Rs Aggarwal 2020 2021 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 2020 2021 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 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 9 Maths are prepared by experts and are 100% accurate.

#### Page No 555:

(i)
Here, l = 12 cm, b = 8 cm, h = 4.5 cm
Volume of the cuboid = $l×b×h\phantom{\rule{0ex}{0ex}}$

Total Surface area = 2(lb + lh+ bh)

Lateral surface area = $2\left(l+b\right)×h$

(ii)
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  $l×b×h\phantom{\rule{0ex}{0ex}}$

Total surface area = 2(lb + lh+ bh)

Lateral surface area =

(iii)
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  $l×b×h\phantom{\rule{0ex}{0ex}}$

Total surface area = 2(lb + lh+ bh

Lateral surface area =

(iv)
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =

Total Surface area = 2(lb + lh+ bh)

Lateral surface area =

#### Page No 555:

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:

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:

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 =  = 2 m

Thus, the breadth of the tank is 2 m.

#### Page No 555:

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

#### Page No 555:

Number of planks =

Volume of one plank =

∴ Number of planks =

#### Page No 555:

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

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

∴ Number of bricks = = $\frac{10800000}{1687.5}=6400$

#### Page No 555:

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(lbbhhl)

= 2(8 × 6 + 6 × 2.5 + 2.5 × 8)

= 2 × 83

= 166 m2

#### Page No 555:

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(lb) − 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:

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 =

Also, volume of the mortar =

Total volume of the bricks in the wall = volume of the wall − volume of the mortar
= (18000000 − 1500000) cm3= 16500000 cm3

∴ Number of bricks =

#### Page No 555:

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:

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

Area of the sheet required =

Let h m be the height of the box.
Then area of the sheet = total surface area of the box

∴ The height of the box is 1.5 m.

#### Page No 556:

Length of the cuboid = 16 m
Suppose that the breadth and height of the cuboid are 3x m and 2x m, respectively.

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

#### Page No 556:

Volume of the dining hall =

Volume of air required by each person = 5 m3

∴ Capacity of the dining hall =

#### Page No 556:

Length of the classroom = 10 m
Breadth of the classroom = 6.4 m
Height of the classroom = 5 m
Area of the floor = length $×$ breadth = 10 $×$ 6.4 m2

Now, volume of the classroom=

#### Page No 556:

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

∴ The height of the cuboid is 9 cm.

#### Page No 556:

Volume of the water that falls on the ground =

#### Page No 556:

Here, a = 9 m
Volume of the cube =
Lateral surface area of the cube =
Total surface area of the cube =
∴ Diagonal of the cube =

#### Page No 556:

Suppose that the side of cube is x cm.
Total surface area of cube = 1176 sq cm

i.e., the side of the cube is 14 cm.

∴ Volume of the cube =

#### Page No 556:

Suppose that the side of cube is x cm.
Lateral surface area of the cube = 900 cm2

i.e., the side of the cube is 15 cm.
∴ Volume of the given cube =

#### Page No 556:

Suppose that the side of the given cube is x cm.
Volume of the cube = 512 cm3

∴ Surface area of the cube =

#### Page No 556:

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

Suppose the edge of the new cube = x cm
Then we have:

∴ Lateral surface area of the new cube =

#### Page No 556:

Length of the longest pole = length of the diagonal of the room

#### Page No 556:

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

∴ lbh = 19           .....(1)

Also,

Length of the diagonal = 11 cm

Squaring (1), we get

(l + b + h)2 = 192

⇒ l2b2 + h+ 2(lbbhhl) = 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:

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

#### Page No 556:

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

∴ Surface area of the cuboid, S = 2(abbcca

Volume of the cuboid, Vabc

Now,

$\frac{S}{V}=\frac{2\left(ab+bc+ca\right)}{abc}\phantom{\rule{0ex}{0ex}}⇒\frac{S}{V}=2\left(\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{S}{V}=2\left(\frac{1}{c}+\frac{1}{a}+\frac{1}{b}\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{V}=\frac{2}{S}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$

#### Page No 556:

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

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:

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 =  = 18

Thus, the number of cubes so formed is 18.

