Surface Areas and Volumes

Given,

Diameter of cone (d) = 10.5 cm

Slant height of cone (l) = 10 cm

By formula,

Radius of cone (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}$ = 5.25 cm

By formula,

Curved surface area of cone (C.S.A.) = πrl

Substituting values we get :

$\text{C.S.A.} = \dfrac{22}{7} × 5.25 × 10 \\[1em] = \dfrac{1155}{7} \\[1em] = 165 \text{ cm}^2$

Hence, curved surface area of the cone = 165 cm².

Given,

Diameter of cone (d) = 24 m

Radius of cone (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{24}{2}$ = 12 m

Slant height of the cone (l) = 21 m

Total surface area of the cone = Curved surface area of cone + area of the base

By formula,

Total surface area of the cone (T.S.A.) = πrl + πr² = πr(l + r)

Susbtituting values we get :

$= \dfrac{22}{7} \times 12 \times (12 + 21) \\[1em] = \dfrac{22}{7} \times 12 \times 33 \\[1em] = \dfrac{8712}{7} \\[1em] = 1244.57 \text{ m}^2$

Hence, total surface area of the cone = 1244.57 m².

(i) Let the radius be r cm.

Given,

Curved surface area of cone = 308 cm²

$\Rightarrow πrl = 308 \\[1em] \Rightarrow r = \dfrac{308}{πl} \\[1em] \Rightarrow r = \dfrac{308}{\dfrac{22}{7} \times 14} \\[1em] \Rightarrow r = \dfrac{308}{22 \times 2} \\[1em] \Rightarrow r = \dfrac{308}{44} \\[1em] \Rightarrow r = 7 \text{ cm}$

Hence, radius of the base of cone = 7 cm.

(ii) By formula,

Total surface area of cone (T.S.A.) = πr(l + r)

Substituting values we get :

T.S.A. = $\dfrac{22}{7}$ × 7 × (7 + 14)

= $\dfrac{22}{7}$ × 7 x 21

= 22 × 21

= 462 cm².

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

(i) Given,

Radius of conical tent (r) = 24 m

Height of conical tent (h) = 10 m

By formula,

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

Substituting values we get :

$l = \sqrt{(24)^2 + (10)^2} \\[1em] = \sqrt{576 + 100} \\[1em] = \sqrt{676} = 26 \text{ m}$

Hence, slant height of the conical tent is 26 m.

(ii) Given,

Curved surface area of the cone = πrl

= $\dfrac{22}{7}$ × 24 × 26

= $\dfrac{13728}{7}$ m²

The cost of the canvas required to make the tent, at ₹ 70 per m² = Curved surface area of the cone x ₹ 70

= $\dfrac{13728}{7}$ × 70

= ₹ 137280.

Hence, the cost of the canvas is ₹ 137280.

Given,

Height of tent (h) = 8 m

Radius of tent (r) = 6 m

By formula,

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

Substituting values we get :

$l = \sqrt{(6)^2 + (8)^2} \\[1em] = \sqrt{36 + 64} \\[1em] = \sqrt{100} = 10 m.$

By formula,

Curved surface area of cone = πrl

= 3.14 × 6 m × 10 m

= 188.4 m².

Area of tarpaulin used to make tent will be equal to curved surface area of tent.

So,

Area of the tarpaulin = Width of the tarpaulin × Length of the tarpaulin

⇒ 188.4 = 3 × length of the tarpaulin

⇒ Length of the tarpaulin = $\dfrac{188.4}{3}$ = 62.8 m

Extra length of the material = 20 cm = $\dfrac{20}{100}$ m = 0.2 m

Total length required = 62.8 m + 0.2 m = 63 m.

Hence, the required length of the tarpaulin is 63 m.

Given,

Diameter of conical tomb (d) = 14 m

Radius of conical tomb (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{14}{2}$ m = 7 m.

Slant height of conical tomb (l) = 25 m

By formula,

Curved surface area of conical tomb = πrl

= $\dfrac{22}{7}$ × 7 × 25

= 550 m².

Given,

Cost of white-washing 100 m² area = ₹ 210

Cost of white-washing 550 m² area = ₹ $210 \times \dfrac{550}{100}$ = ₹ 1155.

Hence, the cost of white-washing the conical tomb is ₹ 1155.

