Chapter 17 from Textbook Rs Aggarwal 2021 2022 for Class 10 MATHS

Question 1:

Answer:

$As, volume of a cube = 27 {cm}^{3} \Rightarrow {(edge)}^{3} = 27 \Rightarrow edge = \sqrt[3]{27} \Rightarrow edge = 3 cm The length of the resulting cuboid, l = 3 + 3 = 6 cm, its breadth, b = 3 cm and its height, h = 3 cm Now, the surface area of the resulting cuboid = 2 (l b + b h + h l) = 2 (6 \times 3 + 3 \times 3 + 3 \times 6) = 2 (18 + 9 + 18) = 2 \times 45 = 90 {cm}^{2}$

Question 2:

$As, volume of a cube = 27 {cm}^{3} \Rightarrow {(edge)}^{3} = 27 \Rightarrow edge = \sqrt[3]{27} \Rightarrow edge = 3 cm The length of the resulting cuboid, l = 3 + 3 = 6 cm, its breadth, b = 3 cm and its height, h = 3 cm Now, the surface area of the resulting cuboid = 2 (l b + b h + h l) = 2 (6 \times 3 + 3 \times 3 + 3 \times 6) = 2 (18 + 9 + 18) = 2 \times 45 = 90 {cm}^{2}$

Answer:

$As, volume of hemisphere = 2425 \frac{1}{2} {cm}^{3} \Rightarrow \frac{2}{3} π r^{3} = 2425 \frac{1}{2} \Rightarrow \frac{2}{3} \times \frac{22}{7} \times r^{3} = \frac{4851}{2} \Rightarrow r^{3} = \frac{4851 \times 3 \times 7}{2 \times 2 \times 22} \Rightarrow r^{3} = \frac{441 \times 3 \times 7}{2 \times 2 \times 2} \Rightarrow r^{3} = \frac{21^{3}}{2^{3}} \Rightarrow r = \frac{21}{2} cm$

$So, the curved surface area of the hemisphere = 2 π r^{2} = 2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 693 {cm}^{2}$

Question 3:

$As, volume of hemisphere = 2425 \frac{1}{2} {cm}^{3} \Rightarrow \frac{2}{3} π r^{3} = 2425 \frac{1}{2} \Rightarrow \frac{2}{3} \times \frac{22}{7} \times r^{3} = \frac{4851}{2} \Rightarrow r^{3} = \frac{4851 \times 3 \times 7}{2 \times 2 \times 22} \Rightarrow r^{3} = \frac{441 \times 3 \times 7}{2 \times 2 \times 2} \Rightarrow r^{3} = \frac{21^{3}}{2^{3}} \Rightarrow r = \frac{21}{2} cm$

$So, the curved surface area of the hemisphere = 2 π r^{2} = 2 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 693 {cm}^{2}$

Answer:

$As, the total surface area of the solid hemisphere = 462 {cm}^{2} \Rightarrow 3 π r^{2} = 462 \Rightarrow 3 \times \frac{22}{7} \times r^{2} = 462 \Rightarrow r^{2} = \frac{462 \times 7}{3 \times 22} \Rightarrow r^{2} = 49 \Rightarrow r^{2} = \sqrt{49} \Rightarrow r = 7 cm$

$Now, the volume of the solid hemisphere = \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 = \frac{2156}{3} {cm}^{3} = 718 \frac{2}{3} {cm}^{3}$
= 718.67 cm³

Question 4:

$As, the total surface area of the solid hemisphere = 462 {cm}^{2} \Rightarrow 3 π r^{2} = 462 \Rightarrow 3 \times \frac{22}{7} \times r^{2} = 462 \Rightarrow r^{2} = \frac{462 \times 7}{3 \times 22} \Rightarrow r^{2} = 49 \Rightarrow r^{2} = \sqrt{49} \Rightarrow r = 7 cm$

$Now, the volume of the solid hemisphere = \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 = \frac{2156}{3} {cm}^{3} = 718 \frac{2}{3} {cm}^{3}$
= 718.67 cm³

Answer:

(i)
$We have, the height of the cone, h = 24 m, the base diameter of the cone, d = 14 m Also, the base radius of the cone, r = \frac{d}{2} = \frac{14}{2} = 7 m The slant height of the cone, l = \sqrt{h^{2} + r^{2}} = \sqrt{24^{2} + 7^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25 m The curved surface area of the tent = π r l = \frac{22}{7} \times 7 \times 25 = 550 m^{2} \Rightarrow The area of cloth required to make the tent = 550 m^{2} \Rightarrow The length of the cloth = \frac{550}{5} = 110 m So, the cost of cloth used = 110 \times 25 = ₹ 2750$

(ii)
The ratio of radius and height of a solid right-circular cone is 5:12.
Let radius, r = 5x and height, h=12x.
Volume = 314 cm³.
$\Rightarrow \frac{1}{3} π r^{2} h = 314 \Rightarrow \frac{1}{3} \times 3.14 \times {(5 x)}^{2} \times (12 x) = 314 \Rightarrow x^{3} = \frac{314 \times 3}{3.14 \times 5 \times 5 \times 12} \Rightarrow x^{3} = 1 \Rightarrow x = 1$
So, radius r = 5 cm and height h = 12 cm.
Using Pythagoras Theorem, slant height is given by $l = \sqrt{r^{2} + h^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 cm$
Total Surface Area of Cone $= π r (r + l) = 3.14 \times 5 \times (5 + 13) = 3.14 \times 5 \times 18 = 282.6 {cm}^{2}$ .

Question 5:

(i)
$We have, the height of the cone, h = 24 m, the base diameter of the cone, d = 14 m Also, the base radius of the cone, r = \frac{d}{2} = \frac{14}{2} = 7 m The slant height of the cone, l = \sqrt{h^{2} + r^{2}} = \sqrt{24^{2} + 7^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25 m The curved surface area of the tent = π r l = \frac{22}{7} \times 7 \times 25 = 550 m^{2} \Rightarrow The area of cloth required to make the tent = 550 m^{2} \Rightarrow The length of the cloth = \frac{550}{5} = 110 m So, the cost of cloth used = 110 \times 25 = ₹ 2750$

(ii)
The ratio of radius and height of a solid right-circular cone is 5:12.
Let radius, r = 5x and height, h=12x.
Volume = 314 cm³.
$\Rightarrow \frac{1}{3} π r^{2} h = 314 \Rightarrow \frac{1}{3} \times 3.14 \times {(5 x)}^{2} \times (12 x) = 314 \Rightarrow x^{3} = \frac{314 \times 3}{3.14 \times 5 \times 5 \times 12} \Rightarrow x^{3} = 1 \Rightarrow x = 1$
So, radius r = 5 cm and height h = 12 cm.
Using Pythagoras Theorem, slant height is given by $l = \sqrt{r^{2} + h^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 cm$
Total Surface Area of Cone $= π r (r + l) = 3.14 \times 5 \times (5 + 13) = 3.14 \times 5 \times 18 = 282.6 {cm}^{2}$ .

Answer:

$Let r and R be the base radii, h and H be the heights, v and V be the volumes of the two given cones . We have, \frac{2 r}{2 R} = \frac{4}{5} o r \frac{r}{R} = \frac{4}{5} . . . . . (i) and \frac{v}{V} = \frac{1}{4} \Rightarrow \frac{(\frac{1}{3} π r^{2} h)}{(\frac{1}{3} π R^{2} H)} = \frac{1}{4} \Rightarrow \frac{r^{2} h}{R^{2} H} = \frac{1}{4} \Rightarrow {(\frac{r}{R})}^{2} \times \frac{h}{H} = \frac{1}{4} \Rightarrow {(\frac{4}{5})}^{2} \times \frac{h}{H} = \frac{1}{4} [Using (i)] \Rightarrow (\frac{16}{25}) \times \frac{h}{H} = \frac{1}{4} \Rightarrow \frac{h}{H} = \frac{1 \times 25}{4 \times 16} \Rightarrow \frac{h}{H} = \frac{25}{64} ∴ h : H = 25 : 64$

So, the ratio of their heights is 25:64.

Question 6:

$Let r and R be the base radii, h and H be the heights, v and V be the volumes of the two given cones . We have, \frac{2 r}{2 R} = \frac{4}{5} o r \frac{r}{R} = \frac{4}{5} . . . . . (i) and \frac{v}{V} = \frac{1}{4} \Rightarrow \frac{(\frac{1}{3} π r^{2} h)}{(\frac{1}{3} π R^{2} H)} = \frac{1}{4} \Rightarrow \frac{r^{2} h}{R^{2} H} = \frac{1}{4} \Rightarrow {(\frac{r}{R})}^{2} \times \frac{h}{H} = \frac{1}{4} \Rightarrow {(\frac{4}{5})}^{2} \times \frac{h}{H} = \frac{1}{4} [Using (i)] \Rightarrow (\frac{16}{25}) \times \frac{h}{H} = \frac{1}{4} \Rightarrow \frac{h}{H} = \frac{1 \times 25}{4 \times 16} \Rightarrow \frac{h}{H} = \frac{25}{64} ∴ h : H = 25 : 64$

So, the ratio of their heights is 25:64.

Answer:

Let r, h and l be the base radius, the height and the slant height of the conical mountain, respectively.

$As, the area of the base = 1.54 {km}^{2} \Rightarrow π r^{2} = 1.54 \Rightarrow \frac{22}{7} \times r^{2} = 1.54 \Rightarrow r^{2} = \frac{1.54 \times 7}{22} \Rightarrow r^{2} = 0.49 \Rightarrow r = \sqrt{0.49} \Rightarrow r = 0.7 km$

$Now, h = \sqrt{l^{2} - r^{2}} = \sqrt{2 . 5^{2} - 0 . 7^{2}} = \sqrt{6.25 - 0.49} = \sqrt{5.76} = 2.4 km$

So, the height of the mountain is 2.4 km.

Question 7:

Let r, h and l be the base radius, the height and the slant height of the conical mountain, respectively.

$As, the area of the base = 1.54 {km}^{2} \Rightarrow π r^{2} = 1.54 \Rightarrow \frac{22}{7} \times r^{2} = 1.54 \Rightarrow r^{2} = \frac{1.54 \times 7}{22} \Rightarrow r^{2} = 0.49 \Rightarrow r = \sqrt{0.49} \Rightarrow r = 0.7 km$

$Now, h = \sqrt{l^{2} - r^{2}} = \sqrt{2 . 5^{2} - 0 . 7^{2}} = \sqrt{6.25 - 0.49} = \sqrt{5.76} = 2.4 km$

So, the height of the mountain is 2.4 km.

Answer:

Let r and h be the base radius and the height of the solid cylinder, respectively.

$We have, (r + h) = 37 m As, the total surface area of the cylinder = 1628 m^{2} \Rightarrow 2 π r (r + h) = 1628 \Rightarrow 2 \times \frac{22}{7} \times r \times 37 = 1628 \Rightarrow r = \frac{1628 \times 7}{2 \times 22 \times 37} \Rightarrow r = 7 m So, h = (37 - 7) = 30 m$

$Now, the volume of the solid cylinder = π r^{2} h = \frac{22}{7} \times 7 \times 7 \times 30 = 4620 m^{3}$

Question 8:

Let r and h be the base radius and the height of the solid cylinder, respectively.

$We have, (r + h) = 37 m As, the total surface area of the cylinder = 1628 m^{2} \Rightarrow 2 π r (r + h) = 1628 \Rightarrow 2 \times \frac{22}{7} \times r \times 37 = 1628 \Rightarrow r = \frac{1628 \times 7}{2 \times 22 \times 37} \Rightarrow r = 7 m So, h = (37 - 7) = 30 m$

$Now, the volume of the solid cylinder = π r^{2} h = \frac{22}{7} \times 7 \times 7 \times 30 = 4620 m^{3}$

Answer:

Let the radii of the given sphere and the new sphere be r and R, respectively.

$We have, R = 2 r and the surface area of the given sphere = 2464 {cm}^{2} i . e . 4 π r^{2} = 2464 {cm}^{2} The surface area of the new sphere = 4 π R^{2} = 4 π {(2 r)}^{2} = 4 π (4 r^{2}) = 4 (4 π r^{2}) = 4 \times 2464 = 9856 {cm}^{2}$

So, the surface area of the new sphere is 9856 cm².

Question 9:

Let the radii of the given sphere and the new sphere be r and R, respectively.

$We have, R = 2 r and the surface area of the given sphere = 2464 {cm}^{2} i . e . 4 π r^{2} = 2464 {cm}^{2} The surface area of the new sphere = 4 π R^{2} = 4 π {(2 r)}^{2} = 4 π (4 r^{2}) = 4 (4 π r^{2}) = 4 \times 2464 = 9856 {cm}^{2}$

So, the surface area of the new sphere is 9856 cm².

Answer:

We have,
the radii of bases of the cone and cylinder, r = 15 m,
the height of the cylinder, h = 5.5 m,
the height of the tent = 8.25 m
Also, the height of the cone, H = $8.25 - 5.5 = 2.75 m$

$The slant height of the cone, l = \sqrt{r^{2} + H^{2}} = \sqrt{15^{2} + 2 . 75^{2}} = \sqrt{225 + 7.5625} = \sqrt{232.5625} = 15.25 m$

$The area of the canvas used in making the tent = CSA of cylinder + CSA of cone = 2 π r h + π r l = π r (2 h + l) = \frac{22}{7} \times 15 (2 \times 5.5 + 15.25) = \frac{22}{7} \times 15 (11 + 15.25) = \frac{22}{7} \times 15 \times 26.25 = 1237.5 m^{2}$

As, the width of the canvas = 1.5 m

$So, the length of the canvas = \frac{237.5}{1.5} = 825 m$

Hence, the length of the tent used for making the tent is 825 m.

Question 10:

We have,
the radii of bases of the cone and cylinder, r = 15 m,
the height of the cylinder, h = 5.5 m,
the height of the tent = 8.25 m
Also, the height of the cone, H = $8.25 - 5.5 = 2.75 m$

$The slant height of the cone, l = \sqrt{r^{2} + H^{2}} = \sqrt{15^{2} + 2 . 75^{2}} = \sqrt{225 + 7.5625} = \sqrt{232.5625} = 15.25 m$

$The area of the canvas used in making the tent = CSA of cylinder + CSA of cone = 2 π r h + π r l = π r (2 h + l) = \frac{22}{7} \times 15 (2 \times 5.5 + 15.25) = \frac{22}{7} \times 15 (11 + 15.25) = \frac{22}{7} \times 15 \times 26.25 = 1237.5 m^{2}$

As, the width of the canvas = 1.5 m

$So, the length of the canvas = \frac{237.5}{1.5} = 825 m$

Hence, the length of the tent used for making the tent is 825 m.

Answer:

Radius of the cylinder = 14 m
Radius of the base of the cone = 14 m
Height of the cylinder (h) = 3 m
Total height of the tent = 13.5 m
Surface area of the cylinder $= 2 π r h = (2 \times \frac{22}{7} \times 14 \times 3) m^{2} = 264 m^{2}$

Height of the cone $= Total height - Height of cone = (13.5 - 3) m = 10.5 m$

Surface area of the cone = $π r \sqrt{r^{2} + h^{2}}$

$= π r \sqrt{r^{2} + h^{2}} = (\frac{22}{7} \times 14 \times \sqrt{14^{2} + 10 . 5^{2}}) m^{2} = (44 \times \sqrt{196 + 110.25}) m^{2} = (44 \times \sqrt{306.25}) m^{2} = (44 \times 17.5) m^{2} = 770 m^{2}$

Total surface area $= (264 + 770) m^{2} = 1034 m^{2}$

∴ Cost of cloth $= Rs 1034 \times 80 = Rs 82720$

Question 11:

Radius of the cylinder = 14 m
Radius of the base of the cone = 14 m
Height of the cylinder (h) = 3 m
Total height of the tent = 13.5 m
Surface area of the cylinder $= 2 π r h = (2 \times \frac{22}{7} \times 14 \times 3) m^{2} = 264 m^{2}$

Height of the cone $= Total height - Height of cone = (13.5 - 3) m = 10.5 m$

Surface area of the cone = $π r \sqrt{r^{2} + h^{2}}$

$= π r \sqrt{r^{2} + h^{2}} = (\frac{22}{7} \times 14 \times \sqrt{14^{2} + 10 . 5^{2}}) m^{2} = (44 \times \sqrt{196 + 110.25}) m^{2} = (44 \times \sqrt{306.25}) m^{2} = (44 \times 17.5) m^{2} = 770 m^{2}$

Total surface area $= (264 + 770) m^{2} = 1034 m^{2}$

∴ Cost of cloth $= Rs 1034 \times 80 = Rs 82720$

Answer:

Radius of the cylinder = 52.5 m
Radius of the base of the cone = 52.5 m
Slant height (l) of the cone = 53 m
Height of the cylinder (h) = 3 m
Curved surface area of the cylindrical portion $= 2 π r h = (2 \times \frac{22}{7} \times 52.5 \times 3) m^{2} = 990 m^{2}$
Curved surface area of the conical portion $= π r l = (\frac{22}{7} \times 52.5 \times 53) m^{2} = 8745 m^{2}$

Thus, the total area of canvas required for making the tent $= (8745 + 990) m^{2} = 9735 m^{2}$

Question 12:

Radius of the cylinder = 52.5 m
Radius of the base of the cone = 52.5 m
Slant height (l) of the cone = 53 m
Height of the cylinder (h) = 3 m
Curved surface area of the cylindrical portion $= 2 π r h = (2 \times \frac{22}{7} \times 52.5 \times 3) m^{2} = 990 m^{2}$
Curved surface area of the conical portion $= π r l = (\frac{22}{7} \times 52.5 \times 53) m^{2} = 8745 m^{2}$

Thus, the total area of canvas required for making the tent $= (8745 + 990) m^{2} = 9735 m^{2}$

Answer:

Radius of the cylinder = 2.5 m
Height of the cylinder = 21 m

Curved surface area of the cylinder $= 2 πrh = 2 \times \frac{22}{7} \times 2.5 \times 21 = 330 m^{2}$
Radius of the cone = 2.5 m
Slant height of the cone = 8 m
Curved surface area of the cone $= πrl = \frac{22}{7} \times 2.5 \times 8 = 62.86 m^{2}$

Area of circular base $= {πr}^{2} = \frac{22}{7} \times 2.5 \times 2.5 = 19.643 m^{2}$
∴ Total surface area of rocket $= 330 + 62.86 + 19.643 = 412.503 m^{2}$

Question 13:

Radius of the cylinder = 2.5 m
Height of the cylinder = 21 m

Curved surface area of the cylinder $= 2 πrh = 2 \times \frac{22}{7} \times 2.5 \times 21 = 330 m^{2}$
Radius of the cone = 2.5 m
Slant height of the cone = 8 m
Curved surface area of the cone $= πrl = \frac{22}{7} \times 2.5 \times 8 = 62.86 m^{2}$

Area of circular base $= {πr}^{2} = \frac{22}{7} \times 2.5 \times 2.5 = 19.643 m^{2}$
∴ Total surface area of rocket $= 330 + 62.86 + 19.643 = 412.503 m^{2}$

Answer:

$We have, Radius of cone = Radius of hemisphere = r = 3.5 cm or AD = BD = CD = 3.5 cm, Total height of the solid, OC = 9.5 cm \Rightarrow OD + CD = 9.5 \Rightarrow OD + 3.5 = 9.5 \Rightarrow OD = 6 cm \Rightarrow Height of cone, h = 6 cm$

$Now, Volume of solid = Volume of cone + Volume of hemisphere = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} (h + 2 r) = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times (6 + 2 \times 3.5) = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times (6 + 7) = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 13 = \frac{500.5}{3} \approx 166.83 {cm}^{3}$

So, the volume of the solid is 166.83 cm³.

Question 14:

$We have, Radius of cone = Radius of hemisphere = r = 3.5 cm or AD = BD = CD = 3.5 cm, Total height of the solid, OC = 9.5 cm \Rightarrow OD + CD = 9.5 \Rightarrow OD + 3.5 = 9.5 \Rightarrow OD = 6 cm \Rightarrow Height of cone, h = 6 cm$

$Now, Volume of solid = Volume of cone + Volume of hemisphere = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} (h + 2 r) = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times (6 + 2 \times 3.5) = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times (6 + 7) = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 13 = \frac{500.5}{3} \approx 166.83 {cm}^{3}$

So, the volume of the solid is 166.83 cm³.

Answer:

The radius of hemisphere as well as that of base of cone is r = 3.5 cm.
The height of toy = 15.5 cm.
The height of hemisphere, h = 3.5 cm and height of cone = 15.5 $-$ 3.5 = 12 cm.
Using Pythagoras Theorem, slant height of cone is $l = \sqrt{{(3.5)}^{2} + {(12)}^{2}} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 cm$ .
Therefore, the Total surface area of toy = Curved surface of cone + Curved surface of hemisphere
$= π r l + 2 π r^{2} = \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times {(3.5)}^{2} = \frac{275}{2} + 77 = \frac{275 + 154}{2} = \frac{429}{2} = 214.5 {cm}^{2} .$

Question 15:

The radius of hemisphere as well as that of base of cone is r = 3.5 cm.
The height of toy = 15.5 cm.
The height of hemisphere, h = 3.5 cm and height of cone = 15.5 $-$ 3.5 = 12 cm.
Using Pythagoras Theorem, slant height of cone is $l = \sqrt{{(3.5)}^{2} + {(12)}^{2}} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5 cm$ .
Therefore, the Total surface area of toy = Curved surface of cone + Curved surface of hemisphere
$= π r l + 2 π r^{2} = \frac{22}{7} \times 3.5 \times 12.5 + 2 \times \frac{22}{7} \times {(3.5)}^{2} = \frac{275}{2} + 77 = \frac{275 + 154}{2} = \frac{429}{2} = 214.5 {cm}^{2} .$

Answer:

$We have, Base radius of cone = Base radius of hemisphere = r = \frac{7}{2} = 3.5 cm, As, the volume of the toy = 231 {cm}^{3} \Rightarrow Volume of cone + Volume of hemisphere = 231 \Rightarrow \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = 231 \Rightarrow \frac{1}{3} π r^{2} (h + 2 r) = 231 \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times (h + 2 \times 3.5) = 231 \Rightarrow \frac{38.5}{3} \times (h + 7) = 231 \Rightarrow h + 7 = \frac{231 \times 3}{38.5} \Rightarrow h + 7 = 18 \Rightarrow h = 18 - 7 \Rightarrow h = 11 cm So, the height of the toy = h + r = 11 + 3.5 = 14.5 cm$

Question 16:

$We have, Base radius of cone = Base radius of hemisphere = r = \frac{7}{2} = 3.5 cm, As, the volume of the toy = 231 {cm}^{3} \Rightarrow Volume of cone + Volume of hemisphere = 231 \Rightarrow \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = 231 \Rightarrow \frac{1}{3} π r^{2} (h + 2 r) = 231 \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times (h + 2 \times 3.5) = 231 \Rightarrow \frac{38.5}{3} \times (h + 7) = 231 \Rightarrow h + 7 = \frac{231 \times 3}{38.5} \Rightarrow h + 7 = 18 \Rightarrow h = 18 - 7 \Rightarrow h = 11 cm So, the height of the toy = h + r = 11 + 3.5 = 14.5 cm$

Answer:

$We have, the base radius of the cylindrical container, R = 6 cm, the height of the container, H = 15 cm, Let the base radius and the height of the ice - cream cone be r and h, respectively . Also, h = 4 r Now, the volume of the cylindrical container = π R^{2} H = \frac{22}{7} \times 6 \times 6 \times 15 = \frac{11880}{7} {cm}^{3} \Rightarrow the volume of the ice - cream distributed to 10 children = \frac{11880}{7} {cm}^{3} \Rightarrow 10 \times Volume of a ice - cream cone = \frac{11880}{7} \Rightarrow 10 \times (Volume of the cone + Volume of the hemisphere) = \frac{11880}{7} \Rightarrow 10 \times (\frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3}) = \frac{11880}{7} \Rightarrow 10 \times (\frac{1}{3} π r^{2} \times 4 r + \frac{2}{3} π r^{3}) = \frac{11880}{7} \Rightarrow 10 \times (\frac{4}{3} π r^{3} + \frac{2}{3} π r^{3}) = \frac{11880}{7} \Rightarrow 10 \times (\frac{6}{3} π r^{3}) = \frac{11880}{7} \Rightarrow 10 \times 2 \times \frac{22}{7} \times r^{3} = \frac{11880}{7} \Rightarrow r^{3} = \frac{11880 \times 7}{7 \times 10 \times 2 \times 22} \Rightarrow r^{3} = 27 \Rightarrow r = \sqrt[3]{27} ∴ r = 3 cm$

So, the radius of the ice-cream cone is 3 cm.

Question 17:

$We have, the base radius of the cylindrical container, R = 6 cm, the height of the container, H = 15 cm, Let the base radius and the height of the ice - cream cone be r and h, respectively . Also, h = 4 r Now, the volume of the cylindrical container = π R^{2} H = \frac{22}{7} \times 6 \times 6 \times 15 = \frac{11880}{7} {cm}^{3} \Rightarrow the volume of the ice - cream distributed to 10 children = \frac{11880}{7} {cm}^{3} \Rightarrow 10 \times Volume of a ice - cream cone = \frac{11880}{7} \Rightarrow 10 \times (Volume of the cone + Volume of the hemisphere) = \frac{11880}{7} \Rightarrow 10 \times (\frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3}) = \frac{11880}{7} \Rightarrow 10 \times (\frac{1}{3} π r^{2} \times 4 r + \frac{2}{3} π r^{3}) = \frac{11880}{7} \Rightarrow 10 \times (\frac{4}{3} π r^{3} + \frac{2}{3} π r^{3}) = \frac{11880}{7} \Rightarrow 10 \times (\frac{6}{3} π r^{3}) = \frac{11880}{7} \Rightarrow 10 \times 2 \times \frac{22}{7} \times r^{3} = \frac{11880}{7} \Rightarrow r^{3} = \frac{11880 \times 7}{7 \times 10 \times 2 \times 22} \Rightarrow r^{3} = 27 \Rightarrow r = \sqrt[3]{27} ∴ r = 3 cm$

So, the radius of the ice-cream cone is 3 cm.

Answer:

Diameter of the hemisphere = 21 cm
Therefore, radius of the hemisphere = 10.5 cm

Volume of the hemisphere $= \frac{2}{3} {πr}^{3} = \frac{2}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5 = 2425.5 {cm}^{3}$

Height of the cylinder $= Total height of the vessel-Radius of the hemisphere = 14.5 - 10.5 = 4 cm$
Volume of the cylinder $= {πr}^{2} h = \frac{22}{7} \times 10.5 \times 10.5 \times 4 = 1386 {cm}^{3}$

Total volume $= V olume of hemispherical part + V olume of cylinder = 1386 + 2425.5 = 3811.5 {cm}^{3}$

Question 18:

Diameter of the hemisphere = 21 cm
Therefore, radius of the hemisphere = 10.5 cm

Volume of the hemisphere $= \frac{2}{3} {πr}^{3} = \frac{2}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5 = 2425.5 {cm}^{3}$

Height of the cylinder $= Total height of the vessel-Radius of the hemisphere = 14.5 - 10.5 = 4 cm$
Volume of the cylinder $= {πr}^{2} h = \frac{22}{7} \times 10.5 \times 10.5 \times 4 = 1386 {cm}^{3}$

Total volume $= V olume of hemispherical part + V olume of cylinder = 1386 + 2425.5 = 3811.5 {cm}^{3}$

Answer:

$We have, the total height of the toy = 90 cm and the radius of the toy, r = \frac{42}{2} = 21 cm Also, the height of the cylinder, h = 90 - 42 = 48 cm Now, the total surface area of the toy = CSA of cylinder + 2 \times CSA of a hemisphere = 2 π r h + 2 \times 2 π r^{2} = 2 π r (h + 2 r) = 2 \times \frac{22}{7} \times 21 \times (48 + 2 \times 21) = 44 \times 3 \times (48 + 42) = 44 \times 3 \times 90 = 11880 {cm}^{2} So, the cost of painting the toy = \frac{11880 \times 70}{100} = Rs 83, 16$

Question 19:

$We have, the total height of the toy = 90 cm and the radius of the toy, r = \frac{42}{2} = 21 cm Also, the height of the cylinder, h = 90 - 42 = 48 cm Now, the total surface area of the toy = CSA of cylinder + 2 \times CSA of a hemisphere = 2 π r h + 2 \times 2 π r^{2} = 2 π r (h + 2 r) = 2 \times \frac{22}{7} \times 21 \times (48 + 2 \times 21) = 44 \times 3 \times (48 + 42) = 44 \times 3 \times 90 = 11880 {cm}^{2} So, the cost of painting the toy = \frac{11880 \times 70}{100} = Rs 83, 16$

Answer:

$We have, the total height of the capsule = 14 mm and the radius of the capsule, r = \frac{5}{2} mm Also, the height of the cylinder, h = 14 - (2 \times \frac{5}{2}) = 14 - 5 = 9 mm Now, the surface area of the capsule = CSA of the cylinder + 2 \times CSA of a hemisphere = 2 π r h + 2 \times 2 π r^{2} = 2 π r (h + 2 r) = 2 \times \frac{22}{7} \times \frac{5}{2} \times (9 + 2 \times \frac{5}{2}) = \frac{22}{7} \times 5 \times (9 + 5) = \frac{22}{7} \times 5 \times 14 = 220 {mm}^{2}$

So, the surface area of the medicine capsule is 220 mm².

Question 20:

$We have, the total height of the capsule = 14 mm and the radius of the capsule, r = \frac{5}{2} mm Also, the height of the cylinder, h = 14 - (2 \times \frac{5}{2}) = 14 - 5 = 9 mm Now, the surface area of the capsule = CSA of the cylinder + 2 \times CSA of a hemisphere = 2 π r h + 2 \times 2 π r^{2} = 2 π r (h + 2 r) = 2 \times \frac{22}{7} \times \frac{5}{2} \times (9 + 2 \times \frac{5}{2}) = \frac{22}{7} \times 5 \times (9 + 5) = \frac{22}{7} \times 5 \times 14 = 220 {mm}^{2}$

So, the surface area of the medicine capsule is 220 mm².

