 cm, 0.25 cm thick and 15 cm long?

Answer

Given,

External diameter of hollow cylinder = 20 cm

So, external radius (R) = $\dfrac{20}{2}$ = 10 cm.

Thickness = 0.25 cm

Hence, the internal radius (r) = (10 - 0.25) = 9.75 cm.

Length of cylinder (h) = 15 cm

By formula,

$\text{Volume of material} = πh(R^2 - r^2) \\[1em] = π \times 15 \times [10^2 - (9.75)^2] \\[1em] = π \times 15 \times [100 - 95.0625] \\[1em] = π \times 15 \times 4.9375.$

Given,

Diameter of the solid cylinder = 2 cm

so, radius (a) = $\dfrac{2}{2}$ = 1 cm

Let h be the length of the solid cylinder,

Volume = πa²h = π(1)²h = πh cm³.

In order for recasting, volume of solid cylinder must be equal to volume of material of hollow cylinder .

∴ πh = 15 π x 4.9375

⇒ h = 15 x 4.9375

⇒ h = 74.0625 cm.

Hence, the length of solid cylinder = 74.06 cm.

A cylinder has a diameter of 20 cm. The area of curved surface is 100 cm². Find :

(i) the height of the cylinder correct to one decimal place.

(ii) the volume of the cylinder correct to one decimal place.

Answer

Given,

The diameter of the cylinder = 20 cm

So, the radius (r) = $\dfrac{20}{2}$ = 10 cm

Given, the curved surface area = 100 cm².

Let height = h cm

(i) By formula,

Curved surface area of cylinder = 2πrh

∴ 2πrh = 100

⇒ 2 x $\dfrac{22}{7}$ x 10 x h = 100

⇒ h = $\dfrac{100 \times 7}{22 \times 10 \times 2}$

⇒ h = $\dfrac{700}{440}$ = 1.6 cm.

Hence, the height of cylinder = 1.6 cm.

(ii) By formula,

Volume of the cylinder = πr²h

= $\dfrac{22}{7}$ x 10 x 10 x 1.6

= 502.9 cm³.

Hence, the volume of cylinder = 502.9 cm³.

A metal pipe has a bore (inner diameter) of 5 cm. The pipe is 5 mm thick all round. Find the weight, in kilogram, of 2 metres of the pipe if 1 cm³ of the metal weights 7.7 g.

Answer

Given,

Inner diameter = 5 cm and thickness of pipe = 5 mm = 0.5 cm.

Inner radius of the pipe (r) = $\dfrac{5}{2}$ = 2.5 cm.

External radius of the pipe (R) = Inner radius of the pipe + Thickness of the pipe

= 2.5 cm + 0.5 cm = 3 cm

Length of the pipe (h) = 2 m = 200 cm.

Volume of the pipe = External Volume - Internal Volume

= πR²h - πr²h

= π(R² - r²)h

= π(R + r)(R - r) h

= $\dfrac{22}{7}$(3 + 2.5)(3 - 2.5) x 200

= $\dfrac{22}{7} \times$(5.5) x (0.5) x 200 = 1728.6 cm³.

Given, 1 cm³ of the metal weight 7.7 g,

∴ Weight of the pipe = (1728.6 x 7.7) g = $\dfrac{(1728.6 \times 7.7)}{1000}$ kg = 13.31 kg.

Hence, weight of the pipe = 13.31 kg.

A cylindrical container with diameter of base 42 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 22 cm x 14 cm x 10.5 cm. Find the rise in the level of the water when the solid is submerged.

Answer

Given,

Diameter of base = 42 cm and radius = $\dfrac{42}{2}$ = 21 cm.

Let rise in level of water be h cm.

Since, water rises on submerging of a rectangular solid.

So, the increase in volume of water = volume of rectangular solid.

∴ πr²h = 22 x 14 x 10.5

$\dfrac{22}{7} \times 21^2 \times h$ = 3234

h = $\dfrac{3234 \times 7}{22 \times 21^2} = \dfrac{22638}{9702} = 2\dfrac{1}{3}$ cm

Hence, the water level will be raised to a level of $2\dfrac{1}{3}$ cm when the solid is submerged.

