Rs Aggarwal 2021 2022 for Class 9 Maths Chapter 15 - Volumes And Surface Area Of Solids

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Rs Aggarwal 2021 2022 Solutions for Class 9 Maths Chapter 15 Volumes And Surface Area Of Solids are provided here with simple step-by-step explanations. These solutions for Volumes And Surface Area Of Solids are extremely popular among Class 9 students for Maths Volumes And Surface Area Of Solids Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 9 Maths Chapter 15 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

Page No 555:

Answer:

(i)
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = $l \times b \times h$
                                  $= (12 \times 8 \times 4.5) {cm}^{3} = 432 {cm}^{3}$

Total Surface area = 2(lb + lh+ bh)
                              $= 2 (12 \times 8 + 12 \times 4.5 + 8 \times 4.5) {cm}^{2} = 2 (96 + 54 + 36) {cm}^{2} = 2 \times 186 {cm}^{2} = 372 cm^{2}$

Lateral surface area = $2 (l + b) \times h$
                              $= [2 (12 + 8) \times 4.5] {cm}^{2} = [2 (20) \times 4.5] {cm}^{2} = 40 \times 4.5 {cm}^{2} = 180 {cm}^{2}$

(ii)
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid = $l \times b \times h$
                                $= (26 \times 14 \times 6.5) m^{3} = 2366 m^{3}$

Total surface area = 2(lb + lh+ bh)
                             $= 2 (26 \times 14 + 26 \times 6.5 + 6.5 \times 14) m^{2} = 2 (364 + 169 + 91) m^{2} = 2 \times 624 m^{2} = 1248 m^{2}$

Lateral surface area = $2 (l + b) \times h$
                            $= [2 (26 + 14) \times 6.5] m^{2} = [2 \times 40 \times 6.5] m^{2} = 520 m^{2}$

(iii)
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid = $l \times b \times h$
                    $= (15 \times 6 \times 0.5) m^{3} = 45 m^{3}$

Total surface area = 2(lb + lh+ bh)
                          $= 2 (15 \times 6 + 15 \times 0.5 + 6 \times 0.5) m^{2} = 2 (90 + 7.5 + 3) m^{2} = 2 \times 100.5 m^{2} = 201 m^{2}$

Lateral surface area = $2 (l + b) \times h$
                             $= [2 (15 + 6) \times 0.5] m^{2} = [2 \times 21 \times 0.5] m^{2} = 21 m^{2}$

(iv)
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid = $l \times b \times h$
                                $= (24 \times 0.25 \times 6) m^{3} = 36 m^{3}$

Total Surface area = 2(lb + lh+ bh)
                            $= 2 (24 \times 0.25 + 24 \times 6 + 0.25 \times 6) m^{2} = 2 (6 + 144 + 1.5) m^{2} = 2 \times 151.5 m^{2} = 303 m^{2}$

Lateral surface area = $2 (l + b) \times h$
                             $= [2 (24 + 0.25) \times 6] m^{2} = [2 \times 24.25 \times 6] m^{2} = 291 m^{2}$

Page No 555:

Answer:

Volume of each matchbox = 4 × 2.5 × 1.5 = 15 cm³

∴ Volume of 12 matchboxes = 12 × 15 = 180 cm³

Thus, the volume of a packet containing 12 such matchboxes is 180 cm³.

Page No 555:

Answer:

Volume of water in the tank = Length × Breadth × Height = 6 × 5 × 4.5 = 135 m³

∴ Volume of water in litres = 135 × 1000 = 135000 L (1 m³ = 1000 L)

Thus, the water tank can hold 135000 L of water.

Page No 555:

Answer:

Capacity of the tank = 50000 L = $\frac{50000}{1000}$ = 50 m³ (1000 L = 1 m³)

Length of the tank = 10 m

Height (or depth) of the tank = 2.5 m

Now,

Volume of the cuboidal tank = Length × Breadth × Height

∴ Breadth of the tank = $\frac{Volume of the tank}{Length \times Height} = \frac{50}{10 \times 2.5} = \frac{50}{25}$ = 2 m

Thus, the breadth of the tank is 2 m.

Page No 555:

Answer:

Volume of the godown = 40 m × 25 m × 15 m = 15000 m³

Volume of each wooden crate = 1.5 m × 1.25 m × 0.5 m = 0.9375 m³

∴ Maximum number of wooden crates that can be stored in the godown

$= \frac{Volume of the godown}{Volume of each wooden crate} = \frac{15000}{0.9375} = 16000$

Page No 555:

Answer:

Number of planks = $\frac{volume of the pit in {cm}^{3}}{volume of 1 plank in {cm}^{3}}$

Volume of one plank = $(l \times b \times h) {cm}^{3}$
$= 500 \times 25 \times 10 {cm}^{3} = 125000 {cm}^{3}$
$Volume of the pit = (l \times b \times h) {cm}^{3} Here, l = 20 m = 2000 cm; b = 6 m = 600 cm; h = 80 cm i . e ., volume of the pit = 2000 \times 600 \times 80 {cm}^{3} = 96000000 {cm}^{3}$

∴ Number of planks = $\frac{96000000}{125000} = \frac{96000}{125} = 768$

Page No 555:

Answer:

Length of the wall = 8 m = 800 cm
Breadth of the wall = 22.5 cm
Height of the wall = 6 m = 600 cm
i.e., volume of wall $= 800 \times 22.5 \times 600 {cm}^{3} = 10800000 {cm}^{3}$

Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm
i.e., volume of one brick $= 25 \times 11.25 \times 6 = 1687.5 {cm}^{3}$

∴ Number of bricks = $\frac{volume of the wall}{volume of one brick}$ = $\frac{10800000}{1687.5} = 6400$

Page No 555:

Answer:

Length of the cistern, l = 8 m

Breadth of the cistern, b = 6 m

Height (or depth) of the cistern, h = 2.5 m

∴ Capacity of the cistern

= Volume of the cistern

= l × b × h

= 8 × 6 × 2.5

= 120 m³

Also,

Area of the iron sheet required to make the cistern

= Total surface area of the cistern

= 2(lb + bh + hl)

= 2(8 × 6 + 6 × 2.5 + 2.5 × 8)

= 2 × 83

= 166 m²

Page No 555:

Answer:

Length of the room, l = 9 m

Breadth of the room, b = 8 m

Height of the room, h = 6.5 m

Now,

Area of the walls to be whitewashed

= Curved surface area of the room − Area of the door − 2 × Area of each window

= 2h(l + b) − 2 m × 1.5 m − 2 × 1.5 m × 1 m

= 2 × 6.5 × (9 + 8) − 3 − 3

= 221 − 6

= 215 m²

∴ Cost of whitewashing the walls at Rs 25 per square metre

= Area of the walls to be whitewashed × Rs 25 per square metre

= 215 × 25

= Rs 5,375

Page No 555:

Answer:

Length of the wall = 15 m = 1500 cm
Breadth of the wall = 30 cm
Height of the wall = 4 m = 400 cm
Volume of wall = $= 1500 \times 30 \times 400 {cm}^{3} = 18000000 {cm}^{3}$

$Now, volume of each brick = 22 \times 12.5 \times 7.5 {cm}^{3} = 2062.5 {cm}^{3}$

Also, volume of the mortar = $\frac{1}{12} \times volume of the wall$
$= \frac{18000000}{12} = 1500000 {cm}^{3}$
Total volume of the bricks in the wall = volume of the wall − volume of the mortar
= (18000000 − 1500000) cm³= 16500000 cm³

∴ Number of bricks = $\frac{volume of bricks}{volume of one brick} = \frac{16500000}{2062.5} = 8000 bricks$

Page No 555:

Answer:

The external dimensions of the box are 36 cm, 25 cm and 16.5 cm.

Thickness of the iron = 1.5 cm

∴ Inner length of the box = 36 − 1.5 −1.5 = 33 cm

Inner breadth of the box = 25 − 1.5 −1.5 = 22 cm

Inner height of the box = 16.5 − 1.5 = 15 cm

Now,

Volume of iron in the open box

= Volume of the outer box − Volume of the inner box

= 36 × 25 × 16.5 − 33 × 22 × 15

= 14850 − 10890

= 3960 cm³

It is given that 1 cm³ of iron weighs 15 g.

∴ Weight of the empty box = 3960 × 15 = 59400 g = $\frac{59400}{1000}$ = 59.4 kg (1 kg = 1000 g)

Page No 556:

Answer:

Length of the box =5 m
Breadth of the box =3 m

Area of the sheet required = $\frac{total cost}{cost per metre square}$

Let h m be the height of the box.
Then area of the sheet = total surface area of the box
$= 2 (l b + l h + b h) m^{2} = 2 (5 \times 3 + 5 \times h + 3 \times h) m^{2} = 2 (15 + 8 h) = (30 + 16 h) m^{2}$

$Now, 30 + 16 h = \frac{6480}{120} \Rightarrow 30 + 16 h = 54 \Rightarrow 16 h = 24 \Rightarrow h = 1.5 m$

∴ The height of the box is 1.5 m.

Page No 556:

Answer:

Length of the cuboid = 16 m
Suppose that the breadth and height of the cuboid are 3x m and 2x m, respectively.
$Then 1536 = 16 \times 3 x \times 2 x \Rightarrow 1536 = 16 \times 6 x^{2} \Rightarrow x^{2} = \frac{1536}{96} = 16 \Rightarrow x = \sqrt{16} = 4$

∴ The breadth and height of the cuboid are 12 m and 8 m, respectively.

Page No 556:

Answer:

Volume of the dining hall = $(20 \times 16 \times 4.5) m^{3} = 1440 m^{3}$

Volume of air required by each person = 5 m³

∴ Capacity of the dining hall = $\frac{volume of dining hall}{volume of air required by each person} = \frac{1440}{5} = 288 persons$

Page No 556:

Answer:

Length of the classroom = 10 m
Breadth of the classroom = 6.4 m
Height of the classroom = 5 m
Area of the floor = length $\times$ breadth = 10 $\times$ 6.4 m²
$No . of students = \frac{area of the floor}{area given to one student on the floor} = \frac{10 \times 6.4}{1.6} = \frac{640}{16} = 40 students$

Now, volume of the classroom= $10 \times 6.4 \times 5 m^{3}$

$∴ Air required by each student = \frac{volume of the room}{number of students} = \frac{10 \times 6.4 \times 5}{40} m^{3} = 8 m^{3}$

Page No 556:

Answer:

Length of the cuboid = 14 cm
Breadth of the cuboid = 11 cm
Let the height of the cuboid be x cm.
Surface area of the cuboid = 758 cm²
$Then 758 = 2 (14 \times 11 + 14 \times x + 11 \times x) \Rightarrow 758 = 2 (154 + 14 x + 11 x) \Rightarrow 758 = 2 (154 + 25 x) \Rightarrow 758 = 308 + 50 x \Rightarrow 50 x = 758 - 308 = 450 \Rightarrow x = \frac{450}{50} = 9$

∴ The height of the cuboid is 9 cm.

Page No 556:

Answer:

Volume of the water that falls on the ground = $area of ground \times depth$
$= 20000 \times 0.05 m^{3} = 1000 m^{3}$

Page No 556:

Answer:

Here, a = 9 m
Volume of the cube = $a^{3} = 9^{3} m^{3} = 729 m^{3}$
Lateral surface area of the cube = $4 a^{2} = 4 \times 9^{2} m^{2} = 4 \times 81 m^{2} = 324 m^{2}$
Total surface area of the cube = $6 a^{2} = 6 \times 9^{2} m^{2} = 6 \times 81 m^{2} = 486 m^{2}$
∴ Diagonal of the cube = $\sqrt{3} a = \sqrt{3} \times 9 = 15.57 m$

Page No 556:

Answer:

Suppose that the side of cube is x cm.
Total surface area of cube = 1176 sq cm
$Then 1176 = 6 x^{2} \Rightarrow x^{2} = \frac{1176}{6} = 196 \Rightarrow x = \sqrt{196} = 14$
i.e., the side of the cube is 14 cm.

∴ Volume of the cube = $x^{3} = 14^{3} {cm}^{3} = 2744 {cm}^{3}$

Page No 556:

Answer:

Suppose that the side of cube is x cm.
Lateral surface area of the cube = 900 cm²
$Then 900 = 4 x^{2} \Rightarrow x^{2} = \frac{900}{4} = 225 \Rightarrow x = \sqrt{225} = 15$
i.e., the side of the cube is 15 cm.
∴ Volume of the given cube = $x^{3} {cm}^{3} = 15^{3} {cm}^{3} = 3375 {cm}^{3}$

Page No 556:

Answer:

Suppose that the side of the given cube is x cm.
Volume of the cube = 512 cm³
$Then 512 = x^{3} \Rightarrow x = \sqrt[3]{512} = 8 i . e ., the side of the cube is 8 cm .$

∴ Surface area of the cube = $6 x^{2} {cm}^{2} = 6 \times 8^{2} {cm}^{2} = 384 {cm}^{2}$

Page No 556:

Answer:

Three cubes of metal with edges 3cm, 4cm and 5 cm are melted to form a single cube.