Given,

Radius of conical cap (r) = 7 cm

Height of conical cap (h) = 24 cm

By formula,

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

Substituting values we get :

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

By formula,

Curved surface area of one conical cap = πrl

= $\dfrac{22}{7}$ × 7 × 25

= $\dfrac{3850}{7}$

= 550 cm²

Curved surface area of 10 such conical caps = 10 × 550 cm² = 5500 cm².

Hence, area of sheet required for making 10 caps = 5500 cm².

Given,

Diameter of cone (d) = 40 cm = $\dfrac{40}{100}$ m = 0.4 m

Radius of cone (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{0.4}{2}$ = 0.2 m

Height of cone (h) = 1 m

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

Substituting values we get :

$l = \sqrt{(0.2)^2 + (1)^2} \\[1em] = \sqrt{0.04 + 1} \\[1em] = \sqrt{1.04} = 1.02 \text{ m}.$

By formula,

Curved surface area of each cone = πrl

= 3.14 × 0.2 m × 1.02 m

= 0.64056 m²

Curved surface area of 50 cones = 50 × 0.64056 m² = 32.028 m².

Given,

Cost of painting = ₹ 12 per m².

Cost of painting of 32.028 m² area = ₹ (32.028 × 12)

= ₹ 384.34 (approx.)

Hence, the cost of painting 50 such hollow cones = ₹ 384.34 (approx.)

Given,

Radius of sphere (r) = 10.5 cm

By formula,

Surface area of the sphere (A) = 4πr²

Substituting values we get :

$A = 4 \times \dfrac{22}{7} \times (10.5)^2 \\[1em] = 4 \times \dfrac{22}{7} \times 10.5 \times 10.5 \\[1em] = 4 \times 22 \times 1.5 \times 10.5 \\[1em] = 1386 \text{ cm}^2.$

Hence, surface area of the sphere = 1386 cm².

Given,

Radius of sphere (r) = 5.6 cm

By formula,

Surface area of the sphere (A) = 4πr²

Substituting values we get :

$A = 4 \times \dfrac{22}{7} \times (5.6)^2 \\[1em] = 4 \times \dfrac{22}{7} \times 5.6 \times 5.6 \\[1em] = 4 \times \dfrac{22}{7} \times 31.36 \\[1em] = 4 \times 22 \times 4.48 \\[1em] = 394.24 \text{ cm}^2$

Hence, surface area of sphere = 394.24 cm².

Given,

Radius of sphere (r) = 14 cm

By formula,

Surface area of the sphere (A) = 4πr²

Substituting values we get :

$A = 4 \times \dfrac{22}{7} \times (14)^2 \\[1em] = 4 \times \dfrac{22}{7} \times 14 \times 14 \\[1em] = 4 \times \dfrac{22}{7} \times 196 \\[1em] = 4 \times 22 \times 28 \\[1em] = 2464 \text{ cm}^2.$

Hence, surface area of sphere = 2464 cm².

Given,

Diameter of sphere (d) = 14 cm

Radius of sphere (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{14}{2}$ = 7 cm.

By formula,

Surface area of a sphere (A) = 4πr²

Substituting values we get :

$A = 4 \times \dfrac{22}{7} \times (7)^2 \\[1em] = 4 \times \dfrac{22}{7} \times 7 \times 7 \\[1em] = 4 \times 22 \times 7 \\[1em] = 616 \text{ cm}^2.$

Hence, the surface area of a sphere = 616 cm².

Given,

Diameter of sphere (d) = 21 cm

Radius of sphere (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{21}{2}$ = 10.5 cm

By formula,

Surface area of a sphere (A) = 4πr²

$A = 4 \times \dfrac{22}{7} \times (10.5)^2 \\[1em] = 4 \times \dfrac{22}{7} \times 10.5 \times 10.5 \\[1em] = 4 \times 22 \times 1.5 \times 10.5 \\[1em] = 1386 \text{ cm}^2.$

Hence, the surface area of a sphere = 1386 cm².