A cylindrical container with internal radius of its base 10 cm, contains water up to a height of 7 cm. Find the area of wet surface of the cylinder.

Answer

Given,

Internal radius of the cylindrical container (r) = 10 cm

Height of water (h) = 7 cm

So, the surface area of the wet surface = 2πrh + πr² = πr(2h + r)

= $\dfrac{22}{7}$ x 10 x (2 x 7 + 10)

= $\dfrac{220}{7}$ x 24

= 754.29 cm².

Hence, the area of wet surface of cylinder = 754.29 cm².

Find the total surface area of an open pipe of length 50 cm, external diameter 20 cm and internal diameter 6 cm.

Answer

Given,

Length of the open pipe = 50 cm

External diameter = 20 cm

External radius (R) = $\dfrac{20}{2}$ = 10 cm

Internal diameter = 6 cm

Internal radius (r) = $\dfrac{6}{2}$ = 3 cm

Surface area of pipe open from both sides = 2πRh + 2πrh = 2πh(R + r)

= 2 x $\dfrac{22}{7}$ x 50 x (10 + 3)

= 4085.71 cm²

Area of upper and lower part = 2πR² - 2πr²

= 2 x $\dfrac{22}{7}$ x (10² - 3²)

= 2 x $\dfrac{22}{7}$ x 91

= 572 cm².

Total surface area = 4085.71 + 572 = 4657.71 cm².

Hence, total surface area of open pipe = 4657.71 cm².

The height and the radius of the base of a cylinder are in the ratio 3 : 1. If its volume is 1029 π cm³; find its total surface area.

Answer

Given,

The ratio between height and radius of a cylinder = 3 : 1

Volume = 1029π cm³ ..............(1)

Let the radius of the base = r

Then, it’s height will be = 3r.

By formula,

Volume = πr²h ............. (2)

From (1) and (2), we get

⇒ πr²h = 1029π

⇒ r² × 3r = 1029

⇒ r³ = $\dfrac{1029}{3}$

⇒ r³ = 343

⇒ r = $\sqrt[3]{343}$

⇒ r = 7 cm.

Thus, radius = r = 7 cm and height = 3r = 3 x 7 = 21 cm.

By formula,

Total surface area = 2πr(h + r)

= 2 x $\dfrac{22}{7}$ x 7 x (21 + 7)

= 2 x 22 x 28

= 1232 cm².

Hence, total surface area of cylinder = 1232 cm².

The radius of a solid right circular cylinder decreases by 20% and its height increases by 10%. Find the percentage change in its:

(i) volume (ii) curved surface area

Answer

Let the original dimensions of the solid right circular cylinder be

radius = r cm and height = h cm.

Volume = πr²h.

Curved surface area = 2πrh

Now, after the changes the new dimensions are:

Radius (r') = r - $\dfrac{20}{100} \times r$ = r - 0.2r = 0.8r

Height (h') = h + $\dfrac{10}{100} \times h$ = h + 0.1h = 1.1h

So,

New volume = πr'²h'

= π(0.8r)²(1.1h)

= 0.704 πr²h.

New curved surface area = 2πr'h' = 2π(0.8r)(1.1h)

= 1.76πrh

(i) Decrease in volume = Original volume - New volume

= πr²h - 0.704 πr²h

= 0.296 πr²h

Percentage change in its volume = $\dfrac{\text{Decrease in volume}}{\text{Original volume}}$ x 100 %

= $\dfrac{0.296πr^2h}{πr^2h}$ x 100 %

= 0.296 x 100 % = 29.6 %.

Hence, decrease in volume = 29.6 %.

(ii) Decrease in curved surface area = Original curved surface area - New curved surface area

= 2πrh - 1.76πrh

= 0.24πrh.

Percentage change in its curved surface area = $\dfrac{\text{Decreased CSA}}{\text{Original CSA}}$ x 100 %

= $\dfrac{0.24πrh}{2πrh}$ x 100 %

= 0.12 x 100 %

= 12 %.

Hence, decrease in volume = 12 %.

