Cylinder, Cone & Sphere (Surface Area & Volume)

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.

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 = $\frac{S}{C}$ = S ÷ C.

Hence, Option 3 is the correct option.

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 = $\frac{a^3}{πr^2}$.

Hence, Option 3 is the correct option.

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 = $\frac{πR^2H}{πr^2h} = \frac{R^2H}{r^2h}$

Hence, Option 2 is the correct option.

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.

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 = $\frac{22}{7} \times (2.1)^2 \times 1200$

= 22 × 0.3 × 2.1 × 1200

= 16632 cm³.

Hence, pipe can hold 16632 cm³ of water.

Given,

Diameter of well = 2.8 m

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

Depth of the well = 28 m.

Volume of earth dug out = Volume of well = πr²h

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

= 172.48 m³.

Area of curved surface = 2πrh

= 2 x $\frac{22}{7}$ 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.

Given,

External diameter of hollow cylinder = 20 cm

So, external radius (R) = $\frac{20}{2}$ = 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,

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

Given,

Diameter of the solid cylinder = 2 cm

so, radius (a) = $\frac{2}{2}$ = 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.

Given,

The diameter of the cylinder = 20 cm

So, the radius (r) = $\frac{20}{2}$ = 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 $\frac{22}{7}$ x 10 x h = 100

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

⇒ h = $\frac{700}{440}$ = 1.6 cm.

Hence, the height of cylinder = 1.6 cm.

(ii) By formula,

Volume of the cylinder = πr²h

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

= 502.9 cm³.

Hence, the volume of cylinder = 502.9 cm³.

Given,

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

Inner radius of the pipe (r) = $\frac{5}{2}$ = 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

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

= $\frac{22}{7} \times$(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 = $\frac{(1728.6 \times 7.7)}{1000}$ kg = 13.31 kg.

Hence, weight of the pipe = 13.31 kg.

Given,

Diameter of base = 42 cm and radius = $\frac{42}{2}$ = 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

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

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

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

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)

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

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

= 754.29 cm².

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

Given,

Length of the open pipe = 50 cm

External diameter = 20 cm

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

Internal diameter = 6 cm

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

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

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

= 4085.71 cm²

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

= 2 x $\frac{22}{7}$ x (10² - 3²)

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

= 572 cm².

Total surface area = 4085.71 + 572 = 4657.71 cm².

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

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 = $\frac{1029}{3}$

⇒ 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 $\frac{22}{7}$ x 7 x (21 + 7)

= 2 x 22 x 28

= 1232 cm².

Hence, total surface area of cylinder = 1232 cm².

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 - $\frac{20}{100} \times r$ = r - 0.2r = 0.8r

Height (h') = h + $\frac{10}{100} \times h$ = 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 = $\frac{\text{Decrease in volume}}{\text{Original volume}}$ x 100 %

= $\frac{0.296πr^2h}{πr^2h}$ 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 = $\frac{\text{Decreased CSA}}{\text{Original CSA}}$ x 100 %

= $\frac{0.24πrh}{2πrh}$ x 100 %

= 0.12 x 100 %

= 12 %.

Hence, decrease in volume = 12 %.

Given,

Height of the cylinder box (h) = 35 cm

Base radius of the cylinder box (r) = $\frac{20}{2}$ = 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 $\frac{22}{7}$ x 10(10 + 35)

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

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

Thus,

Area of metal sheet = length x width = 28 x 100 = 2800 cm² = $\frac{2800}{100 \times 100}$ = 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

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

⇒ $\frac{100x - 10x}{100}$ = 2800

⇒ $\frac{90x}{100}$ = 2800

⇒ $\frac{9x}{10}$ = 2800

⇒ $x = \frac{2800 \times 10}{9}$

⇒ 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.

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,

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

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

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

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 $\frac{20}{2}$ = 10 cm.

Wetted surface area = 2πrh + πr²

= πr(2h + r)

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

= $22 \times 27$

= 594 cm².

Hence, wetted surface area = 594 cm².

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

⇒ Circumference = 33 cm

⇒ 2πr = 33

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

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

Volume = πr²h

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

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

= $\frac{26680.5}{7}$ = 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

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

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

Volume = πr²h

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

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

= 5082 cm³.

Hence, volume of cylinder = 5082 cm³.

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

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

Hence, rise in level of water = 2.16 cm.

Given,

Tank's diameter = 2 m

Tank's radius (r) = $\frac{2}{2}$ = 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.

According to question,

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

Total surface area = 3080

⇒ 2πr(h + r) = 3080

⇒ 2πr × 35 = 3080

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

⇒ 220r = 3080

⇒ r = $\frac{3080}{220}$

⇒ r = 14 cm.

⇒ h + r = 35

⇒ h + 14 = 35

⇒ h = 35 - 14

⇒ h = 21 cm.

Volume of cylinder = πr²h

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

= 22 × 2 × 14 × 21

= 12936 cm³.

Hence, volume of cylinder = 12936 cm³.

Given,

Total surface area = 616 cm²

⇒ 2πr(h + r) = 616

⇒ πr(h + r) = $\frac{616}{2}$

⇒ πr(h + r) = 308 ..........(1)

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

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

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

⇒ πr(r + r) = 308

⇒ πr.2r = 308

⇒ 2πr² = 308

⇒ πr² = 154

⇒ $\frac{22}{7} \times r^2$ = 154

⇒ r² = $\frac{154 \times 7}{22}$

⇒ r² = 49

⇒ r = $\sqrt{49}$

⇒ r = 7 cm

⇒ h = 7 cm.

Volume of cylinder = πr²h

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

= 22 × 49

= 1078 cm³.

Hence, volume of cylinder = 1078 cm³.

Given,

Height of large cylindrical vessel (h₁) = 24 cm

Diameter of large cylindrical vessel = 40 cm

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

Height of small cylindrical vessel (h₂) = 10 cm

Diameter of small cylindrical vessel = 8 cm

Radius of small cylindrical vessel (r₂) = $\frac{8}{2}$ = 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

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

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

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

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

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

Hence, diameter of new cylinder = 180 cm.

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

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

Given,

Total surface area of hollow cylinder = 3575 cm².

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

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

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

Hence, thickness of hollow cylinder = 3.5 cm.

(i) According to question,

r₁ : r₂ = 3 : 5

Let, r₁ = 3x and r₂ = 5x.

h₁ : h₂ = 2 : 3

Let, h₁ = 2y and h₂ = 3y.

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

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

(ii)

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

Hence, ratio between volume = 6 : 25.

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

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

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

= 2 × 22 × 0.6 × 9.6

= 253.44 m².

Given,

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

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

$$\begin{aligned}= \frac{15 - 1}{15} \\[1em] = \frac{14}{15}\end{aligned}$$

Let metal sheet used be $x$ m²

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

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

Area of metal sheet in nearest m² = 272 m²

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

By formula,

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

Given,

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

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

Hence, Option 3 is the correct option.

By formula,

Curved surface area = πrl

= $πr\sqrt{h^2 + r^2}$.

Hence, Option 2 is the correct option.

Since, conical toy is made from the rectangular sheet.

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

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

∴ πrl = length × breadth

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

Hence, Option 1 is the correct option.

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.

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

Hence, Option 2 is the correct option.

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

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

Hence, Option 2 is the correct option.

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 = $\frac{1}{3}πr^2h$

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

Hence, volume of cone = 1005.71 cm³.

By formula,

Curved surface area = πrl

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

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.

Given,

Circumference of base = 66 m

⇒ 2πr = 66

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

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

By formula,

Volume of cone = $\frac{1}{3}πr^2h$

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

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

Given,

Radius : Height = 5 : 12

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

By formula,

Volume of cone = $\frac{1}{3}πr^2h$

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

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.

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.

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

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

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 :

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

⇒ r₁ : r₂ = 4 : 1.

Hence, ratio of radii = 4 : 1.

Given,

Diameter of heap of cone = 16.8 m

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

Height (h) = 3.5 m

Volume of cone = $\frac{1}{3}πr^2h$

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

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

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

= 240.24 m².

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

Given,

Diameter of conical tent = 48 m

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

Height of conical tent (h) = 7 m

Let l be the slant height of the conical tent.

By formula,

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

By formula,

Surface area of conical tent (S) = πrl

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

Given,

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

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

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

Thus,

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

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

⇒ Length of canvas = $\frac{132000}{63 \times 1.5} = \frac{132000}{94.5}$ = 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

For larger cone,

Height (h₁) = 8 cm

Radius (r₁) = 6 cm

For smaller cones,

Height (h₂) = 2 cm

Radius (r₂) = $\frac{1}{2}$ = 0.5 cm

Let no. of smaller cones be n.

Volume of larger cone = n × Volume of smaller cones

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

Hence, no. of cones formed = 576.

(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 = $\frac{1}{3}πr^2h$

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

= $\frac{942}{3}$

= 314 cm³.

Hence, volume of circular cone = 314 cm³.

Given,

Area of the base of conical solid = 38.5 cm²

⇒ πr² = 38.5 ............(1)

⇒ $\frac{22}{7}r^2$ = 38.5

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

⇒ r = $\sqrt{12.25}$ = 3.5 cm

Given,

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

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

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

= 137.5 cm².

Hence, curved surface area = 137.5 cm².

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

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

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,

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

Hence, volume of each cone = 221.76 cm³.

(i) Given,

Area of base floor = 154 m²

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

Hence, radius of the floor = 7 m.

(ii) Given,

Volume of tent = 1232 m³

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

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

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

= 550 m².

Let length of canvas be l.

Area of canvas = Curved surface area of cone

⇒ l × b = 550

⇒ l × 2 = 550

⇒ l = $\frac{550}{2}$

⇒ l = 275 m.

Hence, length of canvas required = 275 m.

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) = $\frac{4}{2}$ = 2 cm

Let no. of smaller spheres formed be n.

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

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

Hence, Option 3 is the correct option.

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.

Let radius of two spherical solids be R and r.

Given,

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

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

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

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

Hence, Option 4 is the correct option.

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

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

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

Hence, Option 1 is the correct option.

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

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

Hence, Option 4 is the correct option.

Given,

Volume of the sphere = 38808 cm³

Let the radius of the sphere = r

By formula,

Volume of sphere = $\frac{4}{3}πr^3$

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

∴ Diameter = 2r = 21 x 2 = 42 cm.

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

= 5544 cm².

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

Let the radius of the spherical ball be r cm.