#### Page No 573:

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

#### Page No 573:

Height of soup in the bowl, h = 4 cm

Volume of soup in one bowl =

∴ 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:

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 =

∴ Amount of concrete mixture required to build 14 such pillars

= Volume of concrete mixture used in each pillar × 14

=

= 17.6 m3

#### Page No 573:

(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 = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}×{\left(\frac{7}{2}\right)}^{2}×10$ = 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:

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 =

Surface area of 20 pillars = Surface area of each pillar × 20 =

Rate of cleaning = ₹ 14 per m2

∴ Total cost of cleaning the 20 pillars

= Surface area of 20 pillars × Rate of cleaning

= ₹ 1760

#### Page No 573:

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.

Thus, the height of the cylinder is 1 m.

(ii) Volume of the cylinder = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}×{\left(0.7\right)}^{2}×1$ = 1.54 m3

Thus, the volume of the cylinder is 1.54 m3.

#### Page No 573:

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.

Thus, the radius of the cylinder is 3 cm.

(ii) Volume of the cylinder = $\mathrm{\pi }{r}^{2}h=3.14×{\left(3\right)}^{2}×5$ = 141.3 cm3

Thus, the volume of the cylinder is 141.3 cm3.

#### Page No 573:

Height of the cylinder, h = 1 m

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

∴ Area of the metal sheet needed to make the cylinder

= Total surface area of the cylinder

Thus, the area of the metal sheet needed to make the cylinder is 0.4708 m2.

#### Page No 573:

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 =

It is given that 1 cmof 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:

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

Thus, the total radiating surface in the system is 4.4 m2.

#### Page No 573:

Here, r = 10.5 cm; h = 60 cm

∴ Weight of cylinder = volume of cylinder $×$weight of cylinder per gram

#### Page No 573:

Curved surface area = 1210 cm2
Suppose that the height of cylinder is h cm.
We have r = 10 cm

#### Page No 573:

Let r be the radius and h be the height of the cylinder.
Circumference of its base(circle) = 110 cm.

Curved surface area of a cylinder = 4400 cm2.

Also, Volume of the cylinder =

#### Page No 573:

Suppose that the radius of the base and the height of the cylinder are 2x cm and 3x cm, respectively.

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

#### Page No 573:

Total surface area = 462 cm2

Given: Curved surface area =$\frac{1}{3}$$×$total surface area =

Now, total surface area − curved surface area =

Now, curved surface area = 154 cm2

#### Page No 573:

Curved surface area = $\frac{2}{3}$$×$total surface area =

Now, total surface area − curved surface area =

#### Page No 574:

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.

#### Page No 574:

Given: Diameter of the cylindrical bucket = 28 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

∴ Height of the rectangular tank = 24 cm

#### Page No 574:

Height of the barrel = h = 7 cm
Radius of the barrel = r = 2.5 mm = 0.25 cm
Volume of the barrel =

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., $\frac{1}{5}×1000$ cm3 =
∴ 48000 words would use up a bottle of ink containing one-fifith of a litre.

#### Page No 574:

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 cm3 of gold is drawn into a wire of diameter 0.1 mm.
Here, r

Hence, length of the wire = 127.27 m

#### Page No 574:

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
$=\mathrm{\pi }×\left(2.5{\right)}^{2}×100-\mathrm{\pi }×{\left(1.5\right)}^{2}×100\phantom{\rule{0ex}{0ex}}=\frac{22}{7}×100×\left[6.25-2.25\right]\phantom{\rule{0ex}{0ex}}=\frac{2200}{7}×4=\frac{8800}{7}{\mathrm{cm}}^{3}$
1 cm3 of cast iron weighs 21 g.

∴ Weight of $\frac{8800}{7}{\mathrm{cm}}^{3}$cast iron =

#### Page No 574:

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

#### Page No 574:

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

Thus, the area of metal sheet required to make the cylindrial tank is 7.48 m2.

#### Page No 574:

Radius of the vassel, R = 15 cm

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

∴ Volume of orange juice in the vassel =

Radius of the glass, r = 3 cm

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

∴ Volume of orange juice in each glass =

Number of glasses of orange juice sold by the juiceseller

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:

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

Thus, the height of the embankment is 2.1 m.

#### Page No 574:

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:

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 = $\mathrm{\pi }{R}^{2}H=\mathrm{\pi }{\left(0.7\right)}^{2}×2.1$ 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 = $\mathrm{\pi }{r}^{2}=\mathrm{\pi }{\left(0.0175\right)}^{2}$ m2

Volume of the water flowing out of the pipe in one second = Area of the cross section of the pipe × 2 m = $\mathrm{\pi }{\left(0.0175\right)}^{2}×2$ 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

$\mathrm{\pi }{\left(0.0175\right)}^{2}×2×t$ m3

Now,

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

$⇒t=\frac{1680}{60}$

⇒ t = 28 minutes

Thus, the tank will be filled in 28 minutes.