Find the minimum length in cm and correct to nearest whole number of the thin metal sheet required to make a hollow and closed cylindrical box of diameter 20 cm and height 35 cm. Given that the width of the metal sheet is 1 m. Also, find the cost of the sheet at the rate of Rs. 56 per m.

Find the area of metal sheet required, if 10% of it is wasted in cutting, overlapping, etc.

Answer

Given,

Height of the cylinder box (h) = 35 cm

Base radius of the cylinder box (r) = $\dfrac{20}{2}$ = 10 cm

Width of metal sheet = 1m = 100 cm

Area of metal sheet required = Total surface area of the box

⇒ Length x width = 2πr(r + h)

⇒ Length x 100 = 2 x $\dfrac{22}{7}$ x 10(10 + 35)

⇒ Length x 100 = 2 x $\dfrac{22}{7}$ x 10 x 45

⇒ Length = $\dfrac{2 \times 22 \times 10 \times 45}{7 \times 100}$ = 28.28 cm ≈ 28 cm (correcting to the nearest whole number)

Thus,

Area of metal sheet = length x width = 28 x 100 = 2800 cm² = $\dfrac{2800}{100 \times 100}$ = 0.28 m².

So, the cost of the sheet at the rate of ₹ 56 per m² = ₹ (56 x 0.28) = ₹ 15.68

Let the total sheet required be x.

Then, x - 10 % of x = 2800 cm

⇒ $x - \dfrac{10}{100} \times x$ = 2800

⇒ $\dfrac{100x - 10x}{100}$ = 2800

⇒ $\dfrac{90x}{100}$ = 2800

⇒ $\dfrac{9x}{10}$ = 2800

⇒ $x = \dfrac{2800 \times 10}{9}$

⇒ x = 3111.11 cm² ≈ 3111 cm² (correcting to the nearest whole number)

Hence, length = 28 cm, cost of sheet = ₹15.68 and area of sheet required = 3111 cm².

3080 cm³ of water is required to fill a cylindrical vessel completely and 2310 cm³ of water is required to fill it upto 5 cm below the top. Find :

(i) radius of the vessel.

(ii) height of the vessel.

(iii) wetted surface area of the vessel when it is half-filled with water.

Answer

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

Given,

Volume of cylindrical vessel = 3080 cm³

∴ πr²h = 3080 ...........(1)

Given,

It takes 2310 cm³ of water to fill cylinder upto 5 cm below the top.

∴ πr²(h - 5) = 2310 ...........(2)

Dividing (1) by (2) we get,

$\Rightarrow \dfrac{πr^2h}{πr^2(h - 5)} = \dfrac{3080}{2310} \\[1em] \Rightarrow \dfrac{h}{h - 5} = \dfrac{308}{231} \\[1em] \Rightarrow 231h = 308(h - 5) \\[1em] \Rightarrow 231h = 308h - 1540 \\[1em] \Rightarrow 308h - 231h = 1540 \\[1em] \Rightarrow 77h = 1540 \\[1em] \Rightarrow h = \dfrac{1540}{77} \\[1em] \Rightarrow h = 20 \text{ cm}$

(i) Substituting value of h in equation (1), we get :

$\Rightarrow πr^2 \times 20 = 3080 \\[1em] \Rightarrow \dfrac{22}{7} \times r^2 \times 20 = 3080 \\[1em] \Rightarrow r^2 = \dfrac{3080 \times 7}{20 \times 22} \\[1em] \Rightarrow r^2 = 49 \\[1em] \Rightarrow r = 7 \text{ cm}.$

Hence, radius of vessel = 7 cm.

(ii) From above,

h = 20 cm.

Hence, height of vessel = 20 cm.

(iii) When vessel is half-filled, water will be filled upto $\dfrac{20}{2}$ = 10 cm.

Wetted surface area = 2πrh + πr²

= πr(2h + r)

= $\dfrac{22}{7} \times 7 \times (2 \times 10 + 7)$

= $22 \times 27$

= 594 cm².

Hence, wetted surface area = 594 cm².

Find the volume of the largest cylinder formed when a rectangular piece of paper 44 cm by 33 cm is rolled along its :

(i) shorter side

(ii) longer side.

Answer

(i) When paper is rolled along shorter side then height = 44 cm and circumference = 33 cm.