So, the volume = $\frac{4}{3} πr^3$

Radius of smaller ball = $\frac{r}{2}$ 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.

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

Hence, 8 balls can be made.

Given,

Diameter of bigger ball = 8 cm

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

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

= $\frac{4}{3} \times π (4)^3$

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

Radius of small ball (r) = 1 cm

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

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

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

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

$$\begin{aligned}\Rightarrow \frac{256π}{3} = n \times \frac{4}{3}π \\[1em] \Rightarrow n = \frac{\frac{256π}{3}}{\frac{4}{3}π} \\[1em] \Rightarrow n = \frac{256 \times π \times 3}{4 \times 3 \times π} \\[1em] \Rightarrow n = 64.\end{aligned}$$

Hence, 64 balls can be made.

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 :

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

Hence, r₁ : r₂ = 3 : 1.

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

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

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

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

Let r be the radius of the sphere.

According to question,

Surface area of sphere = Volume of sphere

$$\begin{aligned}\Rightarrow 4πr^2 = \frac{4}{3}πr^3 \\[1em] \Rightarrow r^2 = \frac{r^3}{3} \\[1em] \Rightarrow \frac{r^3}{r^2} = 3 \\[1em] \Rightarrow r = 3\text{ cm}.\end{aligned}$$

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

Hence, diameter of sphere = 6 cm.

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

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

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

= $\frac{1}{2} \times 4πr^2 + πr^2$

= 2πr² + πr²

= 3πr²

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

Hence, total surface area of each hemisphere 28.88 cm².

(i) Given,

Internal diameter = 21 cm

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

By formula,

Internal curved surface area = 2πr².

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

Hence, internal curved surface area = 693 cm².

(ii) Given,

Internal diameter = 28 cm

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

By formula,

External curved surface area = 2πR².

$$\begin{aligned}= 2 \times \frac{22}{7} \times 14 \times 14 \\[1em] = 1232 \text{ cm}^2.\end{aligned}$$

Hence, external curved surface area = 1232 cm².

(iii) By formula,

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

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

Hence, total surface area of hemisphere = 2194.5 cm².

(iv) By formula,

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

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

Hence, volume of vessel = 3323.83 cm³.

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.

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

Calculating ratio between volumes,

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

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

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

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

By formula,

Surface area of sphere = 4πr²

= 4 × 3.1 × 12²

= 1785.6 cm².

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

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

Hence, Option 2 is the correct option.

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

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

Hence, Option 4 is the correct option.

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.

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

Hence, Option 3 is the correct option.

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 × $\frac{1}{3}πr^2h$

⇒ n = $\frac{3πr^2h}{πr^2h}$ = 3.

Hence, Option 1 is the correct option.

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

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

Hence, Option 2 is the correct option.

Given,

Radius of the sphere (R) = 15 cm

So, the volume of sphere melted = $\frac{4}{3}πR^3 = \frac{4}{3} \times π \times 15 \times 15 \times 15$

= 4500π cm³.

Radius of each cone formed (r) = 2.5 cm

Height of each cone (h) = 8 cm

So, volume of each cone = $\frac{1}{3}πr^2h = \frac{1}{3} \times π \times 2.5 \times 2.5 \times 8$

= $\frac{1}{3} \times 50π = \frac{50}{3}π.$

Let no. of cones formed be n.

Volume of sphere = n × Volume of cone

$$\begin{aligned}\Rightarrow 4500π = n \times \frac{50}{3}π \\[1em] \Rightarrow n = \frac{4500π \times 3}{50π} \\[1em] \Rightarrow n = 270.\end{aligned}$$

Hence, no. of cones formed = 270.

Given,

External diameter of the hollow sphere = 8 cm

So, external radius (R) = $\frac{8}{2}$ = 4 cm

Internal diameter of the hollow sphere = 4 cm

So, internal radius (r) = $\frac{4}{2}$ = 2 cm

Diameter of cone = 8 cm

Radius of cone (r₁) = 4 cm

Let height of the cone = h cm.

Since, hollow sphere is melted and recasted into cone.

∴ Volume of sphere = Volume of cone

$$\begin{aligned}\Rightarrow \frac{4}{3}π(R^3 - r^3) = \frac{1}{3}π(r_1)^2h \\[1em] \Rightarrow 4(R^3 - r^3) = r_1^2h \\[1em] \Rightarrow 4 \times [(4)^3 - (2)^3] = (4)^2 \times h \\[1em] \Rightarrow 4 \times [64 - 8] = 16h \\[1em] \Rightarrow h = \frac{4 \times 56}{16} \\[1em] \Rightarrow h = 14 \text{ cm}.\end{aligned}$$

Hence, the height of the cone = 14 cm.

Given,

Height of the solid right circular cone (h) = 32 cm

Internal radius of metallic spherical shell (r) = 3 cm

External radius of metallic spherical shell (R) = 5 cm

Let radius of cone be r₁ cm.

As, metallic spherical shell is recasted into right circular cone.

∴ Volume of spherical shell = Volume of cone.

$$\begin{aligned}\Rightarrow \frac{4}{3}π(R^3 - r^3) = \frac{1}{3}πr_1^2h \\[1em] \Rightarrow 4(R^3 - r^3) = r_1^2h \\[1em] \Rightarrow r_1^2 = \frac{4(R^3 - r^3)}{h} \\[1em] \Rightarrow r_1^2 = \frac{4 \times (5^3 - 3^3)}{32} \\[1em] \Rightarrow r_1^2 = \frac{4 \times (125 - 27)}{32} \\[1em] \Rightarrow r_1^2 = \frac{4 \times 98}{32} \\[1em] \Rightarrow r_1^2 = \frac{49}{4} \\[1em] \Rightarrow r_1 = \sqrt{\frac{49}{4}} \\[1em] \Rightarrow r_1 = \frac{7}{2}.\end{aligned}$$

Diameter = 2r = $2 \times \frac{7}{2}$ = 7 cm.

Hence, diameter = 7 cm.

Let the radius of the smaller cone be r cm.

Given,

Height of smaller cone (h) = 108 cm

Diameter of bigger cone = 40 cm

So, radius (R) = $\frac{40}{2}$ = 20 cm

Height of bigger cone (H) = 9 cm.

According to question,

Volume of big cone = 3 × Volume of each smaller cone.

$$\begin{aligned}\Rightarrow \frac{1}{3}πR^2H = 3 \times \frac{1}{3}πr^2h \\[1em] \Rightarrow R^2H = 3 \times r^2h \\[1em] \Rightarrow (20)^2 \times 9 = 3 \times r^2 \times 108 \\[1em] \Rightarrow r^2 = \frac{20^2 \times 9}{3 \times 108} \\[1em] \Rightarrow r^2 = \frac{3600}{324} \\[1em] \Rightarrow r^2 = \frac{100}{9} \\[1em] \Rightarrow r = \sqrt{\frac{100}{9}} \\[1em] \Rightarrow r = \frac{10}{3} = 3\frac{1}{3}\text{ cm}.\end{aligned}$$

Hence, radius of base of each cone = $3\frac{1}{3}$ cm.

Let radius of sphere be r cm.

Since,

A solid rectangular block of metal is melted and formed into a solid sphere.

∴ Volume of rectangular block = Volume of sphere

$$\begin{aligned}\Rightarrow 49 \times 44 \times 18 = \frac{4}{3}πr^3 \\[1em] \Rightarrow 38808 = \frac{4}{3} \times \frac{22}{7} \times r^3 \\[1em] \Rightarrow r^3 = \frac{38808 \times 21}{4 \times 22} \\[1em] \Rightarrow r^3 = 9261 \\[1em] \Rightarrow r^3 = (21)^3 \\[1em] \Rightarrow r = 21 \text{ cm}.\end{aligned}$$

Hence, the radius of sphere = 21 cm.

Given,

Internal radius of hemispherical bowl (R) = 9 cm

Diameter of each conical container = 3 cm

So, radius (r) = $\frac{3}{2}$ = 1.5 cm and height of conical container(h) = 4 cm

Let no. of conical containers needed be n.

∴ Volume of hemispherical bowl = n × Volume of each conical shaped container

$$\begin{aligned}\Rightarrow \frac{2}{3}πR^3 = n \times \frac{1}{3}πr^2h \\[1em] \Rightarrow n = \frac{\frac{2}{3}πR^3}{\frac{1}{3}πr^2h} \\[1em] \Rightarrow n = \frac{2 \times (9)^3 \times 3}{1 \times (1.5)^2 \times 4 \times 3} \\[1em] \Rightarrow n = \frac{2 \times 729 \times 3}{2.25 \times 4 \times 3} \\[1em] \Rightarrow n = 162.\end{aligned}$$

Hence, 162 containers are necessary to empty the bowl.

(i) Given,

Total area of solid metallic sphere = 1256 cm²

Let R be the radius of metallic sphere.

By formula,

Total area of solid metallic sphere = 4πR²

∴ 4πR² = 1256

$$\begin{aligned}\Rightarrow 4 \times 3.14 \times R^2 = 1256 \\[1em] \Rightarrow R^2 = \frac{1256}{4 \times 3.14} \\[1em] \Rightarrow R^2 = \frac{1256}{12.56} \\[1em] \Rightarrow R^2 = 100 \\[1em] \Rightarrow R = \sqrt{100} = 10 \text{ cm}.\end{aligned}$$

Hence, radius of sphere = 10 cm.

(ii) Given,

Radius of cone (r) = 2.5 cm

Height of cone (h) = 8 cm.

Let no. of cones formed be n.

Since, sphere is recasted into n cones.

∴ Volume of sphere = n × Volume of cones

$$\begin{aligned}\Rightarrow \frac{4}{3}πR^3 = n \times \frac{1}{3}πr^2h \\[1em] \Rightarrow 4R^3 = n \times r^2h \\[1em] \Rightarrow n = \frac{4R^3}{r^2h} \\[1em] \Rightarrow n = \frac{4 \times 10^3}{(2.5)^2 \times 8} \\[1em] \Rightarrow n = \frac{4 \times 1000}{6.25 \times 8} \\[1em] \Rightarrow n = \frac{4000}{50} \\[1em] \Rightarrow n = 80.\end{aligned}$$

Hence, no. of cones formed = 80.

(i) Given,

Surface area of solid metallic sphere = 2464

Let radius of solid metallic sphere be R cm.