Answer:

The height h of cylinder = 20 cm and diameter of its base = 7 cm.
$\Rightarrow$ The radius r of its base = 3.5 cm.
$\Rightarrow$ Curved surface area of cylinder = $2 π r h = 2 \times \frac{22}{7} \times 3.5 \times 20 = 440 {cm}^{2} .$

Now, curved surface area of one hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times {(3.5)}^{2} = 77 {cm}^{2} .$

Total surface area of the article = Curved surface area of cylinder + 2(curved surface area of hemisphere)
$= 440 + 2 (77) = 440 + 154 = 594 {cm}^{2} .$

Question 21:

The height h of cylinder = 20 cm and diameter of its base = 7 cm.
$\Rightarrow$ The radius r of its base = 3.5 cm.
$\Rightarrow$ Curved surface area of cylinder = $2 π r h = 2 \times \frac{22}{7} \times 3.5 \times 20 = 440 {cm}^{2} .$

Now, curved surface area of one hemisphere $= 2 π r^{2} = 2 \times \frac{22}{7} \times {(3.5)}^{2} = 77 {cm}^{2} .$

Total surface area of the article = Curved surface area of cylinder + 2(curved surface area of hemisphere)
$= 440 + 2 (77) = 440 + 154 = 594 {cm}^{2} .$

Answer:

The object is shown in the figure below.

Radius of hemisphere = 2.1 cm
Volume of hemisphere $= \frac{2}{3} {πr}^{3} = \frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 = 19.404 {cm}^{3}$

Height of cone = 4 cm
Volume of cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4 = 18.48 {cm}^{3}$
Volume of the object $= 18.48 + 19.404 = 37.884 c m^{3}$

Volume of cylindrical tub $= {πr}^{2} h = \frac{22}{7} \times 5 \times 5 \times 9.8 = 770 {cm}^{3}$

When the object is immersed in the tub, volume of water equal to the volume of the object is displaced from the tub.

Volume of water left in the tub = $770 - 37.884 = 732.116 c m^{3}$

Question 22:

The object is shown in the figure below.

Radius of hemisphere = 2.1 cm
Volume of hemisphere $= \frac{2}{3} {πr}^{3} = \frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 = 19.404 {cm}^{3}$

Height of cone = 4 cm
Volume of cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4 = 18.48 {cm}^{3}$
Volume of the object $= 18.48 + 19.404 = 37.884 c m^{3}$

Volume of cylindrical tub $= {πr}^{2} h = \frac{22}{7} \times 5 \times 5 \times 9.8 = 770 {cm}^{3}$

When the object is immersed in the tub, volume of water equal to the volume of the object is displaced from the tub.

Volume of water left in the tub = $770 - 37.884 = 732.116 c m^{3}$

Answer:

Volume of the solid left = Volume of cylinder - Volume of cone $= {πr}^{2} h - \frac{1}{3} {πr}^{2} h = \frac{2}{3} \times \frac{22}{7} \times 8 \times 6 \times 6 = 603.428 {cm}^{3}$

The slant length of the cone, $l = \sqrt{r^{2} + h^{2}} = \sqrt{36 + 64} = 10 c m$

Total surface area of final solid = Area of base circle + Curved surface area of cylinder + Curved surface area of cone
$= π r^{2} + 2 π r h + π r l = π r (r + 2 h + l) = \frac{22}{7} \times 6 \times (6 + 16 + 10) = 603.42 {cm}^{2}$

Question 23:

Volume of the solid left = Volume of cylinder - Volume of cone $= {πr}^{2} h - \frac{1}{3} {πr}^{2} h = \frac{2}{3} \times \frac{22}{7} \times 8 \times 6 \times 6 = 603.428 {cm}^{3}$

The slant length of the cone, $l = \sqrt{r^{2} + h^{2}} = \sqrt{36 + 64} = 10 c m$

Total surface area of final solid = Area of base circle + Curved surface area of cylinder + Curved surface area of cone
$= π r^{2} + 2 π r h + π r l = π r (r + 2 h + l) = \frac{22}{7} \times 6 \times (6 + 16 + 10) = 603.42 {cm}^{2}$

Answer:

$We have, the height of the cone = the height of the cylinder = h = 2.8 cm and the radius of the base, r = \frac{4.2}{2} = 2.1 cm The slant height of the cone, l = \sqrt{r^{2} + h^{2}} = \sqrt{2 . 1^{2} + 2 . 8^{2}} = \sqrt{4.41 + 7.84} = \sqrt{12.25} = 3.5 cm Now, the total surface area of the remaining solid = CSA of cylinder + CSA of cone + Area of a base = 2 π r h + π r l + π r^{2} = π r (2 h + l + r) = \frac{22}{7} \times 2.1 \times (2 \times 2.8 + 3.5 + 2.1) = 22 \times 0.3 \times (5.6 + 5.6) = 6.6 \times 11.2 = 73.92 {cm}^{2}$

So, the total surface area of the remaining solid is 73.92 cm².

Question 24:

$We have, the height of the cone = the height of the cylinder = h = 2.8 cm and the radius of the base, r = \frac{4.2}{2} = 2.1 cm The slant height of the cone, l = \sqrt{r^{2} + h^{2}} = \sqrt{2 . 1^{2} + 2 . 8^{2}} = \sqrt{4.41 + 7.84} = \sqrt{12.25} = 3.5 cm Now, the total surface area of the remaining solid = CSA of cylinder + CSA of cone + Area of a base = 2 π r h + π r l + π r^{2} = π r (2 h + l + r) = \frac{22}{7} \times 2.1 \times (2 \times 2.8 + 3.5 + 2.1) = 22 \times 0.3 \times (5.6 + 5.6) = 6.6 \times 11.2 = 73.92 {cm}^{2}$

So, the total surface area of the remaining solid is 73.92 cm².

Answer:

$We have, the height of the cylinder, H = 14 cm, the base radius of cylinder, R = \frac{7}{2} cm, the base radius of each conical holes, r = 2.1 cm and the height of each conical holes, h = 4 cm Volume of the remaining solid = Volume of the cylinder - Volume of 2 conical holes = π R^{2} H - 2 \times \frac{1}{3} π r^{2} h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 14 - \frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4 = 539 - 36.96 = 502.04 {cm}^{3}$

So, the volume of the remaining solid is 502.04 cm³.

Question 25:

$We have, the height of the cylinder, H = 14 cm, the base radius of cylinder, R = \frac{7}{2} cm, the base radius of each conical holes, r = 2.1 cm and the height of each conical holes, h = 4 cm Volume of the remaining solid = Volume of the cylinder - Volume of 2 conical holes = π R^{2} H - 2 \times \frac{1}{3} π r^{2} h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 14 - \frac{2}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 4 = 539 - 36.96 = 502.04 {cm}^{3}$

So, the volume of the remaining solid is 502.04 cm³.

Answer:

$We have, the base radius of the cylinder, R = 3 cm, the height of the cylinder, H = 5 cm, the base radius of the conical hole, r = \frac{3}{2} cm and the height of the conical hole, h = \frac{8}{9} cm Now, Volume of the cylinder, V = π R^{2} H = π \times 3^{2} \times 5 = 45 π {cm}^{3} Also, Volume of the cone removed from the cylinder, v = \frac{1}{3} π r^{2} h = \frac{π}{3} \times {(\frac{3}{2})}^{2} \times (\frac{8}{9}) = \frac{2 π}{3} {cm}^{3} So, the volume of metal left in the cylinder, V' = V - v = 45 π - \frac{2 π}{3} = \frac{133 π}{3} {cm}^{3} ∴ The required ratio = \frac{V'}{v} = \frac{(\frac{133 π}{3})}{(\frac{2 π}{3})} = \frac{133}{2} = 133 : 2$

So, the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 133 : 2.

Question 26:

$We have, the base radius of the cylinder, R = 3 cm, the height of the cylinder, H = 5 cm, the base radius of the conical hole, r = \frac{3}{2} cm and the height of the conical hole, h = \frac{8}{9} cm Now, Volume of the cylinder, V = π R^{2} H = π \times 3^{2} \times 5 = 45 π {cm}^{3} Also, Volume of the cone removed from the cylinder, v = \frac{1}{3} π r^{2} h = \frac{π}{3} \times {(\frac{3}{2})}^{2} \times (\frac{8}{9}) = \frac{2 π}{3} {cm}^{3} So, the volume of metal left in the cylinder, V' = V - v = 45 π - \frac{2 π}{3} = \frac{133 π}{3} {cm}^{3} ∴ The required ratio = \frac{V'}{v} = \frac{(\frac{133 π}{3})}{(\frac{2 π}{3})} = \frac{133}{2} = 133 : 2$

So, the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape is 133 : 2.

Answer:

Diameter of spherical part = 21 cm
Radius of the spherical part = 10.5 cm

Volume of spherical part of the vessel $= \frac{4}{3} {πr}^{3} = \frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5 = 4851 {cm}^{3}$
Diameter of cylinder = 4 cm
Radius of cylinder = 2 cm
Height of cylinder = 7 cm

Volume of the cylindrical part of the vessel $= π r^{2} h = \frac{22}{7} \times 2 \times 2 \times 7 = 88 {cm}^{3}$

Total volume of the vessel $= 4851 + 88 = 4939 c m^{3}$

Question 27:

Diameter of spherical part = 21 cm
Radius of the spherical part = 10.5 cm

Volume of spherical part of the vessel $= \frac{4}{3} {πr}^{3} = \frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5 = 4851 {cm}^{3}$
Diameter of cylinder = 4 cm
Radius of cylinder = 2 cm
Height of cylinder = 7 cm

Volume of the cylindrical part of the vessel $= π r^{2} h = \frac{22}{7} \times 2 \times 2 \times 7 = 88 {cm}^{3}$

Total volume of the vessel $= 4851 + 88 = 4939 c m^{3}$

Answer:

Diameter of the cylindrical part =7 cm
Therefore,radius of the cylindrical part = 3.5 cm

Volume of hemisphere $= \frac{2}{3} {πr}^{3} = \frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 = 89.83 {cm}^{3}$

Volume of the cylinder $= {πr}^{2} h = \frac{22}{7} \times 3.5 \times 3.5 \times 6.5 = 250.25 {cm}^{3}$

Height of cone $= 12.8 - 6.5 = 6.3 c m$

Volume of the cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 6.3 = 80.85 c m^{3}$

Total volume $= 89.83 + 250.25 + 80.85 = 420.93 c m^{3}$

Question 28:

Diameter of the cylindrical part =7 cm
Therefore,radius of the cylindrical part = 3.5 cm

Volume of hemisphere $= \frac{2}{3} {πr}^{3} = \frac{2}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 = 89.83 {cm}^{3}$

Volume of the cylinder $= {πr}^{2} h = \frac{22}{7} \times 3.5 \times 3.5 \times 6.5 = 250.25 {cm}^{3}$

Height of cone $= 12.8 - 6.5 = 6.3 c m$

Volume of the cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 6.3 = 80.85 c m^{3}$

Total volume $= 89.83 + 250.25 + 80.85 = 420.93 c m^{3}$

Answer:

$We have, the edge of the cubical piece, a = 21 cm and the radius of the hemisphere, r = \frac{a}{2} = \frac{21}{2} cm The surface area of the remaining piece = TSA of cube + CSA of hemisphere - Area of circle = 6 a^{2} + 2 π r^{2} - π r^{2} = 6 a^{2} + π r^{2} = 6 \times 21 \times 21 + \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 21 \times 21 (6 + \frac{22}{7 \times 4}) = 21 \times 21 (6 + \frac{11}{14}) = 21 \times 21 (\frac{84 + 11}{14}) = 21 \times 3 (\frac{95}{2}) = 2992.5 {cm}^{2}$

$Also, Volume of the remaining piece = Volume of the cube - Volume of the hemisphere = a^{3} - \frac{2}{3} π r^{3} = 21 \times 21 \times 21 - \frac{2}{3} \times \frac{22}{7} \times (\frac{21}{2}) \times (\frac{21}{2}) \times (\frac{21}{2}) = 21 \times 21 \times 21 \times (1 - \frac{2}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}) = 21 \times 21 \times 21 \times (\frac{1}{1} - \frac{11}{42}) = 21 \times 21 \times 21 \times (\frac{42 - 11}{42}) = 21 \times 21 \times (\frac{31}{2}) = 6835.5 {cm}^{3}$

Question 29:

$We have, the edge of the cubical piece, a = 21 cm and the radius of the hemisphere, r = \frac{a}{2} = \frac{21}{2} cm The surface area of the remaining piece = TSA of cube + CSA of hemisphere - Area of circle = 6 a^{2} + 2 π r^{2} - π r^{2} = 6 a^{2} + π r^{2} = 6 \times 21 \times 21 + \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 21 \times 21 (6 + \frac{22}{7 \times 4}) = 21 \times 21 (6 + \frac{11}{14}) = 21 \times 21 (\frac{84 + 11}{14}) = 21 \times 3 (\frac{95}{2}) = 2992.5 {cm}^{2}$

$Also, Volume of the remaining piece = Volume of the cube - Volume of the hemisphere = a^{3} - \frac{2}{3} π r^{3} = 21 \times 21 \times 21 - \frac{2}{3} \times \frac{22}{7} \times (\frac{21}{2}) \times (\frac{21}{2}) \times (\frac{21}{2}) = 21 \times 21 \times 21 \times (1 - \frac{2}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}) = 21 \times 21 \times 21 \times (\frac{1}{1} - \frac{11}{42}) = 21 \times 21 \times 21 \times (\frac{42 - 11}{42}) = 21 \times 21 \times (\frac{31}{2}) = 6835.5 {cm}^{3}$

Answer:

(i) Disclaimer: It is written cuboid in the question but it should be cube.

From the figure, it can be observed that the maximum diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7 cm.

Radius (r) of hemispherical part = = 3.5 cm

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

− Area of base of hemispherical part

= 6 (Edge)² − = 6 (Edge)² +

(ii)

We have, the edge of the cubical block, a = 10 cm The largest diameter of the hemisphere = a = 10 cm Also, the radius of the hemisphere, r = \frac{10}{2} = 5 cm Now, Total surface area of the solid = TSA of cube + CSA of hemisphere - Area of circle = 6 a^{2} + 2 π r^{2} - π r^{2} = 6 a^{2} + π r^{2} = 6 \times 10 \times 10 + 3.14 \times 5 \times 5 = 600 + 78.5 = 678.5 {cm}^{2} As, the rate of painting the solid = ₹ 5 per 100 {cm}^{2} So, the cost of painting the solid = 678.5 \times \frac{5}{100} \approx ₹ 33.92

hence, the cost of painting the total surface area of the solid is â‚¹33.92.

Question 30:

(i) Disclaimer: It is written cuboid in the question but it should be cube.

From the figure, it can be observed that the maximum diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7 cm.

Radius (r) of hemispherical part = = 3.5 cm

Total surface area of solid = Surface area of cubical part + CSA of hemispherical part

− Area of base of hemispherical part

= 6 (Edge)² − = 6 (Edge)² +

(ii)

We have, the edge of the cubical block, a = 10 cm The largest diameter of the hemisphere = a = 10 cm Also, the radius of the hemisphere, r = \frac{10}{2} = 5 cm Now, Total surface area of the solid = TSA of cube + CSA of hemisphere - Area of circle = 6 a^{2} + 2 π r^{2} - π r^{2} = 6 a^{2} + π r^{2} = 6 \times 10 \times 10 + 3.14 \times 5 \times 5 = 600 + 78.5 = 678.5 {cm}^{2} As, the rate of painting the solid = ₹ 5 per 100 {cm}^{2} So, the cost of painting the solid = 678.5 \times \frac{5}{100} \approx ₹ 33.92

hence, the cost of painting the total surface area of the solid is â‚¹33.92.

Answer:

$We have, the base radius of cone = the base radius of cylinder = the base radius of hemisphere = r = 5 cm, the height of the cylinder, H = 13 cm, the total height of the toy = 30 cm Also, the height of the cone, h = 30 - (13 + 5) = 12 cm The slant heigt of the cone, l = \sqrt{r^{2} + h^{2}} = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 cm Now, the surface area of the toy = CSA of cone + CSA of cylinder + CSA of hemisphere = π r l + 2 π r H + 2 π r^{2} = π r (l + 2 H + 2 r) = \frac{22}{7} \times 5 \times (13 + 2 \times 13 + 2 \times 5) = \frac{22}{7} \times 5 \times (13 + 26 + 10) = \frac{22}{7} \times 5 \times 49 = 770 {cm}^{2}$

So, the surface area of the toy is 770 cm².

Question 31:

$We have, the base radius of cone = the base radius of cylinder = the base radius of hemisphere = r = 5 cm, the height of the cylinder, H = 13 cm, the total height of the toy = 30 cm Also, the height of the cone, h = 30 - (13 + 5) = 12 cm The slant heigt of the cone, l = \sqrt{r^{2} + h^{2}} = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 cm Now, the surface area of the toy = CSA of cone + CSA of cylinder + CSA of hemisphere = π r l + 2 π r H + 2 π r^{2} = π r (l + 2 H + 2 r) = \frac{22}{7} \times 5 \times (13 + 2 \times 13 + 2 \times 5) = \frac{22}{7} \times 5 \times (13 + 26 + 10) = \frac{22}{7} \times 5 \times 49 = 770 {cm}^{2}$

So, the surface area of the toy is 770 cm².

Answer:

$We have, the height of the glass, h = 16 cm and the base radius of cylinder = the base radius of hemisphere, r = \frac{7}{2} cm Now, The apparent capacity of the glass = Volume of the cylinder = π r^{2} h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 16 = 616 {cm}^{3} Also, The actual capacity of the glass = Volume of cylinder - Volume of hemisphere = 616 - \frac{2}{3} π r^{3} = 616 - \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} = 616 - \frac{539}{6} = \frac{3157}{6} {cm}^{3} \approx 526.17 {cm}^{3}$

Question 32:

$We have, the height of the glass, h = 16 cm and the base radius of cylinder = the base radius of hemisphere, r = \frac{7}{2} cm Now, The apparent capacity of the glass = Volume of the cylinder = π r^{2} h = \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 16 = 616 {cm}^{3} Also, The actual capacity of the glass = Volume of cylinder - Volume of hemisphere = 616 - \frac{2}{3} π r^{3} = 616 - \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} = 616 - \frac{539}{6} = \frac{3157}{6} {cm}^{3} \approx 526.17 {cm}^{3}$

Answer:

$We have, the base radius of the conical part, r = \frac{5}{2} = 2.5 cm, the base radius of the cylindrical part, R = \frac{4}{2} = 2 cm, the total height of the toy = 26 cm, the height of the conical part, h = 6 cm Also, the height of the cylindrical part, H = 26 - 6 = 20 cm And, the slant height of the conical part, l = \sqrt{r^{2} + h^{2}} = \sqrt{2 . 5^{2} + 6^{2}} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5 cm Now, The area to be painted by red colour = CSA of cone + Area of base of conical part - Area of base of cylindrical part = π r l + π r^{2} - π R^{2} = \frac{22}{7} \times 2.5 \times 6.5 + \frac{22}{7} \times 2.5 \times 2.5 - \frac{22}{7} \times 2 \times 2 = \frac{22}{7} \times 16.25 + \frac{22}{7} \times 6.25 - \frac{22}{7} \times 4 = \frac{22}{7} \times (16.25 + 6.25 - 4) = \frac{22}{7} \times 18.5 \approx 58.14 {cm}^{2} Also, The area to be painted by white colour = CSA of cylinder + Area of base of cylinder = 2 π R H + π R^{2} = π R (2 H + R) = \frac{22}{7} \times 2 \times (2 \times 20 + 2) = \frac{22}{7} \times 2 \times 42 = 264 {cm}^{2}$

Question 1:

$We have, the base radius of the conical part, r = \frac{5}{2} = 2.5 cm, the base radius of the cylindrical part, R = \frac{4}{2} = 2 cm, the total height of the toy = 26 cm, the height of the conical part, h = 6 cm Also, the height of the cylindrical part, H = 26 - 6 = 20 cm And, the slant height of the conical part, l = \sqrt{r^{2} + h^{2}} = \sqrt{2 . 5^{2} + 6^{2}} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5 cm Now, The area to be painted by red colour = CSA of cone + Area of base of conical part - Area of base of cylindrical part = π r l + π r^{2} - π R^{2} = \frac{22}{7} \times 2.5 \times 6.5 + \frac{22}{7} \times 2.5 \times 2.5 - \frac{22}{7} \times 2 \times 2 = \frac{22}{7} \times 16.25 + \frac{22}{7} \times 6.25 - \frac{22}{7} \times 4 = \frac{22}{7} \times (16.25 + 6.25 - 4) = \frac{22}{7} \times 18.5 \approx 58.14 {cm}^{2} Also, The area to be painted by white colour = CSA of cylinder + Area of base of cylinder = 2 π R H + π R^{2} = π R (2 H + R) = \frac{22}{7} \times 2 \times (2 \times 20 + 2) = \frac{22}{7} \times 2 \times 42 = 264 {cm}^{2}$

Answer:

The volume of solid metallic cuboid is $9 \times 8 \times 2 = 144 m^{3} .$
This cuboid has been recasted into solid cubes of edge 2 m whose volume is given by $2^{3} = 8 m^{3} .$
Therefore, the total number of cubes so formed $= \frac{the volume of solid metallic cuboid}{the volume of solid cubes} = \frac{144}{8} = 18 .$

Question 2:

The volume of solid metallic cuboid is $9 \times 8 \times 2 = 144 m^{3} .$
This cuboid has been recasted into solid cubes of edge 2 m whose volume is given by $2^{3} = 8 m^{3} .$
Therefore, the total number of cubes so formed $= \frac{the volume of solid metallic cuboid}{the volume of solid cubes} = \frac{144}{8} = 18 .$

Answer:

$We have, the base radius of the cone, r = 5 cm and the height of the cone, h = 20 cm Let the radius of the sphere be R . As, Volume of sphere = Volume of cone \Rightarrow \frac{4}{3} π R^{3} = \frac{1}{3} π r^{2} h \Rightarrow R^{3} = \frac{π r^{2} h \times 3}{3 \times 4 π} \Rightarrow R^{3} = \frac{r^{2} h}{4} \Rightarrow R^{3} = \frac{5 \times 5 \times 20}{4} \Rightarrow R^{3} = 125 \Rightarrow R = \sqrt[3]{125} \Rightarrow R = 5 cm \Rightarrow Diameter of the sphere = 2 R = 2 \times 5 = 10 cm$

So, the diameter of the sphere is 10 cm.

Question 3:

$We have, the base radius of the cone, r = 5 cm and the height of the cone, h = 20 cm Let the radius of the sphere be R . As, Volume of sphere = Volume of cone \Rightarrow \frac{4}{3} π R^{3} = \frac{1}{3} π r^{2} h \Rightarrow R^{3} = \frac{π r^{2} h \times 3}{3 \times 4 π} \Rightarrow R^{3} = \frac{r^{2} h}{4} \Rightarrow R^{3} = \frac{5 \times 5 \times 20}{4} \Rightarrow R^{3} = 125 \Rightarrow R = \sqrt[3]{125} \Rightarrow R = 5 cm \Rightarrow Diameter of the sphere = 2 R = 2 \times 5 = 10 cm$

So, the diameter of the sphere is 10 cm.

Answer:

$We have, the radii r_{1} = 6 cm, r_{2} = 8 cm and r_{3} = 10 cm Let the radius of the resulting sphere be R . As, Volume of resulting sphere = Volume of three metallic spheres \Rightarrow \frac{4}{3} π R^{3} = \frac{4}{3} π {r_{1}}^{3} + \frac{4}{3} π {r_{2}}^{3} + \frac{4}{3} π {r_{3}}^{3} \Rightarrow \frac{4}{3} π R^{3} = \frac{4}{3} π ({r_{1}}^{3} + {r_{2}}^{3} + {r_{3}}^{3}) \Rightarrow R^{3} = {r_{1}}^{3} + {r_{2}}^{3} + {r_{3}}^{3} \Rightarrow R^{3} = 6^{3} + 8^{3} + 10^{3} \Rightarrow R^{3} = 216 + 512 + 1000 \Rightarrow R^{3} = 1728 \Rightarrow R = \sqrt[3]{1728} \Rightarrow R = 12 cm$

So, the radius of the resulting sphere is 12 cm.

Question 4:

$We have, the radii r_{1} = 6 cm, r_{2} = 8 cm and r_{3} = 10 cm Let the radius of the resulting sphere be R . As, Volume of resulting sphere = Volume of three metallic spheres \Rightarrow \frac{4}{3} π R^{3} = \frac{4}{3} π {r_{1}}^{3} + \frac{4}{3} π {r_{2}}^{3} + \frac{4}{3} π {r_{3}}^{3} \Rightarrow \frac{4}{3} π R^{3} = \frac{4}{3} π ({r_{1}}^{3} + {r_{2}}^{3} + {r_{3}}^{3}) \Rightarrow R^{3} = {r_{1}}^{3} + {r_{2}}^{3} + {r_{3}}^{3} \Rightarrow R^{3} = 6^{3} + 8^{3} + 10^{3} \Rightarrow R^{3} = 216 + 512 + 1000 \Rightarrow R^{3} = 1728 \Rightarrow R = \sqrt[3]{1728} \Rightarrow R = 12 cm$

So, the radius of the resulting sphere is 12 cm.

Answer:

Radius of the cone = 12 cm
Height of the cone = 24 cm

Volume $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} π \times 12 \times 12 \times 24 = 48 \times 24 \times π {cm}^{3}$

Radius of each ball = 3 cm
Volume of each ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 3 \times 3 \times 3 = 36 π {cm}^{3}$
Total number of balls formed by melting the cone $= \frac{Volume of cone}{Volume of a ball} = \frac{48 \times 24 π}{36 π} = 32$

Question 5:

Radius of the cone = 12 cm
Height of the cone = 24 cm

Volume $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} π \times 12 \times 12 \times 24 = 48 \times 24 \times π {cm}^{3}$

Radius of each ball = 3 cm
Volume of each ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 3 \times 3 \times 3 = 36 π {cm}^{3}$
Total number of balls formed by melting the cone $= \frac{Volume of cone}{Volume of a ball} = \frac{48 \times 24 π}{36 π} = 32$

Answer:

$We have, the internal base radius of spherical shell, r_{1} = 3 cm, the external base radius of spherical shell, r_{2} = 5 cm and the base radius of solid cylinder, r = \frac{14}{2} = 7 cm Let the height of the cylinder be h . As, Volume of solid cylinder = Volume of spherical shell \Rightarrow π r^{2} h = \frac{4}{3} π {r_{2}}^{3} - \frac{4}{3} π {r_{1}}^{3} \Rightarrow π r^{2} h = \frac{4}{3} π ({r_{2}}^{3} - {r_{1}}^{3}) \Rightarrow r^{2} h = \frac{4}{3} ({r_{2}}^{3} - {r_{1}}^{3}) \Rightarrow 7 \times 7 \times h = \frac{4}{3} (5^{3} - 3^{3}) \Rightarrow 49 \times h = \frac{4}{3} (125 - 27) \Rightarrow h = \frac{4}{3} \times \frac{98}{49} ∴ h = \frac{8}{3} cm$

So, the height of the cylinder is $\frac{8}{3}$ cm.

Question 6:

$We have, the internal base radius of spherical shell, r_{1} = 3 cm, the external base radius of spherical shell, r_{2} = 5 cm and the base radius of solid cylinder, r = \frac{14}{2} = 7 cm Let the height of the cylinder be h . As, Volume of solid cylinder = Volume of spherical shell \Rightarrow π r^{2} h = \frac{4}{3} π {r_{2}}^{3} - \frac{4}{3} π {r_{1}}^{3} \Rightarrow π r^{2} h = \frac{4}{3} π ({r_{2}}^{3} - {r_{1}}^{3}) \Rightarrow r^{2} h = \frac{4}{3} ({r_{2}}^{3} - {r_{1}}^{3}) \Rightarrow 7 \times 7 \times h = \frac{4}{3} (5^{3} - 3^{3}) \Rightarrow 49 \times h = \frac{4}{3} (125 - 27) \Rightarrow h = \frac{4}{3} \times \frac{98}{49} ∴ h = \frac{8}{3} cm$

So, the height of the cylinder is $\frac{8}{3}$ cm.

Answer:

Internal diameter of the hemispherical shell = 6 cm
Therefore, internal radius of the hemispherical shell = 3 cm

External diameter of the hemispherical shell = 10 cm
External radius of the hemispherical shell = 5 cm

Volume of hemispherical shell $= \frac{2}{3} π (5^{3} - 3^{3}) = \frac{196}{3} \times \frac{22}{7} = \frac{616}{3} {cm}^{3}$

Radius of cone = 7 cm
Let the height of the cone be h cm.
Volume of cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 h = \frac{154 h}{3} {cm}^{3}$

The volume of the hemispherical shell must be equal to the volume of the cone. Therefore,

$\frac{154 h}{3} = \frac{616}{3} \Rightarrow h = \frac{616}{154} = 4 c m$

Question 7:

Internal diameter of the hemispherical shell = 6 cm
Therefore, internal radius of the hemispherical shell = 3 cm

External diameter of the hemispherical shell = 10 cm
External radius of the hemispherical shell = 5 cm

Volume of hemispherical shell $= \frac{2}{3} π (5^{3} - 3^{3}) = \frac{196}{3} \times \frac{22}{7} = \frac{616}{3} {cm}^{3}$

Radius of cone = 7 cm
Let the height of the cone be h cm.
Volume of cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 h = \frac{154 h}{3} {cm}^{3}$

The volume of the hemispherical shell must be equal to the volume of the cone. Therefore,

$\frac{154 h}{3} = \frac{616}{3} \Rightarrow h = \frac{616}{154} = 4 c m$

Answer:

$We have, the radius of the copper rod, R = \frac{2}{2} = 1 cm, the height of the copper rod, H = 10 cm and the height of the wire, h = 10 m = 1000 cm Let the radius of the wire be r . As, Volume of the wire = Volume of the rod \Rightarrow π r^{2} h = π R^{2} H \Rightarrow r^{2} h = R^{2} H \Rightarrow r^{2} \times 1000 = 1 \times 10 \Rightarrow r^{2} = \frac{10}{1000} \Rightarrow r^{2} = \frac{1}{100} \Rightarrow r = \sqrt{\frac{1}{100}} \Rightarrow r = \frac{1}{10} \Rightarrow r = 0.1 cm \Rightarrow The diameter of the wire = 2 r = 2 \times 0.1 = 0.2 cm ∴ The thickness of the wire = 0.2 cm$

So, the thickness of the wire is 0.2 cm or 2 mm.