∴ Volume of the new cube = sum of the volumes the old cubes
$= (3^{3} + 4^{3} + 5^{3}) {cm}^{3} = (27 + 64 + 125) {cm}^{3} = 216 {cm}^{3}$

Suppose the edge of the new cube = x cm
Then we have:
$Then 216 = x^{3} \Rightarrow x = \sqrt[3]{216} = 6 i . e ., the edge of the new cube is 6 cm .$

∴ Lateral surface area of the new cube = $4 x^{2} {cm}^{2} = 4 \times 6^{2} {cm}^{2} = 144 {cm}^{2}$

Page No 556:

Answer:

Length of the longest pole = length of the diagonal of the room
$= \sqrt{l^{2} + b^{2} + h^{2}} m = \sqrt{10^{2} + 10^{2} + 5^{2}} m = \sqrt{100 + 100 + 25} = \sqrt{225} = 15 m$

Page No 556:

Answer:

Let the length, breadth and height (or depth) of the cuboid be l cm, b cm and h cm, respectively.

∴ l + b + h = 19 .....(1)

Also,

Length of the diagonal = 11 cm

$\Rightarrow \sqrt{l^{2} + b^{2} + h^{2}} = 11 \Rightarrow l^{2} + b^{2} + h^{2} = 121 . . . . . (2)$

Squaring (1), we get

(l + b + h)² = 19²

⇒ l² + b² + h²+ 2(lb + bh + hl) = 361

⇒ 121 + 2(lb + bh + hl) = 361 [Using (2)]

⇒ 2(lb + bh + hl) = 361 − 121 = 240 cm²

Thus, the surface area of the cuboid is 240 cm².

Page No 556:

Answer:

Let the initial edge of the cube be a units.

∴ Initial surface area of the cube = 6a² square units

New edge of the cube = a + 50% of a = $a + \frac{50}{100} a$ = 1.5a units

∴ New surface of the cube = 6(1.5a)² = 13.5a² square units

Increase in surface area of the cube = 13.5a² − 6a² = 7.5a² square units

∴ Percentage increase in the surface area of the cube

$= \frac{Increase in surface area of the cube}{Initial surface area of the cube} \times 100 % = \frac{7.5 a^{2}}{6 a^{2}} \times 100 % = 125 %$

Page No 556:

Answer:

Let the length, breadth and height of the cuboid be a, b and c, respectively.

∴ Surface area of the cuboid, S = 2(ab + bc + ca)

Volume of the cuboid, V = abc

Now,

$\frac{S}{V} = \frac{2 (a b + b c + c a)}{a b c} \Rightarrow \frac{S}{V} = 2 (\frac{a b}{a b c} + \frac{b c}{a b c} + \frac{c a}{a b c}) \Rightarrow \frac{S}{V} = 2 (\frac{1}{c} + \frac{1}{a} + \frac{1}{b}) \Rightarrow \frac{1}{V} = \frac{2}{S} (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

Page No 556:

Answer:

Width of the canal = 30 dm = 3 m (1 m = 10 dm)

Depth of the canal = 12 dm = 1.2 m

Speed of the water flow = 20 km/h = 20000 m/h

∴ Volume of water flowing out of the canal in 1 h = 3 × 1.2 × 20000 = 72000 m³

Height of standing water on field = 9 cm = 0.09 m (1 m = 100 cm)

Assume that water flows out of the canal for 1 h. Then,

Area of the field irrigated

$= \frac{Volume of water flowing out of the canal}{Height of standing water on the field} = \frac{72000}{0.09} = 800000 m^{2}$
$= \frac{800000}{10000} (1 hectare = 10000 m^{2}) = 80 hectare$

Thus, the area of the field irrigated is 80 hectares.

Disclaimer: In this question time is not given, so the question is solved assuming that the water flows out of the canal for 1 hour.

Page No 556:

Answer:

Volume of the solid metallic cuboid = 9 m × 8 m × 2 m = 144 m³

Volume of each solid cube = (Edge)³ = (2)³ = 8 m³

∴ Number of cubes formed = $\frac{Volume of the solid metallic cuboid}{Volume of each solid cube} = \frac{144}{8}$ = 18

Thus, the number of cubes so formed is 18.

Page No 573:

Answer:

Here, r = 28/2 = 14 cm; h = 40 cm

$Curved surface area of the cylinder = 2 π r h = 2 \times \frac{22}{7} \times 14 \times 40 {cm}^{2} = 2 \times 22 \times 2 \times 40 {cm}^{2} = 3520 {cm}^{2}$

$Total surface area of the cylinder = 2 π r h + 2 π r^{2} = (3520 + 2 \times \frac{22}{7} \times 14^{2}) {cm}^{2} = (3520 + 1232) = 4752 {cm}^{2}$

$∴ Volume of the cylinder = π r^{2} h = \frac{22}{7} \times 14^{2} \times 40 {cm}^{3} = 22 \times 14 \times 2 \times 40 {cm}^{3} = 24640 {cm}^{3}$

Page No 573:

Answer:

Radius of the bowl, r = $\frac{7}{2} cm$

Height of soup in the bowl, h = 4 cm

Volume of soup in one bowl = $π r^{2} h = \frac{22}{7} \times {(\frac{7}{2})}^{2} \times 4 = 154 {cm}^{3}$

∴ Amount of soup the hospital has to prepare daily to serve 250 patients

= Volume of soup in one bowl × 250

= 154 × 250

= 38500 cm³

Page No 573:

Answer:

Radius of each pillar, r = 20 cm = 0.2 m (1 m = 100 cm)

Height of each pillar, h = 10 m

Volume of concrete mixture used in each pillar = $π r^{2} h = \frac{22}{7} \times {(0.2)}^{2} \times 10 m^{3}$

∴ Amount of concrete mixture required to build 14 such pillars

= Volume of concrete mixture used in each pillar × 14

= $\frac{22}{7} \times {(0.2)}^{2} \times 10 \times 14$

= 17.6 m³

Page No 573:

Answer:

(i) Length of tin can, l = 5 cm

Breadth of tin can, b = 4 cm

Height of tin can, h = 15 cm

∴ Volume of soft drink in tin can = l × b × h = 5 × 4 × 15 = 300 cm³

(ii) Radius of plastic cylinder, r = $\frac{7}{2}$ cm

Height of plastic cylinder, h = 10 cm

∴ Volume of soft drink in plastic cylinder = $π r^{2} h = \frac{22}{7} \times {(\frac{7}{2})}^{2} \times 10$ = 385 cm³

So, the capacity of the plastic cylinder pack is greater than the capacity of the tin can pack.

Difference in the capacities of the two packs = 385 − 300 = 85 cm³

Thus, the capacity of the plastic cylinder pack is 85 cm³ more than the capacity of the tin can pack.

Page No 573:

Answer:

Radius of each pillar, r = $\frac{50}{2}$ = 25 cm = 0.25 m (1 m = 100 cm)

Height of each pillar, h = 4 m

∴ Surface area of each pillar = $2 π r h = 2 \times \frac{22}{7} \times 0.25 \times 4 {cm}^{2}$

Surface area of 20 pillars = Surface area of each pillar × 20 = $2 \times \frac{22}{7} \times 0.25 \times 4 \times 20 {cm}^{2}$

Rate of cleaning = ₹ 14 per m²

∴ Total cost of cleaning the 20 pillars

= Surface area of 20 pillars × Rate of cleaning

= $2 \times \frac{22}{7} \times 0.25 \times 4 \times 20 \times 14$

= ₹ 1760

Page No 573:

Answer:

Radius of the cylinder, r = 0.7 m

Curved surface area of cylinder = 4.4 m²

(i) Let the height of the cylinder be h m.

$∴ 2 π r h = 4.4 \Rightarrow 2 \times \frac{22}{7} \times 0.7 \times h = 4.4 \Rightarrow h = \frac{4.4 \times 7}{2 \times 22 \times 0.7} = 1 m$

Thus, the height of the cylinder is 1 m.

(ii) Volume of the cylinder = $π r^{2} h = \frac{22}{7} \times {(0.7)}^{2} \times 1$ = 1.54 m³

Thus, the volume of the cylinder is 1.54 m³.

Page No 573:

Answer:

Height of the cylinder, h = 5 cm

Lateral (or curved) surface area of cylinder = 94.2 cm²

(i) Let the radius of the cylinder be r cm.

$∴ 2 π r h = 94.2 {cm}^{2} \Rightarrow 2 \times 3.14 \times r \times 5 = 94.2 \Rightarrow r = \frac{94.2}{2 \times 3.14 \times 5} = 3 cm$
Thus, the radius of the cylinder is 3 cm.

(ii) Volume of the cylinder = $π r^{2} h = 3.14 \times {(3)}^{2} \times 5$ = 141.3 cm³

Thus, the volume of the cylinder is 141.3 cm³.

Page No 573:

Answer:

Height of the cylinder, h = 1 m

Capacity of the cylinder = 15.4 L = 15.4 × 0.001 m³ = 0.0154 m³

$∴ π r^{2} h = 0.0154 m^{3} \Rightarrow \frac{22}{7} \times r^{2} \times 1 = 0.0154 \Rightarrow r = \sqrt{\frac{0.0154 \times 7}{22}} = 0.07 m$
∴ Area of the metal sheet needed to make the cylinder

= Total surface area of the cylinder

$= 2 π r h (r + h) = 2 \times \frac{22}{7} \times 0.07 \times 1 \times (0.07 + 1) = 2 \times \frac{22}{7} \times 0.07 \times 1 \times 1.07 = 0.4708 m^{2}$
Thus, the area of the metal sheet needed to make the cylinder is 0.4708 m².

Page No 573:

Answer:

Inner radius of the wooden pipe, r = $\frac{24}{2}$ = 12 cm

Outer radius of the wooden pipe, R = $\frac{28}{2}$ = 14 cm

Length of the wooden pipe, h = 35 cm

∴ Volume of wood in the pipe = $π (R^{2} - r^{2}) h = \frac{22}{7} \times (14^{2} - 12^{2}) \times 35 = 5720 {cm}^{3}$

It is given that 1 cm³of wood has a mass of 0.6 g.

∴ Mass of the pipe = Volume of wood in the pipe × 0.6 = 5720 × 0.6 = 3432 g = $\frac{3432}{1000}$ = 3.432 kg

Thus, the mass of the pipe is 3.432 kg.

Page No 573:

Answer:

Length of the cylindrical pipe, h = 28 m

Radius of the cylindrical pipe, r = $\frac{5}{2}$ = 2.5 cm = 0.025 m (1 m = 100 cm)

∴ Total radiating surface in the system

= Curved surface area of the cylindrical pipe

$= 2 π r h = 2 \times \frac{22}{7} \times 0.025 \times 28 = 4.4 m^{2}$
Thus, the total radiating surface in the system is 4.4 m².

Page No 573:

Answer:

Here, r = 10.5 cm; h = 60 cm

$Now, volume of the cylinder = π r^{2} h = \frac{22}{7} \times (10.5)^{2} \times 60 {cm}^{3} = 22 \times 10.5 \times 1.5 \times 60 {cm}^{3} = 20790 {cm}^{3}$

∴ Weight of cylinder = volume of cylinder $\times$ weight of cylinder per gram
$= 20790 \times 5 g = 103950 g = 103.95 kg$

Page No 573:

Answer:

Curved surface area = 1210 cm²
Suppose that the height of cylinder is h cm.
We have r = 10 cm
$Now, 1210 = 2 π r h \Rightarrow 1210 = 2 \times \frac{22}{7} \times 10 \times h \Rightarrow h = \frac{1210 \times 7}{2 \times 22 \times 10} = \frac{11 \times 7}{2 \times 2} = 19.25 cm$

$∴ Volume of the cylinder = π r^{2} h = \frac{22}{7} \times 10^{2} \times 19.25 {cm}^{3} = 2200 \times 2.75 {cm}^{3} = 6050 {cm}^{3}$

Page No 573:

Answer:

Let r be the radius and h be the height of the cylinder.
Circumference of its base(circle) = 110 cm.
$\Rightarrow 2 π r = 110 \Rightarrow r = \frac{110}{2 π} \Rightarrow r = \frac{110}{2 \times \frac{22}{7}} \Rightarrow r = \frac{110 \times 7}{2 \times 22} \Rightarrow r = \frac{35}{2} cm$
Curved surface area of a cylinder = 4400 cm².
$\Rightarrow 2 π r h = 4400 \Rightarrow h = \frac{4400}{2 π r} \Rightarrow h = \frac{4400}{2 \times \frac{22}{7} \times \frac{35}{2}} \Rightarrow h = \frac{4400 \times 7 \times 2}{2 \times 22 \times 35} \Rightarrow h = 40 cm$
Also, Volume of the cylinder = $π r^{2} h = \frac{22}{7} \times {(\frac{35}{2})}^{2} \times 40 = \frac{22 \times 35 \times 35 \times 40}{7 \times 2 \times 2} = 38500 {cm}^{3} .$

