Cylinder, Cone and Sphere (Surface Area and Volume)

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.

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 height, is made from a rectangular sheet of paper 22 cm wide. The smallest length of the paper sheet required is :

1. 70 cm

2. 105 cm

3. 140 cm

4. 280 cm

Answer

Since, conical toy is made from the rectangular sheet.

∴ Surface area of cone = Area of rectangular sheet.

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

∴ πrl = length × breadth

$\Rightarrow πr\sqrt{r^2 + h^2} = \text{length × breadth} \\[1em] \Rightarrow \dfrac{22}{7} \times 28 \times \sqrt{28^2 + 21^2} = a \times 22 \\[1em] \Rightarrow 22 \times 4 \times \sqrt{784 + 441} = a \times 22 \\[1em] \Rightarrow 88 \times \sqrt{1225} = a \times 22 \\[1em] \Rightarrow 88 \times 35 = a \times 22 \\[1em] \Rightarrow a = \dfrac{88 \times 35}{22} \\[1em] \Rightarrow a = 4 \times 35 = 140 \text{ cm}.$

Hence, Option 3 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.

A solid metallic sphere of radius 16 cm is melted to form small identical spheres each of diameter 4 cm. The number of spheres formed is :

1. 64

2. 252

3. 512

4. 1024

Answer

Given,

A solid metallic sphere of radius 16 cm is melted to form small identical spheres each of diameter 4 cm.

Larger metallic ball sphere radius (R) = 16 cm

Radius of smaller metallic sphere (r) = $\dfrac{4}{2}$ = 2 cm

Let no. of smaller spheres formed be n.

∴ Volume of larger metallic ball = n × Volume of smaller metallic ball

$\Rightarrow \dfrac{4}{3}πR^3 = n \times \dfrac{4}{3}πr^3 \\[1em] \Rightarrow n = \dfrac{\dfrac{4}{3}πR^3}{\dfrac{4}{3}πr^3} \\[1em] \Rightarrow n = \dfrac{R^3}{r^3} \\[1em] \Rightarrow n = \dfrac{16^3}{2^3} \\[1em] \Rightarrow n = \dfrac{4096}{8} = 512.$

Hence, Option 3 is the correct option.

A hemi-spherical bowl (as shown) has external radius R and internal radius r, the outer surface area of the bowl is :

1. 2πR² + πr²

2. 3πR² - πr²

3. 2πR² + 2πr²

4. 2πR² - πr²

Answer

Outer surface area of the hemi-spherical bowl = Surface area of outer hemisphere + Area of outer cross-section - Area of inner cross-section

= 2πR² + πR² - πr²

= 3πR² - πr².

Hence, Option 2 is the correct option.

The ratio between the volumes of two spherical solids is 27 : 8. The ratio between their curved surface areas is :

1. 27 : 8

2. 8 : 27

3. 3 : 2

4. 9 : 4

Answer

Let radius of two spherical solids be R and r.

Given,

The ratio between the volumes of two spherical solids is 27 : 8.

$\Rightarrow \dfrac{\text{Vol. of 1st spherical solid}}{\text{Vol. of 2nd spherical solid}} = \dfrac{27}{8} \\[1em] \Rightarrow \dfrac{\dfrac{4}{3}πR^3}{\dfrac{4}{3}πr^3} = \dfrac{27}{8} \\[1em] \Rightarrow \dfrac{R^3}{r^3} = \dfrac{27}{8} \\[1em] \Rightarrow \Big(\dfrac{R}{r}\Big)^3 = \Big(\dfrac{3}{2}\Big)^3 \\[1em] \Rightarrow \dfrac{R}{r} = \dfrac{3}{2} \text{ .........(1)}$

The ratio between the curved surface area of two spherical solids :

$\Rightarrow \dfrac{\text{CSA of 1st spherical solid}}{\text{CSA of 2nd spherical solid}} = \dfrac{4πR^2}{4πr^2} \\[1em] = \dfrac{R^2}{r^2} \\[1em] = \Big(\dfrac{R}{r}\Big)^2 \\[1em] = \Big(\dfrac{3}{2}\Big)^2 \\[1em] = \dfrac{9}{4} \\[1em] = 9 : 4.$

Hence, Option 4 is the correct option.

A solid metallic sphere of radius 8 cm is melted and recast into 64 identical solid spheres. The diameter of each smaller sphere formed is :

1. 4 cm

2. 2 cm

3. 8 cm

4. 1 cm

Answer

Given,

Radius of larger metallic sphere (R) = 8 cm

Let radius of each smaller sphere be r cm.