Given,

Diameter of sphere (d) = 3.5 m

Radius of sphere (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{3.5}{2}$ = 1.75 m

By formula,

Surface area of a sphere (A) = 4πr²

Substituting values we get :

$A = 4 \times \dfrac{22}{7} \times (1.75)^2 \\[1em] = 4 \times \dfrac{22}{7} \times 1.75 \times 1.75 \\[1em] = 4 \times 22 \times 0.25 \times 1.75 \\[1em] = 38.5 \text{ m}^2.$

Hence, the surface area of a sphere = 38.5 m².

Given,

Radius of the hemisphere (r) = 10 cm

By formula,

Total Surface area of hemisphere (A) = 3πr²

Substituting values we get :

$A = 3 \times 3.14 \times (10)^2 \\[1em] = 3 \times 3.14 \times 100 \\[1em] = 942 \text{ cm}^2.$

Hence, the total surface area of the hemisphere is 942 cm².

Given,

Radius of the balloon before pumping air (r₁) = 7 cm

Radius of the balloon after pumping air (r₂) = 14 cm

Initial surface area = 4πr₁²

Surface area after pumping air into ballon = 4πr₂²

Ratio = $\dfrac{\text{Initial surface area}}{\text{Surface area after pumping air into ballon}}$

$= \dfrac{4πr_1^2}{4πr_2^2} \\[1em] = \dfrac{r_1^2}{r_2^2} \\[1em] = \Big(\dfrac{r_1}{r_2}\Big)^2 \\[1em] = \Big(\dfrac{7}{14}\Big)^2 \\[1em] = \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{1}{4} \\[1em] = 1 : 4.$

Hence, the ratio of the surface area of the balloons = 1 : 4.

Given,

Inner diameter of hemispherical bowl (d) = 10.5 cm

Inner radius of hemispherical bowl (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}$ = 5.25 cm

Curved surface area of hemispherical bowl = 2πr²

$= 2 \times \dfrac{22}{7} \times (5.25)^2 \\[1em] = 2 \times \dfrac{22}{7} \times 27.5625 \\[1em] = 2 \times 22 \times 3.9375 \\[1em] = 173.25 \text{ cm}^2.$

Given,

The cost of tin-plating 100 cm² of the bowl = ₹ 16

∴ The cost of tin-plating 1 cm² of the bowl = ₹ $\dfrac{16}{100}$

∴ The cost of tin-plating 173.25 cm² area of the bowl = ₹ $\dfrac{16}{100} \times 173.25$ = ₹ 27.72

Hence, the cost of tin-plating the hemispherical bowl is ₹ 27.72.

Given,

The surface area of the sphere = 154 cm²

$\therefore 4πr^2 = 154 \\[1em] \Rightarrow r^2 = \dfrac{154}{4π} \\[1em] \Rightarrow r^2 = \dfrac{154}{4 \times \dfrac{22}{7}} \\[1em] \Rightarrow r^2 = \dfrac{154 \times 7}{4 \times 22} \\[1em] \Rightarrow r^2 = \dfrac{49}{4} \\[1em] \Rightarrow r = \sqrt{\dfrac{49}{4}} \\[1em] \Rightarrow r = \dfrac{7}{2} = 3.5 \text{ cm}$

Hence, the radius of the sphere = 3.5 cm.

Given,

⇒ Diameter of the moon (d) = $\dfrac{1}{4}$ × diameter of the earth (D)

⇒ $\dfrac{d}{2} = \dfrac{1}{4} \times D$

⇒ Radius of the moon (r) = $\dfrac{1}{4}$ × radius of the earth (R)

⇒ r = $\dfrac{1}{4}$ × R

⇒ $\dfrac{r}{R} = \dfrac{1}{4}$ .......(1)

Now,

Surface area of earth = 4πR²

Surface area of moon = 4πr²

The ratio of their surface areas = $\dfrac{\text{Surface area of moon}}{\text{Surface area of earth}}$

$= \dfrac{4πr^2}{4πR^2} \\[1em] = \dfrac{r^2}{R^2} \\[1em] = \Big(\dfrac{r}{R}\Big)^2 \\[1em] = \Big(\dfrac{1}{4}\Big)^2 \\[1em] = \dfrac{1}{16}.$

Hence, the ratio of their surface area = 1 : 16.