Page No 573:

Answer:

Suppose that the radius of the base and the height of the cylinder are 2x cm and 3x cm, respectively.
$Then 1617 = {πr}^{2} h = \frac{22}{7} \times (2 x)^{2} \times 3 x = \frac{22}{7} \times 12 x^{3} \Rightarrow x^{3} = \frac{1617 \times 7}{22 \times 12} = 42.875 \Rightarrow x = \sqrt[3]{42.875} = 3.5 cm$

Hence, radius = 7 cm; height of the cylinder = 10.5 cm

$∴ Total surface area of the cylinder = 2 π r h + 2 π r^{2} = 2 \times \frac{22}{7} (7 \times 10.5 + 7 \times 7) {cm}^{2} = \frac{44}{7} \times (73.5 + 49) {cm}^{2} = 770 {cm}^{2}$

Page No 573:

Answer:

Total surface area = 462 cm²

Given: Curved surface area = $\frac{1}{3}$ $\times$ total surface area = $\frac{1}{3} \times 462 = 154 {cm}^{2}$

Now, total surface area − curved surface area = $2 πrh + 2 {πr}^{2} - 2 πrh$
$\Rightarrow \Rightarrow 462 - 154 = 2 π r^{2} \Rightarrow 308 = 2 \times \frac{22}{7} \times r^{2} \Rightarrow r^{2} = \frac{308 \times 7}{44} = 49 \Rightarrow r = 7 cm$

Now, curved surface area = 154 cm²

$\Rightarrow 2 π r h = 154 \Rightarrow 2 \times \frac{22}{7} \times 7 \times h = 154 \Rightarrow h = \frac{154}{44} = 3.5 cm$

$∴ Volume of the cylinder = π r^{2} h = \frac{22}{7} \times 7^{2} \times 3.5 = 539 {cm}^{3}$

Page No 573:

Answer:

Curved surface area = $\frac{2}{3}$ $\times$ total surface area = $\frac{2}{3} \times 231 = 2 \times 77 = 154 {cm}^{2}$

Now, total surface area − curved surface area = $2 π r h + 2 π r^{2} - 2 π r h$

$Then 231 - 154 = 2 π r^{2} \Rightarrow 2 \times \frac{22}{7} \times r^{2} = 77 \Rightarrow r^{2} = \frac{77 \times 7}{44} = 12.25 \Rightarrow r = 3.5 cm Also, curved surface area = 154 {cm}^{2} \Rightarrow 2 π r h = 154 \Rightarrow 2 \times \frac{22}{7} \times 3.5 \times h = 154 \Rightarrow h = \frac{154 \times 7}{44 \times 3.5} = 7 cm$

$∴ Volume of the cylinder = π r^{2} h = \frac{22}{7} \times (3.5)^{2} \times 7 = 269.5 {cm}^{3}$

Page No 574:

Answer:

Suppose that the curved surface area and the total surface area of the right circular cylinder are x cm² and 2x cm².
Then we have:
2x = 616
x = 308 sq cm
Hence, the curved surface area of the cylinder is 308 sq cm.
Let r cm and h cm be the radius and height of the cylinder, respectively.
$Then 2 π r h + 2 π r^{2} = 616 {cm}^{2} and 2 π r h = 308 {cm}^{2} ∴ 2 π r h + 2 π r^{2} - 2 π r h = 616 - 308 \Rightarrow 2 π r^{2} = 308 \Rightarrow 2 \times \frac{22}{7} \times r^{2} = 308 \Rightarrow r^{2} = \frac{308 \times 7}{44} = 49 \Rightarrow r = 7 cm Now, 2 π r h = 308 {cm}^{2} \Rightarrow 2 \times \frac{22}{7} \times 7 \times h = 308 \Rightarrow h = \frac{308}{44} = 7 cm$

$∴ Volume of the cylinder = π r^{2} h cubic cm = \frac{22}{7} \times 7^{2} \times 7 = 1078 {cm}^{3}$

Page No 574:

Answer:

Given: Diameter of the cylindrical bucket = 28 cm
i.e., radius = 14 cm
Height of the cylindrical bucket, h₁ = 72 cm
Length of the rectangular tank, l = 66 cm
Breadth of the rectangular tank, b = 28 cm
Let the height of the rectangular tank be h cm.
The water from the cylindrical bucket is emptied into the rectangular tank.
i.e., volume of the bucket = volume of the tank
$\Rightarrow π r^{2} h_{1} = l \times b \times h \Rightarrow \frac{22}{7} \times 14^{2} \times 72 = 66 \times 28 \times h \Rightarrow h = \frac{22 \times 14 \times 2 \times 72}{66 \times 28} = 24 cm$

∴ Height of the rectangular tank = 24 cm

Page No 574:

Answer:

Height of the barrel = h = 7 cm
Radius of the barrel = r = 2.5 mm = 0.25 cm
Volume of the barrel = $π r^{2} h$
$= \frac{22}{7} \times 0.25 \times 0.25 \times 7 = 22 \times 0.25 \times 0.25 = 1.375 {cm}^{3}$

i.e., 1.375 cm³ of ink is used for writing 330 words
Now, number of words that could be written with one-fifth of a litre, i.e., $\frac{1}{5} \times 1000$ cm³ = $330 \times \frac{1}{1.375} \times \frac{1}{5} \times 1000 = 48000$
∴ 48000 words would use up a bottle of ink containing one-fifith of a litre.

Page No 574:

Answer:

Let r cm be the radius of the wire and h cm be the height of the wire.
Volume of the gold = 1 cm³
1 cm³of gold is drawn into a wire of diameter 0.1 mm.
Here, r = $0.1 mm = \frac{0.1}{20} cm = \frac{1}{200} cm$
$∴ π r^{2} h = 1 \Rightarrow \frac{22}{7} \times \frac{1}{200} \times \frac{1}{200} \times h = 1 \Rightarrow h = \frac{40000 \times 7}{22} = 12727.27 cm$

Hence, length of the wire = 127.27 m

Page No 574:

Answer:

Internal radius of the pipe = 1.5 cm
External radius of the pipe = (1.5 + 1) cm = 2.5 cm
Height of the pipe = 1 m = 100 cm
Volume of the cast iron = total volume of the pipe − internal volume of the pipe
$= π \times (2.5)^{2} \times 100 - π \times {(1.5)}^{2} \times 100 = \frac{22}{7} \times 100 \times [6.25 - 2.25] = \frac{2200}{7} \times 4 = \frac{8800}{7} {cm}^{3}$
1 cm³ of cast iron weighs 21 g.

∴ Weight of $\frac{8800}{7} {cm}^{3}$ cast iron = $\frac{8800}{7} \times \frac{21}{1000} kg = 88 \times 0.3 = 26.4 kg$

Page No 574:

Answer:

Inner radius of the cylindrical tube, r = 5.2 cm
Height of the cylindrical tube, h = 25 cm
Outer radius of the cylindrical tube, R = (5.2 + 0.8) cm = 6 cm

$∴ Volume of the metal = external volume - internal volume = π R^{2} h - π r^{2} h (where R and r are the outer and inner radii, respectively) = \frac{22}{7} \times 25 \times (6^{2} - 5 . 2^{2}) = \frac{22}{7} \times 25 \times (36 - 27.04) = \frac{22}{7} \times 25 \times 8.96 = 704 {cm}^{3}$

Page No 574:

Answer:

Height of the cylindrical tank, h = 1 m

Radius of the cylindrical tank, r = $\frac{140}{2}$ = 70 cm = 0.7 m (1 m = 100 cm)

∴ Area of the metal sheet required to make the cylindrical tank

= Total surface area of the cylindrical tank

$= 2 π r h (r + h) = 2 \times \frac{22}{7} \times 0.7 \times 1 \times (0.7 + 1) = 2 \times \frac{22}{7} \times 0.7 \times 1 \times 1.7 = 7.48 m^{2}$
Thus, the area of metal sheet required to make the cylindrial tank is 7.48 m².

Page No 574:

Answer:

Radius of the vassel, R = 15 cm

Height of orange juice in the vassel, H = 32 cm

∴ Volume of orange juice in the vassel = $π R^{2} H = π \times {(15)}^{2} \times 32 {cm}^{3}$

Radius of the glass, r = 3 cm

Height of orange juice in the glass, h = 8 cm

∴ Volume of orange juice in each glass = $π r^{2} h = π \times {(3)}^{2} \times 8 {cm}^{3}$

Number of glasses of orange juice sold by the juiceseller

$= \frac{Volume of orange juice in the vassel}{Volume of orange juice in each glass} = \frac{π \times {(15)}^{2} \times 32}{π \times {(3)}^{2} \times 8} = 100$
Rate of each glass of orange juice = ₹ 15

∴ Total money received by the juiceseller

= Number of glasses of orange juice sold by the juiceseller × Rate of each glass of orange juice

= 100 × 15

= ₹ 1,500

Thus, the total money received by the juiceseller by selling the juice completely is ₹ 1,500.

Page No 574:

Answer:

Inner radius of the well, r = $\frac{10}{2}$ = 5 m

Depth of the well, h = 8.4 m

Suppose the outer radius of the embankment is R m.

Width of the embankment = 7.5 m

∴ R − r = 7.5 m

⇒ R = 7.5 + 5 = 12.5 m

Let the height of the embankment be H m.

Now,

Volume of earth used to form the embankment = Volume of earth dugged out of the well

$∴ π (R^{2} - r^{2}) H = π r^{2} h \Rightarrow H = \frac{5^{2} \times 8.4}{{(12.5)}^{2} - {(7.5)}^{2}} \Rightarrow H = \frac{210}{100} \Rightarrow H = 2.1 m$
Thus, the height of the embankment is 2.1 m.

Page No 574:

Answer:

Speed of the water = 30 cm/s

Area of the cross section = 5 cm²

Volume of the water flowing out of the pipe in one second = Area of the cross section × 30 cm = 5 × 30 = 150 cm³

Now, 1 minute = 60 seconds

∴ Volume of the water flowing out of the pipe in 60 seconds

= Volume of the water flowing out of the pipe in one second × 60

= 150 × 60

= 9000 cm³

= $\frac{9000}{1000}$ (1 L = 1000 cm³)

= 9 L

Thus, 9 L of water flows out of the given pipe in 1 minute.

Page No 574:

Answer:

Radius of the water tank, R = $\frac{1.4}{2}$ = 0.7 m

Height of the water tank, H = 2.1 m

∴ Capacity of the water tank = $π R^{2} H = π {(0.7)}^{2} \times 2.1$ m³

Speed of the water flow = 2 m/s

Radius of the pipe, r = $\frac{3.5}{2}$ = 1.75 cm = 0.0175 m

Area of the cross section of the pipe = $π r^{2} = π {(0.0175)}^{2}$ m²

Volume of the water flowing out of the pipe in one second = Area of the cross section of the pipe × 2 m = $π {(0.0175)}^{2} \times 2$ m³

Let the time taken to fill the tank be t seconds.

∴ Volume of the water flowing out of the pipe in t seconds

= Volume of the water flowing out of the pipe in one second × t

= $π {(0.0175)}^{2} \times 2 \times t$ m³

Now,

Volume of the water flowing out of the pipe in t seconds = Capacity of the water tank

$∴ π {(0.0175)}^{2} \times 2 \times t = π {(0.7)}^{2} \times 2.1 \Rightarrow t = \frac{{(0.7)}^{2} \times 2.1}{{(0.0175)}^{2} \times 2} \Rightarrow t = 1680 s$
$\Rightarrow t = \frac{1680}{60}$

⇒ t = 28 minutes

Thus, the tank will be filled in 28 minutes.

Page No 574:

Answer:

Volume of the rectangular solid of iron = 32 cm × 22 cm × 14 cm

Radius of the container, r = $\frac{56}{2}$ = 28 cm

Let the rise in the level of water in the container when rectangular solid of iron is submerged in it be h cm.

∴ Volume of the water displaced in the container = $π r^{2} h = π \times {(28)}^{2} \times h$

When the rectangular solid of iron is submerged in the container, then the volume of water displaced in the container is equal to the volume of the rectangular solid of iron.

$∴ \frac{22}{7} \times {(28)}^{2} \times h = 32 \times 22 \times 14 \Rightarrow h = \frac{32 \times 22 \times 14 \times 7}{22 \times {(28)}^{2}} \Rightarrow h = 4 cm$
Thus, the rise in the level of water in the container is 4 cm.

Page No 575:

Answer:

Height of the tube-well, h = 280 m

Radius of the tube-well, r = $\frac{3}{2}$ m

∴ Volume of the tube-well = $π r^{2} h = \frac{22}{7} \times {(\frac{3}{2})}^{2} \times 280$ = 1980 m³

Rate of sinking the tube-well = ₹ 15 per cubic metre

∴ Cost of sinking the tube-well = Volume of the tube-well × Rate of sinking the tube-well = 1980 × 15 = ₹ 29,700

Thus, the cost of sinking the tube-well is ₹ 29,700.