⇒ Circumference = 33 cm

⇒ 2πr = 33

⇒ $2 \times \dfrac{22}{7} \times r = 33$

⇒ r = $\dfrac{33 \times 7}{2 \times 22}$ = 5.25 cm.

Volume = πr²h

= $\dfrac{22}{7} \times (5.25)^2 \times 44$

= $\dfrac{22}{7} \times 27.56 \times 44$

= $\dfrac{26680.5}{7}$ = 3811.5 cm³.

Hence, volume of cylinder = 3811.5 cm³.

(ii) When paper is rolled along longer side then height = 33 cm and circumference = 44 cm.

⇒ Circumference = 44 cm

⇒ 2πr = 44

⇒ $2 \times \dfrac{22}{7} \times r = 44$

⇒ r = $\dfrac{44 \times 7}{2 \times 22}$ = 7 cm.

Volume = πr²h

= $\dfrac{22}{7} \times (7)^2 \times 33$

= $\dfrac{22}{7} \times 49 \times 33$

= 5082 cm³.

Hence, volume of cylinder = 5082 cm³.

A metal cube of side 11 cm is completely submerged in water contained in a cylindrical vessel with diameter 28 cm. Find the rise in the level of water.

Answer

Given,

⇒ Diameter = 28 cm

⇒ Radius (r) = 14 cm

⇒ Side of cube (a) = 11 cm.

Let the rise in level of water be h cm.

So, the volume of water risen = volume of cube

∴ πr²h = a³

$\Rightarrow \dfrac{22}{7} \times (14)^2 \times h = (11)^3 \\[1em] \Rightarrow h = \dfrac{11 \times 11 \times 11 \times 7}{22 \times 14 \times 14} \\[1em] \Rightarrow h = \dfrac{9317}{4312} \\[1em] \Rightarrow h = 2.16 \text{ cm}.$

Hence, rise in level of water = 2.16 cm.

A circular tank of diameter 2 m is dug and the earth removed is spread uniformly all around the tank to form an embarkment 2 m in width and 1.6 m in height. Find the depth of the circular tank.

Answer

Given,

Tank's diameter = 2 m

Tank's radius (r) = $\dfrac{2}{2}$ = 1 m

Width of embarkment (w) = 2 m

External radius (R) = r + w = 1 + 2 = 3 m.

Height of embarkment (h) = 1.6 m

Let depth of tank be h' meters.

Volume of earth removed = Volume of embarkment formed

⇒ πr²h' = π(R² - r²)h

⇒ 1²h' = (3² - 1²) × 1.6

⇒ h' = (9 - 1) × 1.6 m

⇒ h' = 12.8 m

Hence, depth of circular tank = 12.8 m.

The sum of the height and the radius of a solid cylinder is 35 cm and its total surface area is 3080 cm²; find the volume of the cylinder.

Answer

According to question,

h + r = 35 .........(1)

Total surface area = 3080

⇒ 2πr(h + r) = 3080

⇒ 2πr × 35 = 3080

⇒ $70 \times \dfrac{22}{7} \times r$ = 3080

⇒ 220r = 3080

⇒ r = $\dfrac{3080}{220}$

⇒ r = 14 cm.

⇒ h + r = 35

⇒ h + 14 = 35

⇒ h = 35 - 14

⇒ h = 21 cm.

Volume of cylinder = πr²h

= $\dfrac{22}{7} \times (14)^2 \times 21$

= 22 × 2 × 14 × 21

= 12936 cm³.

Hence, volume of cylinder = 12936 cm³.

The total surface area of a solid cylinder is 616 cm². If the ratio between its curved surface area and total surface area is 1 : 2; find the volume of the cylinder.