#### Page No 574:

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 = $\mathrm{\pi }{r}^{2}h=\mathrm{\pi }×{\left(28\right)}^{2}×h$

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.

Thus, the rise in the level of water in the container is 4 cm.

#### Page No 575:

Height of the tube-well, h = 280 m

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

∴ Volume of the tube-well = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}×{\left(\frac{3}{2}\right)}^{2}×280$ = 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 = $2\mathrm{\pi }rh=2×\frac{22}{7}×\frac{3}{2}×280$ = 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:

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 = $\mathrm{\pi }{r}^{2}l=\frac{22}{7}×{\left(0.2\right)}^{2}×l$ cm3

Thus, the length of the copper wire is 125 m.

#### Page No 575:

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

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 = $2\mathrm{\pi }rh=2×\frac{22}{7}×r×10$

Thus, the inner radius of the base is 1.75 m.

(iii) Capacity of the vassel = $\mathrm{\pi }{r}^{2}h=\frac{22}{7}×{\left(1.75\right)}^{2}×10=$96.25 m3

Thus, the capacity of the vassel is 96.25 m3.

#### Page No 575:

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)

Volume of the tube = 176 cm3                (Given)

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:

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

Let V1 and Vbe 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.

h1 = 18 cm

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.

h2 = 30 cm

Now,

$\frac{{V}_{1}}{{V}_{2}}=\frac{\mathrm{\pi }{\left(\frac{30}{2\mathrm{\pi }}\right)}^{2}×18}{\mathrm{\pi }{\left(\frac{18}{2\mathrm{\pi }}\right)}^{2}×30}=\frac{5}{3}$

⇒ V1 : V2 = 5 : 3

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

#### Page No 588:

Radius of the base, r = 5.25 cm

Slant height, l = 10 cm

∴ Curved surface area of the cone = $\mathrm{\pi }rl=\frac{22}{7}×5.25×10$ = 165 cm2

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

#### Page No 588:

Slant height, l = 21 m

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

∴ Total surface area of the cone =

Thus, the total surface area of the cone is .

#### Page No 588:

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 = $\mathrm{\pi }rl=\frac{22}{7}×7×25$ = 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:

Slant height, l = 14 cm

Let the radius of the base be r cm.

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

∴ Total surface area of the cone = $\mathrm{\pi }r\left(r+l\right)=\frac{22}{7}×7×\left(7+14\right)=\frac{22}{7}×7×21$ = 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:

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 = $\mathrm{\pi }rl=\frac{22}{7}×7×25$ = 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:

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 =

The cost of 1 m2 canvas is ₹ 70.

∴ Cost of canvas required to make the tent

= Curved surface area of the conical tent × ₹ 70

$=\frac{22}{7}×24×26×70$

= ₹ 1,37,280

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

#### Page No 588:

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, $l=\sqrt{{r}^{2}+{h}^{2}}=\sqrt{{\left(0.2\right)}^{2}+{1}^{2}}=\sqrt{0.04+1}=\sqrt{1.04}$ = 1.02 m

Curved surface area of each cone = $\mathrm{\pi }rl=3.14×0.2×1.02$ = 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:

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 = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×\frac{22}{7}×{\left(35\right)}^{2}×12$ = 15400 cm3

(ii) Curved surface area of the cone = $\mathrm{\pi }rl=\frac{22}{7}×35×37$ = 4070 cm2

(iii) Total surface area of the cone = $\mathrm{\pi }r\left(r+l\right)=\frac{22}{7}×35×\left(35+37\right)=\frac{22}{7}×35×72$ = 7920 cm2

#### Page No 588:

Height of the cone, h = 6 cm
Slant height of the cone, l = 10 cm

Volume of the cone =

Curved surface area of the cone =

∴ Total surface area =

#### Page No 588:

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

Depth of the conical pit, h = 12 m

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

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

#### Page No 588:

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 = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×3.14×{\left(4.5\right)}^{2}×3.5$ = 74.1825 m3

Now,

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

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

= Curved surface area of the heap of wheat

Thus, the area of canvas cloth required to just cover the heap is approximately 80.541 m2 .