Inner curved surface of the tube-well = $2 π r h = 2 \times \frac{22}{7} \times \frac{3}{2} \times 280$ = 2640 m²

Rate of cementing = ₹ 10 per square metre

∴ Cost of cementing the inner curved surface of the tube-well

= Inner curved surface of the tube-well × Rate of cementing

= 2640 × 10

= ₹ 26,400

Thus, the cost of cementing the inner curved surface of the tube-well is ₹ 26,400.

Page No 575:

Answer:

Mass of copper wire = 13.2 kg = 13.2 × 1000 = 13200 g (1 kg = 1000 g)

Volume of 8.4 g of copper wire = 1 cm³

∴ Volume of 13200 g (or 13.2 kg) of copper wire = $\frac{13200}{8.4}$ cm³

Let the length of the copper wire be l cm.

Radius of the copper wire, r = $\frac{4}{2}$ = 2 mm = 0.2 cm (1 cm = 10 mm)

∴ Volume of the copper wire = $π r^{2} l = \frac{22}{7} \times {(0.2)}^{2} \times l$ cm^{3

$\Rightarrow \frac{22}{7} \times {(0.2)}^{2} \times l = \frac{13200}{8.4} \Rightarrow l = \frac{13200 \times 7}{22 \times {(0.2)}^{2} \times 8.4} \Rightarrow l = 12500 cm$}

$\Rightarrow l = \frac{12500}{100} (1 m = 100 cm) \Rightarrow l = 125 m$
Thus, the length of the copper wire is 125 m.

Page No 575:

Answer:

Total cost of paining the inner curved surface of the cylinderical vassel = ₹ 3,300

Rate of painting = ₹ 30 per m²

(i) Inner curved surface area of the vassel

$= \frac{Total cost of painting the inner curved surface of the cylindrical vassel}{Rate of painting} = \frac{3300}{30} = 110 m^{2}$
Thus, the inner curved surface area of the vassel is 110 m².

(ii) Depth of the vassel, h = 10 m

Let the inner radius of the base be r m.

∴ Inner curved surface area of the vasssel = $2 π r h = 2 \times \frac{22}{7} \times r \times 10$

$\Rightarrow 2 \times \frac{22}{7} \times r \times 10 = 110 \Rightarrow r = \frac{110 \times 7}{2 \times 22 \times 10} \Rightarrow r = 1.75 m$
Thus, the inner radius of the base is 1.75 m.

(iii) Capacity of the vassel = $π r^{2} h = \frac{22}{7} \times {(1.75)}^{2} \times 10 =$ 96.25 m³

Thus, the capacity of the vassel is 96.25 m³.

Page No 575:

Answer:

Let the inner and outer radii of the tube be r cm and R cm, respectively.

Length of the cylindrical tube, h = 14 cm

Outer curved surface of the cylinder − Inner curved surface of the cylinder = 88 cm² (Given)

$∴ 2 π R h - 2 π r h = 88 \Rightarrow 2 \times \frac{22}{7} \times 14 \times (R - r) = 88 \Rightarrow R - r = \frac{88 \times 7}{2 \times 22 \times 14} \Rightarrow R - r = 1 cm . . . . . (1)$
Volume of the tube = 176 cm³ (Given)

$∴ π (R^{2} - r^{2}) h = 176 \Rightarrow \frac{22}{7} \times (R - r) \times (R + r) \times 14 = 176 \Rightarrow R + r = \frac{176 \times 7}{22 \times 1 \times 14} [Using (1)] \Rightarrow R + r = 4 cm . . . . . (2)$
Adding (1) and (2), we get

2R = 5

⇒ R = 2.5 cm

Putting R = 2.5 cm in (2), we get

2.5 + r = 4

⇒ r = 4 − 2.5 = 1.5 cm

Thus, the inner and outer radii of the tube are 1.5 cm and 2.5 cm, respectively.

Page No 575:

Answer:

The dimensions of the rectangular sheet of paper are 30 cm × 18 cm.

Let V₁ and V₂be the volumes of the cylinders formed by rolling the rectangular sheet of paper along its length (i.e. 30 cm) and breadth (i.e. 18 cm), respectively.

Suppose r₁ and h₁ be the radius and height of the cylinder formed by rolling the rectangular sheet of paper along its length, respectively.

$∴ 2 π r_{1} = 30 \Rightarrow r_{1} = \frac{30}{2 π} cm$

h₁ = 18 cm

$∴ V_{1} = π r_{1}^{2} h_{1} = π {(\frac{30}{2 π})}^{2} \times 18 {cm}^{3}$

Also, suppose r₂ and h₂ be the radius and height of the cylinder formed by rolling the rectangular sheet of paper along its breadth, respectively.

$∴ 2 π r_{2} = 18 \Rightarrow r_{2} = \frac{18}{2 π} cm$

h₂ = 30 cm

$∴ V_{2} = π r_{2}^{2} h_{2} = π {(\frac{18}{2 π})}^{2} \times 30 {cm}^{3}$

Now,

$\frac{V_{1}}{V_{2}} = \frac{π {(\frac{30}{2 π})}^{2} \times 18}{π {(\frac{18}{2 π})}^{2} \times 30} = \frac{5}{3}$

⇒ V₁ : V₂ = 5 : 3

Thus, the ratio of the volumes of the two cylinders thus formed is 5 : 3.

Page No 588:

Answer:

Radius of the base, r = 5.25 cm

Slant height, l = 10 cm

∴ Curved surface area of the cone = $π r l = \frac{22}{7} \times 5.25 \times 10$ = 165 cm²

Thus, the curved surface area of the cone is 165 cm².

Page No 588:

Answer:

Slant height, l = 21 m

Radius of the base, r = $\frac{24}{2}$ = 12 m

∴ Total surface area of the cone = $π r (r + l) = \frac{22}{7} \times 12 \times (12 + 21) = \frac{22}{7} \times 12 \times 33 = \frac{8712}{7} {cm}^{2}$

Thus, the total surface area of the cone is $\frac{8712}{7} {cm}^{2}$ .

Page No 588:

Answer:

Radius of the cone, r = 7 cm

Height of the cone, h = 24 cm

∴ Slant height of the cone, $l = \sqrt{r^{2} + h^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625}$ = 25 cm

Area of the sheet required to make one cap = Curved surface area of the cone = $π r l = \frac{22}{7} \times 7 \times 25$ = 550 cm²

∴ Area of the sheet required to make 10 such caps = 550 × 10 = 5500 cm²

Thus, the area of the sheet required to make 10 such jocker caps is 5500 cm².

Page No 588:

Answer:

Slant height, l = 14 cm

Let the radius of the base be r cm.

Curved surface area of the cone = 308 cm² (Given)

$∴ π r l = 308 \Rightarrow \frac{22}{7} \times r \times 14 = 308 \Rightarrow r = \frac{308 \times 7}{22 \times 14} \Rightarrow r = 7 cm$
∴ Total surface area of the cone = $π r (r + l) = \frac{22}{7} \times 7 \times (7 + 14) = \frac{22}{7} \times 7 \times 21$ = 462 cm²

Thus, the radius of the base is 7 cm and the total surface area of the cone is 462 cm².

Page No 588:

Answer:

Slant height of the conical tomb, l = 25 m

Radius of the conical tomb, r = $\frac{14}{2}$ = 7 m

∴ Curved surface area of the conical tomb = $π r l = \frac{22}{7} \times 7 \times 25$ = 550 m²

Rate of whitewashing = ₹ 12 per m²

∴ Cost of whitewashing the conical tomb

= Curved surface area of the conical tomb × Rate of whitewashing

= 550 × 12

= ₹ 6,600

Thus, the cost of whitewashing the conical tomb is ₹ 6,600.

Page No 588:

Answer:

Radius of the conical tent, r = 24 m

Height of the conical tent, h = 10 m

∴ Slant height of the conical tent, $l = \sqrt{r^{2} + h^{2}} = \sqrt{24^{2} + 10^{2}} = \sqrt{576 + 100} = \sqrt{676}$ = 26 m

Curved surface area of the conical tent = $π r l = \frac{22}{7} \times 24 \times 26 m^{2}$

The cost of 1 m² canvas is ₹ 70.

∴ Cost of canvas required to make the tent

= Curved surface area of the conical tent × ₹ 70

$= \frac{22}{7} \times 24 \times 26 \times 70$

= ₹ 1,37,280

Thus, the cost of canvas required to make the tent is ₹ 1,37,280.

Page No 588:

Answer:

Radius of each cone, r = $\frac{40}{2}$ = 20 cm = 0.2 m (1 m = 100 cm)

Height of each cone, h = 1 m

∴ Slant height of each cone, $l = \sqrt{r^{2} + h^{2}} = \sqrt{{(0.2)}^{2} + 1^{2}} = \sqrt{0.04 + 1} = \sqrt{1.04}$ = 1.02 m

Curved surface area of each cone = $π r l = 3.14 \times 0.2 \times 1.02$ = 0.64056 m²

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

Cost of painting = ₹ 25 per m²

∴ Total cost of painting all the cones

= Curved surface area of 50 cones × ₹ 25

= 32.028 × 25

= ₹ 800.70

Thus, the cost of painting all the cones is ₹ 800.70.

Page No 588:

Answer:

Radius of the cone, r = 35 cm

Height of the cone, h = 12 cm

∴ Slant height of the cone, $l = \sqrt{r^{2} + h^{2}} = \sqrt{35^{2} + 12^{2}} = \sqrt{1225 + 144} = \sqrt{1369}$ = 37 cm

(i) Volume of the cone = $\frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times {(35)}^{2} \times 12$ = 15400 cm³

(ii) Curved surface area of the cone = $π r l = \frac{22}{7} \times 35 \times 37$ = 4070 cm²

(iii) Total surface area of the cone = $π r (r + l) = \frac{22}{7} \times 35 \times (35 + 37) = \frac{22}{7} \times 35 \times 72$ = 7920 cm²

Page No 588:

Answer:

Height of the cone, h = 6 cm
Slant height of the cone, l = 10 cm
Radius, r = $\sqrt{l^{2} - h^{2}} = \sqrt{100 - 36} = \sqrt{64} = 8 cm$

Volume of the cone = $π r^{2} h$
                      $= \frac{1}{3} \times 3.14 \times 8^{2} \times 6 = 401.92 {cm}^{3}$

Curved surface area of the cone = $πr l$
                                      $= 3.14 \times 8 \times 10 = 251.2 {cm}^{2}$

∴ Total surface area = $π r l + π r^{2}$
                               $= 251.2 + 3.14 \times 8^{2} = 452.16 {cm}^{2}$

Page No 588:

Answer:

Radius of the conical pit, r = $\frac{3.5}{2}$ m

Depth of the conical pit, h = 12 m

∴ Capacity of the conical pit = $\frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times {(\frac{3.5}{2})}^{2} \times 12$ = 38.5 m³ = 38.5 kL (1 m³ = 1 kilolitre)

Thus, the capacity of the conical pit is 38.5 kL.

Page No 588:

Answer:

Radius of the heap, r = $\frac{9}{2}$ m = 4.5 m

Height of the heap, h = 3.5 m

∴ Volume of the heap of wheat = $\frac{1}{3} π r^{2} h = \frac{1}{3} \times 3.14 \times {(4.5)}^{2} \times 3.5$ = 74.1825 m³

Now,

Slant height of the heap, $l = \sqrt{r^{2} + h^{2}} = \sqrt{{(4.5)}^{2} + {(3.5)}^{2}} = \sqrt{20.25 + 12.25} = \sqrt{32.5}$ ≈ 5.7 m

∴ Area of the canvas cloth required to just cover the heap of wheat

= Curved surface area of the heap of wheat

$= π r l \approx 3.14 \times 4.5 \times 5.7 \approx 80.541 m^{2}$
Thus, the area of canvas cloth required to just cover the heap is approximately 80.541 m².

Page No 588:

Answer:

Area of the canvas = 551 m²

Area of the canvas used in stitching margins and wastage incurred while cutting = 1 m²

∴ Area of the canvas used in making the tent = 551 − 1 = 550 m²

Radius of the tent, r = 7 m

Let the slant height and height of the tent be l m and h m, respectively.

Area of the canvas used in making the tent = 550 m²

∴ π r l = 550 \Rightarrow \frac{22}{7} \times 7 \times l = 550 \Rightarrow l = \frac{550}{22} = 25 m

Now,

Height of the tent,

h = \sqrt{l^{2} - r^{2}} = \sqrt{25^{2} - 7^{2}} = \sqrt{625 - 49} = \sqrt{576}

= 24 m

∴ Volume of the tent =

\frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times {(7)}^{2} \times 24

= 1232 m³

Thus, the volume of the tent that can be made with the given canvas is 1232 m³.