Question 8:

$We have, the radius of the copper rod, R = \frac{2}{2} = 1 cm, the height of the copper rod, H = 10 cm and the height of the wire, h = 10 m = 1000 cm Let the radius of the wire be r . As, Volume of the wire = Volume of the rod \Rightarrow π r^{2} h = π R^{2} H \Rightarrow r^{2} h = R^{2} H \Rightarrow r^{2} \times 1000 = 1 \times 10 \Rightarrow r^{2} = \frac{10}{1000} \Rightarrow r^{2} = \frac{1}{100} \Rightarrow r = \sqrt{\frac{1}{100}} \Rightarrow r = \frac{1}{10} \Rightarrow r = 0.1 cm \Rightarrow The diameter of the wire = 2 r = 2 \times 0.1 = 0.2 cm ∴ The thickness of the wire = 0.2 cm$

So, the thickness of the wire is 0.2 cm or 2 mm.

Answer:

Inner diameter of the bowl = 30 cm
Inner radius of the bowl $= \frac{30 c m}{2} = 15 c m$
Inner volume of the bowl = Volume of liquid $= \frac{2}{3} {πr}^{3} = \frac{2}{3} \times π \times 15^{3} {cm}^{3}$

Radius of each bottle = 2.5 cm
Height = 6 cm

Volume of each bottle $= {πr}^{2} h = π \times \frac{5}{2} \times \frac{5}{2} \times 6 = \frac{75 π}{2} {cm}^{3}$

Total number of bottles required $= \frac{[\frac{2}{3} π \times 15 \times 15 \times 15]}{\frac{75 π}{2}} = \frac{2 π \times 15 \times 15 \times 15 \times 2}{3 \times 75 π} = 15 \times 4 = 60$

Question 9:

Inner diameter of the bowl = 30 cm
Inner radius of the bowl $= \frac{30 c m}{2} = 15 c m$
Inner volume of the bowl = Volume of liquid $= \frac{2}{3} {πr}^{3} = \frac{2}{3} \times π \times 15^{3} {cm}^{3}$

Radius of each bottle = 2.5 cm
Height = 6 cm

Volume of each bottle $= {πr}^{2} h = π \times \frac{5}{2} \times \frac{5}{2} \times 6 = \frac{75 π}{2} {cm}^{3}$

Total number of bottles required $= \frac{[\frac{2}{3} π \times 15 \times 15 \times 15]}{\frac{75 π}{2}} = \frac{2 π \times 15 \times 15 \times 15 \times 2}{3 \times 75 π} = 15 \times 4 = 60$

Answer:

Diameter of sphere = 21 cm
Radius of sphere $= \frac{21}{2} c m$

Volume of sphere $= \frac{4}{3} {πr}^{3} = \frac{4 \times 21 \times 21 \times 21 π}{3 \times 2 \times 2 \times 2} = \frac{21 \times 21 \times 21 π}{3 \times 2} {cm}^{3}$

Diameter of the cone = 3.5 cm
Radius of the cone $= \frac{3.5}{2} = \frac{7}{4} c m$
Height = 3 cm

Volume of each cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} π \times 3 \times {(\frac{7}{4})}^{2} = {(\frac{7}{4})}^{2} π {cm}^{3}$

Total number of cones $= \frac{Volume of sphere}{Volume of a cone} = \frac{\frac{21 \times 21 \times 21 π}{3 \times 2}}{{(\frac{7}{4})}^{2} π} = \frac{21 \times 21 \times 21 \times π \times 4 \times 4}{3 \times 2 \times π \times 7 \times 7} = 504$

Question 10:

Diameter of sphere = 21 cm
Radius of sphere $= \frac{21}{2} c m$

Volume of sphere $= \frac{4}{3} {πr}^{3} = \frac{4 \times 21 \times 21 \times 21 π}{3 \times 2 \times 2 \times 2} = \frac{21 \times 21 \times 21 π}{3 \times 2} {cm}^{3}$

Diameter of the cone = 3.5 cm
Radius of the cone $= \frac{3.5}{2} = \frac{7}{4} c m$
Height = 3 cm

Volume of each cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} π \times 3 \times {(\frac{7}{4})}^{2} = {(\frac{7}{4})}^{2} π {cm}^{3}$

Total number of cones $= \frac{Volume of sphere}{Volume of a cone} = \frac{\frac{21 \times 21 \times 21 π}{3 \times 2}}{{(\frac{7}{4})}^{2} π} = \frac{21 \times 21 \times 21 \times π \times 4 \times 4}{3 \times 2 \times π \times 7 \times 7} = 504$

Answer:

Diameter of cannon ball = 28 cm
Radius of the cannon ball = 14 cm

Volume of ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times {(14)}^{3} {cm}^{3}$

Diameter of base of cone = 35 cm

Radius of base of cone $= \frac{35}{2} c m$
Let the height of the cone be h cm.

Volume of cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} π \times {(\frac{35}{2})}^{2} \times h {cm}^{3}$

From the above results and from the given conditions,
Volume of ball = Volume of cone

$Or, \frac{4}{3} π \times {(14)}^{3} = \frac{1}{3} π \times {(\frac{35}{2})}^{2} \times h \Rightarrow h = \frac{\frac{4}{3} π \times {(14)}^{3}}{\frac{1}{3} π \times {(\frac{35}{2})}^{2}} = \frac{4 \times 14 \times 14 \times 14 \times 2 \times 2}{35 \times 35} = 35.84 cm$

Question 11:

Diameter of cannon ball = 28 cm
Radius of the cannon ball = 14 cm

Volume of ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times {(14)}^{3} {cm}^{3}$

Diameter of base of cone = 35 cm

Radius of base of cone $= \frac{35}{2} c m$
Let the height of the cone be h cm.

Volume of cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} π \times {(\frac{35}{2})}^{2} \times h {cm}^{3}$

From the above results and from the given conditions,
Volume of ball = Volume of cone

$Or, \frac{4}{3} π \times {(14)}^{3} = \frac{1}{3} π \times {(\frac{35}{2})}^{2} \times h \Rightarrow h = \frac{\frac{4}{3} π \times {(14)}^{3}}{\frac{1}{3} π \times {(\frac{35}{2})}^{2}} = \frac{4 \times 14 \times 14 \times 14 \times 2 \times 2}{35 \times 35} = 35.84 cm$

Answer:

Radius of sphere = 3 cm
Radius of first ball = 1.5 cm
Radius of second ball = 2 cm
Let radius of the third ball be r cm.
Volume of third ball = Volume of original sphere - Sum of volumes of other two balls

$= \frac{4}{3} π \times 3^{3} - (\frac{4}{3} π \times {\frac{3}{2}}^{3} + \frac{4}{3} π \times 2^{3}) = \frac{4}{3} π \times 3 \times 3 \times 3 - (\frac{4}{3} π \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} + \frac{4}{3} π \times 2 \times 2 \times 2) = 4 π \times 3 \times 3 - (π \times \frac{3 \times 3}{2} + \frac{4}{3} π \times 2 \times 2 \times 2) = 36 π - (π \frac{9}{2} + \frac{32}{3} π) = (\frac{36 \times 6 - 9 \times 3 - 32 \times 2}{6}) π = (\frac{216 - 27 - 64}{6}) π = \frac{125 π}{6}$

Therefore,

$\frac{4}{3} {πr}^{3} = \frac{125 π}{6} O r, r = \sqrt[3]{\frac{125 \times 3}{4 \times 6}} = \sqrt[3]{\frac{125}{8}} = \frac{5}{2} = 2.5 cm$

Question 12:

Radius of sphere = 3 cm
Radius of first ball = 1.5 cm
Radius of second ball = 2 cm
Let radius of the third ball be r cm.
Volume of third ball = Volume of original sphere - Sum of volumes of other two balls

$= \frac{4}{3} π \times 3^{3} - (\frac{4}{3} π \times {\frac{3}{2}}^{3} + \frac{4}{3} π \times 2^{3}) = \frac{4}{3} π \times 3 \times 3 \times 3 - (\frac{4}{3} π \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} + \frac{4}{3} π \times 2 \times 2 \times 2) = 4 π \times 3 \times 3 - (π \times \frac{3 \times 3}{2} + \frac{4}{3} π \times 2 \times 2 \times 2) = 36 π - (π \frac{9}{2} + \frac{32}{3} π) = (\frac{36 \times 6 - 9 \times 3 - 32 \times 2}{6}) π = (\frac{216 - 27 - 64}{6}) π = \frac{125 π}{6}$

Therefore,

$\frac{4}{3} {πr}^{3} = \frac{125 π}{6} O r, r = \sqrt[3]{\frac{125 \times 3}{4 \times 6}} = \sqrt[3]{\frac{125}{8}} = \frac{5}{2} = 2.5 cm$

Answer:

External diameter of the shell = 24 cm
External radius of the shell = 12 cm
Internal diameter of the shell = 18 cm
Internal radius of the shell = 9 cm

Volume of the shell $= \frac{4}{3} π (12^{3} - 9^{3}) = \frac{4}{3} π (1728 - 729) = \frac{4}{3} π \times (999) = 4 π \times (333) {cm}^{3}$

Height of cylinder = 37 cm
Let radius of cylinder be r cm.

Volume of cylinder $= {πr}^{2} h = 37 {πr}^{2} {cm}^{3}$

Volume of the shell = Volume of cylinder

$O r, 4 π \times (333) = 37 {πr}^{2} \Rightarrow r^{2} = \frac{4 \times 333}{37} = 4 \times 9 \Rightarrow r = \sqrt{4 \times 9} = \sqrt{36} = 6 c m$

So, diameter of the base of the cylinder = 2r = 12 cm.

Question 13:

External diameter of the shell = 24 cm
External radius of the shell = 12 cm
Internal diameter of the shell = 18 cm
Internal radius of the shell = 9 cm

Volume of the shell $= \frac{4}{3} π (12^{3} - 9^{3}) = \frac{4}{3} π (1728 - 729) = \frac{4}{3} π \times (999) = 4 π \times (333) {cm}^{3}$

Height of cylinder = 37 cm
Let radius of cylinder be r cm.

Volume of cylinder $= {πr}^{2} h = 37 {πr}^{2} {cm}^{3}$

Volume of the shell = Volume of cylinder

$O r, 4 π \times (333) = 37 {πr}^{2} \Rightarrow r^{2} = \frac{4 \times 333}{37} = 4 \times 9 \Rightarrow r = \sqrt{4 \times 9} = \sqrt{36} = 6 c m$

So, diameter of the base of the cylinder = 2r = 12 cm.

Answer:

Radius of hemisphere = 9 cm

Volume of hemisphere $= \frac{2}{3} {πr}^{3} = \frac{2}{3} π \times 9 \times 9 \times 9 {cm}^{3}$

Height of cone = 72 cm
Let the radius of the cone be r cm.

Volume of the cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} {πr}^{2} \times 72 {cm}^{3}$

The volumes of the hemisphere and cone are equal.
Therefore,

$\frac{1}{3} {πr}^{2} \times 72 = \frac{2}{3} π \times 9 \times 9 \times 9 r^{2} = \frac{2 \times 9 \times 9 \times 9}{72} = \frac{81}{4} r = \sqrt{\frac{81}{4}} = \frac{9}{2} = 4.5 c m$

The radius of the base of the cone is 4.5 cm.

Question 14:

Radius of hemisphere = 9 cm

Volume of hemisphere $= \frac{2}{3} {πr}^{3} = \frac{2}{3} π \times 9 \times 9 \times 9 {cm}^{3}$

Height of cone = 72 cm
Let the radius of the cone be r cm.

Volume of the cone $= \frac{1}{3} {πr}^{2} h = \frac{1}{3} {πr}^{2} \times 72 {cm}^{3}$

The volumes of the hemisphere and cone are equal.
Therefore,

$\frac{1}{3} {πr}^{2} \times 72 = \frac{2}{3} π \times 9 \times 9 \times 9 r^{2} = \frac{2 \times 9 \times 9 \times 9}{72} = \frac{81}{4} r = \sqrt{\frac{81}{4}} = \frac{9}{2} = 4.5 c m$

The radius of the base of the cone is 4.5 cm.

Answer:

Diameter of the spherical ball= 21 cm

Radius of the ball $= \frac{21}{2} c m$
Volume of spherical ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2} = 11 \times 21 \times 21 = 4851 {cm}^{3}$
Volume of each cube $= 1^{3} = 1 c m^{3}$

Number of cubes = $\frac{V o l u m e o f s p h e r i c a l b a l l}{V o l u m e o f e a c h c u b e} = \frac{4851}{1} = 4851$

Question 15:

Diameter of the spherical ball= 21 cm

Radius of the ball $= \frac{21}{2} c m$
Volume of spherical ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \times \frac{21}{2} = 11 \times 21 \times 21 = 4851 {cm}^{3}$
Volume of each cube $= 1^{3} = 1 c m^{3}$

Number of cubes = $\frac{V o l u m e o f s p h e r i c a l b a l l}{V o l u m e o f e a c h c u b e} = \frac{4851}{1} = 4851$

Answer:

Radius of the sphere = R = 8 cm
Volume of the sphere $= \frac{4}{3} π R^{3} = \frac{4}{3} π \times 8 \times 8 \times 8 = \frac{4}{3} π \times 512 {cm}^{3}$

Radius of each new ball = r = 1 cm
Volume of each ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 1 \times 1 \times 1 = \frac{4}{3} π \times 1 {cm}^{3}$

Total number of new balls that can be made $= \frac{V o l u m e o f s p h e r e}{V o l u m e o f e a c h b a l l} = \frac{\frac{4}{3} π \times 512}{\frac{4}{3} π \times 1} = 512$

Question 16:

Radius of the sphere = R = 8 cm
Volume of the sphere $= \frac{4}{3} π R^{3} = \frac{4}{3} π \times 8 \times 8 \times 8 = \frac{4}{3} π \times 512 {cm}^{3}$

Radius of each new ball = r = 1 cm
Volume of each ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 1 \times 1 \times 1 = \frac{4}{3} π \times 1 {cm}^{3}$

Total number of new balls that can be made $= \frac{V o l u m e o f s p h e r e}{V o l u m e o f e a c h b a l l} = \frac{\frac{4}{3} π \times 512}{\frac{4}{3} π \times 1} = 512$

Answer:

Radius of solid sphere = 3 cm
Volume of the sphere $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 3 \times 3 \times 3 {cm}^{3}$

Radius of each new ball = 0.3 cm
Volume of each new ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times \frac{3}{10} \times \frac{3}{10} \times \frac{3}{10} {cm}^{3}$

Total number of balls $= \frac{\frac{4}{3} π \times 3 \times 3 \times 3}{\frac{4}{3} π \times \frac{3}{10} \times \frac{3}{10} \times \frac{3}{10}} = 1000$

Question 17:

Radius of solid sphere = 3 cm
Volume of the sphere $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 3 \times 3 \times 3 {cm}^{3}$

Radius of each new ball = 0.3 cm
Volume of each new ball $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times \frac{3}{10} \times \frac{3}{10} \times \frac{3}{10} {cm}^{3}$

Total number of balls $= \frac{\frac{4}{3} π \times 3 \times 3 \times 3}{\frac{4}{3} π \times \frac{3}{10} \times \frac{3}{10} \times \frac{3}{10}} = 1000$

Answer:

Diameter of sphere = 42 cm
Radius of sphere = 21 cm

Volume of sphere $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 21 \times 21 \times 21 {cm}^{3}$

Diameter of wire = 2.8 cm
Radius of wire = 1.4 cm
Let the length of the wire be l cm.
Volume of the wire $= {πr}^{2} l = π \times 1.4 \times 1.4 \times l$

The volume of the sphere is equal to the volume of the wire.
â€‹Therefore,
$π \times 1.4 \times 1.4 \times l = \frac{4}{3} π \times 21 \times 21 \times 21 l = \frac{4 \times 21 \times 21 \times 21}{3 \times 1.4 \times 1.4} = 6300 cm = 63 m$

So, the wire is 63 m long.

Question 18:

Diameter of sphere = 42 cm
Radius of sphere = 21 cm

Volume of sphere $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 21 \times 21 \times 21 {cm}^{3}$

Diameter of wire = 2.8 cm
Radius of wire = 1.4 cm
Let the length of the wire be l cm.
Volume of the wire $= {πr}^{2} l = π \times 1.4 \times 1.4 \times l$

The volume of the sphere is equal to the volume of the wire.
â€‹Therefore,
$π \times 1.4 \times 1.4 \times l = \frac{4}{3} π \times 21 \times 21 \times 21 l = \frac{4 \times 21 \times 21 \times 21}{3 \times 1.4 \times 1.4} = 6300 cm = 63 m$

So, the wire is 63 m long.

Answer:

Diameter of sphere = 18 cm
Radius of the sphere = 9 cm

Volume of sphere $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 9 \times 9 \times 9 {cm}^{3}$

Length of wire = 108 m = 10800 cm
Let radius of the wire be r cm.
Volume of the wire $= π r^{2} l = π \times r^{2} \times 10800 {cm}^{3}$

The volume of the sphere and the wire are the same.
Therefore,

$π \times r^{2} \times 10800 = \frac{4}{3} π \times 9 \times 9 \times 9 \Rightarrow r^{2} = \frac{4 \times 9 \times 9 \times 9}{3 \times 10800} = \frac{4 \times 9 \times 9}{3 \times 4 \times 3} = 0.09 \Rightarrow r = \sqrt{0.09} = 0.3 cm Thus, d = 2 r = 2 \times 0.3 cm = 0.6 cm$

The diameter of the wire is 0.6 cm.

Question 19:

Diameter of sphere = 18 cm
Radius of the sphere = 9 cm

Volume of sphere $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times 9 \times 9 \times 9 {cm}^{3}$

Length of wire = 108 m = 10800 cm
Let radius of the wire be r cm.
Volume of the wire $= π r^{2} l = π \times r^{2} \times 10800 {cm}^{3}$

The volume of the sphere and the wire are the same.
Therefore,

$π \times r^{2} \times 10800 = \frac{4}{3} π \times 9 \times 9 \times 9 \Rightarrow r^{2} = \frac{4 \times 9 \times 9 \times 9}{3 \times 10800} = \frac{4 \times 9 \times 9}{3 \times 4 \times 3} = 0.09 \Rightarrow r = \sqrt{0.09} = 0.3 cm Thus, d = 2 r = 2 \times 0.3 cm = 0.6 cm$

The diameter of the wire is 0.6 cm.

Answer:

$We have, the radius of the hemispherical bowl, R = 9 cm and the internal base radius of the cylindrical vessel, r = 6 cm Let the height of the water in the cylindrical vessel be h . As, Volume of water in the cylindrical vessel = Volume of hemispherical bowl \Rightarrow π r^{2} h = \frac{2}{3} π R^{3} \Rightarrow r^{2} h = \frac{2}{3} R^{3} \Rightarrow 6 \times 6 \times h = \frac{2}{3} \times 9 \times 9 \times 9 \Rightarrow h = \frac{2}{3} \times \frac{9 \times 9 \times 9}{6 \times 6} \Rightarrow h = \frac{27}{2} ∴ h = 13.5 cm$

So, the height of the water in the cylindrical vessel is 13.5 cm.

Question 20:

$We have, the radius of the hemispherical bowl, R = 9 cm and the internal base radius of the cylindrical vessel, r = 6 cm Let the height of the water in the cylindrical vessel be h . As, Volume of water in the cylindrical vessel = Volume of hemispherical bowl \Rightarrow π r^{2} h = \frac{2}{3} π R^{3} \Rightarrow r^{2} h = \frac{2}{3} R^{3} \Rightarrow 6 \times 6 \times h = \frac{2}{3} \times 9 \times 9 \times 9 \Rightarrow h = \frac{2}{3} \times \frac{9 \times 9 \times 9}{6 \times 6} \Rightarrow h = \frac{27}{2} ∴ h = 13.5 cm$

So, the height of the water in the cylindrical vessel is 13.5 cm.

Answer:

$We have, the radius of the hemispherical tank, r = \frac{3}{2} m Volume of the hemispherical tank = \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} = \frac{99}{14} m^{3} Now, Volume of half tank = \frac{1}{2} \times \frac{99}{14} = \frac{99}{28} m^{3} = \frac{99}{28} kL = \frac{99000}{28} L As, the rate of water emptied by the pipe = \frac{25}{7} L / s So, the time taken to empty half the tank = \frac{(\frac{99000}{28})}{(\frac{25}{7})} = \frac{99000}{25 \times 4} = 990 s = \frac{990}{60} \min = 16.5 \min = 16 \min 30 s$

So, the time taken to empty half the tank is 16 minutes and 30 seconds.

Question 21:

$We have, the radius of the hemispherical tank, r = \frac{3}{2} m Volume of the hemispherical tank = \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{3}{2} = \frac{99}{14} m^{3} Now, Volume of half tank = \frac{1}{2} \times \frac{99}{14} = \frac{99}{28} m^{3} = \frac{99}{28} kL = \frac{99000}{28} L As, the rate of water emptied by the pipe = \frac{25}{7} L / s So, the time taken to empty half the tank = \frac{(\frac{99000}{28})}{(\frac{25}{7})} = \frac{99000}{25 \times 4} = 990 s = \frac{990}{60} \min = 16.5 \min = 16 \min 30 s$

So, the time taken to empty half the tank is 16 minutes and 30 seconds.

Answer:

$We have, the length of the roof, l = 44 m, the width of the roof, b = 20 m, the height of the cylindrical tank, H = 3.5 m and the base radius of the cylindrical tank, R = \frac{4}{2} = 2 m Let the height of the rainfall be h . Now, Volume of rainfall = Volume of cylindrical tank \Rightarrow l b h = π R^{2} H \Rightarrow 44 \times 20 \times h = \frac{22}{7} \times 2 \times 2 \times 3.5 \Rightarrow h = \frac{22}{7} \times \frac{2 \times 2 \times 3.5}{44 \times 20} \Rightarrow h = \frac{1}{20} m \Rightarrow h = \frac{100}{20} cm ∴ h = 5 cm$

So, the height of the rainfall is 5 cm.

Question 22:

$We have, the length of the roof, l = 44 m, the width of the roof, b = 20 m, the height of the cylindrical tank, H = 3.5 m and the base radius of the cylindrical tank, R = \frac{4}{2} = 2 m Let the height of the rainfall be h . Now, Volume of rainfall = Volume of cylindrical tank \Rightarrow l b h = π R^{2} H \Rightarrow 44 \times 20 \times h = \frac{22}{7} \times 2 \times 2 \times 3.5 \Rightarrow h = \frac{22}{7} \times \frac{2 \times 2 \times 3.5}{44 \times 20} \Rightarrow h = \frac{1}{20} m \Rightarrow h = \frac{100}{20} cm ∴ h = 5 cm$

So, the height of the rainfall is 5 cm.

Answer:

$We have, the length of the roof, l = 22 m, the width of the roof, b = 20 m, the base radius of the cylindrical vessel, R = \frac{2}{2} = 1 m and the height of the cylindrical vessel, H = 3.5 m Let the height of the rainfall be h . Now, Volume of rainfall = Volume of rain water collected in the cylindrical vessel \Rightarrow l b h = \frac{4}{5} \times Volume of cylindrical vessel \Rightarrow 22 \times 20 \times h = \frac{4}{5} \times π R^{2} H \Rightarrow 440 h = \frac{4}{5} \times \frac{22}{7} \times 1 \times 1 \times 3.5 \Rightarrow h = \frac{4}{5} \times \frac{22}{7} \times \frac{3.5}{440} \Rightarrow h = 0.02 m ∴ h = 2 cm$

So, the height of the rainfall is 2 cm.

Question 23:

$We have, the length of the roof, l = 22 m, the width of the roof, b = 20 m, the base radius of the cylindrical vessel, R = \frac{2}{2} = 1 m and the height of the cylindrical vessel, H = 3.5 m Let the height of the rainfall be h . Now, Volume of rainfall = Volume of rain water collected in the cylindrical vessel \Rightarrow l b h = \frac{4}{5} \times Volume of cylindrical vessel \Rightarrow 22 \times 20 \times h = \frac{4}{5} \times π R^{2} H \Rightarrow 440 h = \frac{4}{5} \times \frac{22}{7} \times 1 \times 1 \times 3.5 \Rightarrow h = \frac{4}{5} \times \frac{22}{7} \times \frac{3.5}{440} \Rightarrow h = 0.02 m ∴ h = 2 cm$

So, the height of the rainfall is 2 cm.

Answer:

$We have, height of cone, h = 60 cm, the base radius of cone, r = 30 cm, the height of cylinder, H = 180 cm and the base radius of the cylinder, R = 60 cm Now, Volume of water left in the cylinder = Volume of cylinder - Volume of cone = π R^{2} H - \frac{1}{3} π r^{2} h = \frac{22}{7} \times 60 \times 60 \times 180 - \frac{1}{3} \times \frac{22}{7} \times 30 \times 30 \times 60 = \frac{22}{7} \times 30 \times 30 \times 60 (2 \times 2 \times 3 - \frac{1}{3}) = \frac{22}{7} \times 54000 (12 - \frac{1}{3}) = \frac{22}{7} \times 54000 \times \frac{35}{3} = 1980000 {cm}^{3} = \frac{1980000}{1000000} m^{3} = 1.98 m^{3}$

So, the volume of water left in the cylinder is 1.98 m³.

Question 24:

$We have, height of cone, h = 60 cm, the base radius of cone, r = 30 cm, the height of cylinder, H = 180 cm and the base radius of the cylinder, R = 60 cm Now, Volume of water left in the cylinder = Volume of cylinder - Volume of cone = π R^{2} H - \frac{1}{3} π r^{2} h = \frac{22}{7} \times 60 \times 60 \times 180 - \frac{1}{3} \times \frac{22}{7} \times 30 \times 30 \times 60 = \frac{22}{7} \times 30 \times 30 \times 60 (2 \times 2 \times 3 - \frac{1}{3}) = \frac{22}{7} \times 54000 (12 - \frac{1}{3}) = \frac{22}{7} \times 54000 \times \frac{35}{3} = 1980000 {cm}^{3} = \frac{1980000}{1000000} m^{3} = 1.98 m^{3}$

So, the volume of water left in the cylinder is 1.98 m³.

Answer:

$We have, the internal radius of the cylindrical pipe, r = \frac{2}{2} = 1 cm and the base radius of cylindrical tank, R = 40 cm, Also, the rate of water flow, h = 0.4 m / s = 40 cm / s Let the rise in level of water be H . Now, The volume of water flowing out of the cylindrical pipe in 1 \sec = π r^{2} h = π \times 1 \times 1 \times 40 = 40 π {cm}^{3} So, the volume of water flowing out of the cylindrical pipe in half an hour (30 \min) = 40 π \times 60 \times 30 = 72000 π {cm}^{3} As, Volume of water in the cylindrical tank = Volume of standing water in cylindrical pipe for half an hour \Rightarrow π R^{2} H = 72000 π \Rightarrow R^{2} H = 72000 \Rightarrow 40 \times 40 \times H = 72000 \Rightarrow H = \frac{72000}{40 \times 40} ∴ H = 45 cm$

So, the rise in level of water in the tank in half an hour is 45 cm.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Question 25:

$We have, the internal radius of the cylindrical pipe, r = \frac{2}{2} = 1 cm and the base radius of cylindrical tank, R = 40 cm, Also, the rate of water flow, h = 0.4 m / s = 40 cm / s Let the rise in level of water be H . Now, The volume of water flowing out of the cylindrical pipe in 1 \sec = π r^{2} h = π \times 1 \times 1 \times 40 = 40 π {cm}^{3} So, the volume of water flowing out of the cylindrical pipe in half an hour (30 \min) = 40 π \times 60 \times 30 = 72000 π {cm}^{3} As, Volume of water in the cylindrical tank = Volume of standing water in cylindrical pipe for half an hour \Rightarrow π R^{2} H = 72000 π \Rightarrow R^{2} H = 72000 \Rightarrow 40 \times 40 \times H = 72000 \Rightarrow H = \frac{72000}{40 \times 40} ∴ H = 45 cm$

So, the rise in level of water in the tank in half an hour is 45 cm.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Answer:

$We have, Speed of the water flowing through the pipe, H = 6 km / h = \frac{600000 cm}{3600 s} = \frac{500}{3} cm / s, Radius of the pipe, R = \frac{14}{2} = 7 cm, Length of the rectangular tank, l = 60 m = 6000 cm, Breadth of the rectangular tank, b = 22 m = 2200 cm and Rise in the level of water in the tank, h = 7 cm Now, Volume of the water in the rectangular tank = l b h = 6000 \times 2200 \times 7 = 92400000 {cm}^{3} Also, Volume of the water flowing through the pipe in 1 s = π R^{2} H = \frac{22}{7} \times 7 \times 7 \times \frac{500}{3} = \frac{77000}{3} {cm}^{3} So, The time taken = \frac{Volume of the water in the rectangular tank}{Volume of the water flowing through the pipe in 1 s} = \frac{92400000}{(\frac{77000}{3})} = \frac{92400 \times 3}{77} = 3600 s = 1 hr$

So, the time in which the level of water in the tank will rise by 7 cm is 1 hour.