Given,

The inner radius of the bowl (r) = 5 cm

Thickness of steel = 0.25 cm

Outer radius of the bowl (R) = inner radius of the bowl + thickness of steel = 5 cm + 0.25 cm = 5.25 cm

Outer curved surface area of the hemisphere = 2πR²

= 2 × $\dfrac{22}{7}$ × (5.25)²

= $\dfrac{44}{7} \times 27.5625$

= 44 × 3.9375

= 173.25 cm².

Hence, the outer curved surface area of the bowl is 173.25 cm².

From figure,

Radius of cylinder = radius of sphere = r

Height of cylinder = Diameter of sphere = 2r

(i) By formula,

Surface area of sphere = 4πr²

Hence, surface area of sphere = 4πr².

(ii) By formula,

Curved surface area of cylinder = 2πrh

= 2πr × 2r

= 4πr².

Hence, curved surface area of cylinder = 4πr².

(iii) Ratio = $\dfrac{\text{Surface area of sphere}}{\text{C.S.A. of cylinder}} = \dfrac{4πr^2}{4πr^2} = \dfrac{1}{1}$.

Hence, the ratio between these two surface area is 1 : 1.

Given,

Radius of the cone (r) = 6 cm

Height of the cone (h) = 7 cm

By formula,

Volume of the cone (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times \dfrac{22}{7} \times 6^2 \times 7 \\[1em] = \dfrac{1}{3} \times 22 \times 36 \\[1em] = 22 \times 12 \\[1em] = 264 \text{ cm}^3.$

Hence, volume of cone = 264 cm³.

Given,

Radius of the cone (r) = 3.5 cm

Height of the cone (h) = 12 cm

By formula,

Volume of the cone (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times 12 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 12.25 \times 12 \\[1em] = 22 \times 1.75 \times 4 \\[1em] = 154 \text{ cm}^3.$

Hence, volume of cone = 154 cm³.

Given,

Radius of the conical vessel (r) = 7 cm

Slant height of the conical vessel (l) = 25 cm

Let height of cone = h cm

We know that,

⇒ l² = r² + h²

⇒ h² = l² - r²

⇒ h = $\sqrt{l^2 - r^2}$

Substituting values we get :

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

By formula,

Capacity of the conical vessel (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times \dfrac{22}{7} \times 7^2 \times 24 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 49 \times 24 \\[1em] = 22 \times 7 \times 8 \\[1em] = 1232 \text{ cm}^3.\\[1em] = 1232 \times \Big(\dfrac{1}{1000}\Big) \text{ l} \quad [∵ 1000 \text{ cm}^3 = 1 \text{ litre] }\\[1em] = \text{1.232} \text{ l}.$

Hence, capacity of the conical vessel = 1.232 l.

Given,

Height of the conical vessel (h) = 12 cm

Slant height of the conical vessel (l) = 13 cm

Let radius of the vessel = r cm

We know that,

⇒ l² = r² + h²

⇒ r² = l² - h²

⇒ r = $\sqrt{l^2 - h^2}$

Substituting values we get :

$r = \sqrt{(13)^2 - (12)^2} \\[1em] = \sqrt{169 - 144} \\[1em] = \sqrt{25} = 5 \text{ cm}$

By formula,

Capacity of the conical vessel (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times \dfrac{22}{7} \times 5^2 \times 12 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 300 \\[1em] = \dfrac{6600}{21} \\[1em] = \dfrac{2200}{7} \\[1em] = \dfrac{2200}{7} \times \dfrac{1}{1000} \text{ l} \quad [∵ 1000 \text{ cm}^3 = 1 l] \\[1em] = \dfrac{22}{70} \text{ l} \\[1em] = \dfrac{11}{35} \text{ l}.$

Hence, capacity of the conical vessel = $\dfrac{11}{35}$ l.

Given,

Height of the cone (h) = 15 cm

Let radius of base of cone be r cm.

Volume of cone = 1570 cm³

Substituting values we get :

$\therefore \dfrac{1}{3}πr^2h = 1570 \\[1em] \Rightarrow r^2 = \dfrac{1570 \times 3}{πh} \\[1em] \Rightarrow r^2 = \dfrac{1570 \times 3}{3.14 \times 15} \\[1em] \Rightarrow r^2 = \dfrac{4710}{47.1} \\[1em] \Rightarrow r^2 = 100 \\[1em] \Rightarrow r = \sqrt{100} \\[1em] \Rightarrow r = 10 \text{ cm}.$

Hence, radius of the base = 10 cm.