Page No 589:

Answer:

Radius of the conical tent, r = 7 m
Height of the conical tent, h = 24 m
$Now, l = \sqrt{r^{2} + h^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25 m$

$Curved surface area of the cone = π r l = \frac{22}{7} \times 7 \times 25 = 550 m^{2} Here, area of the cloth = curved surface area of the cone = 550 m^{2}$

Width of the cloth = 2.5 m
∴ Length of the cloth = $\frac{area of the cloth}{width of the cloth} = \frac{550}{2.5} = 220 m$

Page No 589:

Answer:

Let the heights of the first and second cones be h and 3h, respectively .
Also, let the radius of the first and second cones be 3r and r, respectively.

∴ Ratio of volumes of the cones = $\frac{volume of the first cone}{volume of the second cone}$

$= \frac{\frac{1}{3} π \times {(3 r)}^{2} \times h}{\frac{1}{3} π \times r^{2} \times 3 h} = \frac{9 r^{2} h}{3 r^{2} h} = 3 : 1$

Page No 589:

Answer:

Suppose that the respective radii and height of the cone and the cylinder are r and h.
Then ratio of curved surface areas = 8 : 5
Let the curved surfaces areas be 8x and 5x.

$i . e ., 2 π r h = 8 x and π r l = 5 x \Rightarrow π r \sqrt{h^{2} + r^{2}} = 5 x Hence 4 π^{2} r^{2} h^{2} = 64 x^{2} and π^{2} r^{2} (h^{2} + r^{2}) = 25 x^{2} ∴ Ratio of curved surface areas = \frac{4 π^{2} r^{2} h^{2}}{π^{2} r^{2} (h^{2} + r^{2})} = \frac{64}{25} \Rightarrow \frac{4 π^{2} r^{2} h^{2}}{π^{2} r^{2} (h^{2} + r^{2})} = \frac{64}{25} \Rightarrow \frac{4 h^{2}}{(h^{2} + r^{2})} = \frac{64}{25} \Rightarrow 25 h^{2} = 16 h^{2} + 16 r^{2} \Rightarrow 9 h^{2} = 16 r^{2} \Rightarrow \frac{r^{2}}{h^{2}} = \frac{9}{16} \Rightarrow \frac{r}{h} = \frac{3}{4}$

∴ The ratio of the radius and height of the cone is 3 : 4.

Page No 589:

Answer:

Let the cone which is being melted be denoted by cone 1 and let the cone into which cone 1 is being melted be denoted by cone 2.

Height of cone 1 = 3.6 cm
Radius of the base of cone 1 = 1.6 cm
Radius of the base of cone 2 = 1.2 cm
Let h cm be the height of cone 2.
The volumes of both the cones should be equal.

$i . e ., \frac{1}{3} π \times 1 . 6^{2} \times 3.6 = \frac{1}{3} π \times 1 . 2^{2} \times h \Rightarrow h = \frac{1.6 \times 1.6 \times 3.6}{1.2 \times 1.2} = 6.4 cm$

∴ Height of cone 2 = 6.4 cm

Page No 589:

Answer:

Radius of the tent, r = 52.5 m
Height of the cylindrical portion of the tent, H = 3 m
Slant height of the conical portion of the tent, l = 53 m
The tent is a combination of a cylindrical and a conical portion.
i.e., area of the canvas = curved surface area of the cone + curved surface area of the cylinder
                         = $π r l + 2 π r H$
                          $= \frac{22}{7} \times \frac{105}{2} (53 + 2 \times 3) = \frac{22}{7} \times \frac{105}{2} \times (53 + 6) = 9735 m^{2}$

But area of the canvas = length $\times$ breadth

∴ Length of the canvas = $\frac{area}{breadth} = \frac{9735}{5} = 1947 m$

Page No 589:

Answer:

Height of the cylindrical portion of the iron pillar, h = 2.8 m = 280 cm
Radius of the cylindrical portion of the iron pillar, r = 20 cm
Height of the cone which is surmounted on the cylindrical portion, H = 42 cm
Now, volume of the pillar = volume of the cylindrical portion + volume of the conical portion
                         = $π r^{2} h + \frac{1}{3} π r^{2} H$
                                $= \frac{22}{7} \times 10^{2} (280 + \frac{1}{3} \times 42) = \frac{22}{7} \times (280 + 14) = \frac{22}{7} \times 100 \times 294 = 92400 {cm}^{3}$

∴ Weight of the pillar = volume of the pillar $\times$ weight per cubic cm
                                 $= 92400 \times \frac{7.5}{1000} = 693 kg$

Page No 589:

Answer:

Height of the cylinder = 10 cm
Radius of the cylinder = 6 cm
The respective heights and radii of the cone and the cylinder are the same.
∴ Volume of the remaining solid = volume of the cylinder − volume of the cone
$= π r^{2} h - \frac{1}{3} π r^{2} h = \frac{2}{3} π r^{2} h = \frac{2}{3} \times 3.14 \times 6^{2} \times 10 = 753.6 {cm}^{3}$

Page No 589:

Answer:

Radius of the cylindrical pipe = 2.5 mm = 0.25 cm
Water flowing per minute = 10 m = 1000 cm
Volume of water flowing per minute through the cylindrical pipe = $π (0.25)^{2} 1000 {cm}^{3}$ = 196.4 cm³
Radius of the the conical vessel = 40 cm
Depth of the vessel = 24 cm
Volume of the vessel = $\frac{1}{3} π (20)^{2} 24 = 10057.1 {cm}^{3}$
Let the time taken to fill the conical vessel be x min.
Volume of water flowing per minute through the cylindrical pipe $\times$ x = volume of the conical vessel
$\Rightarrow x = \frac{10057.1}{196.4} = 51 \min 12 \sec$
∴ The cylindrical pipe would take 51 min 12 sec to fill the conical vessel.

Page No 589:

Answer:

Radius of the conical tent, r = 5 m

Area of the base of the conical tent = $π r^{2} = \frac{22}{7} \times 5^{2} = \frac{550}{7} m^{2}$

Average area occupied by a student on the ground = $\frac{5}{7}$ m²

∴ Number of students who can sit in the tent

$= \frac{Area of the base of the conical tent}{Average area occupied by a student on the ground} = \frac{\frac{550}{7}}{\frac{5}{7}} = 110$
Thus, the number of students who can sit in the tent is 110.

Let the slant height of the tent be l m.

Curved surface area of the tent = 165 m²

$∴ π r l = 165 \Rightarrow \frac{22}{7} \times 5 \times l = 165 \Rightarrow l = \frac{165 \times 7}{22 \times 5} \Rightarrow l = 10.5 m$
Let the height of the tent be h m.

$h = \sqrt{l^{2} - r^{2}} = \sqrt{{(10.5)}^{2} - 5^{2}} = \sqrt{110.25 - 25} = \sqrt{85.25}$ ≈ 9.23 m

∴Volume of the tent = $\frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times {(5)}^{2} \times 9.23$ ≈ 241.74 m³

Thus, the volume of the conical tent is approximately 241.74 m³.

Page No 600:

Answer:

(i) Radius of the sphere = 3.5 cm
Now, volume = $\frac{4}{3} π r^{3}$
                     $= \frac{4}{3} \times \frac{22}{7} \times 3.5 \times 3.5 \times 3.5 = 179.67 {cm}^{3}$

∴ Surface area = $4 π r^{2}$
                       $= 4 \times \frac{22}{7} \times 3.5 \times 3.5 = 154 {cm}^{2}$

(ii) Radius of the sphere=4.2 cm
Now, volume = $\frac{4}{3} π r^{3}$
                      $= \frac{4}{3} \times \frac{22}{7} \times 4.2 \times 4.2 \times 4.2 = 310.46 {cm}^{3}$

∴ Surface area = $4 π r^{2}$
                         $= 4 \times \frac{22}{7} \times 4.2 \times 4.2 = 221.76 {cm}^{2}$

(iii) Radius of sphere=5 m
Now, volume = $\frac{4}{3} π r^{3}$
                    = $\frac{4}{3} \times \frac{22}{7} \times 5^{3} = 523.81 {cm}^{3}$

∴ Surface area = $4 π r^{2}$
                       $= 4 \times \frac{22}{7} \times 5^{2} = 314.29 {cm}^{2}$

Page No 600:

Answer:

Volume of the sphere = 38808 cm³
Suppose that r cm is the radius of the given sphere.

$∴ \frac{4}{3} π r^{3} = 38808 \Rightarrow r^{3} = \frac{38808 \times 3 \times 7}{4 \times 22} = 9261 \Rightarrow r = \sqrt[3]{9261} = 21 cm$

∴ Surface area of the sphere = $4 π r^{2}$
$= 4 \times \frac{22}{7} \times 21 \times 21 = 5544 {cm}^{2}$

Page No 601:

Answer:

Volume of the sphere = 606.375 m³

$Then \frac{4}{3} π r^{3} = 606.375 \Rightarrow r^{3} = \frac{606.375 \times 3 \times 7}{4 \times 22} = 144.703 \Rightarrow r = 5.25 m$

∴ Surface area = $4 {πr}^{2}$
$= 4 \times \frac{22}{7} \times 5.25 \times 5.25 = 346.5 m^{2}$

Page No 601:

Answer:

Let the radius of the sphere be r cm.

Surface area of the sphere = 154 cm²

$∴ 4 π r^{2} = 154 \Rightarrow 4 \times \frac{22}{7} \times r^{2} = 154 \Rightarrow r = \sqrt{\frac{154 \times 7}{4 \times 22}} = \sqrt{12.25} \Rightarrow r = 3.5 cm$
∴ Volume of the sphere = $\frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times {(3.5)}^{3} \approx 179.67 m^{3}$

Thus, the volume of the sphere is approximately 179.67 m³.

Page No 601:

Answer:

Surface area of the sphere = (576π) cm²
^{Suppose that r cm is the radius of the sphere.

$Then 4 π r^{2} = 576 π \Rightarrow r^{2} = \frac{576}{4} = 144 \Rightarrow r = 12 cm$

$∴ Volume of the sphere = \frac{4}{3} \times π \times 12 \times 12 \times 12 {cm}^{3} = 2304 π {cm}^{3}$}

Page No 601:

Answer:

Here, l = 12 cm, b = 11 cm and h = 9 cm

$Volume of the cuboid = l \times b \times h = 12 \times 11 \times 9 = 1188 {cm}^{3}$

Radius of one lead shot = 3 mm= $\frac{0.3}{2} cm$

$Volume of one lead shot = \frac{4}{3} \times \frac{22}{7} \times {(\frac{0.3}{2})}^{3} = \frac{11 \times 9}{7000} = 0.014 {cm}^{3}$

$∴ Number of lead shots = \frac{volume of the cuboid}{volume of one lead shot} = \frac{1188}{0.014} = 84857.14 \approx 84857$

Page No 601:

Answer:

Radius of the sphere = 8 cm

Volume of the sphere = $\frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times 8^{3} = 2145.52 {cm}^{3}$
Radius of one lead ball = 1 cm
Volume of one lead ball = $\frac{4}{3} \times \frac{22}{7} \times 1^{3} = 4.19 {cm}^{3}$

∴ Number of lead balls = $\frac{volume of the sphere}{volume of one lead ball} = \frac{2145.52}{4.19} = 512.05 \approx 512$

Page No 601:

Answer:

Radius of the solid sphere = 3 cm

Volume of the solid sphere = $\frac{4}{3} π r^{3}$
= $\frac{4}{3} \times \frac{22}{7} \times 3^{3} {cm}^{3}$

Diameter of the spherical ball = 0.6 cm
Radius of the spherical ball = 0.3 cm

Volume of the spherical ball = $\frac{4}{3} \times \frac{22}{7} \times {(0.3)}^{3} {cm}^{3}$

Now, number of small spherical balls = $\frac{volume of the sphere}{volume of the spherical ball}$
$= \frac{\frac{4}{3} π \times 27}{\frac{4}{3} π \times (0 . 3^{3})} = 1000$

∴ The number of small balls thus obtained is 1000

Page No 601:

Answer:

Radius of the metallic sphere, r = 10.5 cm
Volume of the sphere = $\frac{4}{3} π r^{3} = \frac{4}{3} π (10.5)^{3} {cm}^{3}$
Radius of each smaller cone, r₂ = 3.05 cm
Height of each smaller cone = 3 cm
$Volume of each smaller cone = \frac{1}{3} π {r_{2}}^{2} h = \frac{1}{3} π (3.05)^{2} \times 3 {cm}^{3}$
Number of cones obtained = $\frac{volume of the sphere}{volume of each smaller cone}$
$= \frac{\frac{4}{3} π r^{3}}{\frac{1}{3} π {r_{2}}^{2} h} = \frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times 3} = 126.006 \approx 126$

∴ 126 cones are obtained.