Question 26:

$We have, Speed of the water flowing through the pipe, H = 6 km / h = \frac{600000 cm}{3600 s} = \frac{500}{3} cm / s, Radius of the pipe, R = \frac{14}{2} = 7 cm, Length of the rectangular tank, l = 60 m = 6000 cm, Breadth of the rectangular tank, b = 22 m = 2200 cm and Rise in the level of water in the tank, h = 7 cm Now, Volume of the water in the rectangular tank = l b h = 6000 \times 2200 \times 7 = 92400000 {cm}^{3} Also, Volume of the water flowing through the pipe in 1 s = π R^{2} H = \frac{22}{7} \times 7 \times 7 \times \frac{500}{3} = \frac{77000}{3} {cm}^{3} So, The time taken = \frac{Volume of the water in the rectangular tank}{Volume of the water flowing through the pipe in 1 s} = \frac{92400000}{(\frac{77000}{3})} = \frac{92400 \times 3}{77} = 3600 s = 1 hr$

So, the time in which the level of water in the tank will rise by 7 cm is 1 hour.

Answer:

Width of the canal = 5.4 m
Depth of the canal = 1.8 m
Height of the standing water needed for irrigation = 10 cm = 0.1 m
Speed of the flowing water = 25 km/h = $\frac{25000}{60} = \frac{1250}{3}$ m/min
Volume of water flowing out of the canal in 1 min
= Area of opening of canal × $\frac{1250}{3}$
= $5.4 \times 1.8 \times \frac{1250}{3}$
= 4050 m³
∴ Volume of water flowing out of the canal in 40 min = 40 × 4050 m³ = 162000 m³
Now,
$Area of irrigation = \frac{Volume of water flowing out from canal in 40 \min}{Height of the standing water needed for irrigation} = \frac{162000}{0.1} = 1620000 m^{2} = 162 hectare (∵ 1 hectare = 10000 m^{2})$
Thus, the area irrigated in 40 minutes is 162 hectare.

Question 27:

Width of the canal = 5.4 m
Depth of the canal = 1.8 m
Height of the standing water needed for irrigation = 10 cm = 0.1 m
Speed of the flowing water = 25 km/h = $\frac{25000}{60} = \frac{1250}{3}$ m/min
Volume of water flowing out of the canal in 1 min
= Area of opening of canal × $\frac{1250}{3}$
= $5.4 \times 1.8 \times \frac{1250}{3}$
= 4050 m³
∴ Volume of water flowing out of the canal in 40 min = 40 × 4050 m³ = 162000 m³
Now,
$Area of irrigation = \frac{Volume of water flowing out from canal in 40 \min}{Height of the standing water needed for irrigation} = \frac{162000}{0.1} = 1620000 m^{2} = 162 hectare (∵ 1 hectare = 10000 m^{2})$
Thus, the area irrigated in 40 minutes is 162 hectare.

Answer:

$We have, the radius of the cylindrical tank, R = \frac{12}{2} = 6 m = 600 cm, the depth of the tank, H = 2.5 m = 250 cm, the radius of the cylindrical pipe, r = \frac{25}{2} = 12.5 cm, speed of the flowing water, h = 3.6 km / h = \frac{360000 m}{3600 s} = 100 cm / s Now, Volume of water flowing out from the pipe in a hour = π r^{2} h = \frac{22}{7} \times 12.5 \times 12.5 \times 100 {cm}^{3} Also, Volume of the tank = π R^{2} H = \frac{22}{7} \times 600 \times 600 \times 250 {cm}^{3} So, the time taken to fill the tank = \frac{Volume of the tank}{Volume of water flowing out from the pipe in a hour} = \frac{\frac{22}{7} \times 600 \times 600 \times 250}{\frac{22}{7} \times 12.5 \times 12.5 \times 100} = 5760 s = \frac{5760}{3600} = 1.6 h Also, the cost of water = 0.07 \times \frac{22}{7} \times 6 \times 6 \times 2.5 = ₹ 19.80$

Question 28:

$We have, the radius of the cylindrical tank, R = \frac{12}{2} = 6 m = 600 cm, the depth of the tank, H = 2.5 m = 250 cm, the radius of the cylindrical pipe, r = \frac{25}{2} = 12.5 cm, speed of the flowing water, h = 3.6 km / h = \frac{360000 m}{3600 s} = 100 cm / s Now, Volume of water flowing out from the pipe in a hour = π r^{2} h = \frac{22}{7} \times 12.5 \times 12.5 \times 100 {cm}^{3} Also, Volume of the tank = π R^{2} H = \frac{22}{7} \times 600 \times 600 \times 250 {cm}^{3} So, the time taken to fill the tank = \frac{Volume of the tank}{Volume of water flowing out from the pipe in a hour} = \frac{\frac{22}{7} \times 600 \times 600 \times 250}{\frac{22}{7} \times 12.5 \times 12.5 \times 100} = 5760 s = \frac{5760}{3600} = 1.6 h Also, the cost of water = 0.07 \times \frac{22}{7} \times 6 \times 6 \times 2.5 = ₹ 19.80$

Answer:

$We have, Radius of cylindrical pipe, r = \frac{7}{2} cm and The rate of flow of water = 192.5 L / \min = \frac{192.5 L}{1 \min} = \frac{192.5 \times 1000 {cm}^{3}}{1 \min} (As, 1 L = 1000 {cm}^{3}) = 192500 {cm}^{3} / \min \Rightarrow The volume of water flowing out from the cylindrical pipe in 1 \min = 192500 {cm}^{3} Now, the rate of flow of water in the pipe = \frac{The volume of water flowing out from the cylindrical pipe in 1 \min}{π r^{2}} = \frac{192500}{(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2})} = \frac{192500 \times 2}{77} = 5000 cm / \min = \frac{5000 \times 60}{1 \times 100000} km / hr = 3 km / hr$

So, the rate of flow of water in the pipe is 3 km/hr.

Question 29:

$We have, Radius of cylindrical pipe, r = \frac{7}{2} cm and The rate of flow of water = 192.5 L / \min = \frac{192.5 L}{1 \min} = \frac{192.5 \times 1000 {cm}^{3}}{1 \min} (As, 1 L = 1000 {cm}^{3}) = 192500 {cm}^{3} / \min \Rightarrow The volume of water flowing out from the cylindrical pipe in 1 \min = 192500 {cm}^{3} Now, the rate of flow of water in the pipe = \frac{The volume of water flowing out from the cylindrical pipe in 1 \min}{π r^{2}} = \frac{192500}{(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2})} = \frac{192500 \times 2}{77} = 5000 cm / \min = \frac{5000 \times 60}{1 \times 100000} km / hr = 3 km / hr$

So, the rate of flow of water in the pipe is 3 km/hr.

Answer:

$We have, the radius of spherical marble, r = \frac{1.4}{2} = 0.7 cm and the radius of the cylindrical vessel, R = \frac{7}{2} cm = 3.5 cm Let the rise in the level of water in the vessel be H . Now, Volume of water rised in the cylindrical vessel = Volume of 150 spherical marbles \Rightarrow π R^{2} H = 150 \times \frac{4}{3} π r^{3} \Rightarrow R^{2} H = 200 r^{3} \Rightarrow 3.5 \times 3.5 \times H = 200 \times 0.7 \times 0.7 \times 0.7 \Rightarrow H = \frac{200 \times 0.7 \times 0.7 \times 0.7}{3.5 \times 3.5} ∴ H = 5.6 cm$

So, the rise in the level of water in the vessel is 5.6 cm.

Disclaimer: The diameter of the spherical marbles should be 1.4 cm instead 14 cm. The has been corrected above.

Question 30:

$We have, the radius of spherical marble, r = \frac{1.4}{2} = 0.7 cm and the radius of the cylindrical vessel, R = \frac{7}{2} cm = 3.5 cm Let the rise in the level of water in the vessel be H . Now, Volume of water rised in the cylindrical vessel = Volume of 150 spherical marbles \Rightarrow π R^{2} H = 150 \times \frac{4}{3} π r^{3} \Rightarrow R^{2} H = 200 r^{3} \Rightarrow 3.5 \times 3.5 \times H = 200 \times 0.7 \times 0.7 \times 0.7 \Rightarrow H = \frac{200 \times 0.7 \times 0.7 \times 0.7}{3.5 \times 3.5} ∴ H = 5.6 cm$

So, the rise in the level of water in the vessel is 5.6 cm.

Disclaimer: The diameter of the spherical marbles should be 1.4 cm instead 14 cm. The has been corrected above.

Answer:

Diameter of each marble = 1.4 cm
Radius of each marble = 0.7 cm
Volume of each marble $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times {(0.7)}^{3} {cm}^{3}$

The water rises as a cylindrical column.
Volume of cylindrical column filled with water $= {πr}^{2} h = π \times {(\frac{7}{2})}^{2} \times 5.6 {cm}^{3}$

Total number of marbles

$= \frac{Volume of cylindrical water column}{Volume of marble} = \frac{π \times {(\frac{7}{2})}^{2} \times 5.6}{\frac{4}{3} π \times {(0.7)}^{3}} = \frac{7 \times 7 \times 5.6 \times 3}{2 \times 2 \times 4 \times 0.7 \times 0.7 \times 0.7} = 150$

Question 31:

Diameter of each marble = 1.4 cm
Radius of each marble = 0.7 cm
Volume of each marble $= \frac{4}{3} {πr}^{3} = \frac{4}{3} π \times {(0.7)}^{3} {cm}^{3}$

The water rises as a cylindrical column.
Volume of cylindrical column filled with water $= {πr}^{2} h = π \times {(\frac{7}{2})}^{2} \times 5.6 {cm}^{3}$

Total number of marbles

$= \frac{Volume of cylindrical water column}{Volume of marble} = \frac{π \times {(\frac{7}{2})}^{2} \times 5.6}{\frac{4}{3} π \times {(0.7)}^{3}} = \frac{7 \times 7 \times 5.6 \times 3}{2 \times 2 \times 4 \times 0.7 \times 0.7 \times 0.7} = 150$

Answer:

$We have, Radius of well, R = \frac{10}{2} = 5 m, Depth of the well, H = 14 m and Width of the embankment = 5 m, Also, the outer radius of the embankment, r = R + 5 = 5 + 5 = 10 m And, the inner radius of the embankment = R = 5 m Let the height of the embankment be h . Now, Volume of the embankment = Volume of the earth taken out \Rightarrow Volume of the embankment = Volume of the well \Rightarrow (π r^{2} - π R^{2}) h = π R^{2} H \Rightarrow π (r^{2} - R^{2}) h = π R^{2} H \Rightarrow (r^{2} - R^{2}) h = R^{2} H \Rightarrow (10^{2} - 5^{2}) h = 5 \times 5 \times 14 \Rightarrow (100 - 25) h = 25 \times 14 \Rightarrow 75 h = 25 \times 14 \Rightarrow h = \frac{25 \times 14}{75} ∴ h = \frac{14}{3} m$

So, the height of the embankment is $\frac{14}{3}$ m.

Value: We must lanour hard to make maximum use of the available resources.

Disclaimer: The answer provided in the textbook is incorrect. It has been corrected above.

Question 32:

$We have, Radius of well, R = \frac{10}{2} = 5 m, Depth of the well, H = 14 m and Width of the embankment = 5 m, Also, the outer radius of the embankment, r = R + 5 = 5 + 5 = 10 m And, the inner radius of the embankment = R = 5 m Let the height of the embankment be h . Now, Volume of the embankment = Volume of the earth taken out \Rightarrow Volume of the embankment = Volume of the well \Rightarrow (π r^{2} - π R^{2}) h = π R^{2} H \Rightarrow π (r^{2} - R^{2}) h = π R^{2} H \Rightarrow (r^{2} - R^{2}) h = R^{2} H \Rightarrow (10^{2} - 5^{2}) h = 5 \times 5 \times 14 \Rightarrow (100 - 25) h = 25 \times 14 \Rightarrow 75 h = 25 \times 14 \Rightarrow h = \frac{25 \times 14}{75} ∴ h = \frac{14}{3} m$

So, the height of the embankment is $\frac{14}{3}$ m.

Value: We must lanour hard to make maximum use of the available resources.

Disclaimer: The answer provided in the textbook is incorrect. It has been corrected above.

Answer:

$We have, Length of the fiel, l = 35 m, Width of the field, b = 22 m, Depth of the well, H = 8 m and Radius of the well, R = \frac{14}{2} = 7 m, Let the rise in the level of the field be h . Now, Volume of the earth on remaining part of the field = Volume of earth dug out \Rightarrow Area of the remaining field \times h = Volume of the well \Rightarrow (Area of the field - Area of base of the well) \times h = π R^{2} H \Rightarrow (l b - π R^{2}) \times h = π R^{2} H \Rightarrow (35 \times 22 - \frac{22}{7} \times 7 \times 7) \times h = \frac{22}{7} \times 7 \times 7 \times 8 \Rightarrow (770 - 154) \times h = 1232 \Rightarrow 616 \times h = 1232 \Rightarrow h = \frac{1232}{616} ∴ h = 2 m$

So, the rise in the level of the field is 2 m.

Question 33:

$We have, Length of the fiel, l = 35 m, Width of the field, b = 22 m, Depth of the well, H = 8 m and Radius of the well, R = \frac{14}{2} = 7 m, Let the rise in the level of the field be h . Now, Volume of the earth on remaining part of the field = Volume of earth dug out \Rightarrow Area of the remaining field \times h = Volume of the well \Rightarrow (Area of the field - Area of base of the well) \times h = π R^{2} H \Rightarrow (l b - π R^{2}) \times h = π R^{2} H \Rightarrow (35 \times 22 - \frac{22}{7} \times 7 \times 7) \times h = \frac{22}{7} \times 7 \times 7 \times 8 \Rightarrow (770 - 154) \times h = 1232 \Rightarrow 616 \times h = 1232 \Rightarrow h = \frac{1232}{616} ∴ h = 2 m$

So, the rise in the level of the field is 2 m.

Answer:

$We have, Diameter of the coppe wire, d = 6 mm = 0.6 cm, Radius of the copper wire, r = \frac{0.6}{2} = 0.3 cm, Length of the cylinder, H = 18 cm, Radius of the cylinder, R = \frac{49}{2} cm The number of rotations of the wire on the cylinder = \frac{Length of the cylinder, H}{Diameter of the copper wire, d} = \frac{18}{0.6} = 30 The circumference of the base of the cylinder = 2 π R = 2 \times \frac{22}{7} \times \frac{49}{2} = 154 cm So, the length of the wire, h = 30 \times 154 = 4620 cm = 46.2 m Now, the volume of the wire = π r^{2} h = \frac{22}{7} \times 0.3 \times 0.3 \times 4620 = 1306.8 {cm}^{3} Also, the weight of the wire = Volume of the wire \times Density of the wire = 1306.8 \times 8.8 = 11499.84 g \approx 11.5 kg$

Question 34:

$We have, Diameter of the coppe wire, d = 6 mm = 0.6 cm, Radius of the copper wire, r = \frac{0.6}{2} = 0.3 cm, Length of the cylinder, H = 18 cm, Radius of the cylinder, R = \frac{49}{2} cm The number of rotations of the wire on the cylinder = \frac{Length of the cylinder, H}{Diameter of the copper wire, d} = \frac{18}{0.6} = 30 The circumference of the base of the cylinder = 2 π R = 2 \times \frac{22}{7} \times \frac{49}{2} = 154 cm So, the length of the wire, h = 30 \times 154 = 4620 cm = 46.2 m Now, the volume of the wire = π r^{2} h = \frac{22}{7} \times 0.3 \times 0.3 \times 4620 = 1306.8 {cm}^{3} Also, the weight of the wire = Volume of the wire \times Density of the wire = 1306.8 \times 8.8 = 11499.84 g \approx 11.5 kg$

Answer:

$We have, In ∆ ABC, ∠ B = 90 °, AB = l_{1} = 15 cm and BC = l_{2} = 20 cm Let OD = OB = r, AO = h_{1} and CO = h_{2} Using Pythagoras theorem, AC = \sqrt{{AB}^{2} + {BC}^{2}} = \sqrt{15^{2} + 20^{2}} = \sqrt{225 + 400} = \sqrt{625} \Rightarrow h = 25 cm As, ar (∆ ABC) = \frac{1}{2} \times AC \times BO = \frac{1}{2} \times AB \times BC \Rightarrow AC \times BO = AB \times BC \Rightarrow 25 r = 15 \times 20 \Rightarrow r = \frac{15 \times 20}{25} \Rightarrow r = 12 cm Now, Volume of the double cone so formed = Volume of cone 1 + Volume of cone 2 = \frac{1}{3} π r^{2} h_{1} + \frac{1}{3} π r^{2} h_{2} = \frac{1}{3} π r^{2} (h_{1} + h_{2}) = \frac{1}{3} π r^{2} h = \frac{1}{3} \times 3.14 \times 12 \times 12 \times 25 = 3768 {cm}^{3} Also, Surace area of the solid so formed = CAS of cone 1 + CSA of cone 2 = π r l_{1} + π r l_{2} = πr (l_{1} + l_{2}) = \frac{22}{7} \times 12 \times (15 + 20) = \frac{22}{7} \times 12 \times 35 = 1320 {cm}^{2}$

Question 35:

$We have, In ∆ ABC, ∠ B = 90 °, AB = l_{1} = 15 cm and BC = l_{2} = 20 cm Let OD = OB = r, AO = h_{1} and CO = h_{2} Using Pythagoras theorem, AC = \sqrt{{AB}^{2} + {BC}^{2}} = \sqrt{15^{2} + 20^{2}} = \sqrt{225 + 400} = \sqrt{625} \Rightarrow h = 25 cm As, ar (∆ ABC) = \frac{1}{2} \times AC \times BO = \frac{1}{2} \times AB \times BC \Rightarrow AC \times BO = AB \times BC \Rightarrow 25 r = 15 \times 20 \Rightarrow r = \frac{15 \times 20}{25} \Rightarrow r = 12 cm Now, Volume of the double cone so formed = Volume of cone 1 + Volume of cone 2 = \frac{1}{3} π r^{2} h_{1} + \frac{1}{3} π r^{2} h_{2} = \frac{1}{3} π r^{2} (h_{1} + h_{2}) = \frac{1}{3} π r^{2} h = \frac{1}{3} \times 3.14 \times 12 \times 12 \times 25 = 3768 {cm}^{3} Also, Surace area of the solid so formed = CAS of cone 1 + CSA of cone 2 = π r l_{1} + π r l_{2} = πr (l_{1} + l_{2}) = \frac{22}{7} \times 12 \times (15 + 20) = \frac{22}{7} \times 12 \times 35 = 1320 {cm}^{2}$

Answer:

Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylindrical tank, V_c = $π r^{2} H$ = $π \times {(1)}^{2} \times 5 = 5 π m$

Length of the park (l) = 25 m
Breadth of park (b) = 20 m
height of standing water in the park = hâ€‹
Volume of water in the park = lbh = $25 \times 20 \times h$

Now water from the tank is used to irrigate the park. So,
Volume of cylindrical tank = Volume of water in the park

$\Rightarrow 5 π = 25 \times 20 \times h \Rightarrow \frac{5 π}{25 \times 20} = h \Rightarrow h = \frac{π}{100} m \Rightarrow h = 0.0314 m$

Through recycling of water, better use of the natural resource occurs without wastage. It helps in reducing and preventing pollution.
It thus helps in conserving water. This keeps the greenery alive in urban areas like in parks gardens etc.

Question 1:

Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylindrical tank, V_c = $π r^{2} H$ = $π \times {(1)}^{2} \times 5 = 5 π m$

Length of the park (l) = 25 m
Breadth of park (b) = 20 m
height of standing water in the park = hâ€‹
Volume of water in the park = lbh = $25 \times 20 \times h$

Now water from the tank is used to irrigate the park. So,
Volume of cylindrical tank = Volume of water in the park

$\Rightarrow 5 π = 25 \times 20 \times h \Rightarrow \frac{5 π}{25 \times 20} = h \Rightarrow h = \frac{π}{100} m \Rightarrow h = 0.0314 m$

Through recycling of water, better use of the natural resource occurs without wastage. It helps in reducing and preventing pollution.
It thus helps in conserving water. This keeps the greenery alive in urban areas like in parks gardens etc.

Answer:

$We have, Height of the frustum, h = 14 cm, Base radii, R = \frac{16}{2} = 8 cm and r = \frac{12}{2} = 6 cm The capacity of the glass = Volume of the frustum = \frac{1}{3} π h (R^{2} + r^{2} + r R) = \frac{1}{3} \times \frac{22}{7} \times 14 \times (8^{2} + 6^{2} + 8 \times 6) = \frac{1}{3} \times 22 \times 2 \times (64 + 36 + 48) = \frac{44}{3} \times 148 = \frac{6512}{3} {cm}^{3} \approx 2170.67 {cm}^{3}$

So, the capacity of the glass is 2170.67 cm³.

Question 2:

$We have, Height of the frustum, h = 14 cm, Base radii, R = \frac{16}{2} = 8 cm and r = \frac{12}{2} = 6 cm The capacity of the glass = Volume of the frustum = \frac{1}{3} π h (R^{2} + r^{2} + r R) = \frac{1}{3} \times \frac{22}{7} \times 14 \times (8^{2} + 6^{2} + 8 \times 6) = \frac{1}{3} \times 22 \times 2 \times (64 + 36 + 48) = \frac{44}{3} \times 148 = \frac{6512}{3} {cm}^{3} \approx 2170.67 {cm}^{3}$

So, the capacity of the glass is 2170.67 cm³.

Answer:

$We have, Height, h = 8 cm, Base radii, R = 18 cm and r = 12 cm Also, the slant height, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(18 - 12)}^{2} + 8^{2}} = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 cm Now, Total surface area of the solid frustum = π (R + r) l + π R^{2} + π r^{2} = 3.14 \times (18 + 12) \times 10 + 3.14 \times 18^{2} + 3.14 \times 12^{2} = 3.14 \times 30 \times 10 + 3.14 \times 324 + 3.14 \times 144 = 3.14 \times (300 + 324 + 144) = 3.14 \times 768 = 2411.52 {cm}^{2}$

So, the total surface area of the solid frustum is 2411.52 cm².

Question 3:

$We have, Height, h = 8 cm, Base radii, R = 18 cm and r = 12 cm Also, the slant height, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(18 - 12)}^{2} + 8^{2}} = \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 cm Now, Total surface area of the solid frustum = π (R + r) l + π R^{2} + π r^{2} = 3.14 \times (18 + 12) \times 10 + 3.14 \times 18^{2} + 3.14 \times 12^{2} = 3.14 \times 30 \times 10 + 3.14 \times 324 + 3.14 \times 144 = 3.14 \times (300 + 324 + 144) = 3.14 \times 768 = 2411.52 {cm}^{2}$

So, the total surface area of the solid frustum is 2411.52 cm².

Answer:

$We have, Height, h = 24 cm, Upper base radius, R = 14 cm and lower base radius, r = 7 cm Also, the slant height, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(14 - 7)}^{2} + 24^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25 cm (i) Volume of the bucket = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 24 \times (14^{2} + 7^{2} + 14 \times 7) = \frac{22}{7} \times 8 \times (196 + 49 + 98) = \frac{176}{7} \times 343 = 8624 {cm}^{3} So, the volume of water which can completely fill the bucket is 8624 {cm}^{3} . (ii) Surface area of the bucket = π (R + r) l + π r^{2} = \frac{22}{7} \times (14 + 7) \times 25 + \frac{22}{7} \times 7 \times 7 = \frac{22}{7} \times 21 \times 25 + 22 \times 7 = 22 \times 3 \times 25 + 22 \times 7 = 1650 + 154 = 1804 {cm}^{2} So, the area of the metal sheet used to make the bucket is 1804 {cm}^{2} .$

Question 4:

$We have, Height, h = 24 cm, Upper base radius, R = 14 cm and lower base radius, r = 7 cm Also, the slant height, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(14 - 7)}^{2} + 24^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25 cm (i) Volume of the bucket = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 24 \times (14^{2} + 7^{2} + 14 \times 7) = \frac{22}{7} \times 8 \times (196 + 49 + 98) = \frac{176}{7} \times 343 = 8624 {cm}^{3} So, the volume of water which can completely fill the bucket is 8624 {cm}^{3} . (ii) Surface area of the bucket = π (R + r) l + π r^{2} = \frac{22}{7} \times (14 + 7) \times 25 + \frac{22}{7} \times 7 \times 7 = \frac{22}{7} \times 21 \times 25 + 22 \times 7 = 22 \times 3 \times 25 + 22 \times 7 = 1650 + 154 = 1804 {cm}^{2} So, the area of the metal sheet used to make the bucket is 1804 {cm}^{2} .$

Answer:

$We have, Height, h = 24 cm, Upper radius, R = 20 cm and Lower radius, r = 8 cm Now, Volume of the container = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 24 \times (20^{2} + 8^{2} + 20 \times 8) = \frac{176}{7} \times (400 + 64 + 160) = \frac{176}{7} \times 624 = \frac{109824}{7} {cm}^{3} = \frac{109.824}{7} L (As, 1000 {cm}^{3} = 1 L) So, the cost of the milk in the container = \frac{109.824}{7} \times 21 = 329.472 \approx ₹ 329.47$

Hence, the cost of milk which can completely fill the container is â‚¹329.47.

Question 5:

$We have, Height, h = 24 cm, Upper radius, R = 20 cm and Lower radius, r = 8 cm Now, Volume of the container = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 24 \times (20^{2} + 8^{2} + 20 \times 8) = \frac{176}{7} \times (400 + 64 + 160) = \frac{176}{7} \times 624 = \frac{109824}{7} {cm}^{3} = \frac{109.824}{7} L (As, 1000 {cm}^{3} = 1 L) So, the cost of the milk in the container = \frac{109.824}{7} \times 21 = 329.472 \approx ₹ 329.47$

Hence, the cost of milk which can completely fill the container is â‚¹329.47.

Answer:

Let $r = 8 cm, R = 20 cm, h = 16 cm .$
$\Rightarrow l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(20 - 8)}^{2} + 16^{2}} = \sqrt{144 + 256} = \sqrt{400} \Rightarrow l = 20 cm .$
(i)
Total metal sheet required to make the container = Curved surface area of frustum + Area of the base
$= π (r + R) l + π r^{2} = π (8 + 20) 20 + π {(8)}^{2} = 560 π + 64 π = 624 π = 1961.14 {cm}^{2} .$
The cost for 100 cm² of sheet = Rs. 10.
The cost of 1 cm² of sheet = $\frac{10}{100} = Rs . 0.1$ .
The cost of 1961.14 cm² of sheet = $1961.14 \times 0.1 = Rs . 196.11$
DISCLAIMER: The answer calculated above is not matching with the answer provided in the textbook.

(ii)
The volume of frustum =
$\frac{π h}{3} [r^{2} + R^{2} + r R] = \frac{22}{7} \times \frac{16}{3} [8^{2} + 20^{2} + 160] = \frac{22}{7} \times \frac{16}{3} [64 + 400 + 160] = \frac{22}{7} \times \frac{16}{3} \times 624 = \frac{73216}{7} {cm}^{3} .$
we know that 1 l=1000 cm³ or 1 cm³ = 0.001 l.
$\Rightarrow$ Volume= $\frac{73216}{7} \times 0.001 = \frac{73216}{7} \times \frac{1}{1000} = \frac{73.216}{7} l .$
Cost of milk is Rs. 35 per litre.
Hence, the cost at which this frustum can be filled = $\frac{73.216}{7} \times 35 = Rs 366.08 .$

Question 6:

Let $r = 8 cm, R = 20 cm, h = 16 cm .$
$\Rightarrow l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(20 - 8)}^{2} + 16^{2}} = \sqrt{144 + 256} = \sqrt{400} \Rightarrow l = 20 cm .$
(i)
Total metal sheet required to make the container = Curved surface area of frustum + Area of the base
$= π (r + R) l + π r^{2} = π (8 + 20) 20 + π {(8)}^{2} = 560 π + 64 π = 624 π = 1961.14 {cm}^{2} .$
The cost for 100 cm² of sheet = Rs. 10.
The cost of 1 cm² of sheet = $\frac{10}{100} = Rs . 0.1$ .
The cost of 1961.14 cm² of sheet = $1961.14 \times 0.1 = Rs . 196.11$
DISCLAIMER: The answer calculated above is not matching with the answer provided in the textbook.