Given,

Height of the cone (h) = 9 cm

Let radius of base of cone be r cm.

Volume of cone = 48π cm³

Substituting values we get :

$\Rightarrow \dfrac{1}{3}πr^2h = 48π \\[1em] \Rightarrow r^2 = \dfrac{48π \times 3}{πh} \\[1em] \Rightarrow r^2 = 48 \times \dfrac{3}{h} \\[1em] \Rightarrow r^2 = 48 \times \dfrac{3}{9} \\[1em] \Rightarrow r^2 = \dfrac{144}{9} \\[1em] \Rightarrow r^2 = 16 \\[1em] \Rightarrow r = \sqrt{16} \\[1em] \Rightarrow r = 4 \text{ cm}.$

By formula,

Diameter = 2 × radius = 2 × 4 cm = 8 cm.

Hence, the diameter of base of cone is 8 cm.

Given,

Diameter of the conical pit (d) = 3.5 m

Radius of the conical pit (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{3.5}{2}$ = 1.75 m

Depth of the conical pit (h) = 12 m

By formula,

Volume of conical pit (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times \dfrac{22}{7} \times (1.75) ^2 \times 12 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 3.0625 \times 12 \\[1em] = 22 \times 0.4375 \times 4 \\[1em] = 38.5 \text{ m}^3.$

As,

1 m³ = 1000 Litres = 1 kilo litres

38.5 m³ = 38.5 × 1 kilo litres = 38.5 kl.

Hence, the capacity of the conical pit is 38.5 kilo litres.

Given,

Volume of cone = 9856 cm³

Diameter of the cone (d) = 28 cm

Radius of the cone (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{28}{2}$ = 14 cm.

(i) Let height of cone be h cm.

Volume of the cone = 9856 cm³

Substituting values we get :

$\Rightarrow \dfrac{1}{3}πr^2h = 9856 \\[1em] \Rightarrow h = 9856 \times \dfrac{3}{πr^2} \\[1em] \Rightarrow h = 9856 \times \dfrac{3}{\dfrac{22}{7} \times 14^2} \\[1em] \Rightarrow h = 9856 \times \dfrac{3 \times 7}{22 \times 196} \\[1em] \Rightarrow h = \dfrac{206976}{4312} \\[1em] \Rightarrow h = 48 \text{ cm}.$

Hence, the height of the cone = 48 cm.

(ii) Let slant height of the cone be l cm.

By formula,

$\Rightarrow l = \sqrt{r^2 + h^2} \\[1em] \Rightarrow l = \sqrt{(14)^2 + (48)^2} \\[1em] \Rightarrow l = \sqrt{196 + 2304} \\[1em] \Rightarrow l = \sqrt{2500} = 50 \text{ cm}.$

Hence, the slant height of the cone = 50 cm.

(iii) By formula,

Curved surface area of cone = πrl

Substituting values we get :

⇒ Curved surface area of cone = $\dfrac{22}{7}$ × 14 × 50

= 22 x 2 x 50

= 44 x 50

= 2200 cm².

Hence, curved surface area of cone = 2200 cm².

On revolving,

Triangle around side 12 cm it becomes a cone with height 12 cm.

From figure,

Radius of the cone (r) = 5 cm

Height of the cone (h) = 12 cm

Volume of the cone (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times π \times 5^2 \times 12 \\[1em] = \dfrac{1}{3} \times π \times 25 \times 12 \\[1em] = 100π\text{ cm}^3$

Hence, the volume of the cone is 100π cm³.

On revolving,

Triangle around side 5 cm it becomes a cone with height 12 cm.

From figure,

Radius of the cone (r) = 12 cm

Height of the cone (h) = 5 cm

By formula,

Volume of the cone (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times π \times 12^2 \times 5 \\[1em] = \dfrac{1}{3} \times π \times 144 \times 5 \\[1em] = \dfrac{1}{3} \times π \times 720 \\[1em] = 240π \text{ cm}^3.$

Volume of the cone = 240π cm³

Volume of the cone in question 7 = 100π cm³

Ratio = Volume of the cone in question 7 : Volume of the cone in question 8

= 100π : 240π

= 5 : 12

Hence, the volume of the cone is 240π cm³ and the required ratio is 5 : 12.