By formula,

Surface area of solid metallic sphere = 4πR²

∴ 4πR² = 2464

⇒ $4 \times \frac{22}{7} \times R^2 = 2464$

⇒ $R^2 = \frac{2464 \times 7}{22 \times 4}$

⇒ $R^2 = \frac{17248}{88}$

⇒ R² = 196

⇒ R = $\sqrt{196}$

⇒ R = 14 cm.

Hence, the radius of sphere = 14 cm.

(ii) Given,

Radius of cones (r) = 3.5 cm

Height of cones (h) = 7 cm.

Let no. of cones formed be n.

Since,

Sphere is melted and recasted into cones.

∴ Volume of sphere = n × Volume of cone

$$\begin{aligned}\Rightarrow \frac{4}{3}πR^3 = \frac{1}{3}πr^2h \\[1em] \Rightarrow 4R^3 = n \times r^2h \\[1em] \Rightarrow n = \frac{4R^3}{r^2h} \\[1em] \Rightarrow n = \frac{4 \times (14)^3}{(3.5)^2 \times 7} \\[1em] \Rightarrow n = \frac{4 \times 2744}{12.25 \times 7} \\[1em] \Rightarrow n = \frac{10976}{85.75} \\[1em] \Rightarrow n = 128.\end{aligned}$$

Hence, no. of cones formed = 128.

Given,

Radius of cone (r) = 2 cm

Height of cone (h) = 2 cm

From figure,

Radius of hemisphere = Radius of cone = 2 cm.

⇒ Volume of body = Volume of cone + Volume of hemisphere

⇒ Volume of body = $\frac{1}{3}πr^2h + \frac{2}{3}πr^3$

$$\begin{aligned}= \frac{πr^2}{3}(h + 2r) \\[1em] = \frac{π \times 2^2}{3} \times (2 + 2 × 2) \\[1em] = \frac{4π}{3} \times (2 + 4) \\[1em] = \frac{4π}{3} \times 6 \\[1em] = \frac{24π}{3} \\[1em] = 8π \text{ cm}^3.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

Radius of cylindrical part (r) = 1 cm

Height of cylindrical part (h) = 1 cm

Height of conical part (H) = 3 cm

From figure,

Radius of conical part = Radius of cylindrical part (r) = 1 cm

Volume of body = Volume of conical part + Volume of cylindrical part

$$\begin{aligned}= \frac{1}{3}πr^2H + πr^2h \\[1em] = \frac{1}{3}π \times 1^2 \times 3 + π \times 1^2 \times 1 \\[1em] = π + π \\[1em] = 2π \text{ cm}^3.\end{aligned}$$

Hence, Option 3 is the correct option.

From figure,

Total surface area of remaining solid = Curved surface area of cylinder + Area of upper surface of cylinder + Curved surface area of cone

= 2πrh + πr² + πrl.

Hence, Option 2 is the correct option.

From figure,

Total wetted surface area of body = Curved surface area of cylinder + Area of upper surface of cylinder + Curved surface area of cone

= 2πrh + πr² + πrl.

Hence, Option 1 is the correct option.

From figure,

Radius of sphere = Radius of cylinder = r

Height of cylinder (h) = 2r

Volume of air left = Volume of cylinder - Volume of sphere

$$\begin{aligned}= πr^2h - \frac{4}{3}πr^3 \\[1em] = πr^2.2r - \frac{4}{3}πr^3 \\[1em] = 2πr^3 - \frac{4}{3}πr^3 \\[1em] = \frac{6πr^3 - 4πr^3}{3} \\[1em] = \frac{2πr^3}{3} \\[1em] = \frac{2}{3}πr^3.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

Height of the cone (h) = 15 cm

Diameter of the cone = 7 cm

Radius of cone (r) = $\frac{7}{2}$ = 3.5 cm

So, radius of the hemisphere (R) = 3.5 cm

From figure,

Volume of the solid (V) = Volume of the cone + Volume of the hemisphere

$$\begin{aligned}V = \frac{1}{3}πr^2h + \frac{2}{3}πR^3 \\[1em] = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 15 + \frac{2}{3} \times \frac{22}{7} \times (3.5)^3 \\[1em] = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 15 + \frac{2}{3} \times \frac{22}{7} \times 42.875 \\[1em] = 22 \times 1.75 \times 5 + \frac{1886.5}{21} \\[1em] = 192.5 + 89.833 \\[1em] = 282.33\text{ cm}^3.\end{aligned}$$

Hence, volume of solid formed = 282.33 cm³.

From figure,

Radius of hemisphere (r) = Radius of cone (R) = 3.5 m.

Volume of hemisphere = $\frac{2}{3}πr^3$

= $\frac{2}{3} \times \frac{22}{7} \times (3.5)^3$

= $\frac{2}{3} \times 22 \times 0.5 \times 3.5 \times 3.5$

= $\frac{269.5}{3}$ m³.

Let height of the cone = h.

By formula,

Volume of conical part = $\frac{1}{3}πR^2h$ ........(1)

Given,

Volume of cone = $\frac{2}{3}$ Volume of hemisphere

= $\frac{2}{3} \times \frac{269.5}{3} = \frac{539}{9}$ m³ ...........(2)

From (1) and (2), we get :

$$\begin{aligned}\Rightarrow \frac{1}{3}πR^2h = \frac{539}{9} \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times h = \frac{539}{9} \\[1em] \Rightarrow h = \frac{539 \times 7 \times 3}{9 \times (3.5)^2 \times 22} \\[1em] \Rightarrow h = \frac{11319}{2425.5} \\[1em] \Rightarrow h = 4.67 \text{ m}.\end{aligned}$$

By formula,

⇒ l² = R² + h²

⇒ l² = (3.5)² + (4.67)²

⇒ l² = 12.25 + 21.81

⇒ l² = 34.06

⇒ l = $\sqrt{34.06}$

⇒ l = 5.83 cm.

Surface area of buoy = Surface area of hemisphere + Surface area of cone = 2πr² + πRl

$$\begin{aligned}= 2 \times \frac{22}{7} \times (3.5)^2 + \frac{22}{7} \times 3.5 \times 5.83 \\[1em] = 2 \times 22 \times 0.5 \times 3.5 + 22 \times 0.5 \times 5.83 \\[1em] = 77 + 64.13 \\[1em] = 141.13 \text{ m}^2.\end{aligned}$$

Hence, height = 4.67 and surface area of buoy = 141.13 m².

Given,

Dimensions of rectangular solid are:

l = 42 cm, b = 30 cm and h = 20 cm.

Conical cavity’s diameter = 14 cm

So, its radius (r) = 7 cm

Depth (h) = 24 cm

(i) Total surface area of cuboid = 2(lb + bh + lh)

= 2 (42 x 30 + 30 x 20 + 20 x 42)

= 2 (1260 + 600 + 840)

= 2 (2700)

= 5400 cm²

Area of circular base of conical cavity = πr² = $\frac{22}{7}$ x 7 x 7 = 154 cm².

By formula,

⇒ l² = r² + h²

⇒ l² = (7)² + (24)²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$

⇒ l = 25 cm.

Area of curved surface area of cone = πrl

= $\frac{22}{7} \times 7 \times 25$ = 22 x 25 = 550 cm²

Surface area of remaining part = Surface area of rectangular solid + Surface area of cone - Area of base of conical cavity

= 5400 + 550 - 154

= 5796 cm².

Hence, surface area of remaining part = 5796 cm².

(ii) Volume of the rectangular solid = lbh

= (42 x 30 x 20) cm³

= 25200 cm³

Radius of conical cavity (r) = 7 cm

Depth (h) = 24 cm

Volume of cone = $\frac{1}{3}πr^2h$

= $\frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24$

= 22 × 7 × 8

= 1232 cm³.

Volume of remaining solid = Volume of rectangular solid - Volume of cone

= 25200 - 1232

= 23968 cm³.

Hence, volume of remaining solid = 23968 cm³.

(iii) Volume of material drilled out = Volume of cone = 1232 cm³.

Weight of material drilled out = 1232 × 7 = 8624 g = 8.624 kg.

Hence, weight of material drilled out = 8.624 kg.

It’s known that, the diameter of the largest hemisphere that can be placed on the face of a cube of side 7 cm will be 7 cm.

So, it’s radius (r) = $\frac{7}{2}$ = 3.5 cm

Curved surface area = 2πr²

= $2 \times \frac{22}{7} \times (3.5)^2$ = 77 cm².

Area of base of hemisphere = πr²

= $\frac{22}{7} \times (3.5)^2$

= 38.5 cm².

From figure,

Surface area of the resulting solid = Surface area of the cube – Area of the base of the hemisphere + Curved surface area of hemisphere

= 6(side)² - 38.5 + 77

= 6(7)² + 38.5

= 294 + 38.5

= 332.5 cm².

Hence, surface area of resulting solid = 332.5 cm³.

Given,

Height of cone (H) = 8 cm

Radius of cone (R) = 5 cm

Radius of sphere (r) = 0.5 cm.

Let no. of lead shots dropped in vessel be n.

According to question,

One-fourth of the water flows out after dropping lead shots.

∴ $\frac{1}{4}$ x Volume of cone = n × Volume of each sphere

$$\begin{aligned}\Rightarrow \frac{1}{4} \times \frac{1}{3}πR^2H = n \times \frac{4}{3}πr^3 \\[1em] \Rightarrow \frac{R^2H}{4} = n \times 4r^3 \\[1em] \Rightarrow n = \frac{R^2H}{16r^3} \\[1em] \Rightarrow n = \frac{5^2 \times 8}{16 \times (0.5)^3} \\[1em] \Rightarrow n = \frac{25 \times 8}{16 \times 0.125} \\[1em] \Rightarrow n = \frac{200}{2} \\[1em] \Rightarrow n = 100.\end{aligned}$$

Hence, no. of shots dropped = 100.

Given,

Circumference = 198 cm

∴ 2πr = 198

⇒ r = $\frac{198}{2π} = \frac{198}{2 \times \frac{22}{7}} = \frac{198 \times 7}{44} = \frac{1386}{44}$ = 31.5 cm.

By formula,

$$\begin{aligned}\text{Capacity of bowl} = \frac{2}{3}πr^3 \\[1em] = \frac{2}{3} \times \frac{22}{7} \times (31.5)^3 \\[1em] = \frac{2 \times 22 \times 31255.875}{21} \\[1em] = \frac{1375258.5}{21} \\[1em] = 65488.5 \text{ cm}^3.\end{aligned}$$

Hence, capacity of bowl = 65488.5 cm³.