Given,

A solid metallic sphere of radius 8 cm is melted and recast into 64 identical solid spheres.

∴ Volume of larger metallic sphere = 64 × Volume of smaller metallic sphere

$\Rightarrow \dfrac{4}{3}πR^3 = 64 \times \dfrac{4}{3}πr^3 \\[1em] \Rightarrow 64 = \dfrac{\dfrac{4}{3}πR^3}{\dfrac{4}{3}πr^3} \\[1em] \Rightarrow 64 = \dfrac{R^3}{r^3} \\[1em] \Rightarrow 64 = \dfrac{8^3}{r^3} \\[1em] \Rightarrow r^3 = \dfrac{8^3}{64} \\[1em] \Rightarrow r^3 = \dfrac{512}{64} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = \sqrt[3]{8} = 2 \text{ cm}.$

Diameter of smaller sphere = 2 × 2 = 4 cm.

Hence, Option 1 is the correct option.

r₁, r₂ and r₃ are the radii of three metallic spheres. If these spheres are melted to form a single solid sphere, the radius of sphere formed is :

1. $\sqrt{r_1^3 + r_2^3 + r_3^3}$

2. r₁ + r₂ + r₃

3. $r_1^3 + r_2^3 + r_3^3$

4. $\sqrt[3]{r_1^3 + r_2^3 + r_3^3}$

Answer

Given,

r₁, r₂ and r₃ are the radii of three metallic spheres. These spheres are melted to form a single solid sphere.

Let the radius of sphere formed be R.

∴ Volume of sphere formed = Sum of volume of three smaller spheres

$\Rightarrow \dfrac{4}{3}πR^3 = \dfrac{4}{3}πr_1^3 + \dfrac{4}{3}πr_2^3 + \dfrac{4}{3}πr_3^3 \\[1em] \Rightarrow \dfrac{4}{3}πR^3 = \dfrac{4}{3}π(r_1^3 + r_2^3 + r_3^3) \\[1em] \Rightarrow R^3 = r_1^3 + r_2^3 + r_3^3 \\[1em] \Rightarrow R = \sqrt[3]{r_1^3 + r_2^3 + r_3^3}.$

Hence, Option 4 is the correct option.

The volume of a sphere is 38808 cm³; find its diameter and the surface area.

Answer

Given,

Volume of the sphere = 38808 cm³

Let the radius of the sphere = r

By formula,

Volume of sphere = $\dfrac{4}{3}πr^3$

$\Rightarrow \dfrac{4}{3} πr^3 = 38808 \\[1em] \Rightarrow \dfrac{4}{3} \times \dfrac{22}{7} \times r^3 = 38808 \\[1em] \Rightarrow r^3 = \dfrac{(38808 \times 7 \times 3)}{(4 \times 22)} \\[1em] \Rightarrow r^3 = \dfrac{814968}{88} \\[1em] \Rightarrow r^3 = 9261 \\[1em] \Rightarrow r = \sqrt[3]{9261} \\[1em] \Rightarrow r = 21 \text{ cm}.$

∴ Diameter = 2r = 21 x 2 = 42 cm.

Surface area = 4πr² = $4 \times \dfrac{22}{7} \times 21 \times 21$

= 5544 cm².

Hence, diameter of ball = 42 cm and surface area = 5544 cm².

A spherical ball of lead has been melted and made into identical smaller balls with radius equal to half the radius of the original one. How many such balls can be made?

Answer

Let the radius of the spherical ball be r cm.

So, the volume = $\dfrac{4}{3} πr^3$

Radius of smaller ball = $\dfrac{r}{2}$ cm

According to question,

The volume of large spherical balls = Volume of all small balls

Let no. of small spherical balls that can be made be n.

$\therefore \dfrac{4}{3}πr^3 = n \times \dfrac{4}{3}π\Big(\dfrac{r}{2}\Big)^3 \\[1em] \Rightarrow n = \dfrac{\dfrac{4}{3}πr^3}{\dfrac{4}{3}π\Big(\dfrac{r}{2}\Big)^3} \\[1em] \Rightarrow n = \dfrac{\dfrac{4}{3}πr^3 \times 2^3}{\dfrac{4}{3}πr^3} \\[1em] \Rightarrow n = 2^3 = 8.$

Hence, 8 balls can be made.