Given,

Diameter of the conical heap (d) = 10.5 m

Radius of the conical heap (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}$ = 5.25 m

Height of the conical heap (h) = 3 m

By formula,

Volume of the conical heap (V) = $\dfrac{1}{3}πr^2h$

Substituting values we get :

$V = \dfrac{1}{3} \times \dfrac{22}{7} \times (5.25)^2 \times 3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 27.5625 \times 3 \\[1em] = \dfrac{1}{3} \times \dfrac{22}{7} \times 82.6875 \\[1em] = \dfrac{1819.125}{21} \\[1em] = \text{86.625 m}^3$

By formula,

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

Substituting values we get :

$l = \sqrt{(5.25)^2 + (3)^2} \\[1em] = \sqrt{27.5625 + 9} \\[1em] = \sqrt{36.5625} \\[1em] = 6.047 \text{ m}$

The area of the canvas required to cover the heap of wheat (A) = Curved surface area of conical heap = πrl

Substituting values we get :

$A = \dfrac{22}{7} \times 5.25 \times 6.047 \\[1em] = \dfrac{698.4285}{7} \\[1em] = 99.775 \text{ m}^2.$

Hence, the volume of the conical heap is 86.625 m³ and the area of the canvas required is 99.775 m².

Given,

Radius of the sphere (r) = 7 cm

By formula,

Volume of the sphere (V) = $\dfrac{4}{3}πr^3$

Substituting values we get :

$V = \dfrac{4}{3} \times \dfrac{22}{7} \times 7^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 343 \\[1em] = \dfrac{88}{21} \times 343 \\[1em] = \dfrac{88}{3} \times 49 \\[1em] = \dfrac{4312}{3} \\[1em] = 1437 \dfrac{1}{3} \text{ cm}^3.$

Hence, the volume of sphere = $1437 \dfrac{1}{3}$ cm³.

Given,

Radius of the sphere (r) = 0.63 m

By formula,

Volume of the sphere = $\dfrac{4}{3}πr^3$

Substituting values we get :

$= \dfrac{4}{3} \times \dfrac{22}{7} \times (0.63)^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 0.250047 \\[1em] = \dfrac{88}{21} \times 0.250047 \\[1em] = \dfrac{22.004}{21} \\[1em] = \text{1.05 m}^3$

Hence, the volume of sphere = 1.05 m³.

Given,

Diameter of the spherical ball (d) = 28 cm

Radius of the spherical ball (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{28}{2}$ cm = 14 cm.

By formula,

The amount of water displaced = volume of the sphere (V) = $\dfrac{4}{3}πr^3$

Substituting values we get :

$V = \dfrac{4}{3} \times \dfrac{22}{7} \times (14)^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 2744 \\[1em] = \dfrac{88}{21} \times 2744 \\[1em] = \dfrac{241472}{21} \\[1em] = 11498\dfrac{2}{3} \text{ cm}^3.$

Hence, the amount of water displaced = $11498\dfrac{2}{3}$ cm³.

Given,

Diameter of the spherical ball (d) = 0.21 m

Radius of the spherical ball (r) = Radius = $\dfrac{\text{Diameter}}{2} = \dfrac{0.21}{2}$ m = 0.105 m

By formula,

The amount of water displaced = volume of a sphere (V) = $\dfrac{4}{3}πr^3$

Substituting values we get :

$V = \dfrac{4}{3} \times \dfrac{22}{7} \times (0.105)^3 \\[1em] = \dfrac{88}{21} \times 0.001157625 \\[1em] = \dfrac{88}{21} \times 0.001157625 \\[1em] = \dfrac{0.101871}{21} \\[1em] = 0.004851\text{ m}^3.$

Hence, the amount of water displaced = 0.004851 m³.

Given,

The diameter of the metallic ball (d) = 4.2 cm

Radius of the metallic ball (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{4.2}{2}$ cm = 2.1 cm

The density of the metal = 8.9 g per cm³

We know that,

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

Mass = Density × Volume ...........(1)

By formula,

Volume of the metallic ball (V) = $\dfrac{4}{3}πr^3$

Substituting values we get :

$V = \dfrac{4}{3} \times \dfrac{22}{7} \times (2.1)^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 9.261 \\[1em] = \dfrac{88}{21} \times 9.261 \\[1em] = \dfrac{814.968}{21} \\[1em] = \text{38.808 cm}^3$

Substituting value in equation (1), we get :

Mass of the metallic ball = 8.9 × 38.808

= 345.39 g

Hence, the mass of the ball is 345.39 g.