The max height and radius of cone, inside a hemisphere of radius r cm can be r cm.

Volume of cone = $\frac{1}{3}πr^2h$

= $\frac{1}{3} \times π \times r^2 \times r$

= $\frac{1}{3}πr^3 \text{ cm}^3.$

Hence, maximum volume of cone that can be carved out of solid hemisphere of radius r cm = $\frac{1}{3}πr^3$ cm³.

Let height of each cone be h cm.

Given,

Cones are melted and recasted into a sphere.

∴ Volume of cone with radii r₁ + Volume of cone with radii r₂ = Volume of sphere

$$\begin{aligned}\Rightarrow \frac{1}{3}πr_1^2h + \frac{1}{3}πr_2^2h = \frac{4}{3}πR^3 \\[1em] \Rightarrow \frac{πh}{3}(r_1^2 + r_2^2) = \frac{π}{3} \times 4R^3 \\[1em] \Rightarrow h(r_1^2 + r_2^2) = 4R^3 \\[1em] \Rightarrow h = \frac{4R^3}{r_1^2 + r_2^2}.\end{aligned}$$

Hence, h = $\frac{4R^3}{r_1^2 + r_2^2}.$

Let radius and height of cone and hemisphere be a cm.

Volume of cone = $\frac{1}{3}π(\text{radius})^2\text{height} = \frac{1}{3}πa^3.$

Volume of hemisphere = $\frac{2}{3}π(\text{radius})^3 = \frac{4}{3}πa^3.$

$$\begin{aligned}\text{Ratio} = \frac{\text{Vol. of cone}}{\text{Vol. of hemisphere}} \\[1em] = \frac{\frac{1}{3}πa^3}{\frac{2}{3}πa^3} \\[1em] = \frac{1}{2} \\[1em] = 1 : 2.\end{aligned}$$

Hence, ratio between volumes = 1 : 2.

Radii of cylinder = Radii of cone = r (let)

Let height of cylinder be h₁ and height of cone be h₂.

Given,

Volume of a solid cylinder and a solid cone are same.

⇒ πr²h₁ = $\frac{1}{3}$ πr²h₂

⇒ h₁ = $\frac{1}{3}$ h₂

⇒ $\frac{h_1}{h_2} = \frac{1}{3}$

⇒ h₁ : h₂ = 1 : 3.

Hence, Option 3 is the correct option.

From figure,

Height of cone (h) = r

Height of cylinder (H) = r

Radius of hemispherical portion = Radius of cylindrical portion = Radius of conical portion = r.

From figure,

Volume of body = Volume of conical portion + Volume of cylindrical portion + Volume of hemispherical portion

$$\begin{aligned}= \frac{1}{3}πr^2h + πr^2H + \frac{2}{3}πr^3 \\[1em] = \frac{1}{3}πr^2.r + πr^2.r + \frac{2}{3}πr^3 \\[1em] = \frac{1}{3}πr^3 + πr^3 + \frac{2}{3}πr^3 \\[1em] = \frac{πr^3 + 3πr^3 + 2πr^3}{3} \\[1em] = \frac{6πr^3}{3}\\[1em] = 2πr^3.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

Radius of small metallic sphere (r) = 1 cm

Eight identical metallic spheres, each of radius 1 cm are melted and recast into a solid sphere.

Let radius of solid sphere formed be R.

∴ Volume of large metallic sphere = 8 × Volume of each smaller sphere

$$\begin{aligned}\Rightarrow \frac{4}{3}πR^3 = 8 \times \frac{4}{3}πr^3 \\[1em] \Rightarrow R^3 = 8r^3 \\[1em] \Rightarrow R^3 = 8 \times 1^3 \\[1em] \Rightarrow R^3 = 8 \\[1em] \Rightarrow R^3 = 2^3 \\[1em] \Rightarrow R = 2 \text{ cm}.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

Ratio between the heights of two solid cones is 2 : 3.

Let height of first (h₁) and second cone (h₂) be 2a and 3a.

Ratio between the radii of two solid cones is 9 : 8.

Let radii of first (r₁) and second cone (r₂) be 9x and 8x.

$$\begin{aligned}\frac{\text{Vol. of 1st cone}}{\text{Vol. of 2nd cone}} = \frac{\frac{1}{3}πr_1^2h_1}{\frac{1}{3}πr_2^2h_2} \\[1em] = \frac{r_1^2h_1}{r_2^2h_2} \\[1em] = \frac{(9x)^2 \times 2a}{(8x)^2 \times 3a} \\[1em] = \frac{81x^2 \times 2a}{64x^2 \times 3a} \\[1em] = \frac{162x^2a}{192x^2a} \\[1em] = \frac{27}{32} \\[1em] = 27 : 32.\end{aligned}$$

Hence, Option 1 is the correct option.

Before change :

Radius of cylinder = r

Height of cylinder = h

Volume of cylinder = πr²h

After change :

Radius of cylinder = 2r

Height of cylinder = h

Volume of cylinder = π(2r)²h = 4πr²h

Difference in volume = Volume after change - Volume before change
= 4πr²h - πr²h
= 3πr²h.

Percentage increase in volume

$$\begin{aligned}= \frac{\text{Difference in vol.}}{\text{Vol. of cylinder before change}} \times 100 \\[1em] = \frac{3πr^2h}{πr^2h} \times 100 \\[1em] = 3 \times 100 \\[1em] = 300%\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

Height of the cylinder (H) = 10 cm

Radius of the base of cylinder (R) = 6 cm

Height of the cone (h) = 10 cm

Radius of the base of cone (r) = 6 cm

Volume of remaining part (V) = Volume of cylinder - Volume of cone

$$\begin{aligned}V = πR^2H - \frac{1}{3}πr^2h \\[1em] = \frac{22}{7} \times 6^2 \times 10 - \frac{1}{3} \times \frac{22}{7} \times 6^2 \times 10 \\[1em] = \frac{22}{7} \times 6^2 \times 10 \times \Big(1 - \frac{1}{3}\Big) \\[1em] = \frac{22}{7} \times 36 \times 10 \times \frac{2}{3} \\[1em] = \frac{22 \times 12 \times 10 \times 2}{7} \\[1em] = \frac{5280}{7} \\[1em] = 754\frac{2}{7} \text{ cm}^3.\end{aligned}$$

Hence, volume of remaining solid = $754\frac{2}{7}$ cm³.

Given,

Height of the cylinder (H) = 16 cm

Radius of the base of cylinder (R) = 12 cm

Height of the cone (h) = 8 cm

Radius of the base of cone (r) = 6 cm

(i) Volume of remaining part (V) = Volume of cylinder - Volume of cone

$$\begin{aligned}V = πR^2H - \frac{1}{3}πr^2h \\[1em] = \frac{22}{7} \times 12^2 \times 16 - \frac{1}{3} \times \frac{22}{7} \times 6^2 \times 8 \\[1em] = \frac{22}{7} \times [(12^2 \times 16) - (\frac{1}{3} \times 6^2 \times 8)] \\[1em] = \frac{22}{7} \times \Big(2304 - \frac{288}{3}\Big) \\[1em] = \frac{22}{7} \times \frac{6912 - 288}{3} \\[1em] = \frac{22}{7} \times \frac{6624}{3} \\[1em] = \frac{22}{7} \times 2208 \\[1em] = \frac{48576}{7} \\[1em] = 6939.43 \text{ cm}^3.\end{aligned}$$

Hence, volume of remaining part = 6939.43 cm³.

(ii) By formula,

⇒ l² = r² + h²

⇒ l² = 6² + 8²

⇒ l² = 36 + 64

⇒ l² = 100

⇒ l² = 10²

⇒ l = 10 cm.

Thus,

Total surface area of remaining solid (T) = Curved surface area of cylinder + Curved surface area of cone + Base area of cylinder + Area of circular ring on upper side of cylinder

(T) = 2πRH + πrl + πR² + π(R² - r²)

$$\begin{aligned}= π(2RH + rl + R^2 + R^2 - r^2) \\[1em] = π(2RH + rl + 2R^2 - r^2) \\[1em] = \frac{22}{7} \times (2 \times 12 \times 16 + 6 \times 10 + 2 \times 12^2 - 6^2) \\[1em] = \frac{22}{7} \times (384 + 60 + 288 - 36) \\[1em] = \frac{22}{7} \times 696 \\[1em] = \frac{15312}{7} \\[1em] = 2187.43 \text{ cm}^2.\end{aligned}$$

Hence, total surface area of remaining solid = 2187.43 cm².

Given,

Radius of the cylindrical part of the tent (R) = $\frac{105}{2}$ m

Radius of conical part (r) = $\frac{105}{2}$ m.

Slant height (l) = 80 m

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

$$\begin{aligned}= (2 \times \frac{22}{7} \times \frac{105}{2} \times 4) + (\frac{22}{7} \times \frac{105}{2} \times 80) \\[1em] = \frac{18480}{14} + \frac{184800}{14} \\[1em] = 1320 + 13200 \\[1em] = 14520 \text{ m}^2.\end{aligned}$$

Width of canvas used = 1.5 m

Length of canvas = $\frac{\text{Area of canvas}}{\text{Width of canvas}} = \frac{14520}{1.5} = 9680 \text{ m}$

Hence,

Total cost of canvas at the rate of ₹ 15 per meter = 9680 x 15 = ₹ 145200.

Hence, total area of canvas required = 9680 m² and cost = ₹ 145200.

Given,

Height of the cylindrical part (H) = 8 m

Height of the conical part (h) = (13 - 8) m = 5 m

Diameter of base = 24 m

From figure,

Radius of cone and cylinder are equal (r) = $\frac{24}{2}$ = 12 m.

By formula,

⇒ (l)² = r² + h²

⇒ l² = 12² + 5²

⇒ l² = 144 + 25

⇒ l² = 169

⇒ l = $\sqrt{169}$

⇒ l = 13 m.

(i) Total surface area of the tent = 2πrH + πrl = πr(2H + l)

= $\frac{22}{7}$ x 12 x (2 x 8 + 13)

= $\frac{264}{7}$ (16 + 13)

= $\frac{7656}{7}$ m²

= 1093.71 m².

Hence, total surface area of tent = 1093.71 m².