#### Page No 588:

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

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 = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×\frac{22}{7}×{\left(7\right)}^{2}×24$ = 1232 m3

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

#### Page No 589:

Radius of the conical tent, r = 7 m
Height of the conical tent, h = 24 m

Width of the cloth = 2.5 m
∴ Length of the cloth =

#### Page No 589:

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 =

#### Page No 589:

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.

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

#### Page No 589:

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.

∴ Height of cone 2 = 6.4 cm

#### Page No 589:

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
=

But area of the canvas = length $×$ breadth

∴ Length of the canvas =

#### Page No 589:

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
=

∴ Weight of the pillar = volume of the pillar $×$ weight per cubic cm

#### Page No 589:

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

#### Page No 589:

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 = = 196.4 cm3
Radius of the the conical vessel = 40 cm
Depth of the vessel = 24 cm
Volume of the vessel =
Let the time taken to fill the conical vessel be x min.
Volume of water flowing per minute through the cylindrical pipe $×$ x = volume of the conical vessel

∴ The cylindrical pipe would take 51 min 12 sec to fill the conical vessel.

#### Page No 589:

Radius of the conical tent, r = 5 m

Area of the base of the conical tent =

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

∴ Number of students who can sit in the tent

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

Let the height of the tent be h m.

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

∴Volume of the tent = $\frac{1}{3}\mathrm{\pi }{r}^{2}h=\frac{1}{3}×\frac{22}{7}×{\left(5\right)}^{2}×9.23$ ≈ 241.74 m3

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

#### Page No 600:

(i) Radius of the sphere = 3.5 cm
Now, volume =

∴ Surface area = $4\mathrm{\pi }{r}^{2}$

(ii) Radius of the sphere=4.2 cm
Now, volume =

∴ Surface area =

Now, volume =
=

∴ Surface area =

#### Page No 600:

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

∴ Surface area of the sphere =

#### Page No 601:

Volume of the sphere = 606.375 m3

∴ Surface area =

#### Page No 601:

Let the radius of the sphere be r cm.

Surface area of the sphere = 154 cm2

∴ Volume of the sphere =

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

#### Page No 601:

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

#### Page No 601:

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

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

#### Page No 601:

Radius of the sphere = 8 cm

Volume of the sphere =
Volume of one lead ball =

∴ Number of lead balls =

#### Page No 601:

Radius of the solid sphere = 3 cm

Volume of the solid sphere =
=

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

Volume of the spherical ball =

Now, number of small spherical balls =

∴ The number of small balls thus obtained is 1000

#### Page No 601:

Radius of the metallic sphere, r = 10.5 cm
Volume of the sphere =
Radius of each smaller cone, r2 = 3.05 cm
Height of each smaller cone = 3 cm

Number of cones obtained =
$=\frac{\frac{4}{3}\mathrm{\pi }{r}^{\mathit{3}}}{\frac{1}{3}\mathrm{\pi }{{r}_{2}}^{\mathit{2}}h}\phantom{\rule{0ex}{0ex}}=\frac{4×10.5×10.5×10.5}{3.5×3.5×3}\phantom{\rule{0ex}{0ex}}=126.006\approx 126$

∴ 126 cones are obtained.

#### Page No 601:

Diameter of each sphere, d=12 cm
Radius of each sphere, r=6 cm

Diameter of base of  the cylinder, D=8 cm
Radius of base of cylinder, R=4 cm
Height of the cylinder, h = 90 cm

Number of spheres=

∴ Five spheres can be made.

#### Page No 601:

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

∴ Length of the wire = 36 m

#### Page No 601:

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.

∴ Diameter of the wire = 0.6 cm

#### Page No 601:

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

∴ The diameter of the cone is 15.6 cm.

#### Page No 601:

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

∴ The height of the cone is 35.84 cm.

#### Page No 601:

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

∴ The radius of the third ball is 2.5 cm.

#### Page No 601:

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

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

#### Page No 601:

Suppose that the radii of the spheres are r and R.
We have:
$\frac{4\mathrm{\pi }{r}^{2}}{4{\mathrm{\pi R}}^{2}}=\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\sqrt{\frac{1}{4}}=\frac{1}{2}$

Now, ratio of the volumes =

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

#### Page No 601:

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

∴ The radius of the ball is 9 cm.