Answer

Given,

Total surface area = 616 cm²

⇒ 2πr(h + r) = 616

⇒ πr(h + r) = $\dfrac{616}{2}$

⇒ πr(h + r) = 308 ..........(1)

Ratio between its curved surface area and total surface area = 1 : 2

$\Rightarrow \dfrac{\text{Curved surface area}}{\text{Total surface area}} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{2πrh}{2πr(h + r)} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{h}{h + r} = \dfrac{1}{2} \\[1em] \Rightarrow 2h = h + r \\[1em] \Rightarrow 2h - h = r \\[1em] \Rightarrow h = r.$

Substituting value of h in equation (1), we get :

⇒ πr(r + r) = 308

⇒ πr.2r = 308

⇒ 2πr² = 308

⇒ πr² = 154

⇒ $\dfrac{22}{7} \times r^2$ = 154

⇒ r² = $\dfrac{154 \times 7}{22}$

⇒ r² = 49

⇒ r = $\sqrt{49}$

⇒ r = 7 cm

⇒ h = 7 cm.

Volume of cylinder = πr²h

= $\dfrac{22}{7} \times (7)^2 \times 7$

= 22 × 49

= 1078 cm³.

Hence, volume of cylinder = 1078 cm³.

A cylindrical vessel of height 24 cm and diameter 40 cm is full of water. Find the exact number of small cylindrical bottles, each of height 10 cm and diameter 8 cm, which can be filled with this water.

Answer

Given,

Height of large cylindrical vessel (h₁) = 24 cm

Diameter of large cylindrical vessel = 40 cm

Radius of large cylindrical vessel (r₁) = $\dfrac{40}{2}$ = 20 cm.

Height of small cylindrical vessel (h₂) = 10 cm

Diameter of small cylindrical vessel = 8 cm

Radius of small cylindrical vessel (r₂) = $\dfrac{8}{2}$ = 4 cm.

Let no. of small cylindrical bottles which can be filled be n.

Volume of large cylindrical vessel = n × Volume of small cylindrical vessel

$\therefore \dfrac{1}{3}πr_1^2h_1 = n \times \dfrac{1}{3}πr_2^2h_2 \\[1em] \Rightarrow n = \dfrac{r_1^2h_1}{r_2^2h_2} \\[1em] \Rightarrow n = \dfrac{20^2 \times 24}{4^2 \times 10} \\[1em] \Rightarrow n = \dfrac{400 \times 24}{16 \times 10} \\[1em] \Rightarrow n = 60.$

Hence, no. of small cylindrical bottles which an be filled = 60.

Two solid cylinders, one with diameter 60 cm and height 30 cm and the other with radius 30 cm and height 60 cm, are melted and recasted into a third solid cylinder of height 10 cm. Find the diameter of the cylinder formed.

Answer

For new cylinder formed,

Let Radius = r

and

Volume = V

Height (h) = 10 cm (Given)

For 1st cylinder melted,

Diameter (d) = 60 cm

Radius (r₁) = 30 cm

Height (h₁) = 30 cm

Volume = V₁

For 2nd cylinder melted,

Radius (r₂) = 30 cm

Height (h₂) = 60 cm

Volume = V₂

Volume of new cylinder formed will be equal to the sum of two cylinders melted,

V = V₁ + V₂

$\Rightarrow \dfrac{1}{3}πr^2h = \dfrac{1}{3}πr_1^2h_1 + \dfrac{1}{3}πr_2^2h_2 \\[1em] \Rightarrow r^2h = r_1^2h_1 + r_2^2h_2 \\[1em] \Rightarrow r^2 \times 10 = (30)^2 \times 30 + (30)^2 \times 60 \\[1em] \Rightarrow r^2 \times 10 = 27000 + 54000 \\[1em] \Rightarrow r^2 = \dfrac{81000}{10} \\[1em] \Rightarrow r^2 = 8100 \\[1em] \Rightarrow r = \sqrt{8100} = 90 \text{ cm.}$

Diameter = 2r = 2 x 90 = 180 cm.

Hence, diameter of new cylinder = 180 cm.

The total surface area of a hollow cylinder, which is open from both the sides, is 3575 cm²; area of its base ring is 357.5 cm² and its height is 14 cm. Find the thickness of the cylinder.

Answer

Let external radius be R cm and internal radius be r cm.

Given,

Area of base of ring = 357.5 cm²

∴ π(R² - r²) = 357.5

$\Rightarrow \dfrac{22}{7} \times (R^2 - r^2) = 357.5 \\[1em] \Rightarrow (R^2 - r^2) = \dfrac{357.5 \times 7}{22} \\[1em] \Rightarrow (R^2 - r^2) = 113.75 \space ..........(1)$

Given,

Total surface area of hollow cylinder = 3575 cm².