Page No 601:

Answer:

Diameter of each sphere, d=12 cm
Radius of each sphere, r=6 cm
$Volume of each sphere = \frac{4}{3} π r^{3} = \frac{4}{3} π (6)^{3} {cm}^{3}$
Diameter of base of the cylinder, D=8 cm
Radius of base of cylinder, R=4 cm
Height of the cylinder, h = 90 cm
$Volume of the cylinder = π R^{2} h = π (4)^{2} \times 90 {cm}^{3}$

Number of spheres= $\frac{volume of the cylinder}{volume of the sphere}$
$= \frac{π R^{2} h}{\frac{4}{3} π r^{3}} = \frac{4^{2} \times 90 \times 3}{4 \times 6^{3}} = \frac{12 \times 90}{216} = 5$

∴ Five spheres can be made.

Page No 601:

Answer:

Let the length of the wire be h cm.
Radius of the wire, r = 1 mm = 0.1 cm
Radius of the sphere, R = 3 cm
Now, volume of the sphere = volume of the cylindrical wire
$\Rightarrow \frac{4}{3} π R^{3} = π r^{2} h \Rightarrow 4 \times 3^{2} = {(0.1)}^{2} \times h \Rightarrow h = \frac{4 \times 9}{0.1 \times 0.1} = 3600 cm = 36 m$

∴ Length of the wire = 36 m

Page No 601:

Answer:

Radius of the copper sphere, R = 9 cm
Length of the wire, h = 108 m = 10800 cm
Volume of the sphere = volume of the wire
Suppose that r cm is the radius of the wire.

$Then \frac{4}{3} π R^{3} = π r^{2} h \Rightarrow \frac{4}{3} \times 9^{3} = r^{2} \times 10800 \Rightarrow r^{2} = \frac{4 \times 729}{3 \times 10800} = \frac{4 \times 81}{3 \times 1200} = \frac{9}{100} \Rightarrow r = \frac{3}{10} = 0.3 cm$

∴ Diameter of the wire = 0.6 cm

Page No 601:

Answer:

Radius of the sphere, r = 7.8 cm
Height of the cone, h = 31.2 cm
Suppose that R cm is the radius of the cone.
Now, volume of the sphere = volume of the cone
$\Rightarrow \frac{4}{3} π r^{3} = \frac{1}{3} π R^{2} h \Rightarrow 4 \times {(7.8)}^{3} = R^{2} \times 31.2 \Rightarrow R^{2} = \frac{4 \times 7.8 \times 7.8 \times 7.8}{31.2} = 60.84 \Rightarrow R = 7.8 cm$

∴ The diameter of the cone is 15.6 cm.

Page No 601:

Answer:

Radius of the spherical cannonball, R = 14 cm
Radius of the base of the cone, r = 17.5 cm
Let h cm be the height of the cone.
Now, volume of the sphere = volume of the cone
$\Rightarrow \frac{4}{3} π R^{3} = \frac{1}{3} π r^{2} h \Rightarrow 4 \times 14^{3} = {(17.5)}^{2} \times h \Rightarrow h = \frac{4 \times 14 \times 14 \times 14}{17.5 \times 17.5} = 35.84 cm$

∴ The height of the cone is 35.84 cm.

Page No 601:

Answer:

Radius of the original spherical ball = 3 cm
Suppose that the radius of third ball is r cm.
Then volume of the original spherical ball = volume of the three spherical balls
$\Rightarrow \frac{4}{3} π \times 3^{3} = \frac{4}{3} π \times 1 . 5^{3} + \frac{4}{3} π \times 2^{3} + \frac{4}{3} π \times r^{3} \Rightarrow 27 = 3.375 + 8 + r^{3} \Rightarrow r^{3} = 27 - 11.375 = 15.625 \Rightarrow r = 2.5 cm$

∴ The radius of the third ball is 2.5 cm.

Page No 601:

Answer:

Suppose that the radii are r and 2r.
Now, ratio of the surface areas = $\frac{4 π r^{2}}{4 π {(2 r)}^{2}} = \frac{r^{2}}{4 r^{2}} = \frac{1}{4}$
= 1:4

∴ The ratio of their surface areas is 1 : 4.

Page No 601:

Answer:

Suppose that the radii of the spheres are r and R.
We have:
$\frac{4 π r^{2}}{4 {πR}^{2}} = \frac{1}{4} \Rightarrow \frac{r}{R} = \sqrt{\frac{1}{4}} = \frac{1}{2}$

Now, ratio of the volumes = $\frac{\frac{4}{3} π r^{3}}{\frac{4}{3} π R^{3}} = {(\frac{r}{R})}^{3} = {(\frac{1}{2})}^{3} = \frac{1}{8}$

∴ The ratio of the volumes of the spheres is 1 : 8.

Page No 601:

Answer:

Suppose that the radius of the ball is r cm.
Radius of the cylindrical tub =12 cm
Depth of the tub= 20 cm
Now, volume of the ball = volume of water raised in the cylinder
$\Rightarrow \frac{4}{3} π r^{3} = π \times 12^{2} \times 6.75 \Rightarrow r^{3} = \frac{144 \times 6.75 \times 3}{4} = 36 \times 6.5 \times 3 = 729 \Rightarrow r = 9 cm$

∴ The radius of the ball is 9 cm.

Page No 601:

Answer:

Let h cm be the increase in the level of water.
Radius of the cylindrical bucket = 15 cm
Height up to which water is being filled = 20 cm
Radius of the spherical ball = 9 cm
Now, volume of the sphere = increased in volume of the cylinder
$\Rightarrow \frac{4}{3} π \times 9^{3} = π \times 15^{2} \times h \Rightarrow h = \frac{4 \times 729}{3 \times 15 \times 15} = \frac{4 \times 27}{25} = 4.32 cm$

∴ The increase in the level of water is 4.32 cm.

Page No 602:

Answer:

Outer radius of the spherical shell = 6 cm
Inner radius of the spherical shell = 4 cm

$Volume of metal contained in the shell = \frac{4}{3} \times \frac{22}{7} (6^{3} - 4^{3}) = \frac{88}{21} \times (216 - 64) = \frac{88}{21} \times 152 = 636.95 {cm}^{3}$

$∴ Outer surface area = 4 \times \frac{22}{7} \times 6 \times 6 = 452.57 {cm}^{2}$

Page No 602:

Answer:

Internal radius of the hollow spherical shell, r= 8 cm
External radius of the hollow spherical shell, R= 9 cm

Volume of the shell = $\frac{4}{3} π (R^{3} - r^{3})$
$= \frac{4}{3} π (9^{3} - 8^{3}) = \frac{4}{3} \times \frac{22}{7} \times (729 - 512) = \frac{4 \times 22 \times 217}{21} = \frac{88 \times 31}{3} = \frac{2728}{3} {cm}^{3}$

Weight of the shell = volume of the shell $\times$ density per cubic cm
= $\frac{2728}{3} \times 4.5 \approx 4092 g = 4.092 kg$

∴ Weight of the shell = 4.092 kg

Page No 602:

Answer:

Radius of the hemisphere = 9 cm
Height of the right circular cone = 72 cm
Suppose that the radius of the base of the cone is r cm.
Volume of the hemisphere = volume of the cone

$\Rightarrow \frac{2}{3} π \times 9^{3} = \frac{1}{3} π \times r^{2} \times 72 \Rightarrow r^{2} = \frac{2 \times 9 \times 9 \times 9}{72} = \frac{81}{4} \Rightarrow r = \frac{9}{2} = 4.5 cm$

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

Page No 602:

Answer:

Internal radius of the hemispherical bowl = 9 cm
Radius of a cylindrical shaped bottle = 1.5 cm
Height of a bottle = 4 cm

Number of bottles required to empty the bowl = $\frac{volume of the hemispherical bowl}{volume of a cylindrical shaped bottle}$
$= \frac{\frac{2}{3} π \times 9^{3}}{π \times 1 . 5^{2} \times 4} = \frac{2 \times 9 \times 9 \times 9}{3 \times 1.5 \times 1.5 \times 4} = 54$

∴ 54 bottles are required to empty the bowl.

Page No 602:

Answer:

Internal radius of the hemispherical bowl = 4 cm
Thickness of a the bowl = 0.5 cm
Now, external radius of the bowl = (4 + 0.5 ) cm = 4.5 cm

Now, volume of steel used in making the bowl = volume of the shell
$= \frac{2}{3} π (4 . 5^{3} - 4^{3}) = \frac{2}{3} \times \frac{22}{7} \times (91.125 - 64) = \frac{2}{3} \times \frac{22}{7} \times 27.125 = 56.83 {cm}^{3}$

∴ 56.83 cm^₃ of steel is used in making the bowl .

Page No 602:

Answer:

Inner radius of the bowl, r = 5 cm

Let the outer radius of the bowl be R cm.

Thickness of the bowl = 0.25 cm (Given)

∴ R − r = 0.25 cm

⇒ R = 0.25 + r = 0.25 + 5 = 5.25 cm

∴ Outer curved surface area of the bowl = $2 π r^{2} = 2 \times \frac{22}{7} \times {(5.25)}^{2}$ = 173.25 cm²

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

Page No 602:

Answer:

Inner radius of the bowl, r = $\frac{10.5}{2}$ = 5.25 cm

∴ Inner curved surface area of the bowl = $2 π r^{2} = 2 \times \frac{22}{7} \times {(5.25)}^{2}$ = 173.25 cm²

Rate of tin-plating = ₹ 32 per 100 cm²

∴ Cost of tin-plating the bowl on the inside

= Inner curved surface area of the bowl × Rate of tin-plating

$= 173.25 \times \frac{32}{100}$

= ₹ 55.44

Thus, the cost of tin-plating the bowl on the inside is ₹ 55.44.

Page No 602:

Answer:

Let the radius of the moon and earth be r units and R units, respectively.

∴ 2r = $\frac{1}{4}$ × 2R (Given)

⇒ $r = \frac{R}{4}$ .....(1)

$∴ \frac{Volume of the moon}{Volume of the earth} = \frac{\frac{4}{3} π r^{3}}{\frac{4}{3} π R^{3}} = \frac{{(\frac{R}{4})}^{3}}{R^{3}} = \frac{1}{64}$ [Using (1)]

Thus, the volume of the moon is $\frac{1}{64}$ of the volume of the earth.

Page No 602:

Answer:

Let the radius of the solid hemisphere be r units.

Numerical value of surface area of the solid hemisphere = $3 π r^{2}$

Numercial value of volume of the solid hemisphere = $\frac{2}{3} π r^{3}$

It is given that the volume and surface area of the solid hemisphere are numerically equal.

$∴ \frac{2}{3} π r^{3} = 3 π r^{2} \Rightarrow 2 r = 9 units$
Thus, the diameter of the hemisphere is 9 units.

Page No 604:

Answer:

(c) 810 cm³

$Volume of the cuboid = l \times b \times h = 15 \times 12 \times 4.5 {cm}^{3} = 810 {cm}^{3}$

Page No 604:

Answer:

(b) 552 cm²

$Total surface area of the cuboid = 2 (l b + b h + l h) {cm}^{2} = 2 (12 \times 9 + 8 \times 9 + 12 \times 8) {cm}^{2} = 2 (108 + 72 + 96) {cm}^{2} = 552 {cm}^{2}$

Page No 605:

Answer:

Length of the cuboid, l = 15 m

Breadth of the cuboid, b = 6 m

Height of the cuboid, h = 5 dm = 0.5 m (1 m = 10 dm)

∴ Lateral surface area of the cuboid = 2h(l + b) = 2 × 0.5 × (15 + 6) = 2 × 0.5 × 21 = 21 cm²

Hence, the correct answer is option (b).

Page No 605:

Answer:

(c) 36 kg
Length = 9 m
Breadth = 40 cm = 0.4 m
Height = 20 cm = 0.2 m
∴ Weight of the beam = volume of the beam $\times$ weight of iron per cubic metre
$= 9 \times 0.4 \times 0.2 \times 50 = 36 kg$

Page No 605:

Answer:

(a) 15 m

Length of longest rod = diagonal of the room
= diagonal of a cuboid
$= \sqrt{l^{2} + b^{2} + h^{2}} = \sqrt{100 + 100 + 25} m = \sqrt{225} m = 15 m$

Page No 605:

Answer:

(d) 11.2 cm
Maximum length of the pencil = diagonal of the box
$= \sqrt{l^{2} + b^{2} + h^{2}} = \sqrt{8^{2} + 6^{2} + 5^{2}} cm = \sqrt{64 + 36 + 25} cm = \sqrt{125} cm = 11.2 cm$

Page No 605:

Answer:

(b) 192

Number of planks = $\frac{volume of the pit}{volume of 1 plank}$
$= \frac{40 \times 12 \times 16}{4 \times 5 \times 2} = \frac{7680}{40} = 192$

Page No 605:

Answer:

(a) 480

Length of the pit = 20 m
Breadth of the pit = 6 m
Height of the pit = 50 cm = 0.5 m

Length of the plank = 5m
Breadth of the plank = 25 cm = 0.25 m
Height of the plank = 10 cm = 0.1 m

∴ Number of planks = $\frac{volume of the pit}{volume of 1 plank}$

$= \frac{20 \times 6 \times 0.5}{5 \times 0.25 \times 0.1} = \frac{60 \times 1000}{125} = 480$

Page No 605:

Answer:

(c) 6400

Length of the wall = 8 m = 800 cm
Breadth of the wall = 6 m = 600 cm
Height of the wall = 22.5 cm

Length of the brick = 25 cm
Breadth of the brick = 11.25 cm
Height of the brick = 6 cm

∴ Number of bricks required = $\frac{volume of the wall}{volume of 1 brick}$

$= \frac{800 \times 600 \times 22.5}{25 \times 11.25 \times 6} = \frac{10800000}{1687.5} = 6400$

Page No 605:

Answer:

(b) 270

Number of persons = $\frac{volume of the hall}{volume of air required by 1 person}$

$= \frac{20 \times 15 \times 4.5}{5} = 20 \times 3 \times 4.5 = 270$

∴ 270 persons can be accommodated.