(ii)
The volume of frustum =
$\frac{π h}{3} [r^{2} + R^{2} + r R] = \frac{22}{7} \times \frac{16}{3} [8^{2} + 20^{2} + 160] = \frac{22}{7} \times \frac{16}{3} [64 + 400 + 160] = \frac{22}{7} \times \frac{16}{3} \times 624 = \frac{73216}{7} {cm}^{3} .$
we know that 1 l=1000 cm³ or 1 cm³ = 0.001 l.
$\Rightarrow$ Volume= $\frac{73216}{7} \times 0.001 = \frac{73216}{7} \times \frac{1}{1000} = \frac{73.216}{7} l .$
Cost of milk is Rs. 35 per litre.
Hence, the cost at which this frustum can be filled = $\frac{73.216}{7} \times 35 = Rs 366.08 .$

Answer:

Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm

Using the formula for height of a frustum:

Height = h =
$= \sqrt{l^{2} - {(R - r)}^{2}} = \sqrt{10^{2} - {(33 - 27)}^{2}} = \sqrt{100 - {(6)}^{2}} = \sqrt{100 - 36} = \sqrt{64} = 8 cm$

Capacity of the frustum
$= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 8 (33^{2} + 27^{2} + 33 \times 27) = \frac{22 \times 8}{3 \times 7} \times 2709 = 22704 {cm}^{3}$

Surface area of the frustum

$= π R^{2} + π r^{2} + π l (R + r) = π [R^{2} + r^{2} + l (R + r)] = \frac{22}{7} [33^{2} + 27^{2} + 10 (33 + 27)] = \frac{22}{7} [1089 + 729 + 10 (60)] = \frac{22 \times 2418}{7} = 7599.43 {cm}^{2}$

Question 7:

Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm

Using the formula for height of a frustum:

Height = h =
$= \sqrt{l^{2} - {(R - r)}^{2}} = \sqrt{10^{2} - {(33 - 27)}^{2}} = \sqrt{100 - {(6)}^{2}} = \sqrt{100 - 36} = \sqrt{64} = 8 cm$

Capacity of the frustum
$= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 8 (33^{2} + 27^{2} + 33 \times 27) = \frac{22 \times 8}{3 \times 7} \times 2709 = 22704 {cm}^{3}$

Surface area of the frustum

$= π R^{2} + π r^{2} + π l (R + r) = π [R^{2} + r^{2} + l (R + r)] = \frac{22}{7} [33^{2} + 27^{2} + 10 (33 + 27)] = \frac{22}{7} [1089 + 729 + 10 (60)] = \frac{22 \times 2418}{7} = 7599.43 {cm}^{2}$

Answer:

Greater diameter of the frustum = 56 cm
Greater radius of the frustum = R = 28 cm
Smaller diameter of the frustum = 42 cm
Radius of the smaller end of the frustum = r = 21 cm
Height of the frustum = h = 15 cm

Capacity of the frustum
$= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 15 (28^{2} + 21^{2} + 28 \times 21) = \frac{22 \times 5}{7} \times 1813 = 28490 {cm}^{3} = 28.49 litres$

Question 8:

Greater diameter of the frustum = 56 cm
Greater radius of the frustum = R = 28 cm
Smaller diameter of the frustum = 42 cm
Radius of the smaller end of the frustum = r = 21 cm
Height of the frustum = h = 15 cm

Capacity of the frustum
$= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 15 (28^{2} + 21^{2} + 28 \times 21) = \frac{22 \times 5}{7} \times 1813 = 28490 {cm}^{3} = 28.49 litres$

Answer:

Greater radius of the frustum = R = 20 cm
Smaller radius of the frustum = r = 8 cm
Height of the frustum = h = 16 cm

Slant height, l, of the frustum
$= \sqrt{h^{2} + (R^{2} - r^{2})} = \sqrt{16^{2} + {(20 - 8)}^{2}} = \sqrt{256 + {(12)}^{2}} = \sqrt{256 + 144} = \sqrt{400} = 20 cm$

Surface area of the frustum
$= {πr}^{2} + πl (R + r) = π [r^{2} + l (R + r)] = \frac{22}{7} [8^{2} + 20 (20 + 8)] = \frac{22}{7} [64 + 20 (28)] = \frac{22 \times 624}{7} = 1961.14 {cm}^{2}$

100 cm² of metal sheet costs Rs 15.
So, total cost of the metal sheet used for fabrication
$= \frac{1961.14}{100} \times 15 = Rs 294.171$

Question 9:

Greater radius of the frustum = R = 20 cm
Smaller radius of the frustum = r = 8 cm
Height of the frustum = h = 16 cm

Slant height, l, of the frustum
$= \sqrt{h^{2} + (R^{2} - r^{2})} = \sqrt{16^{2} + {(20 - 8)}^{2}} = \sqrt{256 + {(12)}^{2}} = \sqrt{256 + 144} = \sqrt{400} = 20 cm$

Surface area of the frustum
$= {πr}^{2} + πl (R + r) = π [r^{2} + l (R + r)] = \frac{22}{7} [8^{2} + 20 (20 + 8)] = \frac{22}{7} [64 + 20 (28)] = \frac{22 \times 624}{7} = 1961.14 {cm}^{2}$

100 cm² of metal sheet costs Rs 15.
So, total cost of the metal sheet used for fabrication
$= \frac{1961.14}{100} \times 15 = Rs 294.171$

Answer:

Greater diameter of the bucket = 30 cm
Radius of the bigger end of the bucket = R = 15 cm
Diameter of the smaller end of the bucket = 10 cm
Radius of the smaller end of the bucket = r = 5 cm
Height of the bucket = 24 cm

Slant height, l
$= \sqrt{h^{2} + {(R - r)}^{2}} = \sqrt{24^{2} + {(15 - 5)}^{2}} = \sqrt{576 + {(10)}^{2}} = \sqrt{576 + 100} = \sqrt{676} = 26 cm$

Capacity of the frustum

$= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times 3.14 \times 24 (15^{2} + 5^{2} + 15 \times 5) = 3.14 \times 8 \times 325 = 8164 {cm}^{3} = 8.164 litres$

A litre of milk cost Rs 20.
So, total cost of filling the bucket with milk $= 8.164 \times 20 = Rs 163.28$

Surface area of the bucket
$= π r^{2} + π l (R + r) = π [5^{2} + 26 (15 + 5)] = 3.14 [25 + 26 (20)] = 3.14 [25 + 520] = 1711.3 {cm}^{2}$

Cost of 100 cm²of metal sheet is Rs 10.
So, cost of metal used for making the bucket $= \frac{1711.3}{100} \times 10 = Rs 171.13$

Question 10:

Greater diameter of the bucket = 30 cm
Radius of the bigger end of the bucket = R = 15 cm
Diameter of the smaller end of the bucket = 10 cm
Radius of the smaller end of the bucket = r = 5 cm
Height of the bucket = 24 cm

Slant height, l
$= \sqrt{h^{2} + {(R - r)}^{2}} = \sqrt{24^{2} + {(15 - 5)}^{2}} = \sqrt{576 + {(10)}^{2}} = \sqrt{576 + 100} = \sqrt{676} = 26 cm$

Capacity of the frustum

$= \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times 3.14 \times 24 (15^{2} + 5^{2} + 15 \times 5) = 3.14 \times 8 \times 325 = 8164 {cm}^{3} = 8.164 litres$

A litre of milk cost Rs 20.
So, total cost of filling the bucket with milk $= 8.164 \times 20 = Rs 163.28$

Surface area of the bucket
$= π r^{2} + π l (R + r) = π [5^{2} + 26 (15 + 5)] = 3.14 [25 + 26 (20)] = 3.14 [25 + 520] = 1711.3 {cm}^{2}$

Cost of 100 cm²of metal sheet is Rs 10.
So, cost of metal used for making the bucket $= \frac{1711.3}{100} \times 10 = Rs 171.13$

Answer:

$We have, Height, h = 14 cm, Radius of upper end, R = \frac{35}{2} = 17.5 cm and Radius of lower end, r = \frac{30}{2} = 15 cm Now, Volume of the container = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 14 \times (17 . 5^{2} + 15^{2} + 17.5 \times 15) = \frac{44}{3} \times (306.25 + 225 + 262.5) = \frac{44}{3} \times 793.75 = \frac{34925}{3} {cm}^{3} So, the mass of the oil that is completely filled in the container = \frac{34925}{3} \times 1.2 = 13970 kg = 13.97 kg ∴ The cost of the oil in the container = 40 \times 13.97 = ₹ 558.80$

Question 11:

$We have, Height, h = 14 cm, Radius of upper end, R = \frac{35}{2} = 17.5 cm and Radius of lower end, r = \frac{30}{2} = 15 cm Now, Volume of the container = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 14 \times (17 . 5^{2} + 15^{2} + 17.5 \times 15) = \frac{44}{3} \times (306.25 + 225 + 262.5) = \frac{44}{3} \times 793.75 = \frac{34925}{3} {cm}^{3} So, the mass of the oil that is completely filled in the container = \frac{34925}{3} \times 1.2 = 13970 kg = 13.97 kg ∴ The cost of the oil in the container = 40 \times 13.97 = ₹ 558.80$

Answer:

$We have, Radius of upper end, R = 28 cm and Radius of lower end, r = 21 cm Let the height of the bucket be h . Now, Volume of water the bucket can hold = 28.49 L \Rightarrow Volume of bucket = 28490 {cm}^{3} (As, 1 L = 1000 {cm}^{3}) \Rightarrow \frac{1}{3} π h (R^{2} + r^{2} + R r) = 28490 \Rightarrow \frac{1}{3} \times \frac{22}{7} \times h \times (28^{2} + 21^{2} + 28 \times 21) = 28490 \Rightarrow \frac{22 h}{21} \times 49 \times (16 + 9 + 12) = 28490 \Rightarrow \frac{22 h}{3} \times 7 \times 37 = 28490 \Rightarrow h = \frac{28490 \times 3}{22 \times 7 \times 37} ∴ h = 15 cm$

So, the height of the bucket is 15 cm.

Question 12:

$We have, Radius of upper end, R = 28 cm and Radius of lower end, r = 21 cm Let the height of the bucket be h . Now, Volume of water the bucket can hold = 28.49 L \Rightarrow Volume of bucket = 28490 {cm}^{3} (As, 1 L = 1000 {cm}^{3}) \Rightarrow \frac{1}{3} π h (R^{2} + r^{2} + R r) = 28490 \Rightarrow \frac{1}{3} \times \frac{22}{7} \times h \times (28^{2} + 21^{2} + 28 \times 21) = 28490 \Rightarrow \frac{22 h}{21} \times 49 \times (16 + 9 + 12) = 28490 \Rightarrow \frac{22 h}{3} \times 7 \times 37 = 28490 \Rightarrow h = \frac{28490 \times 3}{22 \times 7 \times 37} ∴ h = 15 cm$

So, the height of the bucket is 15 cm.

Answer:

$We have, Height, h = 15 cm, Radius of the upper end, R = 14 cm, Radius of lower end = r, As, Volume of the bucket = 5390 {cm}^{3} \Rightarrow \frac{1}{3} π h (R^{2} + r^{2} + R r) = 5390 \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 15 \times (14^{2} + r^{2} + 14 r) = 5390 \Rightarrow \frac{110}{7} \times (196 + r^{2} + 14 r) = 5390 \Rightarrow 196 + r^{2} + 14 r = \frac{5390 \times 7}{110} \Rightarrow 196 + r^{2} + 14 r = 343 \Rightarrow r^{2} + 14 r - 147 = 0 \Rightarrow r^{2} + 21 r - 7 r - 147 = 0 \Rightarrow r (r + 21) - 7 (r + 21) = 0 \Rightarrow (r + 21) (r - 7) = 0 \Rightarrow r + 21 = 0 or r - 7 = 0 \Rightarrow r = - 21 or r = 7 As, r cannot be negative ∴ r = 7 cm$

So, the value of r is 7 cm.

Question 13:

$We have, Height, h = 15 cm, Radius of the upper end, R = 14 cm, Radius of lower end = r, As, Volume of the bucket = 5390 {cm}^{3} \Rightarrow \frac{1}{3} π h (R^{2} + r^{2} + R r) = 5390 \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 15 \times (14^{2} + r^{2} + 14 r) = 5390 \Rightarrow \frac{110}{7} \times (196 + r^{2} + 14 r) = 5390 \Rightarrow 196 + r^{2} + 14 r = \frac{5390 \times 7}{110} \Rightarrow 196 + r^{2} + 14 r = 343 \Rightarrow r^{2} + 14 r - 147 = 0 \Rightarrow r^{2} + 21 r - 7 r - 147 = 0 \Rightarrow r (r + 21) - 7 (r + 21) = 0 \Rightarrow (r + 21) (r - 7) = 0 \Rightarrow r + 21 = 0 or r - 7 = 0 \Rightarrow r = - 21 or r = 7 As, r cannot be negative ∴ r = 7 cm$

So, the value of r is 7 cm.

Answer:

Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm

Surface area of the frustum

$= π R^{2} + π r^{2} + π l (R + r) = π [R^{2} + r^{2} + l (R + r)] = [33^{2} + 27^{2} + 10 (33 + 27)] π = [1089 + 729 + 10 (60)] π = 2418 \times 3.14 = 7592.52 {cm}^{2}$

Question 14:

Greater radius = R = 33 cm
Smaller radius = r = 27 cm
Slant height = l = 10 cm

Surface area of the frustum

$= π R^{2} + π r^{2} + π l (R + r) = π [R^{2} + r^{2} + l (R + r)] = [33^{2} + 27^{2} + 10 (33 + 27)] π = [1089 + 729 + 10 (60)] π = 2418 \times 3.14 = 7592.52 {cm}^{2}$

Answer:

For the lower portion of the tent:
Diameter of the base= 20 m
Radius, R, of the base = 10 m
Diameter of the top end of the frustum = 6 m
Radius of the top end of the frustum = r = 3 m

Height of the frustum = h = 24 m

Slant height = l
$= \sqrt{h^{2} + {(R - r)}^{2}} = \sqrt{24^{2} + {(10 - 3)}^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25 m$

For the conical part:
Radius of the cone's base = r = 3 m
Height of the cone = Total height - Height of the frustum = 28-24 = 4 m

Slant height, L, of the cone = $\sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 m$

Total quantity of canvas = Curved surface area of the frustum + curved surface area of the conical top
$= (π l (R + r)) + π L r = π (l (R + r) + L r) = \frac{22}{7} (25 \times 13 + 5 \times 3) = \frac{22}{7} (325 + 15) = 1068.57 m^{2}$

Question 15:

For the lower portion of the tent:
Diameter of the base= 20 m
Radius, R, of the base = 10 m
Diameter of the top end of the frustum = 6 m
Radius of the top end of the frustum = r = 3 m

Height of the frustum = h = 24 m

Slant height = l
$= \sqrt{h^{2} + {(R - r)}^{2}} = \sqrt{24^{2} + {(10 - 3)}^{2}} = \sqrt{576 + 49} = \sqrt{625} = 25 m$

For the conical part:
Radius of the cone's base = r = 3 m
Height of the cone = Total height - Height of the frustum = 28-24 = 4 m

Slant height, L, of the cone = $\sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 m$

Total quantity of canvas = Curved surface area of the frustum + curved surface area of the conical top
$= (π l (R + r)) + π L r = π (l (R + r) + L r) = \frac{22}{7} (25 \times 13 + 5 \times 3) = \frac{22}{7} (325 + 15) = 1068.57 m^{2}$

Answer:

For the frustum:
Upper diameter = 14 m
Upper Radius, r = 7 m
Lower diameter = 26 m
Lower Radius, R = 13 m

Height of the frustum= h = 8 m
Slant height l =
$\sqrt{h^{2} + {(R - r)}^{2}} = \sqrt{8^{2} + {(13 - 7)}^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 m$

For the conical part:
Radius of the base = r = 7 m
Slant height = L =12 m

Total surface area of the tent = Curved area of frustum + Curved area of the cone
$= πl (R + r) + πrL = \frac{22}{7} \times 10 (13 + 7) + \frac{22}{7} \times 7 \times 12 = \frac{22}{7} (200 + 84) = 892.57 m^{2}$

Question 16:

For the frustum:
Upper diameter = 14 m
Upper Radius, r = 7 m
Lower diameter = 26 m
Lower Radius, R = 13 m

Height of the frustum= h = 8 m
Slant height l =
$\sqrt{h^{2} + {(R - r)}^{2}} = \sqrt{8^{2} + {(13 - 7)}^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 m$

For the conical part:
Radius of the base = r = 7 m
Slant height = L =12 m

Total surface area of the tent = Curved area of frustum + Curved area of the cone
$= πl (R + r) + πrL = \frac{22}{7} \times 10 (13 + 7) + \frac{22}{7} \times 7 \times 12 = \frac{22}{7} (200 + 84) = 892.57 m^{2}$

Answer:

$We have, Perimeter of upper end, C = 48 cm, Perimeter of lower end, c = 36 cm and Height, h = 11 cm Let the radius of upper end be R and the radius of lower end be r . As, C = 48 cm \Rightarrow 2 π R = 48 \Rightarrow R = \frac{48}{2 π} \Rightarrow R = \frac{24}{π} cm Similarly, c = 36 cm \Rightarrow r = \frac{36}{2 π} \Rightarrow r = \frac{18}{π} cm And, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(\frac{24}{π} - \frac{18}{π})}^{2} + 11^{2}} = \sqrt{{(\frac{6}{π})}^{2} + 11^{2}} = \sqrt{{(\frac{6 \times 7}{22})}^{2} + 11^{2}} = \sqrt{{(\frac{21}{11})}^{2} + 11^{2}} = \sqrt{\frac{441 + 121 \times 121}{121}} = \sqrt{\frac{441 + 14641}{121}} = \frac{\sqrt{15082}}{11} cm Now, Volume of the frustum = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times π \times 11 \times [{(\frac{24}{π})}^{2} + {(\frac{18}{π})}^{2} + (\frac{24}{π}) \times (\frac{18}{π})] = \frac{11 π}{3} \times [\frac{576}{π^{2}} + \frac{324}{π^{2}} + \frac{432}{π^{2}}] = \frac{11 π}{3} \times \frac{1332}{π^{2}} = \frac{11}{3} \times \frac{1332}{π} = \frac{11}{3} \times \frac{1332 \times 7}{22} = 1554 {cm}^{3} Also, Curved surface area of the frustum = π (R + r) l = \frac{22}{7} \times (\frac{24}{π} + \frac{18}{π}) \times \frac{\sqrt{15082}}{11} = \frac{22}{7} \times \frac{42}{π} \times \frac{\sqrt{15082}}{11} = \frac{22}{7} \times \frac{42 \times 7}{22} \times \frac{\sqrt{15082}}{11} \approx 42 \times 11.164436 \approx 468.91 {cm}^{2}$

Question 17:

$We have, Perimeter of upper end, C = 48 cm, Perimeter of lower end, c = 36 cm and Height, h = 11 cm Let the radius of upper end be R and the radius of lower end be r . As, C = 48 cm \Rightarrow 2 π R = 48 \Rightarrow R = \frac{48}{2 π} \Rightarrow R = \frac{24}{π} cm Similarly, c = 36 cm \Rightarrow r = \frac{36}{2 π} \Rightarrow r = \frac{18}{π} cm And, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(\frac{24}{π} - \frac{18}{π})}^{2} + 11^{2}} = \sqrt{{(\frac{6}{π})}^{2} + 11^{2}} = \sqrt{{(\frac{6 \times 7}{22})}^{2} + 11^{2}} = \sqrt{{(\frac{21}{11})}^{2} + 11^{2}} = \sqrt{\frac{441 + 121 \times 121}{121}} = \sqrt{\frac{441 + 14641}{121}} = \frac{\sqrt{15082}}{11} cm Now, Volume of the frustum = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times π \times 11 \times [{(\frac{24}{π})}^{2} + {(\frac{18}{π})}^{2} + (\frac{24}{π}) \times (\frac{18}{π})] = \frac{11 π}{3} \times [\frac{576}{π^{2}} + \frac{324}{π^{2}} + \frac{432}{π^{2}}] = \frac{11 π}{3} \times \frac{1332}{π^{2}} = \frac{11}{3} \times \frac{1332}{π} = \frac{11}{3} \times \frac{1332 \times 7}{22} = 1554 {cm}^{3} Also, Curved surface area of the frustum = π (R + r) l = \frac{22}{7} \times (\frac{24}{π} + \frac{18}{π}) \times \frac{\sqrt{15082}}{11} = \frac{22}{7} \times \frac{42}{π} \times \frac{\sqrt{15082}}{11} = \frac{22}{7} \times \frac{42 \times 7}{22} \times \frac{\sqrt{15082}}{11} \approx 42 \times 11.164436 \approx 468.91 {cm}^{2}$

Answer:

$We have, Radius of solid cone, R = CP = 10 cm, Let the height of the solid cone be, AP = H, the radius of the smaller cone, QD = r and the height of the smaller cone be, AQ = h . Also, AQ = \frac{AP}{2} i . e . h = \frac{H}{2} or H = 2 h . . . . . (i) Now, in ∆ AQD and ∆ APC, ∠ QAD = ∠ PAC (Common angle) ∠ AQD = ∠ APC = 90 ° So, by AA criteria ∆ AQD ~ APC \Rightarrow \frac{AQ}{AP} = \frac{QD}{PC} \Rightarrow \frac{h}{H} = \frac{r}{R} \Rightarrow \frac{h}{2 h} = \frac{r}{R} [Using (i)] \Rightarrow \frac{1}{2} = \frac{r}{R} \Rightarrow R = 2 r . . . . . (ii) As, Volume of smaller cone = \frac{1}{3} π r^{2} h And, Volume of solid cone = \frac{1}{3} π R^{2} H = \frac{1}{3} π {(2 r)}^{2} \times (2 h) [Using (i) and (ii)] = \frac{8}{3} π r^{2} h So, Volume of frustum = Volume of solid cone - Volume of smaller cone = \frac{8}{3} π r^{2} h - \frac{1}{3} π r^{2} h = \frac{7}{3} π r^{2} h Now, the ratio of the volumes of the two parts = \frac{Volume of the smaller cone}{Volume of the frustum} = \frac{(\frac{1}{3} π r^{2} h)}{(\frac{7}{3} π r^{2} h)} = \frac{1}{7} = 1 : 7$

So, the ratio of the volume of the two parts of the cone is 1 : 7.

Question 18:

$We have, Radius of solid cone, R = CP = 10 cm, Let the height of the solid cone be, AP = H, the radius of the smaller cone, QD = r and the height of the smaller cone be, AQ = h . Also, AQ = \frac{AP}{2} i . e . h = \frac{H}{2} or H = 2 h . . . . . (i) Now, in ∆ AQD and ∆ APC, ∠ QAD = ∠ PAC (Common angle) ∠ AQD = ∠ APC = 90 ° So, by AA criteria ∆ AQD ~ APC \Rightarrow \frac{AQ}{AP} = \frac{QD}{PC} \Rightarrow \frac{h}{H} = \frac{r}{R} \Rightarrow \frac{h}{2 h} = \frac{r}{R} [Using (i)] \Rightarrow \frac{1}{2} = \frac{r}{R} \Rightarrow R = 2 r . . . . . (ii) As, Volume of smaller cone = \frac{1}{3} π r^{2} h And, Volume of solid cone = \frac{1}{3} π R^{2} H = \frac{1}{3} π {(2 r)}^{2} \times (2 h) [Using (i) and (ii)] = \frac{8}{3} π r^{2} h So, Volume of frustum = Volume of solid cone - Volume of smaller cone = \frac{8}{3} π r^{2} h - \frac{1}{3} π r^{2} h = \frac{7}{3} π r^{2} h Now, the ratio of the volumes of the two parts = \frac{Volume of the smaller cone}{Volume of the frustum} = \frac{(\frac{1}{3} π r^{2} h)}{(\frac{7}{3} π r^{2} h)} = \frac{1}{7} = 1 : 7$

So, the ratio of the volume of the two parts of the cone is 1 : 7.

Answer:

$We have, Height of the given cone, H = 20 cm Let the radius of the given cone be R, the height of the smaller cone be h and the radius of the smaller cone be r . Now, in ∆ AQD and ∆ APC, ∠ QAD = ∠ PAC (Common angle) ∠ AQD = ∠ APC = 90 ° So, by AA criteria ∆ AQD ~ ∆ APC \Rightarrow \frac{AQ}{AP} = \frac{QD}{PC} \Rightarrow \frac{h}{H} = \frac{r}{R} . . . . . (i) Volume of smaller cone = \frac{1}{8} \times Volume of the given cone \Rightarrow \frac{Volume of smaller cone}{Volume of the given cone} = \frac{1}{8} \Rightarrow \frac{(\frac{1}{3} π r^{2} h)}{(\frac{1}{3} π R^{2} H)} = \frac{1}{8} \Rightarrow {(\frac{r}{R})}^{2} \times (\frac{h}{H}) = \frac{1}{8} \Rightarrow {(\frac{h}{H})}^{2} \times (\frac{h}{H}) = \frac{1}{8} [Using (i)] \Rightarrow {(\frac{h}{H})}^{3} = \frac{1}{8} \Rightarrow \frac{h}{H} = \sqrt[3]{\frac{1}{8}} \Rightarrow \frac{h}{20} = \frac{1}{2} \Rightarrow h = \frac{20}{2} \Rightarrow h = 10 cm ∴ PQ = H - h = 20 - 10 = 10 cm$

So, the section is made at the height of 10 cm above the base.

Question 19:

$We have, Height of the given cone, H = 20 cm Let the radius of the given cone be R, the height of the smaller cone be h and the radius of the smaller cone be r . Now, in ∆ AQD and ∆ APC, ∠ QAD = ∠ PAC (Common angle) ∠ AQD = ∠ APC = 90 ° So, by AA criteria ∆ AQD ~ ∆ APC \Rightarrow \frac{AQ}{AP} = \frac{QD}{PC} \Rightarrow \frac{h}{H} = \frac{r}{R} . . . . . (i) Volume of smaller cone = \frac{1}{8} \times Volume of the given cone \Rightarrow \frac{Volume of smaller cone}{Volume of the given cone} = \frac{1}{8} \Rightarrow \frac{(\frac{1}{3} π r^{2} h)}{(\frac{1}{3} π R^{2} H)} = \frac{1}{8} \Rightarrow {(\frac{r}{R})}^{2} \times (\frac{h}{H}) = \frac{1}{8} \Rightarrow {(\frac{h}{H})}^{2} \times (\frac{h}{H}) = \frac{1}{8} [Using (i)] \Rightarrow {(\frac{h}{H})}^{3} = \frac{1}{8} \Rightarrow \frac{h}{H} = \sqrt[3]{\frac{1}{8}} \Rightarrow \frac{h}{20} = \frac{1}{2} \Rightarrow h = \frac{20}{2} \Rightarrow h = 10 cm ∴ PQ = H - h = 20 - 10 = 10 cm$

So, the section is made at the height of 10 cm above the base.

Answer:

$We have, Height of the solid metallic cone, H = 20 cm, Height of the frustum, h = \frac{20}{2} = 10 cm and Radius of the wire = \frac{1}{24} cm Let the length of the wire be l, EG = r and BD = R . In ∆ AEG, \tan 30 ° = \frac{EG}{AG} \Rightarrow \frac{1}{\sqrt{3}} = \frac{r}{H - h} \Rightarrow \frac{1}{\sqrt{3}} = \frac{r}{20 - 10} \Rightarrow r = \frac{10}{\sqrt{3}} cm Also, in ∆ ABD, \tan 30 ° = \frac{BD}{AD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{R}{H} \Rightarrow \frac{1}{\sqrt{3}} = \frac{R}{20} \Rightarrow R = \frac{20}{\sqrt{3}} cm Now, Volume of the wire = Volume of the frustum \Rightarrow π {(\frac{1}{24})}^{2} l = \frac{1}{3} π h (R^{2} + r^{2} + R r) \Rightarrow \frac{l}{576} = \frac{1}{3} \times 10 \times [{(\frac{20}{\sqrt{3}})}^{2} + {(\frac{10}{\sqrt{3}})}^{2} + (\frac{20}{\sqrt{3}}) (\frac{10}{\sqrt{3}})] \Rightarrow l = \frac{576}{3} \times 10 \times [\frac{400}{3} + \frac{100}{3} + \frac{200}{3}] \Rightarrow l = \frac{576}{3} \times 10 \times \frac{700}{3} \Rightarrow l = 448000 cm ∴ l = 4480 m$

So, the length of the wire is 4480 m.

Question 20:

$We have, Height of the solid metallic cone, H = 20 cm, Height of the frustum, h = \frac{20}{2} = 10 cm and Radius of the wire = \frac{1}{24} cm Let the length of the wire be l, EG = r and BD = R . In ∆ AEG, \tan 30 ° = \frac{EG}{AG} \Rightarrow \frac{1}{\sqrt{3}} = \frac{r}{H - h} \Rightarrow \frac{1}{\sqrt{3}} = \frac{r}{20 - 10} \Rightarrow r = \frac{10}{\sqrt{3}} cm Also, in ∆ ABD, \tan 30 ° = \frac{BD}{AD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{R}{H} \Rightarrow \frac{1}{\sqrt{3}} = \frac{R}{20} \Rightarrow R = \frac{20}{\sqrt{3}} cm Now, Volume of the wire = Volume of the frustum \Rightarrow π {(\frac{1}{24})}^{2} l = \frac{1}{3} π h (R^{2} + r^{2} + R r) \Rightarrow \frac{l}{576} = \frac{1}{3} \times 10 \times [{(\frac{20}{\sqrt{3}})}^{2} + {(\frac{10}{\sqrt{3}})}^{2} + (\frac{20}{\sqrt{3}}) (\frac{10}{\sqrt{3}})] \Rightarrow l = \frac{576}{3} \times 10 \times [\frac{400}{3} + \frac{100}{3} + \frac{200}{3}] \Rightarrow l = \frac{576}{3} \times 10 \times \frac{700}{3} \Rightarrow l = 448000 cm ∴ l = 4480 m$

So, the length of the wire is 4480 m.

Answer:

$We have, Radius of open side, R = 10 cm, Radius of upper base, r = 4 cm and Slant height, l = 15 cm Now, The area of material used = π (R + r) l + π r^{2} = \frac{22}{7} \times (10 + 4) \times 15 + \frac{22}{7} \times 4 \times 4 = \frac{22}{7} \times 14 \times 15 + \frac{22}{7} \times 4 \times 4 = \frac{22}{7} \times (14 \times 15 + 4 \times 4) = \frac{22}{7} \times (210 + 16) = \frac{22}{7} \times 226 = \frac{4972}{7} {cm}^{2} \approx 710.28 {cm}^{2}$

So, the area of material used for making the fez is 710.28 cm².

Question 21:

$We have, Radius of open side, R = 10 cm, Radius of upper base, r = 4 cm and Slant height, l = 15 cm Now, The area of material used = π (R + r) l + π r^{2} = \frac{22}{7} \times (10 + 4) \times 15 + \frac{22}{7} \times 4 \times 4 = \frac{22}{7} \times 14 \times 15 + \frac{22}{7} \times 4 \times 4 = \frac{22}{7} \times (14 \times 15 + 4 \times 4) = \frac{22}{7} \times (210 + 16) = \frac{22}{7} \times 226 = \frac{4972}{7} {cm}^{2} \approx 710.28 {cm}^{2}$

So, the area of material used for making the fez is 710.28 cm².

Answer:

$We have, Height of the cylindrical portion, h = 10 cm, Height of the frustum of cone portion, H = 22 - 10 = 12 cm, Radius of the cylindical portion = Radius of smaller end of frustum portion, r = \frac{8}{2} = 4 cm and Radius of larger end of frustum portion, R = \frac{18}{2} = 9 cm Also, the slant height of the frustum, l = \sqrt{{(R - r)}^{2} + H^{2}} = \sqrt{{(9 - 4)}^{2} + 12^{2}} = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 cm Now, The area of the tin sheet required = CSA of frustum of cone + CSA of cylinder = π (R + r) l + 2 π r h = \frac{22}{7} \times (9 + 4) \times 13 + 2 \times \frac{22}{7} \times 4 \times 10 = \frac{22}{7} \times 13 \times 13 + \frac{22}{7} \times 80 = \frac{22}{7} \times (169 + 80) = \frac{22}{7} \times 249 \approx 782.57 {cm}^{2}$

So, the area of the tin sheet required to make the funnel is 782.57 cm².