(ii) According to question,

Area of canvas used in stitching = $\frac{10}{100} \times$ Total area of canvas

⇒ Total area of canvas required = Total surface area of tent + Area of canvas used in stitching

⇒ Total area of canvas = $\frac{7656}{7} + \frac{1}{10}$ Total area of canvas

⇒ Total area of canvas - $\frac{1}{10}$ Total area of canvas = $\frac{7656}{7}$

⇒ $\frac{9}{10}$ Total area of canvas = $\frac{7656}{7}$

⇒ Total area of canvas = $\frac{7656}{7} \times \frac{10}{9}$

⇒ Total area of canvas = 1215.23 m².

Hence, the total area of canvas required = 1215.23 m².

Given,

Diameter of base of cylindrical container = 42 cm

Radius of base of cylindrical container (r) = $\frac{42}{2}$ = 21 cm.

Let rise in level of water be h cm.

∴ Volume of solid submerged = Volume of increased water

⇒ 22 cm × 14 cm × 10.5 cm = πr²h

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

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

⇒ h = $\frac{22638}{9702} = 2\frac{1}{3}$ cm.

Hence, water rises by $2\frac{1}{3}$ cm in container.

Given,

Diameter of spherical marbles = 1.4 cm

Radius (r) = $\frac{1.4}{2}$ = 0.7 cm.

Diameter of beaker = 7 cm

Radius of beaker (R) = $\frac{7}{2}$ = 3.5 cm

Increase in water level (h) = 5.6 cm.

Let n marbles are dropped.

⇒ Volume of water increased in beaker = n × Volume of one marble (sphere)

⇒ πR²h = $n \times \frac{4}{3}πr^3$

$$\begin{aligned}\Rightarrow R^2h = n \times \frac{4}{3}r^3\\[1em] \Rightarrow n = \frac{3R^2h}{4r^3} \\[1em] \Rightarrow n = \frac{3 \times (3.5)^2 \times 5.6}{4 \times (0.7)^3} \\[1em] \Rightarrow n = \frac{3 \times 12.25 \times 5.6}{4 \times 0.343} \\[1em] \Rightarrow n = 150.\end{aligned}$$

Hence, 150 marbles are dropped in the beaker.

Given,

Diameter of cylindrical vessel = 7 cm

Radius of cylindrical vessel (R) = $\frac{7}{2}$ = 3.5 cm

Height of cylindrical vessel (H) = 8 cm

Diameter of base of cone = $3\frac{1}{2} = \frac{7}{2}$ cm

Radius of base of cone (r) = $\frac{7}{4}$ cm

Height of cone (h) = 8 cm

Volume of cylindrical vessel = $πR^2H = \frac{22}{7} \times (3.5)^2 \times 8$

= $22 \times 0.5 \times 3.5 \times 8$

= 308 cm³.

Volume of original cone = $\frac{1}{3}πr^2h$

= $\frac{1}{3} \times \frac{22}{7} \times \Big(\frac{7}{4}\Big)^2 \times 8$

= $\frac{1}{3} \times 22 \times \frac{1}{4} \times \frac{7}{4} \times 8$

= $\frac{308}{12}$ cm³

Volume of water required to fill the vessel = Volume of cylindrical vessel - Volume of original cone

$$\begin{aligned}= 308 - \frac{308}{12} \\[1em] = \frac{12 \times 308 - 308}{12} \\[1em] = \frac{3388}{12} \\[1em] = 282.33 \text{ cm}^3.\end{aligned}$$

Given,

Radius of new cone (r₁) = 2 cm

Height of new cone (h₁) = $\frac{7}{4}$ cm

Volume of new cone = $\frac{1}{3}πr_1^2h_1$

$$\begin{aligned}= \frac{1}{3} \times \frac{22}{7} \times 2^2 \times \frac{7}{4} \\[1em] = \frac{1}{3} \times 22 \times 4 \times \frac{1}{4} \\[1em] = \frac{22}{3} \text{ cm}^3\end{aligned}$$

Volume of water which comes down = Volume of original cone - Volume of new cone

= $\frac{308}{12} - \frac{22}{3}$

= $\frac{77}{3} - \frac{22}{3} = \frac{55}{3}$ cm³.

Let drop in height of water be h₂ cm.

Drop in volume of water = $\frac{55}{3}$ cm³

$$\begin{aligned}\therefore πR^2h_2 = \frac{55}{3} \\[1em] \Rightarrow \frac{22}{7} \times (3.5)^2 \times h_2 = \frac{55}{3} \\[1em] \Rightarrow 22 \times 0.5 \times 3.5 \times h_2 = \frac{55}{3} \\[1em] \Rightarrow 38.5 \times h_2 = \frac{55}{3} \\[1em] \Rightarrow h_2 = \frac{55}{115.5} \\[1em] \Rightarrow h_2 = \frac{10}{21} \text{ cm}. \\[1em]\end{aligned}$$

Hence, volume of water required to fill the vessel = 282.33 cm³ and drop in level of water = $\frac{10}{21}$ cm.

(i) Given,

Radius of base of cylindrical can (R) = 3.5 cm

Since,

When a sphere is placed in the can, the water just covers the sphere.

∴ Height (H) = 2R = 7 cm.

Total surface area = 2πRH + πR²

= πR(2H + R)

= $\frac{22}{7} \times 3.5 \times (2 \times 7 + 3.5)$

= 22 × 0.5 × (14 + 3.5)

= 192.5 cm².

Hence, the total surface area of the can in contact with water when the sphere is in it = 192.5 cm².

(ii) Let the depth of the water be h cm in the can.

Volume of water = Volume of cylinder - Volume of sphere

$$\begin{aligned}\Rightarrow πR^2h = πR^2H - \frac{4}{3}πR^3 \\[1em] \Rightarrow πR^2h = πR^2\Big(H - \frac{4}{3}πR\Big) \\[1em] \Rightarrow h = H - \frac{4}{3}R \\[1em] \Rightarrow h = 7 - \frac{4}{3} \times 3.5 \\[1em] \Rightarrow h = 7 - \frac{14}{3} \\[1em] \Rightarrow h = \frac{21 - 14}{3} \\[1em] \Rightarrow h = \frac{7}{3} = 2\frac{1}{3}\text{ cm}.\end{aligned}$$

Hence, depth of water in the can before the sphere was put into the can = $2\frac{1}{3}$ cm.

Given,

For Cylinder :

Height (H) = 20 cm, Radius (R) = 3.5 cm

For Hemisphere :

Radius (r) = Radius of cylinder (R) = 3.5 cm

When flat surface is downward, then height of water (h) = 10 cm.

So, when circular surface part will be downward then,

Let height upto which water fills be h₁.

Volume of cylinder (upto height h₁) = Volume of cylinder (upto height h) + Volume of hemisphere

$$\begin{aligned}\Rightarrow πR^2h_1 = πR^2h + \frac{2}{3}πr^3 \\[1em] \Rightarrow πR^2h_1 = πR^2h + \frac{2}{3}πR^3 [As, r = R] \\[1em] \Rightarrow πR^2h_1 = πR^2\Big(h + \frac{2}{3}R\Big) \\[1em] \Rightarrow h_1 = h + \frac{2}{3}R \\[1em] \Rightarrow h_1 = 10 + \frac{2}{3}\times 3.5 \\[1em] \Rightarrow h_1 = 10 + \frac{7}{3} \\[1em] \Rightarrow h_1 = \frac{30 + 7}{3} \\[1em] \Rightarrow h_1 = \frac{37}{3} = 12\frac{1}{3} = 12.33\text{ cm}.\end{aligned}$$

Hence, the level of water, when cylinder is inverted upside down = 12.33 cm.

From figure,

Total surface area of solid = Curved surface area of cylinder + Curved surface area of cone + Area of cross-section

= 2πrh + πrl + πr²

= 2πr.r + πr.2r + πr²

= 2πr² + 2πr² + πr²

= 5πr².

Hence, Option 2 is the correct option.

Radius of first solid sphere (r) = 10 cm

Radius of second solid sphere (R) = 20 cm

Ratio of volumes of two spheres :

$$\begin{aligned}\Rightarrow \frac{\text{Vol. of 1st sphere}}{\text{Vol. of 2nd sphere}} = \frac{\frac{4}{3}πr^3}{\frac{4}{3}πR^3} \\[1em] = \frac{r^3}{R^3} \\[1em] = \frac{10^3}{20^3} \\[1em] = \frac{1000}{8000} \\[1em] = \frac{1}{8} \\[1em] = 1 : 8.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

Height of the cone (H) = radius of the sphere (r) = 10 cm

Volume of cone and sphere are equal.

Let radius of cone be R cm.

$$\begin{aligned}\therefore \frac{1}{3}πR^2H = \frac{4}{3}πr^3 \\[1em] \Rightarrow R^2H = 4r^3 \\[1em] \Rightarrow R^2 = \frac{4r^3}{H} \\[1em] \Rightarrow R^2 = \frac{4 \times 10^3}{10} \\[1em] \Rightarrow R^2 = 4 \times 10^2 \\[1em] \Rightarrow R^2 = 400 \\[1em] \Rightarrow R = \sqrt{400} = 20 \text{ cm}.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

A solid metal cuboid with dimensions 10 cm, 11 cm and 8 cm is melted and identical solid spheres, each of radius 1 cm, are formed. Let no. of spheres formed be n.

∴ Volume of metal cuboid = n × Volume of sphere

⇒ l × b × h = n × $\frac{4}{3}πr^3$

⇒ 10 × 11 × 8 = n × $\frac{4}{3} \times \frac{22}{7} \times 1^3$

⇒ 880 = n × $\frac{88}{21}$

⇒ n = $\frac{880 \times 21}{88}$

⇒ n = 10 x 21 = 210.

Hence, option 2 is the correct option.

On transforming a solid from one shape to another by any external force, the volume of solid remains same.

Let initial length of wire be l and radius be r.

After extension,

Length of wire = 4l and Radius be R.

Since, volume is equal in both the cases :

∴ πr²h = πR²H

⇒ πr²l = πR².4l

⇒ r² = 4R²

⇒ r = $\sqrt{4R^2}$

⇒ r = 2R

⇒ R = $\frac{r}{2}$

Surface area in first case :

⇒ 2πrh

⇒ 2πrl

Surface area in second case :

⇒ 2πRh

⇒ $2π \times \frac{r}{2} \times 4l$

⇒ 4πrl.

∴ Surface area of solid wires in both the cases are different.

Hence, Option 1 is the correct option.

The toy has the hemisphere, a cylinder and a cone all of the same radius.