Page No 605:

Answer:

(b) 2250 m³

Length of the river = 1.5 m
Breadth of the river = 30 m
Depth of the river = 3 km = 3000 m

$Now, volume of water that runs into the sea = 1.5 \times 30 \times 3000 m^{3} = 135000 m^{3}$

∴ Volume of water that runs into the sea per minute = $\frac{135000}{60} = 2250 m^{3}$

Page No 605:

Answer:

(d) 512 m³

Suppose that a m be the edge of the cube.
We have:
$4 a^{2} = 256 \Rightarrow a^{2} = \frac{256}{4} = 64 \Rightarrow a = 8 m$

∴ Volume of the cube = $a^{3} m^{3} = 8^{3} m^{3} = 512 m^{3}$

Page No 605:

Answer:

(c) 64 cm³

Let a cm be the edge of the cube.
We have:
$6 a^{2} = 96 \Rightarrow a^{2} = 16 \Rightarrow a = 4 cm$

∴ Volume of the cube = $a^{3} {cm}^{3} = 4^{3} {cm}^{3} = 64 {cm}^{3}$

Page No 605:

Answer:

(b) 384 cm²

Suppose that a cm is the edge of the cube.
We have:
$a^{3} = 512 \Rightarrow a = \sqrt[3]{512} = 8 cm$

$∴ Total surface area of cube = 6 a^{2} {cm}^{2} = 6 \times 8 \times 8 {cm}^{2} = 384 {cm}^{2}$

Page No 605:

Answer:

(d) $10 \sqrt{3}$ cm103√ cm
Length of the longest rod = body diagonal of the vessel
$= \sqrt{3} a = \sqrt{3} \times 10 = 10 \sqrt{3} cm$

Page No 606:

Answer:

(b) 384 cm²
We have:
$\sqrt{3} a = 8 \sqrt{3} \Rightarrow a = 8 cm$

∴ Surface area of the cube = $6 a^{2} = 6 \times 8 \times 8 = 384 {cm}^{2}$

Page No 606:

Answer:

Let a be the edge of the cube.
Then the surface area is $6 a^{2} = S (say)$

Now, increased edge = $(a + \frac{50}{100} a)$ = $\frac{150}{100} a = \frac{3}{2} a$

$Then, new surface area = 6 {(\frac{3}{2} a)}^{2} = 6 \times \frac{9}{4} a^{2} = \frac{9}{4} S In crease in surface area = \frac{9}{4} S - S = \frac{5}{4} S ∴ Percentage increase in surface area = \frac{\frac{5}{4} S}{S} = \frac{5}{4} \times 100 % = 125 %$

Page No 606:

Answer:

(b) 144 cm²
Volume of the new cube formed = total volume of the three cubes
Suppose that a cm is the edge of the new cube, then

$a^{3} = 3^{3} + 4^{3} + 5^{3} = 27 + 64 + 125 = 216 = 6 cm$

∴ Lateral surface area of the new cube = $4 a^{2} = 4 \times 6 \times 6 = 144 {cm}^{2}$

Page No 606:

Answer:

(d) 1000 m³

$Area of the land = 2 sq hec = 2000 sq m Amount of rainfall = 5 cm = 0.05 m ∴ Volume of the water = area of the land \times amount of rainfall = 2000 \times 0.05 = 1000 m^{3}$

Page No 606:

Answer:

(c) 1 : 9
Suppose that the edges of the cubes are a and b.
We have:
$\frac{a^{3}}{b^{3}} = \frac{1}{27} \Rightarrow {(\frac{a}{b})}^{3} = \frac{1}{27} \Rightarrow \frac{a}{b} = \frac{1}{3}$

∴ Ratio of the surface areas = $\frac{6 a^{2}}{6 b^{2}} = {(\frac{a}{b})}^{2} = {(\frac{1}{3})}^{2} = \frac{1}{9}$

Page No 606:

Answer:

(d) becomes 8 times

Suppose that the side of the cube is a.
When it is doubled, it becomes 2a.
New volume of the cube = ${(2 a)}^{3} = 8 a^{3}$

Hence, the volume becomes 8 times the original volume.

Page No 606:

Answer:

(b) 396 cm³

$Volume of the cylinder = π r^{2} h = \frac{22}{7} \times 3^{2} \times 14 = 22 \times 9 \times 2 = 396 {cm}^{3}$

Page No 606:

Answer:

(b) 1760 cm²

$Curved surface area of the cylinder = 2 πrh = 2 \times \frac{22}{7} \times 14 \times 20 = 44 \times 40 = 1760 {cm}^{2}$

Page No 606:

Answer:

(c) 20 cm

Curved surface area = 1760 cm²
Suppose that h cm is the height of the cylinder.
Then we have:
$2 πr h = 1760 \Rightarrow 2 \times \frac{22}{7} \times 14 \times h = 1760 \Rightarrow h = \frac{1760 \times 7}{44 \times 14} = 20 cm$

Page No 606:

Answer:

(b) 396 cm³

Curved surface area = 264 cm².
Let r cm be the radius of the cylinder.
Then we have:

$2 π r h = 264 \Rightarrow 2 \times \frac{22}{7} \times r \times 14 = 264 \Rightarrow r = \frac{264 \times 7}{44 \times 14} = 3 cm$

$∴ Volume of the cylinder = \frac{22}{7} \times 3^{2} \times 14 = 22 \times 9 \times 2 = 396 {cm}^{3}$

Page No 606:

Answer:

(c) 6 m

Curved surface area = 264 m²
Volume = 924 m³
Let r m be the radius and h m be the height of the cylinder.
Then we have:
$2 π r h = 264 and π r^{2} h = 924 \Rightarrow r h = \frac{264}{2 π} \Rightarrow h = \frac{264}{2 r \times π} Now, π r^{2} h = π \times r^{2} \times \frac{264}{2 r \times π} = 924 \Rightarrow r = \frac{924 \times 2}{264} \Rightarrow r = 7 m ∴ h = \frac{264 \times 7}{2 \times 7 \times 22} = 6 m$

Page No 606:

Answer:

(c) 10 : 9

Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h.
Then, ratio of the curved surface areas = $\frac{2 π (2 r) (5 h)}{2 π (3 r) (3 h)}$ =10 : 9

Page No 607:

Answer:

(b) 20 : 27

Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h..
$∴ Ratio of their volumes = \frac{π {(2 r)}^{2} \times 5 h}{π {(3 r)}^{2} \times 3 h} = \frac{4 \times 5}{9 \times 3} = \frac{20}{27}$

Page No 607:

Answer:

(d) 770 cm²
We have:
r: h = 2 : 3
$\Rightarrow \frac{r}{h} = \frac{2}{3} \Rightarrow h = \frac{3}{2} r$
Now, volume = 1617 cm³
$\Rightarrow π r^{2} h = 1617 \Rightarrow \frac{22}{7} \times r^{2} \times \frac{3}{2} r = 1617 \Rightarrow r^{3} = \frac{1617 \times 14}{66} = 343 \Rightarrow r = 7 cm$

∴ h = 10.5 cm

$Hence, total surface area = 2 π r h + 2 π r^{2} = \frac{22}{7} (2 \times 7 \times 10.5 + 2 \times 7^{2}) = \frac{22}{7} (147 + 2 \times 49) = \frac{22}{7} \times 245 = 770 {cm}^{2}$

Page No 607:

Answer:

(b) $\sqrt{2} : 1$

Suppose that the heights of two cylinders are h and 2h whose radii are r and R, respectively.
Since the volumes of the cylinders are equal, we have:

$π r^{2} h = π R^{2} \times 2 h \Rightarrow \frac{r^{2}}{R^{2}} = \frac{2}{1} \Rightarrow (\frac{r}{R}) = \sqrt{\frac{2}{1}} \Rightarrow r : R = \sqrt{2} : 1$

Page No 607:

Answer:

(a) 1078 cm³

We have:

$\frac{2 π r h}{2 π r h + 2 π r^{2}} = \frac{1}{2} \Rightarrow 4 π r h = 2 π r h + 2 π r^{2} \Rightarrow 2 π r h = 2 π r^{2} \Rightarrow \frac{r}{h} = \frac{1}{1}$

$Also, 2 π r h + 2 π r^{2} = 616 \Rightarrow 2 π r^{2} + 2 π r^{2} = 616 \Rightarrow 4 π r^{2} = 616 \Rightarrow π r^{2} = 154 \Rightarrow r^{2} = \frac{154}{π} {cm}^{2}$

$∴ Volume of the cylinder = π r^{2} h = π \times \frac{154}{π} \times r = 154 \times \sqrt{\frac{154 \times 7}{22}} = 154 \times 7 = 1078 {cm}^{3}$

Page No 607:

Answer:

(c) halved

Suppose that the new radius is $\frac{1}{2}$ r and the height is 2h.
$∴ Volume = π \times {(\frac{1}{2} r)}^{2} \times 2 h = π \times \frac{r^{2}}{4} \times 2 h = \frac{1}{2} π r^{2} h$

Page No 607:

Answer:

(b) 450

Number of coins = $\frac{volume of cylinder to be formed}{volume of a coin}$

$= \frac{π \times 2.25 \times 2.25 \times 10}{π \times 0.75 \times 0.75 \times 0.2} = \frac{225 \times 225 \times 5}{75 \times 75} = 450$

Page No 607:

Answer:

(d) 9 times

Let the new radius be $\frac{1}{3}$ r.
Suppose that the new height is H.
The volume remains the same.
$i . e ., π r^{2} h = π \times {(\frac{1}{3} r)}^{2} \times H \Rightarrow h = \frac{1}{9} H \Rightarrow H = 9 h ∴ The new height becomes nine times the original height .$

Page No 607:

Answer:

(b) 1320 m²

Area covered by the roller in 1 revolution = 2 $π$ rh
$= 2 \times \frac{22}{7} \times 42 \times 100 = 44 \times 600 = 26400 {cm}^{2} ∴ Area covered in 50 revolutions = 26400 \times 500 {cm}^{2} = 1320 m^{2}$

Page No 607:

Answer:

(b) 112 m

Volume of the lead = volume of the cylindrical wire
Suppose that h m is the length of the wire.
As 2.2 dm³ of lead is to be drawn into a cylindrical wire of diameter 0.50 cm, we have:
$π {(0.25)}^{2} \times h = 2.2 \times 10 \times 10 \times 10 \Rightarrow h = \frac{2200 \times 7}{22 \times 0.0625} = \frac{700}{0.0625} = 11200 cm = 112 m$

Page No 607:

Answer:

(c) 2πrh
The lateral surface area of a cylinder is equal to its curved surface area.
∴ Lateral surface area of a cylinder = $2 π r h$

Page No 607:

Answer:

(b) 550 cm²

$l = \sqrt{r^{2} + h^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = \sqrt{625} = 25 cm ∴ Curved surface area of the cone = π r l = \frac{22}{7} \times 7 \times 25 = 550 {cm}^{2}$

Page No 607:

Answer:

(d) (144π) cm³

Volume of the cone = $\frac{1}{3} π r^{2} h$
$= \frac{1}{3} π \times 6^{2} \times 12 = π \times 36 \times 4 = 144 π {cm}^{3}$

Page No 608:

Answer:

(c) 220 m
Let the length of the required cloth be L m and its breadth be B m.
Here, B = 2.5 m
Radius of the conical tent =7 m
Height of the tent = 24 m
Area of cloth required = curved surface area of the conical tent
$\Rightarrow L \times B = πrl \Rightarrow L \times 2.5 = \frac{22}{7} \times 7 \times \sqrt{7^{2} + 24^{2}} \Rightarrow 2.5 L = 22 \times \sqrt{49 + 576} ∴ L = \frac{22 \times 25}{2.5} = 220 m$

Page No 608:

Answer:

(a) 10 cm
Let r cm be the radius of the cone.
Volume = 1570 cm³

$Then \frac{1}{3} \times 3.14 \times r^{2} \times 15 = 1570 \Rightarrow r^{2} = \frac{1570}{3.14 \times 5} = 100 \Rightarrow r = 10 cm$