Given,

⇒ Diameter of the moon (d) = $\dfrac{1}{4}$ × diameter of the earth (D)

⇒ $\dfrac{\text{d}}{2} = \dfrac{1}{4} \times \dfrac{D}{2}$

⇒ The radius of the moon = $\dfrac{1}{4}$ × radius of the earth

Let radius of moon be r and radius of earth be R.

⇒ r = $\dfrac{1}{4}\times R$

⇒ r = $\dfrac{R}{4}$ .....(1)

Since, earth and moon are spherical.

Volume of the earth (V) = $\dfrac{4}{3}πR^3$

Volume of the moon (v) = $\dfrac{4}{3}πr^3$

Substituting value of r from equation (1) in above equation, we get :

$\Rightarrow v = \dfrac{4}{3}\times π \times \Big(\dfrac{R}{4}\Big)^3 (\text{From equation (1)}) \\[1em] \Rightarrow v = \dfrac{4}{3} \times π \times \dfrac{R}{4} \times \dfrac{R}{4} \times \dfrac{R}{4} \\[1em] \Rightarrow v = \dfrac{1}{64} \times \dfrac{4}{3}π R^3 \\[1em] \Rightarrow v = \dfrac{1}{64} \times V.$

Hence, the volume of the moon is $\dfrac{1}{64}$ times the volume of the earth.

Given,

Diameter of the hemispherical ball (d) = 10.5 cm

Radius of the hemispherical ball (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{10.5}{2}$ cm = 5.25 cm

By formula,

Volume of the hemispherical ball (V) = $\dfrac{2}{3}πr^3$

Substituting values we get :

$V = \dfrac{2}{3} \times \dfrac{22}{7} \times (5.25)^3 \\[1em] = \dfrac{44}{21} \times 144.703 \\[1em] = \dfrac{6366.9375}{21} \\[1em] = 303.1875 \\[1em] = \dfrac{303.1875}{1000} (∵ 1000 \text{ cm}^3 = 1 \text{ L}) \\[1em] = 0.3031875 \\[1em] \approx 0.303 \text{ l}$

Hence, the hemispherical bowl can hold 0.303 litres of milk.

Given,

Inner radius of the tank (r) = 1 m

Thickness of iron = 1 cm = $\dfrac{1}{100}$ m = 0.01 m

Outer radius of the tank (R) = Inner radius + Thickness = 1 m + 0.01 m = 1.01 m

Volume of iron used (V) = Volume of tank with outer radius - Volume of tank with inner radius

$V = \dfrac{2}{3}πR^3 - \dfrac{2}{3}πr^3 \\[1em] = \dfrac{2}{3}π(R^3 - r^3) \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times [(1.01)^3 - (1)^3] \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times [1.030301 - 1 ] \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 0.030301 \\[1em] = \dfrac{44}{21} \times 0.030301 \\[1em] = \dfrac{1.333244}{21} \\[1em] = \text{0.06348 m}^3.$

Hence, volume of iron used to make the tank = 0.06348 m³.

Let the radius of the sphere be r.

By formula,

Surface area of the sphere = 4πr²

Substituting values we get :

$\Rightarrow 4πr^2 = 154 \\[1em] \Rightarrow r^2 = \dfrac{154}{4π} \\[1em] \Rightarrow r^2 = \dfrac{154}{4 \times \dfrac{22}{7}} \\[1em] \Rightarrow r^2 = \dfrac{154 \times 7}{4 \times 22} \\[1em] \Rightarrow r^2 = \dfrac{49}{4} \\[1em] \Rightarrow r = \sqrt{\dfrac{49}{4}} \\[1em] \Rightarrow r = \dfrac{7}{2} \text{ cm}.$

By formula,

Volume of a sphere (V) = $\dfrac{4}{3}πr^3$

$V = \dfrac{4}{3} \times \dfrac{22}{7} \times \Big(\dfrac{7}{2}\Big)^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times \dfrac{343}{8} \\[1em] = \dfrac{11}{3} \times 49 \\[1em] = \dfrac{539}{3} \\[1em] = 179 \dfrac{2}{3} \text{ cm}^3$

Hence, volume of the sphere is $179 \dfrac{2}{3} \text{ cm}^3$.