Metal sheet required = minimum area of metal sheet required

= curved surface area of conical part + curved surface area of cylinder part + curved surface area of hemisphere part

= curved surface area of (conical part + cylindrical part + hemispherical part)

∴ A is false, R is true.

Hence, option 2 is the correct option.

Given, the base radius of two right circular cone of the same height are in ratio 3 : 5.

As we know that volume of cone = $\frac{1}{3}πr^2h$

$$\begin{aligned}\frac{\text{Volume of 1st cone}}{\text{Volume of 2nd cone}} = \frac{\frac{1}{3}πr_1^2h}{\frac{1}{3}πr_2^2h}\\[1em] = \frac{πr_1^2h}{πr_2^2h}\\[1em] = \frac{r_1^2}{r_2^2}\\[1em] = \frac{3^2}{5^2}\\[1em] = \Big(\frac{3}{5}\Big)^2\\[1em] = \frac{9}{25}.\end{aligned}$$

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

Given, radius of solid cone, R = 3 cm

Height of solid cone, H = 3 cm

Radius of solid cylinder, r = 1 cm

Height of solid cylinder, h = 1 cm

By formula,

Volume of cone = $\frac{1}{3}πR^2H$

Volume of cylinder = $πr^2h$.

Given,

Solid cone is recasted into cylinders. Let no. of cylinders formed be n.

∴ Volume of cone = n × Volume of cylinder

$$\begin{aligned}\Rightarrow \frac{1}{3}πR^2H = n × πr^2h\\[1em] \Rightarrow \frac{1}{3}π(3)^2 \times 3 = n × π(1)^2 \times 1\\[1em] \Rightarrow \frac{1}{3} (3)^2 \times 3 = n × 1 \times 1\\[1em] \Rightarrow n = \frac{1}{3} \times 3 \times 3 \times 3\\[1em] \Rightarrow n = 9.\end{aligned}$$

∴ Both A and R are true and R is correct reason for A.

Hence, option 3 is the correct option.

The volume of the remaining wood is the volume of the original solid cylinder minus the volume of the conical cavity drilled out.

Let radius and height of cylinder and conical cavity (both are equal) be r and h units.

Volume of the cylinder : πr²h

Volume of the cone : $\frac{1}{3}$ πr²h

The volume of the remaining wood = Volume of the solid cylinder - Volume of the cone drilled.

= πr²h - $\frac{1}{3}$ πr²h

= $\Big(1 - \frac{1}{3}\Big)$ πr²h

= $\Big(\frac{3 - 1}{3}\Big)$ πr²h

= $\frac{2}{3}$ πr²h.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Given, radius of solid sphere, R = 6 cm

Diameter of new sphere, d = 2 cm

Radius of new sphere, r = $\frac{2}{2}$ cm = 1 cm

Volume of the sphere: $\frac{4}{3}$ πr³

Given,

A solid sphere of radius 6 cm is melted and recast into solid spheres of diameter 2 cm each. Let no. of spheres formed be n.

∴ Volume of sphere = n × Volume of sphere recasted.

⇒ n = $\frac{\text{Volume of sphere melted}}{\text{Volume of each new sphere formed}}$

∴ Statement 2 is true.

$$\begin{aligned}\Rightarrow n = \frac{\frac{4}{3}πR^3}{\frac{4}{3}πr^3}\\[1em] \Rightarrow n = \frac{R^3}{r^3}\\[1em] \Rightarrow n = \frac{6^3}{1^3}\\[1em] \Rightarrow n = \frac{216}{1}\\[1em] \Rightarrow n = 216.\end{aligned}$$

∴ Statement 1 is true.

Thus, both the statements are true.

Hence, option 1 is the correct option.

Given, water in a canal 6 m wide and 2 m deep, is flowing with the speed of 18 km/h.

Volume of water flowing through the canal = Area of cross-section x speed x time

Area of cross-section = width × depth = (6 × 2) m²

Speed = 18 km/h = 18 × $\frac{1000}{3600}$ = $\Big(18 × \frac{5}{18}\Big)$ m/sec

Time = 20 minutes = 20 × 60 sec

∴ Volume = 6 × 2 x 18 × $\frac{5}{18}$ x 20 × 60 m³

Thus, Statement 1 is true, and statement 2 is false.

Hence, option 3 is the correct option.

Given,

Diameter of metallic spheres = 6 cm

Radius of metallic spheres (r) = $\frac{6}{2}$ = 3 cm.

Height of cone (H) = 45 cm

Diameter of cone = 12 cm

Radius of cone (R) = $\frac{12}{2}$ = 6 cm.

Let no. of spheres melted to form a solid metal cone be n.

∴ n × Volume of each sphere = Volume of solid metal cone

$$\begin{aligned}\Rightarrow n \times \frac{4}{3}πr^3 = \frac{1}{3}πR^2H \\[1em] \Rightarrow n = \frac{\frac{1}{3}πR^2H}{\frac{4}{3}πr^3} \\[1em] \Rightarrow n = \frac{πR^2H \times 3}{3 \times4 \times πr^3} \\[1em] \Rightarrow n = \frac{R^2H}{4r^3} \\[1em] \Rightarrow n = \frac{6^2 \times 45}{4 \times 3^3} \\[1em] \Rightarrow n = \frac{36 \times 45}{4 \times 27} \\[1em] \Rightarrow n = \frac{1620}{108} \\[1em] \Rightarrow n = 15.\end{aligned}$$

Hence, 15 solid metallic spheres need to be melted.

Radius of largest sphere from a right circular cylinder will be equal to radius of circular cylinder = 7 cm.

Volume of sphere = $\frac{4}{3}πr^3$

$$\begin{aligned}= \frac{4}{3} \times \frac{22}{7} \times 7^3 \\[1em] = \frac{4}{3} \times 22 \times 7^2 \\[1em] = 1437 \text{ m}^3.\end{aligned}$$

Hence, volume of sphere = 1437 m³.

Given,

Diameter of the cylinder = 12 cm

So, its radius (r) = $\frac{12}{2}$ = 6 cm

Height of the cylinder (h) = 15 cm

Diameter of the cone = 6 cm

So, its radius (R) = $\frac{6}{2}$ = 3 cm

Height of the cone (H) = 12 cm

Radius of the hemisphere = Radius of cone = R = 3 cm

Now,

Let the number of cones be n.

Volume of the cylinder = πr²h

= π × (6)² × 15

= 540π cm².

Volume of an ice-cream cone with ice-cream = Volume of cone + Volume of hemisphere

$$\begin{aligned}= \frac{1}{3}πR^2H + \frac{2}{3} \times π \times R^3 \\[1em] = \frac{1}{3} \times π \times 3^2 \times 12 + \frac{2}{3} \times π \times 3^3 \\[1em] = 36 π + 18 π \\[1em] = 54 π \text{ cm}^3.\end{aligned}$$

Let no. of cones required be n.

Volume of cylinder = n × Volume of each cone

n = $\frac{\text{Vol. of cylinder}}{\text{Vol. of ice-cream cone}} = \frac{540π}{54π}$ = 10.

Hence, the number of cones required = 10.

Given,

Radius of both cone and hemisphere (r) = 8 cm

Height of cone (h) = 8 cm

From figure,

Volume of the solid = Volume of cone + Volume of hemisphere

$$\begin{aligned}= \frac{1}{3}πr^2h + \frac{2}{3} πr^3 \\[1em] = \frac{1}{3} \times π \times 8^2 \times 8 + \frac{2}{3} \times π \times 8^3 \\[1em] = \frac{1}{3}.π.8^3 + \frac{2}{3}.π.8^3 \\[1em] = \Big(\frac{1 + 2}{3}\Big) \times π \times 512 \\[1em] = 512π \text{ cm}^3.\end{aligned}$$

Hence, volume of solid = 512π cm³.

Given,

Diameter of the sphere = 6 cm

So, its radius (R) = $\frac{6}{2}$ = 3 cm

Diameter of cylindrical wire = 0.2 cm

So, the radius of the wire (r) = $\frac{0.2}{2}$ = 0.1 cm

Let length of wire = h

Since, sphere is melted and recasted into a wire.

∴ Volume of sphere = Volume of wire

$$\begin{aligned}\Rightarrow \frac{4}{3}πR^3 = πr^2h \\[1em] \Rightarrow h = \frac{\frac{4}{3}πR^3}{πr^2} \\[1em] \Rightarrow h = \frac{4R^3}{3r^2} \\[1em] \Rightarrow h = \frac{4 \times 3^3}{3 \times (0.1)^2} \\[1em] \Rightarrow h = \frac{4 \times 3^2}{0.01} \\[1em] \Rightarrow h = 4 \times 3^2 \times 100 \\[1em] \Rightarrow h = 3600 \text{ cm} \\[1em] \Rightarrow h = 3600 \times \frac{1}{100} \text{ m} \\[1em] \Rightarrow h = 36 \text{ m}.\end{aligned}$$

Hence, length of the wire = 36 m.

Let edge of the cube = a units.

Then, Volume of the cube = a x a x a = a³

The sphere that exactly fits in the cube will have diameter a units.

Radius of sphere (r) = $\frac{a}{2}$ units.

$$\begin{aligned}\frac{\text{Vol. of cube}}{\text{Vol. of sphere}} = \frac{\text{(side)}^3}{\frac{4}{3}πr^3} \\[1em] = \frac{a^3}{\frac{4}{3}π \times \Big(\frac{a}{2}\Big)^3} \\[1em] = \frac{a^3 \times 3 \times 2^3}{4π \times a^3} \\[1em] = \frac{3 \times 8}{4 \times \frac{22}{7}} \\[1em] = \frac{3 \times 8 \times 7}{4 \times 22} \\[1em] = \frac{21}{11} \\[1em] = 21 : 11.\end{aligned}$$

Hence, ratio of volume of cube to volume of sphere = 21 : 11.

Given,

Diameter of cylindrical portion = 12 cm

Radius of cylindrical portion (r) = $\frac{12}{2}$ = 6 cm

Height of the cylindrical part (H) = 110 cm

Height of the conical part (h) = 9 cm

From figure,

Radius of conical part = Radius of cylindrical portion = r = 6 cm.

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

$$\begin{aligned}= πr^2H + \frac{1}{3}πr^2h \\[1em] = πr^2\Big(H + \frac{1}{3}h\Big) \\[1em] = \frac{355}{113} \times 6^2 \times \Big(110 + \frac{1}{3} \times 9\Big) \\[1em] = \frac{355}{113} \times 36 \times 113 \\[1em] = 12780 \text{ cm}^3.\end{aligned}$$

Given,

Weight of 1 cm³ of iron = 8 gm.