Page No 608:

Answer:

(b) 7546 cm³

$Radius of the cone, r = \sqrt{l^{2} - h^{2}} = \sqrt{28^{2} - 21^{2}} = \sqrt{784 - 441} = \sqrt{343} cm$

$∴ Volume of the cone = \frac{1}{3} π r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 343 \times 21 = 22 \times 343 = 7546 {cm}^{3}$

Page No 608:

Answer:

(c) 550 cm²
Let r cm be the radius of the cone.
Volume of the right circular cone = 1232 cm³
Then we have:
$\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 24 = 1232 \Rightarrow r^{2} = \frac{1232 \times 21}{22 \times 24} \Rightarrow r^{2} = 49 \Rightarrow r = 7 cm$

$∴ Curved surface area of the cone = π r l = \frac{22}{7} \times 7 \times \sqrt{7^{2} + 24^{2}} = 22 \times \sqrt{49 + 576} = 22 \times 25 = 550 {cm}^{2}$

Page No 608:

Answer:

(d) 25 : 64

Suppose that the radii of the cones are 4r and 5r and their heights are h and H, respectively .
It is given that the ratio of the volumes of the two cones is 1:4
Then we have:
$\frac{\frac{1}{3} π {(4 r)}^{2} h}{\frac{1}{3} π {(5 r)}^{2} H} = \frac{1}{4} \Rightarrow \frac{16 r^{2} h}{25 r^{2} H} = \frac{1}{4} \Rightarrow \frac{h}{H} = \frac{25}{64} ∴ h : H = 25 : 64$

Page No 608:

Answer:

(a) 100 %

Suppose that height of the cone becomes 2h and let its radius be r.
Then new volume of the cone = $\frac{1}{3} π r^{2} (2 h) = \frac{2}{3} π r^{2} h = 2 \times volume of the cone$
Increase in volume = $\frac{2}{3} π r^{2} h - \frac{1}{3} π r^{2} h = \frac{1}{3} π r^{2} h$
$∴ Percentage increase = \frac{increase in volume}{initial volume} \times 100 %$ = $\frac{\frac{1}{3} π r^{2} h}{\frac{1}{3} π r^{2} h} \times 100 %$ =100%

Hence, the volume increases by 100%.

Page No 608:

Answer:

(b) 4 : 1
If the slant height of the first cone is l, then the slant height of the second cone will be 2l.
Let the radii of the first and second cones be r and R, respectively.
Then we have:

$π r l = 2 \times (π R \times 2 l) \Rightarrow r = 4 R \Rightarrow \frac{r}{R} = \frac{4}{1}$

Page No 608:

Answer:

(b) 3 : 1

It is given that the right circular cylinder and the right circular cone have the same base and height.
Suppose that the respective radii of bases and heights are equal to r and h.
Then ratio of their volumes = $\frac{π r^{2} h}{\frac{1}{3} π r^{2} h} = \frac{3}{1}$

Page No 608:

Answer:

(d) 1 : 3
It is given that the right circular cylinder and the right circular cone have the same radius and volume.
Suppose that the radii of their bases are equal to r and the respective heights of the cylinder and the cone are h and H.
As the volumes of the cylinder and the cone are the same, we have:
$π r^{2} h = \frac{1}{3} π r^{2} H \Rightarrow \frac{h}{H} = \frac{1}{3}$

Page No 608:

Answer:

(a) 9 : 8

Suppose that the respective radii of the cylinder and the cone are 3r and 4r and their respective heights are 2h and 3h.

$∴ Ratio of the volumes = \frac{π {(3 r)}^{2} \times 2 h}{\frac{1}{3} π {(4 r)}^{2} \times 3 h} = \frac{3 \times 9 \times 2}{16 \times 3} = \frac{9}{8}$

Page No 608:

Answer:

(d) 8 times

Let the new height and radius be 2h and 2r, respectively.

$New volume of the cone = \frac{1}{3} π {(2 r)}^{2} \times 2 h = \frac{8}{3} π r^{2} h = 8 \times initial volume of the cone$

Page No 608:

Answer:

(d) 13500
Radius of the cylinder = 3 cm
Height of the cylinder = 5 cm
Radius of the cone = 1 mm = 0.1 cm
Height of the cone = 1 cm

∴ Number of cones, n = $\frac{volume of the cylinder}{volume of 1 cone}$
$= \frac{π \times 3^{2} \times 5}{\frac{1}{3} π \times 0 . 1^{2} \times 1} = \frac{3 \times 9 \times 5}{0.01} = 13500$

Page No 609:

Answer:

(b) 15 m

Suppose that the height of the cone is h m.
Area of the ground = $11 \times 4 = 44 m^{2}$

$∴ π r^{2} = 44 \Rightarrow r^{2} = \frac{44 \times 7}{22} = 14 Also, \frac{1}{3} π r^{2} h = 220 \Rightarrow \frac{1}{3} \times \frac{22}{7} \times 14 h = 220 \Rightarrow h = \frac{220 \times 21}{22 \times 14} = 15 m$
Hence, the height of the cone is 15 m.

Page No 609:

Answer:

(a) $\frac{32 {πr}^{3}}{3}$

$Volume of the sphere of radius 2 r = \frac{4}{3} π {(2 r)}^{3} = \frac{32}{3} π r^{3}$

Page No 609:

Answer:

(b) 4851 cm³
Volume of the sphere = $\frac{4}{3} π r^{3}$
$= \frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5 = 88 \times 0.5 \times 10.5 \times 10.5 = 4851 {cm}^{3}$

Page No 609:

Answer:

(d) 5544 cm²

Surface area of sphere = $4 π r^{2}$
= $4 \times \frac{22}{7} \times 21 \times 21 = 5544 {cm}^{2}$

Page No 609:

Answer:

(c) 4851 cm³

Surface area = 1386 cm²
Let r cm be the radius of the sphere.
Then we have:
$4 π r^{2} = 1386 \Rightarrow r^{2} = \frac{1386 \times 7}{4 \times 22} = 110.25 \Rightarrow r = 10.5 cm ∴ Volume of the sphere = \frac{4}{3} π r^{3} = \frac{4}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5 = 4851 {cm}^{3}$

Page No 609:

Answer:

(a) (288π) m³

Surface area = (144π) m²
Ler r m be the radius of the sphere.
Then we have:

$4 π r^{2} = 144 π \Rightarrow r^{2} = \frac{144}{4} = 36 \Rightarrow r = 6 m ∴ Volume of the sphere = \frac{4}{3} π r^{3} = \frac{4}{3} π \times 6 \times 6 \times 6 = 288 π m^{3}$

Page No 609:

Answer:

(a) 5544 cm²
Let r cm be the radius of the sphere.
Then we have:

$\frac{4}{3} π r^{3} = 38808 \Rightarrow r^{3} = \frac{38808 \times 3 \times 7}{4 \times 22} = 9261 \Rightarrow r = 21 cm ∴ Curved surface area = 4 π r^{2} = 4 \times \frac{22}{7} \times 21 \times 21 = 5544 {cm}^{3}$

Page No 609:

Answer:

(b) 1 : 4
Suppose that r and R are the radii of the spheres.
Then we have:

$\frac{\frac{4}{3} π r^{3}}{\frac{4}{3} π R^{3}} = \frac{1}{8} \Rightarrow {(\frac{r}{R})}^{3} = \frac{1}{8} \Rightarrow \frac{r}{R} = \frac{1}{2}$
∴ Ratio of surface area of spheres = $\frac{4 π r^{2}}{4 π R^{2}} = {(\frac{r}{R})}^{2} = {(\frac{1}{2})}^{2} = \frac{1}{4}$

Page No 609:

Answer:

(d) 64

Number of balls = $\frac{volume of solid metal ball}{volumeof 1 small ball}$
= $\frac{\frac{4}{3} π \times 8^{3}}{\frac{4}{3} π \times 2^{3}} = \frac{512}{8} = 64$

Page No 609:

Answer:

(b) 2.1 cm

Let r cm be the radius of the sphere.
Volume of the cone = volume of the sphere
$\Rightarrow \frac{1}{3} π \times 2.1 \times 2.1 \times 8.4 = \frac{4}{3} π r^{3} \Rightarrow r^{3} = 2.1 \times 2.1 \times 2.1 \Rightarrow r = 2.1 cm$

Page No 609:

Answer:

(b) 288 m

Let h m be the length of wire.
Volume of the lead ball = volume of the wire
$\Rightarrow \frac{4}{3} π \times 6^{3} = π \times {(0.1)}^{2} h \Rightarrow h = \frac{4 \times 216}{3 \times 0.01} = 28800 cm = 288 m$

Page No 609:

Answer:

(c) 126

Number of cones = $\frac{volume of the sphere}{volume of 1 cone}$
$= \frac{\frac{4}{3} π \times 10.5 \times 10.5 \times 10.5}{\frac{1}{3} π \times 3.5 \times 3.5 \times 3} = 4 \times 3 \times 3 \times 3.5 = 126$

Page No 609:

Answer:

(d) 84000

$Number of lead shots = \frac{volume of cuboid}{volume of 1 lead shot} = \frac{9 \times 11 \times 12}{\frac{4}{3} \times \frac{22}{7} \times 0.15 \times 0.15 \times 0.15} = \frac{9 \times 11 \times 3 \times 3 \times 7}{22 \times 0.15 \times 0.15 \times 0.15} = 84000$

Page No 610:

Answer:

(c) 36 m
Let h m be the length of the wire.
Volume of the sphere = volume of the wire
$\Rightarrow \frac{4}{3} π \times 3^{3} = π \times {(0.1)}^{2} \times h \Rightarrow h = \frac{4 \times 9}{0.001} = 3600 cm = 36 m$

Page No 610:

Answer:

(a) 6.3 cm
Let r cm be the radius of base of the cone.
Volume of the sphere = volume of the cone
$\Rightarrow \frac{4}{3} π \times {(6.3)}^{3} = \frac{1}{3} π \times r^{2} \times 25.2 \Rightarrow r^{2} = \frac{4 \times 6.3 \times 6.3 \times 6.3}{25.2} = 39.69 \Rightarrow r = \sqrt{39.69} = 6.3 cm$

Page No 610:

Answer:

(c) 2.5 cm

Let r cm be the radius of the third ball.
Volume of the original ball = volume of the three balls
$\frac{4}{3} π \times 3^{3} = \frac{4}{3} π \times 1 . 5^{3} + \frac{4}{3} π \times 2^{3} + \frac{4}{3} π r^{3} \Rightarrow 27 = 3.375 + 8 + r^{3} \Rightarrow r^{3} = 27 - 11.375 = 15.625 \Rightarrow r = 2.5 cm$

Page No 610:

Answer:

(a) 1 : 4
Ratio of the surface areas of balloon = $\frac{2 π \times 6^{2}}{2 π \times 12^{2}} = \frac{36}{144} = \frac{1}{4}$

Page No 610:

Answer:

(d) 88 cm²

Suppose that the radii of the spheres are r cm and (7 − r) cm. Then we have:
$\frac{\frac{4}{3} π {(7 - r)}^{3}}{\frac{4}{3} {πr}^{3}} = \frac{64}{27} \Rightarrow \frac{(7 - r)}{r} = \frac{4}{3} \Rightarrow 21 - 3 r = 4 r \Rightarrow 21 = 7 r \Rightarrow r = 3 cm$

Now, the radii of the two spheres are 3 cm and 4 cm.

$∴ Required difference = 4 \times \frac{22}{7} \times 4^{2} - 4 \times \frac{22}{7} \times 3^{2} = 4 \times \frac{22}{7} \times (16 - 9) = 4 \times \frac{22}{7} \times 7 = 88 {cm}^{2}$

Page No 610:

Answer:

(c) 54

Number of bottles = $\frac{volume of bowl}{volume of 1 bottle}$

$= \frac{\frac{2}{3} \times π \times 9^{3}}{π \times 1 . 5^{2} \times 4} = \frac{2 \times 729}{2.25 \times 12} = 54$

Page No 610:

Answer:

(b) 2 : 1

Let the radii of the cone and the hemisphere be r and their respective heights be h and H.
$Then \frac{1}{3} π \times r^{2} \times h = \frac{2}{3} π \times r^{2} \times H \Rightarrow \frac{h}{H} = \frac{2}{1}$

Page No 610:

Answer:

(a) 1 : 2 : 3
The cone, hemisphere and the cylinder stand on equal bases and have the same height.
We know that radius and height of a hemisphere are the same.
Hence, the height of the cone and the cylinder will be the common radius.
i.e., r = h
Ratio of the volumes of the cone, hemisphere and the cylinder:

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

Page No 610:

Answer:

(c) 3 units

We have:

$\frac{4}{3} π r^{3} = 4 π r^{2} \Rightarrow \frac{1}{3} r = 1 \Rightarrow r = 3 units$

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