Question 22:

$We have, Height of the cylindrical portion, h = 10 cm, Height of the frustum of cone portion, H = 22 - 10 = 12 cm, Radius of the cylindical portion = Radius of smaller end of frustum portion, r = \frac{8}{2} = 4 cm and Radius of larger end of frustum portion, R = \frac{18}{2} = 9 cm Also, the slant height of the frustum, l = \sqrt{{(R - r)}^{2} + H^{2}} = \sqrt{{(9 - 4)}^{2} + 12^{2}} = \sqrt{5^{2} + 12^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 cm Now, The area of the tin sheet required = CSA of frustum of cone + CSA of cylinder = π (R + r) l + 2 π r h = \frac{22}{7} \times (9 + 4) \times 13 + 2 \times \frac{22}{7} \times 4 \times 10 = \frac{22}{7} \times 13 \times 13 + \frac{22}{7} \times 80 = \frac{22}{7} \times (169 + 80) = \frac{22}{7} \times 249 \approx 782.57 {cm}^{2}$

So, the area of the tin sheet required to make the funnel is 782.57 cm².

Answer:

Let ABC be a right circular cone of height 3h and base radius r. This cone is cut by two planes such that AQ = QP = PO = h.

Since $∆ ABO ~ ∆ AEP$ (AA Similarity)
$∴ \frac{AO}{AP} = \frac{BO}{EP} \Rightarrow \frac{3 h}{2 h} = \frac{r}{r_{1}} \Rightarrow r_{1} = \frac{2 r}{3} . . . . . (1)$
Also, $∆ ABO ~ ∆ AGQ$ (AA Similarity)
$∴ \frac{AO}{AQ} = \frac{BO}{GQ} \Rightarrow \frac{3 h}{h} = \frac{r}{r_{2}} \Rightarrow r_{2} = \frac{r}{3} . . . . . (2)$
Now,
Volume of cone AGF,
$V_{1} = \frac{1}{3} π {r_{2}}^{2} h = \frac{1}{3} π {(\frac{r}{3})}^{2} h [From (2)] = \frac{1}{27} π r^{2} h$
Voulme of the frustum GFDE,
$V_{2} = \frac{1}{3} π ({r_{1}}^{2} + {r_{2}}^{2} + r_{1} r_{2}) h = \frac{1}{3} π (\frac{4 r^{2}}{9} + \frac{r^{2}}{9} + \frac{2 r^{2}}{9}) h [From (1) and (2)] = \frac{7}{27} π r^{2} h$
Voulme of the frustum EDCB,
$V_{3} = \frac{1}{3} π (r^{2} + {r_{1}}^{2} + r_{1} r) h = \frac{1}{3} π (r^{2} + \frac{4 r^{2}}{9} + \frac{2 r^{2}}{3}) h [From (1) and (2)] = \frac{19}{27} π r^{2} h$
∴ Required ratio = $V_{1} : V_{2} : V_{3} = \frac{1}{27} π r^{2} h : \frac{7}{27} π r^{2} h : \frac{19}{27} π r^{2} h = 1 : 7 : 19$

Question 1:

Let ABC be a right circular cone of height 3h and base radius r. This cone is cut by two planes such that AQ = QP = PO = h.

Since $∆ ABO ~ ∆ AEP$ (AA Similarity)
$∴ \frac{AO}{AP} = \frac{BO}{EP} \Rightarrow \frac{3 h}{2 h} = \frac{r}{r_{1}} \Rightarrow r_{1} = \frac{2 r}{3} . . . . . (1)$
Also, $∆ ABO ~ ∆ AGQ$ (AA Similarity)
$∴ \frac{AO}{AQ} = \frac{BO}{GQ} \Rightarrow \frac{3 h}{h} = \frac{r}{r_{2}} \Rightarrow r_{2} = \frac{r}{3} . . . . . (2)$
Now,
Volume of cone AGF,
$V_{1} = \frac{1}{3} π {r_{2}}^{2} h = \frac{1}{3} π {(\frac{r}{3})}^{2} h [From (2)] = \frac{1}{27} π r^{2} h$
Voulme of the frustum GFDE,
$V_{2} = \frac{1}{3} π ({r_{1}}^{2} + {r_{2}}^{2} + r_{1} r_{2}) h = \frac{1}{3} π (\frac{4 r^{2}}{9} + \frac{r^{2}}{9} + \frac{2 r^{2}}{9}) h [From (1) and (2)] = \frac{7}{27} π r^{2} h$
Voulme of the frustum EDCB,
$V_{3} = \frac{1}{3} π (r^{2} + {r_{1}}^{2} + r_{1} r) h = \frac{1}{3} π (r^{2} + \frac{4 r^{2}}{9} + \frac{2 r^{2}}{3}) h [From (1) and (2)] = \frac{19}{27} π r^{2} h$
∴ Required ratio = $V_{1} : V_{2} : V_{3} = \frac{1}{27} π r^{2} h : \frac{7}{27} π r^{2} h : \frac{19}{27} π r^{2} h = 1 : 7 : 19$

Answer:

$We have, Depth of the river, h = 1.5 m, Width of the river, b = 36 m, Speed of the flowing water, l = 3.5 km / hr = \frac{3.5 \times 1000 m}{60 \min} = \frac{175}{3} m / \min Now, The amount of water that runs into the sea per minute = l b h = \frac{175}{3} \times 36 \times 1.5 = 3150 m^{3} / \min$

So, the amount of water that runs into the sea per minute is 3150 m³.

Question 2:

$We have, Depth of the river, h = 1.5 m, Width of the river, b = 36 m, Speed of the flowing water, l = 3.5 km / hr = \frac{3.5 \times 1000 m}{60 \min} = \frac{175}{3} m / \min Now, The amount of water that runs into the sea per minute = l b h = \frac{175}{3} \times 36 \times 1.5 = 3150 m^{3} / \min$

So, the amount of water that runs into the sea per minute is 3150 m³.

Answer:

$Let the edge of the cube be a . As, Volume of the cube = 729 {cm}^{3} \Rightarrow a^{3} = 729 \Rightarrow a = \sqrt[3]{729} \Rightarrow a = 9 cm Now, Surface area of the cube = 6 a^{2} = 6 \times 9 \times 9 = 486 {cm}^{2}$

So, the surface area of the cube is 486 cm².

Question 3:

$Let the edge of the cube be a . As, Volume of the cube = 729 {cm}^{3} \Rightarrow a^{3} = 729 \Rightarrow a = \sqrt[3]{729} \Rightarrow a = 9 cm Now, Surface area of the cube = 6 a^{2} = 6 \times 9 \times 9 = 486 {cm}^{2}$

So, the surface area of the cube is 486 cm².

Answer:

$We have, Edge of the cube, a = 10 cm and Edge of the cubical box, l = 1 m = 100 cm Now, The number of cubes that can be put in the box = \frac{Volume of the cubical box}{Volume of the cube} = \frac{l^{3}}{a^{3}} = \frac{100^{3}}{10^{3}} = 10^{3} = 1000$

So, the number of cubes that can be put in the cubical box is 1000.

Question 4:

$We have, Edge of the cube, a = 10 cm and Edge of the cubical box, l = 1 m = 100 cm Now, The number of cubes that can be put in the box = \frac{Volume of the cubical box}{Volume of the cube} = \frac{l^{3}}{a^{3}} = \frac{100^{3}}{10^{3}} = 10^{3} = 1000$

So, the number of cubes that can be put in the cubical box is 1000.

Answer:

$We have, Edges of the cubes : a_{1} = 6 cm, a_{2} = 8 cm and a_{3} = 10 cm Let the edge of the new cube so formed be a . As, Volume of the new cube so formed = {a_{1}}^{3} + {a_{2}}^{3} + {a_{3}}^{3} \Rightarrow a^{3} = 6^{3} + 8^{3} + 10^{3} \Rightarrow a^{3} = 216 + 512 + 1000 \Rightarrow a^{3} = 1728 \Rightarrow a = \sqrt[3]{1728} ∴ a = 12 cm$

So, the edge of the new cube so formed is 12 cm.

Question 5:

$We have, Edges of the cubes : a_{1} = 6 cm, a_{2} = 8 cm and a_{3} = 10 cm Let the edge of the new cube so formed be a . As, Volume of the new cube so formed = {a_{1}}^{3} + {a_{2}}^{3} + {a_{3}}^{3} \Rightarrow a^{3} = 6^{3} + 8^{3} + 10^{3} \Rightarrow a^{3} = 216 + 512 + 1000 \Rightarrow a^{3} = 1728 \Rightarrow a = \sqrt[3]{1728} ∴ a = 12 cm$

So, the edge of the new cube so formed is 12 cm.

Answer:

$We have, Length of the resulting cuboid, l = 5 \times 5 = 25 cm, Breadth of the resulting cuboid, b = 5 cm and Height of the resulting cuboid, h = 5 cm Now, Volume of the resulting cuboid = l b h = 25 \times 5 \times 5 = 625 {cm}^{3}$

So, the volume of the resulting cuboid is 625 cm³.

Question 6:

$We have, Length of the resulting cuboid, l = 5 \times 5 = 25 cm, Breadth of the resulting cuboid, b = 5 cm and Height of the resulting cuboid, h = 5 cm Now, Volume of the resulting cuboid = l b h = 25 \times 5 \times 5 = 625 {cm}^{3}$

So, the volume of the resulting cuboid is 625 cm³.

Answer:

$Let the edges of the cubes be a and b . As, \frac{Volume of the first cube}{Volume of the second cube} = \frac{8}{27} \Rightarrow \frac{a^{3}}{b^{3}} = \frac{8}{27} \Rightarrow \frac{a}{b} = \sqrt[3]{\frac{8}{27}} \Rightarrow \frac{a}{b} = \frac{2}{3} . . . . . (i) Now, The ratio of the surface areas of the cubes = \frac{Surface area of the first cube}{Surface area of the second cube} = \frac{6 a^{2}}{6 b^{2}} = {(\frac{a}{b})}^{2} = {(\frac{2}{3})}^{2} [Using (i)] = \frac{4}{9} = 4 : 9$

So, the ratio of the surface areas of the given cubes is 4 : 9.

Question 7:

$Let the edges of the cubes be a and b . As, \frac{Volume of the first cube}{Volume of the second cube} = \frac{8}{27} \Rightarrow \frac{a^{3}}{b^{3}} = \frac{8}{27} \Rightarrow \frac{a}{b} = \sqrt[3]{\frac{8}{27}} \Rightarrow \frac{a}{b} = \frac{2}{3} . . . . . (i) Now, The ratio of the surface areas of the cubes = \frac{Surface area of the first cube}{Surface area of the second cube} = \frac{6 a^{2}}{6 b^{2}} = {(\frac{a}{b})}^{2} = {(\frac{2}{3})}^{2} [Using (i)] = \frac{4}{9} = 4 : 9$

So, the ratio of the surface areas of the given cubes is 4 : 9.

Answer:

$We have, Height = Base radius i . e . h = r As, Volume of the cylinder = 25 \frac{1}{7} {cm}^{3} \Rightarrow π r^{2} h = \frac{176}{7} \Rightarrow \frac{22}{7} \times h^{2} \times h = \frac{176}{7} \Rightarrow h^{3} = \frac{176 \times 7}{7 \times 22} \Rightarrow h^{3} = 8 \Rightarrow h = \sqrt[3]{8} ∴ h = 2 cm$

So, the height of the cylinder is 2 cm.

Question 8:

$We have, Height = Base radius i . e . h = r As, Volume of the cylinder = 25 \frac{1}{7} {cm}^{3} \Rightarrow π r^{2} h = \frac{176}{7} \Rightarrow \frac{22}{7} \times h^{2} \times h = \frac{176}{7} \Rightarrow h^{3} = \frac{176 \times 7}{7 \times 22} \Rightarrow h^{3} = 8 \Rightarrow h = \sqrt[3]{8} ∴ h = 2 cm$

So, the height of the cylinder is 2 cm.

Answer:

$Let the radius of the base and the height of the cylinder be r and h, respectively . We have, r : h = 2 : 3 i . e . \frac{r}{h} = \frac{2}{3} or h = \frac{3 r}{2} . . . . . (i) As, Volume of the cylinder = 12936 {cm}^{3} \Rightarrow π r^{2} h = 12936 \Rightarrow \frac{22}{7} \times r^{2} \times \frac{3 r}{2} = 12936 [Using (i)] \Rightarrow \frac{33}{7} \times r^{3} = 12936 \Rightarrow r^{3} = 12936 \times \frac{7}{33} \Rightarrow r^{3} = 2744 \Rightarrow r = \sqrt[3]{2744} ∴ r = 14 cm$

So, the radius of the base of the cylinder is 14 cm.

Question 9:

$Let the radius of the base and the height of the cylinder be r and h, respectively . We have, r : h = 2 : 3 i . e . \frac{r}{h} = \frac{2}{3} or h = \frac{3 r}{2} . . . . . (i) As, Volume of the cylinder = 12936 {cm}^{3} \Rightarrow π r^{2} h = 12936 \Rightarrow \frac{22}{7} \times r^{2} \times \frac{3 r}{2} = 12936 [Using (i)] \Rightarrow \frac{33}{7} \times r^{3} = 12936 \Rightarrow r^{3} = 12936 \times \frac{7}{33} \Rightarrow r^{3} = 2744 \Rightarrow r = \sqrt[3]{2744} ∴ r = 14 cm$

So, the radius of the base of the cylinder is 14 cm.

Answer:

$Let the radii of the cylinders be r_{1} and r_{2}; and their heights be h_{1} and h_{2} . We have, r_{1} : r_{2} = 2 : 3 or \frac{r_{1}}{r_{2}} = \frac{2}{3} . . . . . (i) and h_{1} : h_{2} = 5 : 3 or \frac{h_{1}}{h_{2}} = \frac{5}{3} . . . . . (ii) Now, The ratio of the volumes of the cylinders = \frac{Volume of the first cylinder}{Volume of the second cylinder} = \frac{π {r_{1}}^{2} h_{1}}{π {r_{2}}^{2} h_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} \times \frac{h_{1}}{h_{2}} = {(\frac{2}{3})}^{2} \times \frac{5}{3} [Using (i) and (ii)] = \frac{20}{27} = 20 : 27$

So, the ratio of the volumes of the given cylinders is 20 : 27.

Question 10:

$Let the radii of the cylinders be r_{1} and r_{2}; and their heights be h_{1} and h_{2} . We have, r_{1} : r_{2} = 2 : 3 or \frac{r_{1}}{r_{2}} = \frac{2}{3} . . . . . (i) and h_{1} : h_{2} = 5 : 3 or \frac{h_{1}}{h_{2}} = \frac{5}{3} . . . . . (ii) Now, The ratio of the volumes of the cylinders = \frac{Volume of the first cylinder}{Volume of the second cylinder} = \frac{π {r_{1}}^{2} h_{1}}{π {r_{2}}^{2} h_{2}} = {(\frac{r_{1}}{r_{2}})}^{2} \times \frac{h_{1}}{h_{2}} = {(\frac{2}{3})}^{2} \times \frac{5}{3} [Using (i) and (ii)] = \frac{20}{27} = 20 : 27$

So, the ratio of the volumes of the given cylinders is 20 : 27.

Answer:

$We have, Radius of wire, r = \frac{1}{2} = 0.5 mm = 0.05 cm Let the length of the wire be l . As, Volume of the wire = 66 {cm}^{3} \Rightarrow π r^{2} l = 66 \Rightarrow \frac{22}{7} \times 0.05 \times 0.05 \times l = 66 \Rightarrow l = 66 \times \frac{7}{22 \times 0.05 \times 0.05} ∴ l = 8400 cm = 84 m$

So, the length of the wire is 84 m.

Question 11:

$We have, Radius of wire, r = \frac{1}{2} = 0.5 mm = 0.05 cm Let the length of the wire be l . As, Volume of the wire = 66 {cm}^{3} \Rightarrow π r^{2} l = 66 \Rightarrow \frac{22}{7} \times 0.05 \times 0.05 \times l = 66 \Rightarrow l = 66 \times \frac{7}{22 \times 0.05 \times 0.05} ∴ l = 8400 cm = 84 m$

So, the length of the wire is 84 m.

Answer:

$We have, Height = 84 cm Let the radius and the slant height of the cone be r and l, respectively . As, Area of the base of the cone = 3850 {cm}^{2} \Rightarrow π r^{2} = 3850 \Rightarrow \frac{22}{7} \times r^{2} = 3850 \Rightarrow r^{2} = 3850 \times \frac{7}{22} \Rightarrow r^{2} = 1225 \Rightarrow r = \sqrt{1225} ∴ r = 35 cm Now, l = \sqrt{h^{2} + r^{2}} = \sqrt{84^{2} + 35^{2}} = \sqrt{7056 + 1225} = \sqrt{8281} = 91 cm$

So, the slant height of the given cone is 91 cm.

Question 12:

$We have, Height = 84 cm Let the radius and the slant height of the cone be r and l, respectively . As, Area of the base of the cone = 3850 {cm}^{2} \Rightarrow π r^{2} = 3850 \Rightarrow \frac{22}{7} \times r^{2} = 3850 \Rightarrow r^{2} = 3850 \times \frac{7}{22} \Rightarrow r^{2} = 1225 \Rightarrow r = \sqrt{1225} ∴ r = 35 cm Now, l = \sqrt{h^{2} + r^{2}} = \sqrt{84^{2} + 35^{2}} = \sqrt{7056 + 1225} = \sqrt{8281} = 91 cm$

So, the slant height of the given cone is 91 cm.

Answer:

$We have, Base radius of the cylinder, r = 8 cm, Height of the cylinder, h = 2 cm and Height of the cone, H = 6 cm Let the base radius of the cone be R . Now, Volume of the cone = Volume of the cylinder \Rightarrow \frac{1}{3} π R^{2} H = π r^{2} h \Rightarrow R^{2} = \frac{3 r^{2} h}{H} \Rightarrow R^{2} = \frac{3 \times 8 \times 8 \times 2}{6} \Rightarrow R^{2} = 64 \Rightarrow R = \sqrt{64} ∴ R = 8 cm$

So, the radius of the base of the cone is 8 cm.

Question 13:

$We have, Base radius of the cylinder, r = 8 cm, Height of the cylinder, h = 2 cm and Height of the cone, H = 6 cm Let the base radius of the cone be R . Now, Volume of the cone = Volume of the cylinder \Rightarrow \frac{1}{3} π R^{2} H = π r^{2} h \Rightarrow R^{2} = \frac{3 r^{2} h}{H} \Rightarrow R^{2} = \frac{3 \times 8 \times 8 \times 2}{6} \Rightarrow R^{2} = 64 \Rightarrow R = \sqrt{64} ∴ R = 8 cm$

So, the radius of the base of the cone is 8 cm.

Answer:

$Let the radius and height of the cone be r and h, respectively . Then, Radius of the cylindrical vessel = r and Height of the cylindrical vessel = h Now, The number of cones = \frac{Volume of the cylindrical vessel}{Volume of a cone} = \frac{π r^{2} h}{(\frac{1}{3} π r^{2} h)} = 3$

So, the number of cones that will be needed to store the water is 3.

Question 14:

$Let the radius and height of the cone be r and h, respectively . Then, Radius of the cylindrical vessel = r and Height of the cylindrical vessel = h Now, The number of cones = \frac{Volume of the cylindrical vessel}{Volume of a cone} = \frac{π r^{2} h}{(\frac{1}{3} π r^{2} h)} = 3$

So, the number of cones that will be needed to store the water is 3.

Answer:

$Let the radius of the sphere be r . As, Volume of the sphere = 4851 {cm}^{3} \Rightarrow \frac{4}{3} π r^{3} = 4851 \Rightarrow \frac{4}{3} \times \frac{22}{7} \times r^{3} = 4851 \Rightarrow r^{3} = 4851 \times \frac{3 \times 7}{4 \times 22} \Rightarrow r^{3} = \frac{9261}{8} \Rightarrow r = \sqrt[3]{\frac{9261}{8}} \Rightarrow r = \frac{21}{2} cm Now, Curved surface area of the sphere = 4 π r^{2} = 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 1386 {cm}^{2}$

So, the curved surface area of the sphere is 1386 cm².

Question 15:

$Let the radius of the sphere be r . As, Volume of the sphere = 4851 {cm}^{3} \Rightarrow \frac{4}{3} π r^{3} = 4851 \Rightarrow \frac{4}{3} \times \frac{22}{7} \times r^{3} = 4851 \Rightarrow r^{3} = 4851 \times \frac{3 \times 7}{4 \times 22} \Rightarrow r^{3} = \frac{9261}{8} \Rightarrow r = \sqrt[3]{\frac{9261}{8}} \Rightarrow r = \frac{21}{2} cm Now, Curved surface area of the sphere = 4 π r^{2} = 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} = 1386 {cm}^{2}$

So, the curved surface area of the sphere is 1386 cm².

Answer:

$Let the radius of the sphere be r . As, Curved surface area of the sphere = 5544 {cm}^{2} \Rightarrow 4 π r^{2} = 5544 \Rightarrow 4 \times \frac{22}{7} \times r^{2} = 5544 \Rightarrow r^{2} = 5544 \times \frac{7}{4 \times 22} \Rightarrow r^{2} = 441 \Rightarrow r = \sqrt{441} \Rightarrow r = 21 cm Now, Volume of the sphere = \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 38808 {cm}^{3}$

So, the volume of the sphere is 38808 cm³.

Question 16:

$Let the radius of the sphere be r . As, Curved surface area of the sphere = 5544 {cm}^{2} \Rightarrow 4 π r^{2} = 5544 \Rightarrow 4 \times \frac{22}{7} \times r^{2} = 5544 \Rightarrow r^{2} = 5544 \times \frac{7}{4 \times 22} \Rightarrow r^{2} = 441 \Rightarrow r = \sqrt{441} \Rightarrow r = 21 cm Now, Volume of the sphere = \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 38808 {cm}^{3}$

So, the volume of the sphere is 38808 cm³.

Answer:

$Let the radii of the two spheres be r and R . As, \frac{Surface area of the first sphere}{Surface area of the second sphere} = \frac{4}{25} \Rightarrow \frac{4 π r^{2}}{4 π R^{2}} = \frac{4}{25} \Rightarrow {(\frac{r}{R})}^{2} = \frac{4}{25} \Rightarrow \frac{r}{R} = \sqrt{\frac{4}{25}} \Rightarrow \frac{r}{R} = \frac{2}{5} . . . . . (i) Now, The ratio of the volumes of the two spheres = \frac{Volume of the first sphere}{Volume of the second sphere} = \frac{(\frac{4}{3} π r^{3})}{(\frac{4}{3} π R^{3})} = {(\frac{r}{R})}^{3} = {(\frac{2}{5})}^{3} [Using (i)] = \frac{8}{125} = 8 : 125$

So, the ratio of the volumes of the given spheres is 8 : 125.

Question 17:

$Let the radii of the two spheres be r and R . As, \frac{Surface area of the first sphere}{Surface area of the second sphere} = \frac{4}{25} \Rightarrow \frac{4 π r^{2}}{4 π R^{2}} = \frac{4}{25} \Rightarrow {(\frac{r}{R})}^{2} = \frac{4}{25} \Rightarrow \frac{r}{R} = \sqrt{\frac{4}{25}} \Rightarrow \frac{r}{R} = \frac{2}{5} . . . . . (i) Now, The ratio of the volumes of the two spheres = \frac{Volume of the first sphere}{Volume of the second sphere} = \frac{(\frac{4}{3} π r^{3})}{(\frac{4}{3} π R^{3})} = {(\frac{r}{R})}^{3} = {(\frac{2}{5})}^{3} [Using (i)] = \frac{8}{125} = 8 : 125$

So, the ratio of the volumes of the given spheres is 8 : 125.

Answer:

$We have, Radius of the solid metallic sphere, R = 8 cm and Radius of the spherical ball, r = 2 cm Now, The number spherical balls obtained = \frac{Volume of the solid metallic sphere}{Volume of a spherical ball} = \frac{(\frac{4}{3} π R^{3})}{(\frac{4}{3} π r^{3})} = {(\frac{R}{r})}^{3} = {(\frac{8}{2})}^{3} = 4^{3} = 64$

So, the number of spherical balls obtained is 64.

Question 18:

$We have, Radius of the solid metallic sphere, R = 8 cm and Radius of the spherical ball, r = 2 cm Now, The number spherical balls obtained = \frac{Volume of the solid metallic sphere}{Volume of a spherical ball} = \frac{(\frac{4}{3} π R^{3})}{(\frac{4}{3} π r^{3})} = {(\frac{R}{r})}^{3} = {(\frac{8}{2})}^{3} = 4^{3} = 64$

So, the number of spherical balls obtained is 64.

Answer:

$We have, Radius of a lead shot, r = \frac{3}{2} = 1.5 mm = 0.15 cm and Dimensions of the cuboid are 9 cm \times 11 cm \times 12 cm Now, The number of the lead shots = \frac{Volume of the cuboid}{Volume of a lead shot} = \frac{9 \times 11 \times 12}{(\frac{4}{3} π r^{3})} = \frac{9 \times 11 \times 12}{(\frac{4}{3} \times \frac{22}{7} \times 0.15 \times 0.15 \times 0.15)} = 84000$

So, the number of lead shots that can be made from the cuboid is 84000.

Question 19:

$We have, Radius of a lead shot, r = \frac{3}{2} = 1.5 mm = 0.15 cm and Dimensions of the cuboid are 9 cm \times 11 cm \times 12 cm Now, The number of the lead shots = \frac{Volume of the cuboid}{Volume of a lead shot} = \frac{9 \times 11 \times 12}{(\frac{4}{3} π r^{3})} = \frac{9 \times 11 \times 12}{(\frac{4}{3} \times \frac{22}{7} \times 0.15 \times 0.15 \times 0.15)} = 84000$

So, the number of lead shots that can be made from the cuboid is 84000.

Answer:

$We have, Radius of the metallic cone, r = 12 cm, Height of the metallic cone, h = 24 cm and Radius of the sphere, R = 2 cm Now, The number of spheres so formed = \frac{Volume of the metallic cone}{Volume of a sphere} = \frac{(\frac{1}{3} π r^{2} h)}{(\frac{4}{3} π R^{3})} = \frac{r^{2} h}{4 R^{3}} = \frac{12 \times 12 \times 24}{4 \times 2 \times 2 \times 2} = 108$

So, the number of spheres so formed is 108.

Question 20:

$We have, Radius of the metallic cone, r = 12 cm, Height of the metallic cone, h = 24 cm and Radius of the sphere, R = 2 cm Now, The number of spheres so formed = \frac{Volume of the metallic cone}{Volume of a sphere} = \frac{(\frac{1}{3} π r^{2} h)}{(\frac{4}{3} π R^{3})} = \frac{r^{2} h}{4 R^{3}} = \frac{12 \times 12 \times 24}{4 \times 2 \times 2 \times 2} = 108$

So, the number of spheres so formed is 108.

Answer:

$We have, Radius of the hemisphere, R = 6 cm and Height of the cone, h = 75 cm Let the radius of the base of the cone be r . Now, Volume of the cone = Volume of the hemisphere \Rightarrow \frac{1}{3} π r^{2} h = \frac{2}{3} π R^{3} \Rightarrow r^{2} = \frac{2 R^{3}}{h} \Rightarrow r^{2} = \frac{2 \times 6 \times 6 \times 6}{75} \Rightarrow r^{2} = 5.76 \Rightarrow r = \sqrt{5.76} ∴ r = 2.4 cm$

So, the radius of the base of the cone is 2.4 cm.

Question 21:

$We have, Radius of the hemisphere, R = 6 cm and Height of the cone, h = 75 cm Let the radius of the base of the cone be r . Now, Volume of the cone = Volume of the hemisphere \Rightarrow \frac{1}{3} π r^{2} h = \frac{2}{3} π R^{3} \Rightarrow r^{2} = \frac{2 R^{3}}{h} \Rightarrow r^{2} = \frac{2 \times 6 \times 6 \times 6}{75} \Rightarrow r^{2} = 5.76 \Rightarrow r = \sqrt{5.76} ∴ r = 2.4 cm$

So, the radius of the base of the cone is 2.4 cm.

Answer:

$We have, Radius of the sphere, R = \frac{18}{2} = 9 cm and Radius of the wire, r = \frac{4}{2} = 2 mm = 0.2 cm Let the length of the wire be l . Now, Volume of the wire = Volume of the copper sphere \Rightarrow π r^{2} l = \frac{4}{3} π R^{3} \Rightarrow l = \frac{4 R^{3}}{3 r^{2}} \Rightarrow l = \frac{4 \times 9 \times 9 \times 9}{3 \times 0.2 \times 0.2} ∴ l = 24300 cm = 243 m$

So, the length of the wire is 243 m.

Question 22:

$We have, Radius of the sphere, R = \frac{18}{2} = 9 cm and Radius of the wire, r = \frac{4}{2} = 2 mm = 0.2 cm Let the length of the wire be l . Now, Volume of the wire = Volume of the copper sphere \Rightarrow π r^{2} l = \frac{4}{3} π R^{3} \Rightarrow l = \frac{4 R^{3}}{3 r^{2}} \Rightarrow l = \frac{4 \times 9 \times 9 \times 9}{3 \times 0.2 \times 0.2} ∴ l = 24300 cm = 243 m$

So, the length of the wire is 243 m.

Answer:

$We have, Height of the frustum, h = 6 cm, Radii of the circular ends, R = 14 cm and r = 6 cm Let the slant height of the frustum be l . Now, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(14 - 6)}^{2} + 6^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 cm$

So, the slant height of the frustum is 10 cm.

Question 23:

$We have, Height of the frustum, h = 6 cm, Radii of the circular ends, R = 14 cm and r = 6 cm Let the slant height of the frustum be l . Now, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(14 - 6)}^{2} + 6^{2}} = \sqrt{8^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10 cm$

So, the slant height of the frustum is 10 cm.