(i) Given,

Inside of dome was white-washed at the cost of ₹ 4989.60.

Cost of white washing = ₹ 20 per square metre.

Inside surface area of the dome

= $\dfrac{\text{Total cost for white-washing the dome inside}}{\text{Rate of white-washing}}$

⇒ $\dfrac{4989.60}{20}$ = 249.48 m²

Hence, inner surface area of the dome = 249.48 m².

(ii) Let 'r' be the radius of a hemispherical dome.

Inner surface area of the hemispherical dome = 2πr²

Substituting values we get :

$\Rightarrow 2πr^2 = 249.48 \\[1em] \Rightarrow r^2 = \dfrac{249.48}{2π} \\[1em] \Rightarrow r^2 = \dfrac{249.48}{2 \times \dfrac{22}{7}} \\[1em] \Rightarrow r^2 = \dfrac{249.48 \times 7}{44} \\[1em] \Rightarrow r^2 = 5.67 \times 7\\[1em] \Rightarrow r^2 = 39.69 \\[1em] \Rightarrow r = \sqrt{39.69} \\[1em] \Rightarrow r = 6.3 \text{ m}$

The volume of the air inside the dome will be the same as the volume of the hemisphere.

Now the volume of the air inside the dome (v) = $\dfrac{2}{3}πr^3$

$v = \dfrac{2}{3} \times \dfrac{22}{7} \times 6.3 \times 6.3 \times 6.3 \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 6.3 \times 39.69 \\[1em] = \dfrac{2}{3} \times \dfrac{22}{7} \times 250.047 \\[1em] = \dfrac{44}{21} \times 250.047 \\[1em] = \dfrac{11002.068}{21} \\[1em] = 523.9 \text{ m}^3$

Hence, the volume of the air inside the dome is 523.9 m³.

By formula,

The volume of a sphere = $\dfrac{4}{3}πr^3$

The volume of 27 solid spheres with radius r = $27 \times \dfrac{4}{3}πr^3 = 36πr^3$ ......(1)

The volume of the new sphere with radius r' = $\dfrac{4}{3}π(r')^3$ .......(2)

(i) Volume of the new sphere = Volume of 27 solid spheres

$\Rightarrow \dfrac{4}{3}π(r')^3 = 36πr^3 (\text{From equation (1) and (2)}) \\[1em] \Rightarrow (r')^3 = 36r^3 \times \dfrac{3}{4} \\[1em] \Rightarrow (r')^3 = 9r^3 \times 3 \\[1em] \Rightarrow (r')^3 = 27r^3 \\[1em] \Rightarrow r' = \sqrt[3]{27r^3} \\[1em] \Rightarrow r' = 3r$

Hence, radius of the new sphere r' = 3r.

(ii) By formula,

Surface area of each iron sphere (S) = 4πr²

Surface area of the new sphere (S') = $4π(r')^2 = 4π(3r)^2 = 36πr^2$

Required ratio = $\dfrac{4πr^2}{36πr^2}$

The ratio of the S and S’ = $\dfrac{1}{9}$

Hence, the ratio of S and S’ is 1 : 9.

Given,

Diameter of the spherical capsule (d) = 3.5 mm

Radius of the spherical capsule (r) = $\dfrac{\text{Diameter}}{2} = \dfrac{3.5}{2}$ mm = 1.75 mm

Medicine needed to fill the capsule = Volume of the spherical capsule

∴ Medicine needed to fill the capsule (v) = $\dfrac{4}{3}πr^3$

$v = \dfrac{4}{3} \times \dfrac{22}{7} \times (1.75)^3 \\[1em] = \dfrac{4}{3} \times \dfrac{22}{7} \times 5.359375 \\[1em] = \dfrac{88}{21} \times 5.359375 \\[1em] = \dfrac{471.625}{21} \\[1em] = 22.46 \text{ mm}^3$

Hence, the medicine needed to fill the capsule = 22.46 mm³.