How many balls each of radius 1 cm can be made by melting a bigger ball whose diameter is 8 cm.

Answer

Given,

Diameter of bigger ball = 8 cm

So, radius of bigger ball (R) = $\dfrac{8}{2}$ = 4 cm.

Volume of bigger ball = $\dfrac{4}{3}πR^3$

= $\dfrac{4}{3} \times π (4)^3$

= $\dfrac{4}{3} \times π \times 64 = \dfrac{256π}{3} \text{ cm}^3$.

Radius of small ball (r) = 1 cm

Volume of each smaller ball = $\dfrac{4}{3}πr^3$

= $\dfrac{4}{3} \times π \times (1)^3 = \dfrac{4}{3}π \text{ cm}^3.$

Let n smaller balls can be made by, melting bigger ball.

Volume of bigger ball = n × Volume of each smaller ball

$\Rightarrow \dfrac{256π}{3} = n \times \dfrac{4}{3}π \\[1em] \Rightarrow n = \dfrac{\dfrac{256π}{3}}{\dfrac{4}{3}π} \\[1em] \Rightarrow n = \dfrac{256 \times π \times 3}{4 \times 3 \times π} \\[1em] \Rightarrow n = 64.$

Hence, 64 balls can be made.

The volume of one sphere is 27 times that of another sphere. Calculate the ratio of their:

(i) radii

(ii) surface areas

Answer

Given,

Volume of first sphere = 27 x volume of second sphere

Let the radius of the first sphere = r₁ and, radius of second sphere = r₂

(i) According to the question, we have :

$\Rightarrow \dfrac{4}{3} πr_1^3 = 27 \times \dfrac{4}{3} πr_2^3 \\[1em] \Rightarrow r_1^3 = 27 \times r_2^3 \\[1em] \Rightarrow \dfrac{r_1^3}{r_2^3} = \dfrac{27}{1} \\[1em] \Rightarrow \Big(\dfrac{r_1}{r_2}\Big)^3 = \Big(\dfrac{3}{1}\Big)^3 \\[1em] \Rightarrow \dfrac{r_1}{r_2} = \dfrac{3}{1}.$

Hence, r₁ : r₂ = 3 : 1.

(ii) Surface area of the first sphere = 4π(r₁)²

Surface area of second sphere = 4π(r₂)²

$\text{Ratio of surface areas} = \dfrac{4πr_1^2}{4πr_2^2} \\[1em] = \dfrac{r_1^2}{r_2^2} = \Big(\dfrac{3}{1}\Big)^2 \\[1em] = \dfrac{9}{1}.$

Hence, the ratio of surface areas = 9 : 1.

If the number of square centimeters on the surface of a sphere is equal to the number of cubic centimeters in its volume, what is the diameter of the sphere ?

Answer

Let r be the radius of the sphere.

According to question,

Surface area of sphere = Volume of sphere

$\Rightarrow 4πr^2 = \dfrac{4}{3}πr^3 \\[1em] \Rightarrow r^2 = \dfrac{r^3}{3} \\[1em] \Rightarrow \dfrac{r^3}{r^2} = 3 \\[1em] \Rightarrow r = 3\text{ cm}.$

Diameter of sphere = 2r = 2 × 3 = 6 cm.

Hence, diameter of sphere = 6 cm.

A solid metal sphere is cut through its center into 2 equal parts. If the diameter of the sphere is $3\dfrac{1}{2}$ cm, find the total surface area of each part correct to two decimal places.

Answer

Diameter of sphere = $3\dfrac{1}{2} = \dfrac{7}{2}$ cm.

Radius of sphere (r) = $\dfrac{\dfrac{7}{2}}{2} = \dfrac{7}{4}$ cm.

Total surface area of each hemisphere = $\dfrac{1}{2}$ Curved surface area of sphere + Area of circular base

= $\dfrac{1}{2} \times 4πr^2 + πr^2$

= 2πr² + πr²

= 3πr²

$= 3 \times \dfrac{22}{7} \times \dfrac{7}{4} \times \dfrac{7}{4} \\[1em] = 3 \times \dfrac{22 \times 7}{16} \\[1em] = 3 \times \dfrac{77}{8} \\[1em] = 28.88 \text{ cm}^2.$

Hence, total surface area of each hemisphere 28.88 cm².

The internal and external diameters of a hollow hemispherical vessel are 21 cm and 28 cm respectively. Find :

(i) internal curved surface area,

(ii) external curved surface area,

(iii) total surface area,

(iv) volume of material of the vessel.