Total weight = 12780 x 8 gm = 102240 gm

= $\frac{102240}{1000}$ kg = 102.24 kg.

Hence, mass of pole = 102.24 kg

Given,

Diameter of cylindrical tank = 2.8 m

So, its radius (R) = $\frac{2.8}{2}$ = 1.4 m

Height (H) = 4.2 m

Volume of water filled in it = πr²h

= $\frac{22}{7}$ x 1.4 x 1.4 x 4.2

= 25.872 m³

= 25.872 × (100)³ cm³

Given,

Diameter of the pipe = 7 cm

Radius (r) = $\frac{7}{2}$ = 3.5 cm

Area of cross section of pipe = πr²

= $\frac{22}{7} \times (3.5)^2$

= 22 × 0.5 × 3.5

= 38.5 cm².

Volume of water discharged per second = 38.5 cm² × 4 m s^-1

= 38.5 × 400 cm s^-1

Let the pipe fill the tank in n seconds.

∴ n × Volume of water discharged per second = Volume of tank

$$\begin{aligned}\Rightarrow n \times 38.5 × 400 = 25.872 × (100)^3 \\[1em] \Rightarrow n = \frac{25.872 \times 100^3}{38.5 \times 400} \\[1em] \Rightarrow n = \frac{25872000}{15400} \\[1em] \Rightarrow n = 1680 \text{ seconds} \\[1em] \Rightarrow n = 1680 \times \frac{1}{60} \text{ minutes} \\[1em] \Rightarrow n = 28\text{ minutes}.\end{aligned}$$

Hence, it takes 28 minutes to fill the tank.

Given,

Rate of water flow = 9 km/hr

= 9 × 10⁵ cm/hr. [As, 1 km = 10⁵ cm.]

1 hour 15 minutes = $1 + \frac{15}{60} = 1 + \frac{1}{4} = \frac{5}{4}$ hours.

Volume of water flowing in $\frac{5}{4}$ hours = Area of cross-section of pipe × Rate of water flow × $\frac{5}{4}$ hours

= 25 cm² × 9 km/hr × $\frac{5}{4}$ hr

= 25 cm² × (9 × 10⁵) cm/hr × $\frac{5}{4}$ hr

= $\frac{1125 \times 10^5}{4}$ cm³.

Let increase in level of water be h cm.

Given,

Length of rectangular cistern = 7.5 m = 750 cm

Breadth of rectangular cistern = 5 m = 500 cm

Volume of water increase in cistern = 750 × 500 × h cm³

We know that,

Volume of water increase in cistern = Volume of water flowing in $\frac{5}{4}$ hours

$$\begin{aligned}\Rightarrow 750 × 500 × h = \frac{1125 \times 10^5}{4} \\[1em] \Rightarrow h = \frac{1125 \times 10^5}{4 \times 750 \times 500} \\[1em] \Rightarrow h = 0.00075 \times 10^5 \\[1em] \Rightarrow h = 75 \text{ cm}.\end{aligned}$$

Hence, there is an increase of 75 cm in level of water in cistern.

(i) Given,

Diameter of cone, cylinder and hemisphere = 10 cm

Radius of cone, cylinder and hemisphere (r) = $\frac{10}{2}$ = 5 cm.

From figure,

Height of cone (h) = 12 cm

By formula,

⇒ l² = r² + h²

⇒ l² = 5² + 12²

⇒ l² = 25 + 144

⇒ l² = 169

⇒ l = $\sqrt{169}$ = 13 cm.

Total surface area = Surface area of cone + Surface area of cylinder + Surface area of hemisphere

= πrl + 2πrh + 2πr²

= πr(l + 2h + 2r)

= $\frac{22}{7} \times 5 \times (13 + 2 \times 12 + 2 \times 5)$

= $\frac{110}{7} \times (13 + 24 + 10)$

= $\frac{110 \times 47}{7}$

= 738.57 cm².

Hence, surface area of figure = 738.57 cm².

(ii) From figure,

Volume of figure = Volume of cone + Volume of cylinder + Volume of hemisphere

$$\begin{aligned}= \frac{1}{3}πr^2h + πr^2h + \frac{2}{3}πr^3 \\[1em] = πr^2\Big(\frac{1}{3}h + h + \frac{2}{3}r\Big) \\[1em] = \frac{22}{7} \times 5^2 \times \Big(\frac{1}{3} \times 12 + 12 + \frac{2}{3} \times 5\Big) \\[1em] = \frac{550}{7} \times \Big(4 + 12 + \frac{10}{3}\Big) \\[1em] = \frac{550}{7} \times \frac{58}{3} \\[1em] = \frac{31900}{21} \\[1em] = 1519.05 \text{ cm}^3.\end{aligned}$$

Hence, volume of figure = 1519.05 cm³.

(iii) By formula,

Density = $\frac{\text{Mass}}{\text{Volume}}$

Given,

Mass = 1.7 kg = 1700 gm

Substituting values we get,

Density = $\frac{1700}{1519.05}$ = 1.12 g/cm³.

Hence, density of material = 1.12 g/cm³.

Given,

Height of cylinder (H) = 6 cm

Radius of cylinder (R) = 3 cm

Height of cone (h) = 4 cm

Radius of cone = Radius of hemisphere = r = 2 cm

Volume of water left in cylinder = Volume of cylinder - Volume of cone - Volume of hemisphere

$$\begin{aligned}= πR^2H - \frac{1}{3}πr^2h - \frac{2}{3}πr^3 \\[1em] = π\Big(R^2H - \frac{1}{3}r^2h - \frac{2}{3}r^3\Big) \\[1em] = \frac{22}{7} \times \Big(3^2 \times 6 - \frac{1}{3} \times 2^2 \times 4 - \frac{2}{3} \times 2^3\Big) \\[1em] = \frac{22}{7} \times \Big(54 - \frac{16}{3} - \frac{16}{3}\Big) \\[1em] = \frac{22}{7} \times \Big(54 - \frac{32}{3}\Big) \\[1em] = \frac{22}{7} \times \Big(\frac{162 - 32}{3}\Big) \\[1em] = \frac{22}{7} \times \frac{130}{3} \\[1em] = \frac{2860}{21} \\[1em] = 136 \text{ cm}^3.\end{aligned}$$

Hence, volume of water left in cylinder = 136 cm³.

Given,

Radius of cylindrical portion = Radius of hemispherical portion = r = 3.5 m.

Height of cylinder (h) = 7 m.

(i) Area of internal surface = Surface area of cylinder + Surface area of hemisphere

= 2πrh + 2πr²

= 2πr(h + r)

= $2 \times \frac{22}{7} \times 3.5 \times (7 + 3.5)$

= 2 × 22 × 0.5 × 10.5

= 231 m².

Hence, the total area of the internal surface = 231 m².

(ii) Internal volume of container = Volume of hemisphere + Volume of cylinder

$$\begin{aligned}= \frac{2}{3}πr^3 + πr^2h \\[1em] = πr^2\Big(\frac{2r}{3} + h\Big) \\[1em] = \frac{22}{7} \times (3.5)^2 \times \Big(\frac{2 \times 3.5}{3} + 7\Big) \\[1em] = 22 \times 0.5 \times 3.5 \times \Big(\frac{7}{3} + 7\Big) \\[1em] = 38.5 \times \frac{28}{3} \\[1em] = 359.33 \text{ m}^3.\end{aligned}$$

Hence, volume of container = 359.33 m³.

Given,

Diameter of base = 168 m

From figure,

Diameter of cylindrical base = Diameter of cone

Radius of cylindrical base = Radius of conical base = r = $\frac{168}{2}$ = 84 m.

Height of cylindrical part (H) = 50 m

Height of conical part (h) = 35 m

By formula,

⇒ l² = r² + h²

⇒ l² = 84² + 35²

⇒ l² = 7056 + 1225

⇒ l² = 8281

⇒ l = $\sqrt{8281}$ = 91 m.

Surface area of exhibition tent = Surface area of cylindrical part + Surface area of conical part

= 2πrH + πrl

$$\begin{aligned}= 2 \times \frac{22}{7} \times 84 \times 50 + \frac{22}{7} \times 84 \times 91 \\[1em] = 26400 + 24024 \\[1em] = 50424 \text{ m}^2.\end{aligned}$$

Area required for folds and stitching = $\frac{20}{100} \times 50424$ = 10084.8 m².

Total area of canvas required = 50424 + 10084.8 = 60508.8 m² ≈ 60509 m² (to the nearest m²).

Hence, area of canvas required = 60509 m².

Given,

Let radius of hemisphere be r cm and height of cylindrical portion be h cm.

Total volume of test tube = Volume of hemisphere + Volume of cylinder

$$\begin{aligned}\Rightarrow \frac{5159}{6} = \frac{2}{3}πr^3 + πr^2h \\[1em] \Rightarrow \frac{5159}{6} = πr^2\Big(\frac{2}{3}r + h\Big) \\[1em] \Rightarrow \frac{5159}{6} = πr^2\Big(\frac{2r + 3h}{3}\Big) \\[1em] \Rightarrow πr^2(2r + 3h) = \frac{5159}{6} \times 3 \\[1em] \Rightarrow πr^2(2r + 3h) = \frac{5159}{2} ..............(1)\end{aligned}$$

Given,

$\frac{4235}{6}$ cm³ of water is required to fill the tube to a level which is 4 cm below the top of the tube.

$$\begin{aligned}\therefore \frac{4235}{6} = \text{Vol. of hemisphere + Vol. of cylinder[upto (h - 4) cm]} \\[1em] \Rightarrow \frac{4235}{6} = \frac{2}{3}πr^3 + πr^2(h - 4) \\[1em] \Rightarrow \frac{4235}{6} = πr^2\Big(\frac{2}{3}r + h - 4\Big) \\[1em] \Rightarrow \frac{4235}{6} = πr^2\Big(\frac{2r + 3h - 12}{3}\Big) \\[1em] \Rightarrow \frac{4235 \times 3}{6} = πr^2(2r + 3h) - 12πr^2 \\[1em] \Rightarrow \frac{4235}{2} = πr^2(2r + 3h) - 12πr^2\end{aligned}$$

Substituting value of πr²(2r + 3h) from equation 1 in above equation.