Answer:

$Let the radius of the shere be R and the edge of the cube be a . As, the sphere is fit inside the cube . So, diameter of the sphere = edge of the cube \Rightarrow 2 R = a . . . . . (i) Now, The ratio of the volume of the cube to that of the sphere = \frac{Volume of the cube}{Volume of the sphere} = \frac{a^{3}}{(\frac{4}{3} π R^{3})} = \frac{{(2 R)}^{3}}{(\frac{4}{3} π R^{3})} [Using (i)] = \frac{3 \times 8 R^{3}}{4 π R^{3}} = \frac{6}{π} = 6 : π$

So, the ratio of the volume of the cube to that of the sphere is 6 : $π$ .

Question 24:

$Let the radius of the shere be R and the edge of the cube be a . As, the sphere is fit inside the cube . So, diameter of the sphere = edge of the cube \Rightarrow 2 R = a . . . . . (i) Now, The ratio of the volume of the cube to that of the sphere = \frac{Volume of the cube}{Volume of the sphere} = \frac{a^{3}}{(\frac{4}{3} π R^{3})} = \frac{{(2 R)}^{3}}{(\frac{4}{3} π R^{3})} [Using (i)] = \frac{3 \times 8 R^{3}}{4 π R^{3}} = \frac{6}{π} = 6 : π$

So, the ratio of the volume of the cube to that of the sphere is 6 : $π$ .

Answer:

$Let the radius of the sphere be r . We have, The radius of the cone = The radius of the cylinder = The radius of the sphere = r and The height of the cylinder = The height of the cone = The height of the sphere = 2 r Now, Volume of the cylinder = π r^{2} (2 r) = 2 π r^{3}, Volume of the cone = \frac{1}{3} π r^{2} (2 r) = \frac{2}{3} π r^{3} and Volume of the sphere = \frac{4}{3} π r^{3} So, The ratio of the volumes of the cylinder, the cone and the sphere = 2 π r^{3} : \frac{2}{3} π r^{3} : \frac{4}{3} π r^{3} = 1 : \frac{1}{3} : \frac{2}{3} = 3 : 1 : 2$

So, the ratio of the volumes of the cylinder, the cone and the sphere is 3 : 1 : 2.

Question 25:

$Let the radius of the sphere be r . We have, The radius of the cone = The radius of the cylinder = The radius of the sphere = r and The height of the cylinder = The height of the cone = The height of the sphere = 2 r Now, Volume of the cylinder = π r^{2} (2 r) = 2 π r^{3}, Volume of the cone = \frac{1}{3} π r^{2} (2 r) = \frac{2}{3} π r^{3} and Volume of the sphere = \frac{4}{3} π r^{3} So, The ratio of the volumes of the cylinder, the cone and the sphere = 2 π r^{3} : \frac{2}{3} π r^{3} : \frac{4}{3} π r^{3} = 1 : \frac{1}{3} : \frac{2}{3} = 3 : 1 : 2$

So, the ratio of the volumes of the cylinder, the cone and the sphere is 3 : 1 : 2.

Answer:

$Let the edge of the cube be a . As, Volume of the cube = 125 {cm}^{3} \Rightarrow a^{3} = 125 \Rightarrow a = \sqrt[3]{125} \Rightarrow a = 5 cm So, Length of the resulting cuboid, l = 2 \times 5 = 10 cm, Breadth of the resulting cuboid, b = 5 cm and Height of the resulting cuboid, h = 5 cm Now, Surface area of the resulting cuboid = 2 (l b + b h + h l) = 2 \times (10 \times 5 + 5 \times 5 + 5 \times 10) = 2 \times (50 + 25 + 50) = 2 \times 125 = 250 {cm}^{2}$

So, the surface area of the resulting cuboid is 250 cm².

Question 26:

$Let the edge of the cube be a . As, Volume of the cube = 125 {cm}^{3} \Rightarrow a^{3} = 125 \Rightarrow a = \sqrt[3]{125} \Rightarrow a = 5 cm So, Length of the resulting cuboid, l = 2 \times 5 = 10 cm, Breadth of the resulting cuboid, b = 5 cm and Height of the resulting cuboid, h = 5 cm Now, Surface area of the resulting cuboid = 2 (l b + b h + h l) = 2 \times (10 \times 5 + 5 \times 5 + 5 \times 10) = 2 \times (50 + 25 + 50) = 2 \times 125 = 250 {cm}^{2}$

So, the surface area of the resulting cuboid is 250 cm².

Answer:

$We have, Edges of the cubes : a_{1} = 3 cm, a_{2} = 4 cm and a_{3} = 5 cm Let the edge of the new cube be a . Now, Volume of the new cube = {a_{1}}^{3} + {a_{2}}^{3} + {a_{3}}^{3} \Rightarrow a^{3} = 3^{3} + 4^{3} + 5^{3} \Rightarrow a^{3} = 27 + 64 + 125 \Rightarrow a^{3} = 216 \Rightarrow a = \sqrt[3]{216} ∴ a = 6 cm$

So, the edge of the new cube so formed is 6 cm.

Question 27:

$We have, Edges of the cubes : a_{1} = 3 cm, a_{2} = 4 cm and a_{3} = 5 cm Let the edge of the new cube be a . Now, Volume of the new cube = {a_{1}}^{3} + {a_{2}}^{3} + {a_{3}}^{3} \Rightarrow a^{3} = 3^{3} + 4^{3} + 5^{3} \Rightarrow a^{3} = 27 + 64 + 125 \Rightarrow a^{3} = 216 \Rightarrow a = \sqrt[3]{216} ∴ a = 6 cm$

So, the edge of the new cube so formed is 6 cm.

Answer:

$We have, Radius of the metallic sphere, R = \frac{8}{2} = 4 cm and Height of the cylindrical wire, h = 12 m = 1200 cm Let the radius of the base be r . Now, Volume of the cylindrical wire = Volume of the metallic sphere \Rightarrow π r^{2} h = \frac{4}{3} π R^{3} \Rightarrow r^{2} = \frac{4 R^{3}}{3 h} \Rightarrow r^{2} = \frac{4 \times 4 \times 4 \times 4}{3 \times 1200} \Rightarrow r^{2} = \frac{16}{225} \Rightarrow r = \sqrt{\frac{16}{225}} \Rightarrow r = \frac{4}{15} cm ∴ The width of the wire = 2 r = 2 \times \frac{4}{15} = \frac{8}{15} cm$

So, the width of the wire is $\frac{8}{15}$ cm.

Question 28:

$We have, Radius of the metallic sphere, R = \frac{8}{2} = 4 cm and Height of the cylindrical wire, h = 12 m = 1200 cm Let the radius of the base be r . Now, Volume of the cylindrical wire = Volume of the metallic sphere \Rightarrow π r^{2} h = \frac{4}{3} π R^{3} \Rightarrow r^{2} = \frac{4 R^{3}}{3 h} \Rightarrow r^{2} = \frac{4 \times 4 \times 4 \times 4}{3 \times 1200} \Rightarrow r^{2} = \frac{16}{225} \Rightarrow r = \sqrt{\frac{16}{225}} \Rightarrow r = \frac{4}{15} cm ∴ The width of the wire = 2 r = 2 \times \frac{4}{15} = \frac{8}{15} cm$

So, the width of the wire is $\frac{8}{15}$ cm.

Answer:

$We have, Width of the cloth, B = 5 m, Radius of the conical tent, r = \frac{14}{2} = 7 m and Height of the conical tent, h = 24 m Let the length of the cloth used for making the tent be L . Also, The slant height of the conical tent, l = \sqrt{r^{2} + h^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25 m Now, The curved surface of the conical tent = π r l = \frac{22}{7} \times 7 \times 25 \Rightarrow The area of the cloth used for making the tent = 550 m^{2} \Rightarrow L B = 550 \Rightarrow L = \frac{550}{B} \Rightarrow L = \frac{550}{5} \Rightarrow L = 110 m So, the cost of the cloth used = 25 \times 110 = ₹ 2750$

So, the cost of the cloth used for making the tent is â‚¹2750.

Question 29:

$We have, Width of the cloth, B = 5 m, Radius of the conical tent, r = \frac{14}{2} = 7 m and Height of the conical tent, h = 24 m Let the length of the cloth used for making the tent be L . Also, The slant height of the conical tent, l = \sqrt{r^{2} + h^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25 m Now, The curved surface of the conical tent = π r l = \frac{22}{7} \times 7 \times 25 \Rightarrow The area of the cloth used for making the tent = 550 m^{2} \Rightarrow L B = 550 \Rightarrow L = \frac{550}{B} \Rightarrow L = \frac{550}{5} \Rightarrow L = 110 m So, the cost of the cloth used = 25 \times 110 = ₹ 2750$

So, the cost of the cloth used for making the tent is â‚¹2750.

Answer:

$We have, Radius of the cylinder = Radius of the hemispher = r = 3.5 cm and Height of the cylinder, h = 10 cm Now, Volume of the toy = Volume of the cylinder - Volume of the two hemispheres = π r^{2} h - 2 \times \frac{2}{3} π r^{3} = π r^{2} (h - \frac{4 r}{3}) = \frac{22}{7} \times 3.5 \times 3.5 \times (10 - \frac{4 \times 3.5}{3}) = 38.5 \times (10 - \frac{14}{3}) = 38.5 \times \frac{16}{3} = \frac{616}{3} {cm}^{3} \approx 205.33 {cm}^{3}$

So, the volume of wood in the toy is $\frac{616}{3}$ cm³ or 205.33 cm³.

Question 30:

$We have, Radius of the cylinder = Radius of the hemispher = r = 3.5 cm and Height of the cylinder, h = 10 cm Now, Volume of the toy = Volume of the cylinder - Volume of the two hemispheres = π r^{2} h - 2 \times \frac{2}{3} π r^{3} = π r^{2} (h - \frac{4 r}{3}) = \frac{22}{7} \times 3.5 \times 3.5 \times (10 - \frac{4 \times 3.5}{3}) = 38.5 \times (10 - \frac{14}{3}) = 38.5 \times \frac{16}{3} = \frac{616}{3} {cm}^{3} \approx 205.33 {cm}^{3}$

So, the volume of wood in the toy is $\frac{616}{3}$ cm³ or 205.33 cm³.

Answer:

$Let the edge of the metal cubes be 3 x, 4 x and 5 x . Let the edge of the single cube be a . As, Diagonal of the single cube = 12 \sqrt{3} cm \Rightarrow a \sqrt{3} = 12 \sqrt{3} \Rightarrow a = 12 cm Now, Volume of the single cube = Sum of the volumes of the metallic cubes \Rightarrow a^{3} = {(3 x)}^{3} + {(4 x)}^{3} + {(5 x)}^{3} \Rightarrow 12^{3} = 27 x^{3} + 64 x^{3} + 125 x^{3} \Rightarrow 1728 = 216 x^{3} \Rightarrow x^{3} = \frac{1728}{216} \Rightarrow x^{3} = 8 \Rightarrow x = \sqrt[3]{8} \Rightarrow x = 2 So, the egdes of the cubes are 3 \times 2 = 6 cm, 4 \times 2 = 8 cm and 5 \times 2 = 10 cm .$

Hence, the edges of the given three metallic cubes are 6 cm, 8 cm and 10 cm.

Question 31:

$Let the edge of the metal cubes be 3 x, 4 x and 5 x . Let the edge of the single cube be a . As, Diagonal of the single cube = 12 \sqrt{3} cm \Rightarrow a \sqrt{3} = 12 \sqrt{3} \Rightarrow a = 12 cm Now, Volume of the single cube = Sum of the volumes of the metallic cubes \Rightarrow a^{3} = {(3 x)}^{3} + {(4 x)}^{3} + {(5 x)}^{3} \Rightarrow 12^{3} = 27 x^{3} + 64 x^{3} + 125 x^{3} \Rightarrow 1728 = 216 x^{3} \Rightarrow x^{3} = \frac{1728}{216} \Rightarrow x^{3} = 8 \Rightarrow x = \sqrt[3]{8} \Rightarrow x = 2 So, the egdes of the cubes are 3 \times 2 = 6 cm, 4 \times 2 = 8 cm and 5 \times 2 = 10 cm .$

Hence, the edges of the given three metallic cubes are 6 cm, 8 cm and 10 cm.

Answer:

$We have, External radius of the hollow sphere, R_{1} = \frac{8}{2} = 4 cm, Internal radius of the hollow sphere, R_{2} = \frac{4}{2} = 2 cm and Base radius of the cone, r = \frac{8}{2} = 4 cm Let the height of the cone be h . Now, Volume of the cone = Volume of the hollow sphere \Rightarrow \frac{1}{3} π r^{2} h = \frac{4}{3} π {R_{1}}^{3} - \frac{4}{3} π {R_{2}}^{3} \Rightarrow \frac{1}{3} π r^{2} h = \frac{4}{3} π ({R_{1}}^{3} - {R_{2}}^{3}) \Rightarrow h = \frac{4}{r^{2}} ({R_{1}}^{3} - {R_{2}}^{3}) \Rightarrow h = \frac{4}{4 \times 4} (4^{3} - 2^{3}) \Rightarrow h = \frac{1}{4} (64 - 8) \Rightarrow h = \frac{1}{4} \times 56 ∴ h = 14 cm$

So, the height of the cone is 14 cm.

Question 32:

$We have, External radius of the hollow sphere, R_{1} = \frac{8}{2} = 4 cm, Internal radius of the hollow sphere, R_{2} = \frac{4}{2} = 2 cm and Base radius of the cone, r = \frac{8}{2} = 4 cm Let the height of the cone be h . Now, Volume of the cone = Volume of the hollow sphere \Rightarrow \frac{1}{3} π r^{2} h = \frac{4}{3} π {R_{1}}^{3} - \frac{4}{3} π {R_{2}}^{3} \Rightarrow \frac{1}{3} π r^{2} h = \frac{4}{3} π ({R_{1}}^{3} - {R_{2}}^{3}) \Rightarrow h = \frac{4}{r^{2}} ({R_{1}}^{3} - {R_{2}}^{3}) \Rightarrow h = \frac{4}{4 \times 4} (4^{3} - 2^{3}) \Rightarrow h = \frac{1}{4} (64 - 8) \Rightarrow h = \frac{1}{4} \times 56 ∴ h = 14 cm$

So, the height of the cone is 14 cm.

Answer:

$We have, Height of the frustum, h = 24 cm, Radius of the open end, R = \frac{42}{2} = 21 cm and Radius of the close end, r = \frac{28}{2} = 14 cm Now, Volume of the bucket = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 24 \times (21^{2} + 14^{2} + 21 \times 14) = \frac{176}{7} \times (441 + 196 + 294) = \frac{176}{7} \times 931 = 23408 {cm}^{3} = 23.408 L (As, 1000 {cm}^{3} = 1 L) ∴ The cost of the milk = 30 \times 23.408 = ₹ 702.24$

So, the cost of the milk which the bucket can hold is â‚¹702.24.

Question 33:

$We have, Height of the frustum, h = 24 cm, Radius of the open end, R = \frac{42}{2} = 21 cm and Radius of the close end, r = \frac{28}{2} = 14 cm Now, Volume of the bucket = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 24 \times (21^{2} + 14^{2} + 21 \times 14) = \frac{176}{7} \times (441 + 196 + 294) = \frac{176}{7} \times 931 = 23408 {cm}^{3} = 23.408 L (As, 1000 {cm}^{3} = 1 L) ∴ The cost of the milk = 30 \times 23.408 = ₹ 702.24$

So, the cost of the milk which the bucket can hold is â‚¹702.24.

Answer:

$We have, Radius of the cylinder = Radius of the cone = r = \frac{4.2}{2} = 2.1 m, Height of the cylinder, H = 4 m and Height of the cone, h = 2.8 m Also, The slant height of the cone, l = \sqrt{r^{2} + h^{2}} = \sqrt{2 . 1^{2} + 2 . 8^{2}} = \sqrt{4.41 + 7.84} = \sqrt{12.25} = 3.5 m Now, The outer surface area of the building = CSA of the cylinder + CSA of the cone = 2 π r H + π r l = π r (2 H + l) = \frac{22}{7} \times 2.1 \times (2 \times 4 + 3.5) = 6.6 \times 11.5 = 75.9 m^{2}$

So, the outer surface area of the building is 75.9 m².

Question 34:

$We have, Radius of the cylinder = Radius of the cone = r = \frac{4.2}{2} = 2.1 m, Height of the cylinder, H = 4 m and Height of the cone, h = 2.8 m Also, The slant height of the cone, l = \sqrt{r^{2} + h^{2}} = \sqrt{2 . 1^{2} + 2 . 8^{2}} = \sqrt{4.41 + 7.84} = \sqrt{12.25} = 3.5 m Now, The outer surface area of the building = CSA of the cylinder + CSA of the cone = 2 π r H + π r l = π r (2 H + l) = \frac{22}{7} \times 2.1 \times (2 \times 4 + 3.5) = 6.6 \times 11.5 = 75.9 m^{2}$

So, the outer surface area of the building is 75.9 m².

Answer:

$We have, Height of the cone, h = 84 cm and Base radius of the cone, r = 21 cm Let the radius of the solid sphere be R . Now, Volume of the solid sphere = Volume of the solid cone \Rightarrow \frac{4}{3} π R^{3} = \frac{1}{3} π r^{2} h \Rightarrow R^{3} = \frac{r^{2} h}{4} \Rightarrow R^{3} = \frac{21 \times 21 \times 84}{4} \Rightarrow R^{3} = 21 \times 21 \times 21 \Rightarrow R = 21 cm ∴ Diameter = 2 R = 2 \times 21 = 42 cm$

So, the diameter of the solid sphere is 42 cm.

Question 35:

$We have, Height of the cone, h = 84 cm and Base radius of the cone, r = 21 cm Let the radius of the solid sphere be R . Now, Volume of the solid sphere = Volume of the solid cone \Rightarrow \frac{4}{3} π R^{3} = \frac{1}{3} π r^{2} h \Rightarrow R^{3} = \frac{r^{2} h}{4} \Rightarrow R^{3} = \frac{21 \times 21 \times 84}{4} \Rightarrow R^{3} = 21 \times 21 \times 21 \Rightarrow R = 21 cm ∴ Diameter = 2 R = 2 \times 21 = 42 cm$

So, the diameter of the solid sphere is 42 cm.

Answer:

$We have, Radius of the hemisphere = Radius of the cone = r = 3.5 cm and Height of the cone = 15.5 - 3.5 = 12 cm Also, The slant height of the cone, l = \sqrt{h^{2} + r^{2}} = \sqrt{12^{2} + 3 . 5^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 cm Now, Total surface area of the toy = CSA of cone + CSA of hemisphere = π r l + 2 π r^{2} = π r (l + 2 r) = \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5) = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5 {cm}^{2}$

So, the total surface area of the toy is 214.5 cm².

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Question 36:

$We have, Radius of the hemisphere = Radius of the cone = r = 3.5 cm and Height of the cone = 15.5 - 3.5 = 12 cm Also, The slant height of the cone, l = \sqrt{h^{2} + r^{2}} = \sqrt{12^{2} + 3 . 5^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 cm Now, Total surface area of the toy = CSA of cone + CSA of hemisphere = π r l + 2 π r^{2} = π r (l + 2 r) = \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5) = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5 {cm}^{2}$

So, the total surface area of the toy is 214.5 cm².

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Answer:

$We have, Height, h = 28 cm, Radius of the upper end, R = 28 cm and Radius of the lower end, r = 7 cm Also, The slant height, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(28 - 7)}^{2} + 28^{2}} = \sqrt{21^{2} + 28^{2}} = \sqrt{441 + 784} = \sqrt{1225} = 35 cm Now, Capacity of the bucket = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 28 \times (28^{2} + 7^{2} + 28 \times 7) = \frac{88}{3} \times (784 + 49 + 196) = \frac{88}{3} \times 1029 = 30184 {cm}^{3} Also, Total surface area of the bucket = π l (R + r) + π r^{2} = \frac{22}{7} \times 35 \times (28 + 7) + \frac{22}{7} \times 7 \times 7 = 110 \times (35) + 154 = 3850 + 154 = 4004 {cm}^{2}$

Disclaimer: The answer of the total surface area given in the textbook is incorrect. The same has been corrected above.

Question 37:

$We have, Height, h = 28 cm, Radius of the upper end, R = 28 cm and Radius of the lower end, r = 7 cm Also, The slant height, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(28 - 7)}^{2} + 28^{2}} = \sqrt{21^{2} + 28^{2}} = \sqrt{441 + 784} = \sqrt{1225} = 35 cm Now, Capacity of the bucket = \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{1}{3} \times \frac{22}{7} \times 28 \times (28^{2} + 7^{2} + 28 \times 7) = \frac{88}{3} \times (784 + 49 + 196) = \frac{88}{3} \times 1029 = 30184 {cm}^{3} Also, Total surface area of the bucket = π l (R + r) + π r^{2} = \frac{22}{7} \times 35 \times (28 + 7) + \frac{22}{7} \times 7 \times 7 = 110 \times (35) + 154 = 3850 + 154 = 4004 {cm}^{2}$

Disclaimer: The answer of the total surface area given in the textbook is incorrect. The same has been corrected above.

Answer:

$We have, Radius of the upper end, R = 20 cm and Radius of the lower end, r = 12 cm Let the height of the bucket be h . As, Volume of the bucket = 12308.8 {cm}^{3} \Rightarrow \frac{1}{3} π h (R^{2} + r^{2} + R r) = 12308.8 \Rightarrow \frac{1}{3} \times 3.14 \times h \times (20^{2} + 12^{2} + 20 \times 12) = 12308.8 \Rightarrow \frac{3.14 h}{3} \times (400 + 144 + 240) = 12308.8 \Rightarrow \frac{3.14 h}{3} \times 784 = 12308.8 \Rightarrow h = \frac{12308.8 \times 3}{3.14 \times 784} ∴ h = 15 cm$

So, the height of the bucket is 15 cm.

Question 38:

$We have, Radius of the upper end, R = 20 cm and Radius of the lower end, r = 12 cm Let the height of the bucket be h . As, Volume of the bucket = 12308.8 {cm}^{3} \Rightarrow \frac{1}{3} π h (R^{2} + r^{2} + R r) = 12308.8 \Rightarrow \frac{1}{3} \times 3.14 \times h \times (20^{2} + 12^{2} + 20 \times 12) = 12308.8 \Rightarrow \frac{3.14 h}{3} \times (400 + 144 + 240) = 12308.8 \Rightarrow \frac{3.14 h}{3} \times 784 = 12308.8 \Rightarrow h = \frac{12308.8 \times 3}{3.14 \times 784} ∴ h = 15 cm$

So, the height of the bucket is 15 cm.

Answer:

$We have, Radius of the upper end, R = 20 cm and Radius of the lower end, r = 8 cm Let the height of the container be h . As, Volume of the container = 10459 \frac{3}{7} {cm}^{3} \Rightarrow \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{73216}{7} \Rightarrow \frac{1}{3} \times \frac{22}{7} \times h \times (20^{2} + 8^{2} + 20 \times 8) = \frac{73216}{7} \Rightarrow \frac{22 h}{21} \times (400 + 64 + 160) = \frac{73216}{7} \Rightarrow \frac{22 h}{21} \times 624 = \frac{73216}{7} \Rightarrow h = \frac{73216 \times 21}{7 \times 22 \times 624} \Rightarrow h = 16 cm Also, The slant height of the container, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(20 - 8)}^{2} + 16^{2}} = \sqrt{12^{2} + 16^{2}} = \sqrt{144 + 256} = \sqrt{400} = 20 cm Now, Total surface area of the container = π l (R + r) + π r^{2} = \frac{22}{7} \times 20 \times (20 + 8) + \frac{22}{7} \times 8 \times 8 = \frac{22}{7} \times 20 \times 28 + \frac{22}{7} \times 64 = \frac{22}{7} \times (560 + 64) = \frac{22}{7} \times 624 = \frac{13728}{7} {cm}^{2} So, the cost of metal sheet = 1.4 \times \frac{13728}{7} = ₹ 2745.60$

Hence, the cost of the metal sheet used for making the milk container is â‚¹2745.60.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Question 39:

$We have, Radius of the upper end, R = 20 cm and Radius of the lower end, r = 8 cm Let the height of the container be h . As, Volume of the container = 10459 \frac{3}{7} {cm}^{3} \Rightarrow \frac{1}{3} π h (R^{2} + r^{2} + R r) = \frac{73216}{7} \Rightarrow \frac{1}{3} \times \frac{22}{7} \times h \times (20^{2} + 8^{2} + 20 \times 8) = \frac{73216}{7} \Rightarrow \frac{22 h}{21} \times (400 + 64 + 160) = \frac{73216}{7} \Rightarrow \frac{22 h}{21} \times 624 = \frac{73216}{7} \Rightarrow h = \frac{73216 \times 21}{7 \times 22 \times 624} \Rightarrow h = 16 cm Also, The slant height of the container, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(20 - 8)}^{2} + 16^{2}} = \sqrt{12^{2} + 16^{2}} = \sqrt{144 + 256} = \sqrt{400} = 20 cm Now, Total surface area of the container = π l (R + r) + π r^{2} = \frac{22}{7} \times 20 \times (20 + 8) + \frac{22}{7} \times 8 \times 8 = \frac{22}{7} \times 20 \times 28 + \frac{22}{7} \times 64 = \frac{22}{7} \times (560 + 64) = \frac{22}{7} \times 624 = \frac{13728}{7} {cm}^{2} So, the cost of metal sheet = 1.4 \times \frac{13728}{7} = ₹ 2745.60$

Hence, the cost of the metal sheet used for making the milk container is â‚¹2745.60.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Answer:

$We have, Radius of the metallic sphere, R = \frac{28}{2} = 14 cm, Radius of the smaller cone, r = \frac{1}{2} \times (4 \frac{2}{3}) = \frac{1}{2} \times \frac{14}{3} = \frac{7}{3} cm and Height of the smaller cone, h = 3 cm Now, The number of cones so formed = \frac{Volume of the metallic sphere}{Volume of a smaller cone} = \frac{(\frac{4}{3} π R^{3})}{(\frac{1}{3} π r^{2} h)} = \frac{4 R^{3}}{r^{2} h} = \frac{4 \times 14 \times 14 \times 14}{(\frac{7}{3} \times \frac{7}{3} \times 3)} = 672$

So, the number of cones so formed is 672.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Question 40:

$We have, Radius of the metallic sphere, R = \frac{28}{2} = 14 cm, Radius of the smaller cone, r = \frac{1}{2} \times (4 \frac{2}{3}) = \frac{1}{2} \times \frac{14}{3} = \frac{7}{3} cm and Height of the smaller cone, h = 3 cm Now, The number of cones so formed = \frac{Volume of the metallic sphere}{Volume of a smaller cone} = \frac{(\frac{4}{3} π R^{3})}{(\frac{1}{3} π r^{2} h)} = \frac{4 R^{3}}{r^{2} h} = \frac{4 \times 14 \times 14 \times 14}{(\frac{7}{3} \times \frac{7}{3} \times 3)} = 672$

So, the number of cones so formed is 672.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.

Answer:

$We have, Internal radius of the cylindrical vessel, R = \frac{10}{2} = 5 cm, Height of the cylindrical vessel, H = 10.5 cm, Radius of the solid cone, r = \frac{7}{2} = 3.5 cm and Height of the solid cone, h = 6 cm (i) Volume of water displaced out of the cylinder = Volume of the solid cone = \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 6 = 77 {cm}^{3} (ii) As, Volume of the cylindrical vessel = π R^{2} H = \frac{22}{7} \times 5 \times 5 \times 10.5 = 825 {cm}^{3} So, the volume of water left in the cylindrical vessel = Volume of the cylindrical vessel - Volume of the solid cone = 825 - 77 = 748 {cm}^{3}$

Question 1:

$We have, Internal radius of the cylindrical vessel, R = \frac{10}{2} = 5 cm, Height of the cylindrical vessel, H = 10.5 cm, Radius of the solid cone, r = \frac{7}{2} = 3.5 cm and Height of the solid cone, h = 6 cm (i) Volume of water displaced out of the cylinder = Volume of the solid cone = \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 6 = 77 {cm}^{3} (ii) As, Volume of the cylindrical vessel = π R^{2} H = \frac{22}{7} \times 5 \times 5 \times 10.5 = 825 {cm}^{3} So, the volume of water left in the cylindrical vessel = Volume of the cylindrical vessel - Volume of the solid cone = 825 - 77 = 748 {cm}^{3}$

Answer:

(a) a cylinder and a cone
A cylindrical pencil sharpened at one end is a combination of a cylinder and a cone.

Question 2:

(a) a cylinder and a cone
A cylindrical pencil sharpened at one end is a combination of a cylinder and a cone.

Answer:

(b) frustum of a cone and a hemisphere
A shuttlecock used for playing badminton is a combination of frustum of a cone and a hemisphere.

Question 3:

(b) frustum of a cone and a hemisphere
A shuttlecock used for playing badminton is a combination of frustum of a cone and a hemisphere.

Answer:

(c) a cylinder and frustum of a cone
A funnel is a combination of a cylinder and frustum of a cone.

Question 4:

(c) a cylinder and frustum of a cone
A funnel is a combination of a cylinder and frustum of a cone.

Answer:

(a) a sphere and a cylinder
A surahi is a combination of a sphere and a cylinder.

Question 5:

(a) a sphere and a cylinder
A surahi is a combination of a sphere and a cylinder.

Answer:

(b) frustum of a cone
The shape of a glass (tumbler) is usually in the form of frustum of a cone.

Question 6:

(b) frustum of a cone
The shape of a glass (tumbler) is usually in the form of frustum of a cone.

Answer:

(c) two cones and a cylinder
The shape of the gilli used in a gilli-danda game is a combination of two cones and a cylinder.

Question 7:

(c) two cones and a cylinder
The shape of the gilli used in a gilli-danda game is a combination of two cones and a cylinder.

Answer:

(a) a hemisphere and a cone
A plumbline (sahul) is a combination of a hemisphere and a cone.

Question 8:

(a) a hemisphere and a cone
A plumbline (sahul) is a combination of a hemisphere and a cone.

Answer:

(d) frustum of a cone
A cone is cut by a plane parallel to its base and the upper part is removed. The part that is left is called frustum of a cone.

Question 9:

(d) frustum of a cone
A cone is cut by a plane parallel to its base and the upper part is removed. The part that is left is called frustum of a cone.