Answer

(i) Given,

Internal diameter = 21 cm

Internal radius (r) = $\dfrac{21}{2}$ cm.

By formula,

Internal curved surface area = 2πr².

$= 2 \times \dfrac{22}{7} \times \dfrac{21}{2} \times \dfrac{21}{2} \\[1em] = \dfrac{19404}{28} \\[1em] = 693 \text{ cm}^2.$

Hence, internal curved surface area = 693 cm².

(ii) Given,

Internal diameter = 28 cm

Internal radius (R) = $\dfrac{28}{2}$ = 14 cm.

By formula,

External curved surface area = 2πR².

$= 2 \times \dfrac{22}{7} \times 14 \times 14 \\[1em] = 1232 \text{ cm}^2.$

Hence, external curved surface area = 1232 cm².

(iii) By formula,

Total surface area of hemisphere = 2πr² + 2πR² + π(R² - r²)

$= 693 + 1232 + \dfrac{22}{7} \times \Big[(14)^2 - \Big(\dfrac{21}{2}\Big)^2\Big] \\[1em] = 1925 + \dfrac{22}{7} \times \Big[196 - \dfrac{441}{4}\Big] \\[1em] = 1925 + \dfrac{22}{7} \times \dfrac{784 - 441}{4} \\[1em] = 1925 + \dfrac{22}{7} \times \dfrac{343}{4} \\[1em] = 1925 + \dfrac{11 \times 49}{2} \\[1em] = 1925 + \dfrac{539}{2} \\[1em] = 1925 + 269.5 \\[1em] = 2194.5 \text{ cm}^2.$

Hence, total surface area of hemisphere = 2194.5 cm².

(iv) By formula,

Volume of hemispherical vessel = $\dfrac{2}{3}π(R^3 - r^3)$

$= \dfrac{2}{3} \times \dfrac{22}{7} \times [(14)^3 - \Big(\dfrac{21}{2}\Big)^3] \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times [2744 - 1157.625] \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 1586.375 \\[1em] = \dfrac{44}{21} \times 1586.375 \\[1em] = 3323.83 \text{ cm}^3.$

Hence, volume of vessel = 3323.83 cm³.

A solid sphere and a solid hemi-sphere have the same total surface area. Find the ratio between their volumes.

Answer

Let radius of sphere be R cm and hemi-sphere be r cm.

Given,

Solid sphere and a solid hemi-sphere have the same total surface area.

$\therefore 4πR^2 = 3πr^2 \\[1em] \Rightarrow \dfrac{R^2}{r^2} = \dfrac{3π}{4π} \\[1em] \Rightarrow \dfrac{R^2}{r^2} = \dfrac{3}{4} \\[1em] \Rightarrow \dfrac{R}{r} = \sqrt{\dfrac{3}{4}} \\[1em] \Rightarrow \dfrac{R}{r} = \dfrac{\sqrt{3}}{2}.$

Calculating ratio between volumes,

$\dfrac{\text{Vol. of sphere}}{\text{Vol. of hemi-sphere}} = \dfrac{\dfrac{4}{3}πR^3}{\dfrac{2}{3}πr^3} \\[1em] = \dfrac{4πR^3 \times 3}{2πr^3 \times 3} \\[1em] = 2 \times \dfrac{R^3}{r^3} \\[1em] = 2 \times \Big(\dfrac{\sqrt{3}}{2}\Big)^3 \\[1em] = 2 \times \dfrac{3\sqrt{3}}{8} \\[1em] = \dfrac{3\sqrt{3}}{4}.$

Hence, ratio between volumes = $3\sqrt{3} : 4$.

Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted into a single solid sphere. Taking π = 3.1, find the surface area of the solid sphere formed.

Answer

Let radius of spheres be r₁, r₂ and r₃ cm respectively.

Let r be the radius of new sphere formed.

Volume of new sphere formed = Total volume of the three spheres melted

$\Rightarrow \dfrac{4}{3}πr^3 = \dfrac{4}{3}π(r_1)^3 + \dfrac{4}{3}π(r_2)^3 + \dfrac{4}{3}π(r_3)^3 \\[1em] \Rightarrow \dfrac{4}{3}πr^3 = \dfrac{4}{3}π(r_1^3 + r_2^3 + r_3^3) \\[1em] \Rightarrow r^3 = r_1^3 + r_2^3 + r_3^3 \\[1em] \Rightarrow r^3 = (6)^3 + (8)^3 + (10)^3 \\[1em] \Rightarrow r^3 = 216 + 512 + 1000 \\[1em] \Rightarrow r^3 = 1728 \\[1em] \Rightarrow r^3 = (12)^3 \\[1em] \Rightarrow r = 12 \text{ cm.}$

By formula,

Surface area of sphere = 4πr²

= 4 × 3.1 × 12²

= 1785.6 cm².