$$\begin{aligned}\Rightarrow \frac{4235}{2} = \frac{5159}{2} - 12πr^2 \\[1em] \Rightarrow 12πr^2 = \frac{5159}{2} - \frac{4235}{2} \\[1em] \Rightarrow 12πr^2 = \frac{924}{2} \\[1em] \Rightarrow r^2 = \frac{924}{2 \times 12π} \\[1em] \Rightarrow r^2 = \frac{924}{24 \times \frac{22}{7}} \\[1em] \Rightarrow r^2 = \frac{924 \times 7}{24 \times 22} \\[1em] \Rightarrow r^2 = \frac{6468}{528} \\[1em] \Rightarrow r^2 = 12.25 \\[1em] \Rightarrow r = \sqrt{12.25} \\[1em] \Rightarrow r = 3.5 \text{ cm}.\end{aligned}$$

Substituting value of r in equation 1, we get :

$$\begin{aligned}\Rightarrow πr^2(2r + 3h) = \frac{5159}{2} \\[1em] \Rightarrow \frac{22}{7} \times (3.5)^2 \times (2 \times 3.5 + 3h) = \frac{5159}{2} \\[1em] \Rightarrow 22 \times 0.5 \times 3.5 \times (7 + 3h) = \frac{5159}{2} \\[1em] \Rightarrow 38.5 \times (7 + 3h) = \frac{5159}{2} \\[1em] \Rightarrow 7 + 3h = \frac{5159}{2 \times 38.5} \\[1em] \Rightarrow 7 + 3h = \frac{5159}{77} \\[1em] \Rightarrow 7 + 3h = 67 \\[1em] \Rightarrow 3h = 67 - 7 \\[1em] \Rightarrow 3h = 60 \\[1em] \Rightarrow h = \frac{60}{3} \\[1em] \Rightarrow h = 20 \text{ cm}.\end{aligned}$$

Hence, radius of cylindrical part = 3.5 cm and height = 20 cm.

Given,

Diameter of cone = 7 cm

Radius of cone (r) = $\frac{7}{2}$ = 3.5 cm

Height of cone (h) = 8 cm

Radius of hemisphere = radius of cone = r = 3.5 cm

Radius of cylindrical vessel (R) = 7 cm

Height of cylindrical vessel (H) = 15 cm.

Volume of solid = Volume of cone + Volume of hemisphere

$$\begin{aligned}= \frac{1}{3}πr^2h + \frac{2}{3}πr^3 = πr^2\Big(\frac{1}{3}h + \frac{2}{3}r) \\[1em] = \frac{22}{7} \times (3.5)^2 \times \Big(\frac{1}{3} \times 8 + \frac{2}{3} \times 3.5\Big) \\[1em] = 22 \times 0.5 \times 3.5 \times \Big(\frac{8}{3} + \frac{7}{3}\Big) \\[1em] = 38.5 \times \frac{15}{3} \\[1em] = 38.5 \times 5 \\[1em] = 192.5 \text{ cm}^3.\end{aligned}$$

By formula,

Volume of cylindrical vessel = πR²H

$$\begin{aligned}= \frac{22}{7} \times 7^2 \times 15 \\[1em] = 22 \times 7 \times 15 \\[1em] = 2310 \text{ cm}^3.\end{aligned}$$

Volume of water required to fill the vessel completely = Volume of cylindrical vessel - Volume of solid

= 2310 - 192.5

= 2117.5 cm³.

Hence, volume of water required to fill the vessel completely = 2117.5 cm³.

Given,

Radius of cone (r) = 2 cm,

Height of cone (h) = 3 cm,

Radius of sphere (R) = 6 cm.

Let n cones be melted to recast into a solid sphere.

∴ n × Volume of each cone = Volume of sphere

$$\begin{aligned}\Rightarrow n \times \frac{1}{3}πr^2h = \frac{4}{3}πR^3 \\[1em] \Rightarrow n = \frac{\frac{4}{3}πR^3}{\frac{1}{3}πr^2h} \\[1em] \Rightarrow n = \frac{4πR^3 \times 3}{πr^2h \times 3} \\[1em] \Rightarrow n = \frac{4R^3}{r^2h} \\[1em] \Rightarrow n = \frac{4 \times 6^3}{2^2 \times 3} \\[1em] \Rightarrow n = \frac{864}{12} \\[1em] \Rightarrow n = 72.\end{aligned}$$

Hence, 72 cones need to be melted to recast into a solid sphere.

Given,

Each person must have 16 m³ of air to breathe.

∴ 77 persons need 77 × 16 m³ = 1232 m³.

Radius of tent (r) = 7 m.

Let height of conical tent be h meters.

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

$$\begin{aligned}\Rightarrow \frac{1}{3}πr^2h = 1232 \\[1em] \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 7^2 \times h = 1232 \\[1em] \Rightarrow \frac{1}{3} \times 22 \times 7 \times h = 1232 \\[1em] \Rightarrow h = \frac{1232 \times 3}{22 \times 7} \\[1em] \Rightarrow h = \frac{3696}{154} \\[1em] \Rightarrow h = 24 \text{ m}.\end{aligned}$$

By formula,

⇒ l² = r² + h²

⇒ l² = 7² + 24²

⇒ l² = 49 + 576

⇒ l² = 625

⇒ l = $\sqrt{625}$ = 25 m.

Curved surface area of tent = πrl

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

= 22 × 25

= 550 m².

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

When the three portions are separated, each becomes a complete solid. So, we add the entire surface area of the cone, the cylinder and the hemisphere.

For the cone :

Radius = r cm and height = h cm.

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

Total surface area of cone = πrl + πr² = $πr\sqrt{h^2 + r^2}$ + πr².

For the cylinder :

Radius = r cm and height = h cm.

Total surface area of cylinder = 2πrh + 2πr².

For the hemisphere :

Radius = r cm.

Total surface area of hemisphere = 2πr² + πr² = 3πr².

∴ Total surface area of the three parts

$$\begin{aligned}= \left(πr\sqrt{h^2 + r^2} + πr^2\right) + \left(2πrh + 2πr^2\right) + 3πr^2 \\[1em] = πr\sqrt{h^2 + r^2} + 2πrh + 6πr^2 \\[1em] = πr\left(\sqrt{h^2 + r^2} + 2h + 6r\right).\end{aligned}$$

Hence, total surface area of the three parts = πr$\left(\sqrt{h^2 + r^2} + 2h + 6r\right)$ cm².

Let the radius of the base of the cone be R = 20 cm and its height be H cm.

The cone is cut at the middle of its height by a plane parallel to the base. So the cut is at a height $\frac{H}{2}$ from the apex.

The upper part is a small cone similar to the whole cone. Since the height is halved, by similar triangles its radius is also halved.

∴ Radius of small cone (r) = $\frac{R}{2}$ = 10 cm and height = $\frac{H}{2}$.

Volume of small cone (upper part) :

$V_1 = \frac{1}{3}πr^2\left(\frac{H}{2}\right) = \frac{1}{3}π(10)^2 \times \frac{H}{2} = \frac{1}{3}π \times 50H.$

Volume of the whole cone :

$V = \frac{1}{3}πR^2H = \frac{1}{3}π(20)^2H = \frac{1}{3}π \times 400H.$

Volume of lower part (frustum) = Volume of whole conce - Volume of small cone

$V_2 = V - V_1 = \frac{1}{3}π \times 400H - \frac{1}{3}π \times 50H = \frac{1}{3}π \times 350H.$

∴ Ratio of the two parts :

$V_1 : V_2 = \frac{1}{3}π \times 50H : \frac{1}{3}π \times 350H = 50 : 350 = 1 : 7.$

Hence, the ratio between the volumes of the two parts (smaller cone : frustum) is 1 : 7.

(i) F is the apex of the cone FDB. P and O lie on the axis with PF = 1 m = 100 cm and OP = 20 cm.

∴ FO = FP + PO = 100 + 20 = 120 cm.

In △FPE and △FOD :

∠PFE = ∠OFD (common angle)

∠FPE = ∠FOD (each = 90°, both perpendicular to the axis)

∴ △FPE ∼ △FOD (by AA axiom)

We know that,

In similar triangles corresponding sides are proportional.

$$\begin{aligned}\Rightarrow \frac{PE}{OD} = \frac{FP}{FO} \\[1em] \Rightarrow \frac{PE}{30} = \frac{100}{120} \\[1em] \Rightarrow PE = \frac{100}{120} \times 30 = 25 \text{ cm}.\end{aligned}$$

Hence, PE = 25 cm.

(ii) From figure,

For cone FBD :

Radius (R) = 30 cm

Height (H) = 120 cm

For cone FAE :

Radius (r) = 25 cm

Height (h) = 100 cm

Volume of part with cross section ABDE (V) = Volume of cone FBD - Volume of cone FAE

$$\begin{aligned}\Rightarrow V = \frac{1}{3} \times \frac{22}{7} \times 30^2 \times 120 - \frac{1}{3} \times \frac{22}{7} \times 25^2 \times 100 \\[1em] \Rightarrow V = \frac{1}{3} \times \frac{22}{7} \times 900 \times 120 - \frac{1}{3} \times \frac{22}{7} \times 625 \times 100 \\[1em] \Rightarrow V = \frac{1}{3} \times \frac{22}{7} \times \left(108000 - 62500\right) \\[1em] \Rightarrow V = \frac{1}{3} \times \frac{22}{7} \times 45500 \\[1em] \Rightarrow V = \frac{22}{3} \times 6500 \\[1em] \Rightarrow V = \frac{143000}{3} = 47666\frac{2}{3} \approx 47666.67 \text{ cm}^3\end{aligned}$$

Hence, volume of part with cross section ABDE = 47666.7 cm³.

(iii) From figure,

For hemisphere BCD,

Radius (R) = 30 cm

Volume of hemisphere BCD = $\frac{2}{3}πR^3$

$$\begin{aligned}\text{Volume of hemisphere BCD} = \frac{2}{3} \times \frac{22}{7} \times (30)^3 \\[1em] = \frac{2}{3} \times \frac{22}{7} \times 27000 \\[1em] = \frac{22}{7} \times 18000 \\[1em] = \frac{396000}{7} = 56571\frac{3}{7} \approx 56571.43 \text{ cm}^3\end{aligned}$$

Volume of whole concrete mixer = Volume of part with cross section ABDE + Volume of hemisphere BCD

= 47666.7 + 56571.4

= 104238.1 cm³

= 104238.1 × 0.001 liters

= 104.238 liters.

Hence, the volume of the whole concrete mixer ≈ 104 litres.