Answer:

(c) remain unaltered
During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

Question 10:

(c) remain unaltered
During conversion of a solid from one shape to another, the volume of the new shape will remain unaltered.

Answer:

(c) circle
In a right circular cone, the cross-section made by a plane parallel to the base is a circle.

Question 11:

(c) circle
In a right circular cone, the cross-section made by a plane parallel to the base is a circle.

Answer:

(b) 21 cm
Volume of the cuboid $= (l \times b \times h)$ = $49 \times 33 \times 24 c m^{3}$
Let the radius of the sphere be r cm.
Volume of the sphere $= \frac{4}{3} π r^{3}$

The volume of the sphere and the cuboid are the same.
Therefore,

$\frac{4}{3} π r^{3} = 49 \times 33 \times 24 \Rightarrow \frac{4}{3} \times \frac{22}{7} \times r^{3} = 49 \times 33 \times 24$
$\Rightarrow r^{3} = 49 \times 33 \times 24 \times \frac{21}{88} \Rightarrow r^{3} = 21 \times 21 \times 21 \Rightarrow r^{3} = {(21)}^{3} \Rightarrow r = 21 cm$

Hence, the radius of the sphere is 21 cm.

Question 12:

(b) 21 cm
Volume of the cuboid $= (l \times b \times h)$ = $49 \times 33 \times 24 c m^{3}$
Let the radius of the sphere be r cm.
Volume of the sphere $= \frac{4}{3} π r^{3}$

The volume of the sphere and the cuboid are the same.
Therefore,

$\frac{4}{3} π r^{3} = 49 \times 33 \times 24 \Rightarrow \frac{4}{3} \times \frac{22}{7} \times r^{3} = 49 \times 33 \times 24$
$\Rightarrow r^{3} = 49 \times 33 \times 24 \times \frac{21}{88} \Rightarrow r^{3} = 21 \times 21 \times 21 \Rightarrow r^{3} = {(21)}^{3} \Rightarrow r = 21 cm$

Hence, the radius of the sphere is 21 cm.

Answer:

Since, the diameter of the base of the largest cone that can be cut out from the cube = Edge of the cube = 4.2 cm

So, the radius of the base of the largest cone = $\frac{4.2}{2}$ = 2.1 cm

Hence, the correct answer is option (a).

Question 13:

Since, the diameter of the base of the largest cone that can be cut out from the cube = Edge of the cube = 4.2 cm

So, the radius of the base of the largest cone = $\frac{4.2}{2}$ = 2.1 cm

Hence, the correct answer is option (a).

Answer:

$We have, Radius of the solid sphere, R = 9 cm and Radius of the solid cylinder, r = 9 cm Let the height of the cylinder be h . Now, Volume of the cylinder = Volume of the sphere \Rightarrow π r^{2} h = \frac{4}{3} π R^{3} \Rightarrow h = \frac{4 R^{3}}{3 r^{2}} \Rightarrow h = \frac{4 \times 9 \times 9 \times 9}{3 \times 9 \times 9} ∴ h = 12 cm$

Hence, the correct answer is option (a).

Question 14:

$We have, Radius of the solid sphere, R = 9 cm and Radius of the solid cylinder, r = 9 cm Let the height of the cylinder be h . Now, Volume of the cylinder = Volume of the sphere \Rightarrow π r^{2} h = \frac{4}{3} π R^{3} \Rightarrow h = \frac{4 R^{3}}{3 r^{2}} \Rightarrow h = \frac{4 \times 9 \times 9 \times 9}{3 \times 9 \times 9} ∴ h = 12 cm$

Hence, the correct answer is option (a).

Answer:

$We have, Length of the rectangular sheet, l = 40 cm, Width of the rectangular sheet, b = 22 cm and Height of the hollow cylinder, h = 40 cm Let the radius of the cylinder be r . As, l = h So, the circumference of base of the cylinder = b \Rightarrow 2 π r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = \frac{22 \times 7}{2 \times 22} ∴ r = 3.5 cm$

Hence, the correct answer is option (a).

Question 15:

$We have, Length of the rectangular sheet, l = 40 cm, Width of the rectangular sheet, b = 22 cm and Height of the hollow cylinder, h = 40 cm Let the radius of the cylinder be r . As, l = h So, the circumference of base of the cylinder = b \Rightarrow 2 π r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = \frac{22 \times 7}{2 \times 22} ∴ r = 3.5 cm$

Hence, the correct answer is option (a).

Answer:

$We have, Radius of the spher, r = \frac{6}{2} = 3 cm, Height of the cylinder, H = 45 cm and Radius of the cylinder, R = \frac{4}{2} = 2 cm Now, The number of sphere that can be made = \frac{Volume of the cylinder}{Volume of the sphere} = \frac{π R^{2} H}{(\frac{4}{3} π r^{3})} = \frac{3 R^{2} H}{4 r^{3}} = \frac{3 \times 2 \times 2 \times 45}{4 \times 3 \times 3 \times 3} = 5$

Hence, the correct answer is option (c).

Question 16:

$We have, Radius of the spher, r = \frac{6}{2} = 3 cm, Height of the cylinder, H = 45 cm and Radius of the cylinder, R = \frac{4}{2} = 2 cm Now, The number of sphere that can be made = \frac{Volume of the cylinder}{Volume of the sphere} = \frac{π R^{2} H}{(\frac{4}{3} π r^{3})} = \frac{3 R^{2} H}{4 r^{3}} = \frac{3 \times 2 \times 2 \times 45}{4 \times 3 \times 3 \times 3} = 5$

Hence, the correct answer is option (c).

Answer:

$Let the radius of the two spheres be r and R . As, \frac{Surface area of the first sphere}{Surface area of the second sphere} = \frac{16}{9} \Rightarrow \frac{4 π R^{2}}{4 π r^{2}} = \frac{16}{9} \Rightarrow {(\frac{R}{r})}^{2} = \frac{16}{9} \Rightarrow \frac{R}{r} = \sqrt{\frac{16}{9}} \Rightarrow \frac{R}{r} = \frac{4}{3} . . . . . (i) Now, The ratio of their volumes = \frac{Volume of the first sphere}{Volume of the second sphere} = \frac{(\frac{4}{3} π R^{3})}{(\frac{4}{3} π r^{3})} = {(\frac{R}{r})}^{3} = {(\frac{4}{3})}^{3} [Using (i)] = \frac{64}{27} = 64 : 27$

Hence, the correct answer is option (a).

Question 17:

$Let the radius of the two spheres be r and R . As, \frac{Surface area of the first sphere}{Surface area of the second sphere} = \frac{16}{9} \Rightarrow \frac{4 π R^{2}}{4 π r^{2}} = \frac{16}{9} \Rightarrow {(\frac{R}{r})}^{2} = \frac{16}{9} \Rightarrow \frac{R}{r} = \sqrt{\frac{16}{9}} \Rightarrow \frac{R}{r} = \frac{4}{3} . . . . . (i) Now, The ratio of their volumes = \frac{Volume of the first sphere}{Volume of the second sphere} = \frac{(\frac{4}{3} π R^{3})}{(\frac{4}{3} π r^{3})} = {(\frac{R}{r})}^{3} = {(\frac{4}{3})}^{3} [Using (i)] = \frac{64}{27} = 64 : 27$

Hence, the correct answer is option (a).

Answer:

$Let the radius of the sphere be r . As, Surface area of the sphere = 616 {cm}^{2} \Rightarrow 4 π r^{2} = 616 \Rightarrow 4 \times \frac{22}{7} \times r^{2} = 616 \Rightarrow r^{2} = \frac{616 \times 7}{4 \times 22} \Rightarrow r^{2} = 49 \Rightarrow r = \sqrt{49} \Rightarrow r = 7 cm ∴ Diameter of the sphere = 2 r = 2 \times 7 = 14 cm$

Hence, the correct answer is option (b).

Question 18:

$Let the radius of the sphere be r . As, Surface area of the sphere = 616 {cm}^{2} \Rightarrow 4 π r^{2} = 616 \Rightarrow 4 \times \frac{22}{7} \times r^{2} = 616 \Rightarrow r^{2} = \frac{616 \times 7}{4 \times 22} \Rightarrow r^{2} = 49 \Rightarrow r = \sqrt{49} \Rightarrow r = 7 cm ∴ Diameter of the sphere = 2 r = 2 \times 7 = 14 cm$

Hence, the correct answer is option (b).

Answer:

$Let the radius of the given sphere and that of the new sphere be r and R, respectively; Also, the volume of the given sphere and that of the new sphere be v and V, respectively . We have, R = 3 r . . . . . (i) Now, \frac{Volume of the new sphere}{Volume of the given sphere} = \frac{(\frac{4}{3} {πR}^{3})}{(\frac{4}{3} {πr}^{3})} = {(\frac{R}{r})}^{3} = {(\frac{3 r}{r})}^{3} [Using (i)] = 3^{3} = 27$

Hence, the correct answer is option (d).

Question 19:

$Let the radius of the given sphere and that of the new sphere be r and R, respectively; Also, the volume of the given sphere and that of the new sphere be v and V, respectively . We have, R = 3 r . . . . . (i) Now, \frac{Volume of the new sphere}{Volume of the given sphere} = \frac{(\frac{4}{3} {πR}^{3})}{(\frac{4}{3} {πr}^{3})} = {(\frac{R}{r})}^{3} = {(\frac{3 r}{r})}^{3} [Using (i)] = 3^{3} = 27$

Hence, the correct answer is option (d).

Answer:

$We have, Height of the frustum, h = 16 cm, Radii of the circular ends, R = \frac{40}{2} = 20 cm and r = \frac{16}{2} = 8 cm Now, The slant height of the frustum, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(20 - 8)}^{2} + 16^{2}} = \sqrt{12^{2} + 16^{2}} = \sqrt{144 + 256} = \sqrt{400} = 20 cm$

Hence, the correct answer is option (a).

Question 20:

$We have, Height of the frustum, h = 16 cm, Radii of the circular ends, R = \frac{40}{2} = 20 cm and r = \frac{16}{2} = 8 cm Now, The slant height of the frustum, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(20 - 8)}^{2} + 16^{2}} = \sqrt{12^{2} + 16^{2}} = \sqrt{144 + 256} = \sqrt{400} = 20 cm$

Hence, the correct answer is option (a).

Answer:

$We have, Radius of the sphere, r = \frac{18}{2} = 9 cm and Radius of the cylindrical vessel, R = \frac{36}{2} = 18 cm Let the rise of water level be H . Now, Volume of the water rised = Volume of the sphere \Rightarrow π R^{2} H = \frac{4}{3} π r^{3} \Rightarrow H = \frac{4 r^{3}}{3 R^{2}} \Rightarrow H = \frac{4 \times 9 \times 9 \times 9}{3 \times 18 \times 18} ∴ H = 3 cm$

Hence, the correct answer is option (a).

Question 21:

$We have, Radius of the sphere, r = \frac{18}{2} = 9 cm and Radius of the cylindrical vessel, R = \frac{36}{2} = 18 cm Let the rise of water level be H . Now, Volume of the water rised = Volume of the sphere \Rightarrow π R^{2} H = \frac{4}{3} π r^{3} \Rightarrow H = \frac{4 r^{3}}{3 R^{2}} \Rightarrow H = \frac{4 \times 9 \times 9 \times 9}{3 \times 18 \times 18} ∴ H = 3 cm$

Hence, the correct answer is option (a).

Answer:

$Let the radii of the smaller and given cones be r and R, respectively; and their heights be h and H, respectively . We have, H = 2 h . . . . . (i) In ∆ AQD and ∆ APC, ∠ QAD = ∠ PAC (Common angle) ∠ AQD = ∠ APC = 90 ° So, by AA criteria ∆ AQD ~ ∆ APC \Rightarrow \frac{AQ}{AP} = \frac{QD}{PC} \Rightarrow \frac{h}{H} = \frac{r}{R} \Rightarrow \frac{h}{2 h} = \frac{r}{R} [Using (i)] \Rightarrow \frac{1}{2} = \frac{r}{R} \Rightarrow R = 2 r . . . . . (ii) Now, The ratio of the volume of the smaller cone to the whole cone = \frac{Volume of the smaller cone}{Volume of the whole cone} = \frac{(\frac{1}{3} π r^{2} h)}{(\frac{1}{3} π R^{2} H)} = {(\frac{r}{R})}^{2} \times (\frac{h}{H}) = {(\frac{r}{2 r})}^{2} \times (\frac{h}{2 h}) [Using (i) and (ii)] = {(\frac{1}{2})}^{2} \times (\frac{1}{2}) = \frac{1}{8} = 1 : 8$

Hence, the correct answer is option (d).

Question 22:

$Let the radii of the smaller and given cones be r and R, respectively; and their heights be h and H, respectively . We have, H = 2 h . . . . . (i) In ∆ AQD and ∆ APC, ∠ QAD = ∠ PAC (Common angle) ∠ AQD = ∠ APC = 90 ° So, by AA criteria ∆ AQD ~ ∆ APC \Rightarrow \frac{AQ}{AP} = \frac{QD}{PC} \Rightarrow \frac{h}{H} = \frac{r}{R} \Rightarrow \frac{h}{2 h} = \frac{r}{R} [Using (i)] \Rightarrow \frac{1}{2} = \frac{r}{R} \Rightarrow R = 2 r . . . . . (ii) Now, The ratio of the volume of the smaller cone to the whole cone = \frac{Volume of the smaller cone}{Volume of the whole cone} = \frac{(\frac{1}{3} π r^{2} h)}{(\frac{1}{3} π R^{2} H)} = {(\frac{r}{R})}^{2} \times (\frac{h}{H}) = {(\frac{r}{2 r})}^{2} \times (\frac{h}{2 h}) [Using (i) and (ii)] = {(\frac{1}{2})}^{2} \times (\frac{1}{2}) = \frac{1}{8} = 1 : 8$

Hence, the correct answer is option (d).

Answer:

$We have, Height of the bucket, h = 40 cm, Radius of the upper end, R = 24 cm and Radius of the lower end, r = 15 cm Now, The slant height, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(24 - 15)}^{2} + 40^{2}} = \sqrt{9^{2} + 40^{2}} = \sqrt{81 + 1600} = \sqrt{1681} = 41 cm$

Hence, the correct answer is option (a).

Question 23:

$We have, Height of the bucket, h = 40 cm, Radius of the upper end, R = 24 cm and Radius of the lower end, r = 15 cm Now, The slant height, l = \sqrt{{(R - r)}^{2} + h^{2}} = \sqrt{{(24 - 15)}^{2} + 40^{2}} = \sqrt{9^{2} + 40^{2}} = \sqrt{81 + 1600} = \sqrt{1681} = 41 cm$

Hence, the correct answer is option (a).

Answer:

$Let the radius of the hemisphere or the radius of the cone be r and the slant height of the cone be l . Now, Surface area of the hemisphere = Surface area of the cone \Rightarrow 2 π r^{2} = π r l \Rightarrow \frac{π r^{2}}{π r l} = \frac{1}{2} \Rightarrow \frac{r}{l} = \frac{1}{2} ∴ r : l = 1 : 2$

Hence, the correct answer is option (a).

Question 24:

$Let the radius of the hemisphere or the radius of the cone be r and the slant height of the cone be l . Now, Surface area of the hemisphere = Surface area of the cone \Rightarrow 2 π r^{2} = π r l \Rightarrow \frac{π r^{2}}{π r l} = \frac{1}{2} \Rightarrow \frac{r}{l} = \frac{1}{2} ∴ r : l = 1 : 2$

Hence, the correct answer is option (a).

Answer:

$Let the base radius and height of the original cylinder be r and h, respectively . Also, The radius of the new cylinder, R = \frac{r}{2} and its height, H = h . Now, The ratio of the volume of the new cylinder to that of the original cylinder = \frac{Volume of the new cylinder}{Voilume of the original cylinder} = \frac{π r^{2} h}{π R^{2} H} = \frac{π r^{2} h}{π {(\frac{r}{2})}^{2} h} = \frac{4 π r^{2} h}{π r^{2} h} = \frac{4}{1} = 4 : 1$

Hence, the correct answer is option (d).

Question 25:

$Let the base radius and height of the original cylinder be r and h, respectively . Also, The radius of the new cylinder, R = \frac{r}{2} and its height, H = h . Now, The ratio of the volume of the new cylinder to that of the original cylinder = \frac{Volume of the new cylinder}{Voilume of the original cylinder} = \frac{π r^{2} h}{π R^{2} H} = \frac{π r^{2} h}{π {(\frac{r}{2})}^{2} h} = \frac{4 π r^{2} h}{π r^{2} h} = \frac{4}{1} = 4 : 1$

Hence, the correct answer is option (d).

Answer:

(c) 363
The edge of the cubical ice-cream brick = a = 22 cm
Volume of the cubical ice-cream brick $= {(a)}^{3}$
$= (22 \times 22 \times 22) {cm}^{3}$

Radius of an ice-cream cone = 2 cm
Height of an ice-cream cone = 7 cm
Volume of each ice-cream cone $= \frac{1}{3} π r^{2} h$
$= (\frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times 7) {cm}^{3}$

Number of ice-cream cones $= \frac{Volume of the cubical ice cream brick}{Volume of each ice cream cone}$

$= \frac{22 \times 22 \times 22 \times 3 \times 7}{22 \times 2 \times 2 \times 7} = 363$
Hence, the number of ice-cream cones is 363.

Question 26:

(c) 363
The edge of the cubical ice-cream brick = a = 22 cm
Volume of the cubical ice-cream brick $= {(a)}^{3}$
$= (22 \times 22 \times 22) {cm}^{3}$

Radius of an ice-cream cone = 2 cm
Height of an ice-cream cone = 7 cm
Volume of each ice-cream cone $= \frac{1}{3} π r^{2} h$
$= (\frac{1}{3} \times \frac{22}{7} \times 2 \times 2 \times 7) {cm}^{3}$

Number of ice-cream cones $= \frac{Volume of the cubical ice cream brick}{Volume of each ice cream cone}$

$= \frac{22 \times 22 \times 22 \times 3 \times 7}{22 \times 2 \times 2 \times 7} = 363$
Hence, the number of ice-cream cones is 363.

Answer:

(c) 11200
Volume of wall = $270 \times 300 \times 350 {cm}^{3}$
$\frac{1}{8} th$ of the wall is covered with mortar.
So,
Volume of the wall filled with bricks $= (\frac{7}{8} \times 270 \times 300 \times 350) {cm}^{3}$
Volume of each brick $= (\frac{225}{10} \times \frac{1125}{100} \times \frac{875}{100}) {cm}^{3}$
   $= (\frac{9 \times 225 \times 35}{32}) {cm}^{3}$

Number of bricks used to construct the wall $= \frac{Volume of the wall composed of bricks}{Volulme of each brick}$

   $= \frac{7 \times 270 \times 300 \times 350 \times 32}{8 \times 9 \times 225 \times 35}$
= 11200

Hence, the number of bricks used to construct the wall is 11200.

Question 27:

(c) 11200
Volume of wall = $270 \times 300 \times 350 {cm}^{3}$
$\frac{1}{8} th$ of the wall is covered with mortar.
So,
Volume of the wall filled with bricks $= (\frac{7}{8} \times 270 \times 300 \times 350) {cm}^{3}$
Volume of each brick $= (\frac{225}{10} \times \frac{1125}{100} \times \frac{875}{100}) {cm}^{3}$
   $= (\frac{9 \times 225 \times 35}{32}) {cm}^{3}$

Number of bricks used to construct the wall $= \frac{Volume of the wall composed of bricks}{Volulme of each brick}$

   $= \frac{7 \times 270 \times 300 \times 350 \times 32}{8 \times 9 \times 225 \times 35}$
= 11200

Hence, the number of bricks used to construct the wall is 11200.

Answer:

(a) 2 cm
Let the diameter of each sphere be d cm.
Let r and R be the radii of the sphere and the cylinder, respectively,
and h be the height of the cylinder.

$As R = \frac{Diameter}{2}, R = \frac{2}{2} cm = 1 cm h = 16 cm$

Therefore,
$12 \times \frac{4}{3} π r^{3} = π R^{2} h \Rightarrow 12 \times \frac{4}{3} r^{3} = R^{2} h$

$\Rightarrow 12 \times \frac{4}{3} {(\frac{d}{2})}^{3} = {(1)}^{2} \times 16 \Rightarrow 16 \times \frac{d^{3}}{8} = 16$ ^{$\Rightarrow d^{3} = 8 \Rightarrow d = \pm 2 Since d cannot be negative, thus, d = 2$}

Hence, the diameter of each sphere is 2 cm.

Question 28:

(a) 2 cm
Let the diameter of each sphere be d cm.
Let r and R be the radii of the sphere and the cylinder, respectively,
and h be the height of the cylinder.

$As R = \frac{Diameter}{2}, R = \frac{2}{2} cm = 1 cm h = 16 cm$

Therefore,
$12 \times \frac{4}{3} π r^{3} = π R^{2} h \Rightarrow 12 \times \frac{4}{3} r^{3} = R^{2} h$

$\Rightarrow 12 \times \frac{4}{3} {(\frac{d}{2})}^{3} = {(1)}^{2} \times 16 \Rightarrow 16 \times \frac{d^{3}}{8} = 16$ ^{$\Rightarrow d^{3} = 8 \Rightarrow d = \pm 2 Since d cannot be negative, thus, d = 2$}

Hence, the diameter of each sphere is 2 cm.

Answer:

(b) 32.7 litres
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.

Then, $R = \frac{44}{2} cm = 22 cm, r = \frac{24}{2} cm = 12 cm, h = 35 cm$

Capacity of the bucket = Volume of the frustum of the cone
$= \frac{1}{3} π h [R^{2} + r^{2} + R r] {cm}^{3}$
$= \frac{1}{3} \times \frac{22}{7} \times 35 \times [{(22)}^{2} + {(12)}^{2} + (22 \times 12)] {cm}^{3} = (\frac{110}{3} \times 892) {cm}^{3} = (\frac{110 \times 892}{3 \times 1000}) litres = 32.7 litres$

Hence, the capacity of the bucket is 32.7 litres.

Question 29:

(b) 32.7 litres
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.

Then, $R = \frac{44}{2} cm = 22 cm, r = \frac{24}{2} cm = 12 cm, h = 35 cm$

Capacity of the bucket = Volume of the frustum of the cone
$= \frac{1}{3} π h [R^{2} + r^{2} + R r] {cm}^{3}$
$= \frac{1}{3} \times \frac{22}{7} \times 35 \times [{(22)}^{2} + {(12)}^{2} + (22 \times 12)] {cm}^{3} = (\frac{110}{3} \times 892) {cm}^{3} = (\frac{110 \times 892}{3 \times 1000}) litres = 32.7 litres$

Hence, the capacity of the bucket is 32.7 litres.

Answer:

(d) 4950 cm²
Let the radii of the top and bottom of the bucket be R and r and let its slant height be l.
Then, $R = 28 cm, r = 7 cm, l = 45 cm$
Curved surface area of the bucket $= π l (R + r)$
$= \frac{22}{7} \times 45 \times (28 + 7) {cm}^{2} = 4950 {cm}^{2}$

Hence, the curved surface area of the bucket is 4950 cm².

Question 30:

(d) 4950 cm²
Let the radii of the top and bottom of the bucket be R and r and let its slant height be l.
Then, $R = 28 cm, r = 7 cm, l = 45 cm$
Curved surface area of the bucket $= π l (R + r)$
$= \frac{22}{7} \times 45 \times (28 + 7) {cm}^{2} = 4950 {cm}^{2}$

Hence, the curved surface area of the bucket is 4950 cm².

Answer:

(b) 16 : 9
Let the radii of the spheres be R and r.
Then, ratio of their volumes
$= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}}$

Therefore,

$\frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}} = \frac{64}{27} \Rightarrow \frac{R^{3}}{r^{3}} = \frac{64}{27}$

$\Rightarrow {(\frac{R}{r})}^{3} = {(\frac{4}{3})}^{3} \Rightarrow \frac{R}{r} = \frac{4}{3}$
Hence, the ratio of their surface areas $= \frac{4 π R^{2}}{4 {πr}^{2}}$

$= \frac{R^{2}}{r^{2}} = {(\frac{R}{r})}^{2} = {(\frac{4}{3})}^{2} = \frac{16}{9} = 16 : 9$

Question 31:

(b) 16 : 9
Let the radii of the spheres be R and r.
Then, ratio of their volumes
$= \frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}}$

Therefore,

$\frac{\frac{4}{3} π R^{3}}{\frac{4}{3} {πr}^{3}} = \frac{64}{27} \Rightarrow \frac{R^{3}}{r^{3}} = \frac{64}{27}$

$\Rightarrow {(\frac{R}{r})}^{3} = {(\frac{4}{3})}^{3} \Rightarrow \frac{R}{r} = \frac{4}{3}$
Hence, the ratio of their surface areas $= \frac{4 π R^{2}}{4 {πr}^{2}}$

$= \frac{R^{2}}{r^{2}} = {(\frac{R}{r})}^{2} = {(\frac{4}{3})}^{2} = \frac{16}{9} = 16 : 9$

Answer:

(a) 142296
Since $\frac{1}{8} th$ of the cube remains unfulfilled,
volume of the cube = $22 \times 22 \times 22 {cm}^{3}$

Space filled in the cube $= (\frac{7}{8} \times 22 \times 22 \times 22) {cm}^{3}$
$= (7 \times 1331) {cm}^{3}$

Radius of each marble $= \frac{0.5}{2} cm$
$= \frac{5}{20} cm = \frac{1}{4} cm$

Volume of each marble $= \frac{4}{3} π r^{3}$
$= (\frac{4}{3} \times \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}) {cm}^{3} = (\frac{11}{24 \times 7}) {cm}^{3}$

Therefore, number of marbles required $= (\frac{7 \times 1331 \times 24 \times 7}{11})$
= 142296

Question 32:

(a) 142296
Since $\frac{1}{8} th$ of the cube remains unfulfilled,
volume of the cube = $22 \times 22 \times 22 {cm}^{3}$

Space filled in the cube $= (\frac{7}{8} \times 22 \times 22 \times 22) {cm}^{3}$
$= (7 \times 1331) {cm}^{3}$

Radius of each marble $= \frac{0.5}{2} cm$
$= \frac{5}{20} cm = \frac{1}{4} cm$

Volume of each marble $= \frac{4}{3} π r^{3}$
$= (\frac{4}{3} \times \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}) {cm}^{3} = (\frac{11}{24 \times 7}) {cm}^{3}$

Therefore, number of marbles required $= (\frac{7 \times 1331 \times 24 \times 7}{11})$
= 142296

Answer:

(b) 14 cm
Let the internal and external radii of the spherical shell be r and R, respectively.
Volume of the spherical shell $= \frac{4}{3} π (R^{3} - r^{3})$
$= \frac{4}{3} π [{(4)}^{3} - {(2)}^{3}] {cm}^{3} [Since r \Rightarrow \frac{4}{2} = 2 cm, R = \frac{8}{2} = 4 cm] = (\frac{4}{3} π \times 56) {cm}^{3}$
Radius of the cone = $\frac{8}{2} = 4 cm$
Volume of the cone $= \frac{1}{3} π r^{2} h$
$= (\frac{1}{3} π \times 4 \times 4 \times h) {cm}^{3}$
Therefore,
$\frac{1}{3} π \times 4 \times 4 \times h = \frac{4}{3} π \times 56 \Rightarrow \frac{16}{3} \times h = \frac{4}{3} π \times 56 \Rightarrow h = \frac{4 \times 56 \times 3}{3 \times 16} \Rightarrow h = 14 cm$

Question 33:

(b) 14 cm
Let the internal and external radii of the spherical shell be r and R, respectively.
Volume of the spherical shell $= \frac{4}{3} π (R^{3} - r^{3})$
$= \frac{4}{3} π [{(4)}^{3} - {(2)}^{3}] {cm}^{3} [Since r \Rightarrow \frac{4}{2} = 2 cm, R = \frac{8}{2} = 4 cm] = (\frac{4}{3} π \times 56) {cm}^{3}$
Radius of the cone = $\frac{8}{2} = 4 cm$
Volume of the cone $= \frac{1}{3} π r^{2} h$
$= (\frac{1}{3} π \times 4 \times 4 \times h) {cm}^{3}$
Therefore,
$\frac{1}{3} π \times 4 \times 4 \times h = \frac{4}{3} π \times 56 \Rightarrow \frac{16}{3} \times h = \frac{4}{3} π \times 56 \Rightarrow h = \frac{4 \times 56 \times 3}{3 \times 16} \Rightarrow h = 14 cm$

Answer:

(d) 0.36 cm³
Radius of the capsule
$= \frac{0.5}{2} cm$
= 0.25 cm

Let the length of the cylindrical part of the capsule be x cm.

Then,

$0.25 + x + 0.25 = 2 \Rightarrow 0.5 + x = 2 \Rightarrow x = 1.5$

Hence, the capacity of the capsule $= 2 \times (Volume of the hemisphere) + (Volume of the cylinder)$
$= (2 \times \frac{2}{3} π r)$ ³+πr²h
$= (\frac{4}{3} \times \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}) + (\frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times 1.5) [Since 0.25 = \frac{1}{4}]$
$= (\frac{11}{168} + \frac{33}{112}) cm$ ³
$= \frac{121}{336} cm$ ³= 0.36 cm³

Question 34:

(d) 0.36 cm³
Radius of the capsule
$= \frac{0.5}{2} cm$
= 0.25 cm

Let the length of the cylindrical part of the capsule be x cm.

Then,

$0.25 + x + 0.25 = 2 \Rightarrow 0.5 + x = 2 \Rightarrow x = 1.5$

Hence, the capacity of the capsule $= 2 \times (Volume of the hemisphere) + (Volume of the cylinder)$
$= (2 \times \frac{2}{3} π r)$ ³+πr²h
$= (\frac{4}{3} \times \frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}) + (\frac{22}{7} \times \frac{1}{4} \times \frac{1}{4} \times 1.5) [Since 0.25 = \frac{1}{4}]$
$= (\frac{11}{168} + \frac{33}{112}) cm$ ³
$= \frac{121}{336}$

RS Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 17 - Volumes And Surface Areas Of Solids

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