A cone and a sphere have equal volumes. Their radii are also equal each being 10 cm; the height of the cone is :

1. 70 cm

2. 40 cm

3. 50 cm

4. 20 cm

Answer

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

Given,

A cone and a sphere have equal volumes and their radius are equal.

∴ Volume of cone = Volume of sphere

$\Rightarrow \dfrac{1}{3}πr^2h = \dfrac{4}{3}πr^3 \\[1em] \Rightarrow h = \dfrac{4 \times 3πr^3}{3πr^2} \\[1em] \Rightarrow h = 4r \\[1em] \Rightarrow h = 4 \times 10 = 40 \text{ cm}.$

Hence, Option 2 is the correct option.

A sphere and a cone have equal volumes and equal radii. The ratio between the radius and height of the cone is :

1. 3 : 4

2. 4 : 3

3. 4 : 1

4. 1 : 4

Answer

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

Given,

A cone and a sphere have equal volumes and their radius are equal.

∴ Volume of cone = Volume of sphere

$\Rightarrow \dfrac{1}{3}πr^2h = \dfrac{4}{3}πr^3 \\[1em] \Rightarrow h = \dfrac{4 \times 3πr^3}{3πr^2} \\[1em] \Rightarrow h = 4r \\[1em] \Rightarrow \dfrac{r}{h} = \dfrac{1}{4} \\[1em] \Rightarrow r : h = 1 : 4.$

Hence, Option 4 is the correct option.

The radius and the height of a cone are in the ratio 2 : 1. The ratio between the volumes of a sphere and this cone (both having equal radii) is :

1. 1 : 4

2. 4 : 1

3. 8 : 1

4. 1 : 8

Answer

Given,

The radius and the height of a cone are in the ratio 2 : 1.

Let radius of cone be 2x and height of cone be x.

Let radius f sphere be r.

Given,

Radius of cone and sphere are equal.

∴ r = 2x.

$\Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{\dfrac{4}{3}πr^3}{\dfrac{1}{3}πr^2h} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{4 \times 3 \times πr^3}{3 \times πr^2h} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{4r}{h} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{4 \times 2x}{x} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{8x}{x} \\[1em] \Rightarrow \dfrac{\text{Vol. of sphere}}{\text{Vol. of cone}} = \dfrac{8}{1} = 8 : 1.$

Hence, Option 3 is the correct option.

A solid metallic cylinder is melted and formed into identical solid cones each having same radius and same height as that of the cylinder. The number of cones formed is :

1. 3

2. 6

3. 9

4. 2

Answer

Given,

A solid metallic cylinder is melted and formed into identical solid cones.

Let no. of cones formed be n.

As, radius and height of cones formed is equal to the cylinder.

Let radius of cylinder and cone be r units, height of cylinder and cone be h units.

∴ Volume of cylinder = n × Volume of each cone

⇒ πr²h = n × $\dfrac{1}{3}πr^2h$

⇒ n = $\dfrac{3πr^2h}{πr^2h}$ = 3.

Hence, Option 1 is the correct option.

A solid metallic cone of radius 10 cm and height 12 cm is melted to form a wire of uniform cross-section 1 cm². The length of wire formed is :

1. 40 × π cm

2. 4 × π m

3. 400 × π m

4. 2 × π m

Answer

Given,

A solid metallic cone of radius (r) 10 cm and height (h) 12 cm is melted to form a wire of uniform cross-section 1 cm².

Let length of wire formed be l cm.

∴ Volume of cone = Volume of wire

$\Rightarrow \dfrac{1}{3}πr^2h = \text{Area of cross-section × length} \\[1em] \Rightarrow \dfrac{1}{3}π \times 10^2 \times 12 = 1 \times l \\[1em] \Rightarrow l = 4 \times π \times 10^2 \\[1em] \Rightarrow l = 400 π \text{ cm} \\[1em] \Rightarrow l = \dfrac{400π}{100} \text{ m} \\[1em] \Rightarrow l = 4 × π \text{ m}.$