$\Rightarrow 2πRh + 2πrh + 2π(R^2 - r^2) = 3575 \\[1em] \Rightarrow 2πh(R + r) + 2π \times 113.75 = 3575 \space ..........\text{[From (1)]} \\[1em] \Rightarrow 2 \times \dfrac{22}{7} \times 14 (R + r) + 2 \times \dfrac{22}{7} \times 113.75 = 3575 \\[1em] \Rightarrow 88(R + r) + 715 = 3575 \\[1em] \Rightarrow 88(R + r) = 3575 - 715 \\[1em] \Rightarrow 88(R + r) = 2860 \\[1em] \Rightarrow (R + r) = \dfrac{2860}{88} \\[1em] \Rightarrow (R + r) = 32.5 \space ..........(2)$

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

$\Rightarrow \dfrac{R^2 - r^2}{R + r} = \dfrac{113.75}{32.5} \\[1em] \Rightarrow \dfrac{(R - r)(R + r)}{(R + r)} = 3.5 \\[1em] \Rightarrow (R - r) = 3.5 \text{ cm}.$

Hence, thickness of hollow cylinder = 3.5 cm.

Two right circular solid cylinders have radii in the ratio 3 : 5 and heights in the ratio 2 : 3. Find the ratio between their :

(i) curved surface areas.

(ii) volumes.

Answer

(i) According to question,

r₁ : r₂ = 3 : 5

Let, r₁ = 3x and r₂ = 5x.

h₁ : h₂ = 2 : 3

Let, h₁ = 2y and h₂ = 3y.

$\dfrac{\text{CSA of 1st cylinder}}{\text{CSA of 2nd cylinder}} = \dfrac{2πr_1h_1}{2πr_2h_2} \\[1em] = \dfrac{r_1h_1}{r_2h_2} \\[1em] = \dfrac{3x \times 2y}{5x \times 3y} \\[1em] = \dfrac{6xy}{15xy} \\[1em] = \dfrac{2}{5} \\[1em] = 2 : 5.$

Hence, ratio between curved surface area = 2 : 5.

(ii)

$\dfrac{\text{Vol. of 1st cylinder}}{\text{Vol. of 2nd cylinder}} = \dfrac{πr_1^2h_1}{πr_2^2h_2} \\[1em] = \dfrac{r_1^2h_1}{r_2^2h_2} \\[1em] = \dfrac{(3x)^2 \times 2y}{(5x)^2 \times 3y} \\[1em] = \dfrac{18x^2y}{75x^2y} \\[1em] = 6 : 25.$

Hence, ratio between volume = 6 : 25.

A closed cylindrical tank, made of thin iron-sheet, has diameter = 8.4 m and height 5.4 m. How much metal sheet, to the nearest m², is used in making this tank, if $\dfrac{1}{15}$ of the sheet actually used was wasted in making the tank ?

Answer

Radius of cylindrical tank = $\dfrac{8.4}{2}$ = 4.2 m.

Total surface area of cylindrical tank = 2πr(h + r)

= 2 × $\dfrac{22}{7}$ × 4.2 × (5.4 + 4.2)

= 2 × 22 × 0.6 × 9.6

= 253.44 m².

Given,

$\dfrac{1}{15}$ of the sheet actually used was wasted in making the tank.

∴ Fraction of sheet used in making the tank = $1 - \dfrac{1}{15}$

$= \dfrac{15 - 1}{15} \\[1em] = \dfrac{14}{15}$

Let metal sheet used be $x$ m²

$\dfrac{14}{15}$ of $x$ = TSA of Cylindrical tank

$\Rightarrow \dfrac{14}{15} \times x = 253.44 \\[1em] \Rightarrow x = \dfrac{253.44}{\dfrac{14}{15}} \\[1em] \Rightarrow x = 253.44 \times \dfrac{15}{14} \\[1em] \Rightarrow x \approx 271.54 \text{m}^2$

Area of metal sheet in nearest m² = 272 m²

Hence, total metal (iron) sheet used in nearest m² is 272 m².