#### Page No 601:

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

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

#### Page No 602:

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

#### Page No 602:

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

Volume of the shell =
$=\frac{4}{3}\mathrm{\pi }\left({9}^{3}-{8}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{4}{3}×\frac{22}{7}×\left(729-512\right)\phantom{\rule{0ex}{0ex}}=\frac{4×22×217}{21}\phantom{\rule{0ex}{0ex}}=\frac{88×31}{3}\phantom{\rule{0ex}{0ex}}=\frac{2728}{3}{\mathrm{cm}}^{3}$

Weight of the shell = volume of the shell $×$ density per cubic cm
=

∴ Weight of the shell = 4.092 kg

#### Page No 602:

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

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

#### Page No 602:

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  =
$=\frac{\frac{2}{3}\mathrm{\pi }×{9}^{3}}{\mathrm{\pi }×1.{5}^{2}×4}\phantom{\rule{0ex}{0ex}}=\frac{2×9×9×9}{3×1.5×1.5×4}\phantom{\rule{0ex}{0ex}}=54$

∴ 54 bottles are required to empty the bowl.

#### Page No 602:

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

∴ 56.83 cm3 of steel is used in making the bowl .

#### Page No 602:

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 = $2\mathrm{\pi }{r}^{2}=2×\frac{22}{7}×{\left(5.25\right)}^{2}$ = 173.25 cm2

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

#### Page No 602:

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

∴ Inner curved surface area of the bowl = $2\mathrm{\pi }{r}^{2}=2×\frac{22}{7}×{\left(5.25\right)}^{2}$ = 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

$=173.25×\frac{32}{100}$

= ₹ 55.44

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

#### Page No 602:

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)

[Using (1)]

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

#### Page No 602:

Let the radius of the solid hemisphere be r units.

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

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

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

Thus, the diameter of the hemisphere is 9 units.

(c) 810 cm3

(b) 552 cm2

#### Page No 605:

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(lb) = 2 × 0.5 × (15 + 6) = 2 × 0.5 × 21 = 21 cm2

Hence, the correct answer is option (b).

#### Page No 605:

(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 $×$weight of iron per cubic metre

#### Page No 605:

(a) 15 m

Length of longest rod = diagonal of the room
= diagonal of a cuboid

#### Page No 605:

(d) 11.2 cm
Maximum length of the pencil =  diagonal of the box

#### Page No 605:

(b)  192

Number of planks =

#### Page No 605:

(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 =

$=\frac{20×6×0.5}{5×0.25×0.1}\phantom{\rule{0ex}{0ex}}=\frac{60×1000}{125}\phantom{\rule{0ex}{0ex}}=480$

#### Page No 605:

(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 =

#### Page No 605:

(b) 270

Number of persons  =

∴ 270 persons can be accommodated.

#### Page No 605:

(b) 2250 m3

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

∴ Volume of water that runs into the sea per minute =

#### Page No 605:

(d) 512  m3

Suppose that a m be the edge of the cube.
We have:

∴ Volume of the cube =

#### Page No 605:

(c)  64 cm3

Let a cm be the edge of the cube.
We have:

∴ Volume of the cube =

#### Page No 605:

(b) 384 cm2

Suppose that a cm is the edge of the cube.
We have:

#### Page No 605:

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

#### Page No 606:

(b)  384 cm2
We have:

∴ Surface area of the cube =

#### Page No 606:

Let a be the edge of the cube.
Then the surface area is

Now, increased edge = $\left(a+\frac{50}{100}a\right)$ =

#### Page No 606:

(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

∴ Lateral surface area of the new cube =

(d) 1000 m3

#### Page No 606:

(c) 1 : 9
Suppose that the edges of the cubes are a and b.
We have:
$\frac{{a}^{3}}{{b}^{3}}=\frac{1}{27}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{a}{b}\right)}^{3}=\frac{1}{27}\phantom{\rule{0ex}{0ex}}⇒\frac{a}{b}=\frac{1}{3}$

∴ Ratio of the surface areas =

#### Page No 606:

(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 =

Hence, the volume becomes 8 times the original volume.

(b)  396 cm3

(b)  1760 cm2

#### Page No 606:

(c) 20 cm

Curved surface area = 1760 cm2
Suppose that h cm is the height of the cylinder.
Then we have:

#### Page No 606:

(b)  396 cm3

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

#### Page No 606:

(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:

#### Page No 606:

(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 = $\frac{2\mathrm{\pi }\left(2r\right)\left(5h\right)}{2\mathrm{\pi }\left(3r\right)\left(3h\right)}$=10 : 9

#### Page No 607:

(b)  20 : 27

Suppose that the radii of the cylinders are 2r and 3r and their respective heights are 5h and 3h..

#### Page No 607:

(d) 770 cm2
We have:
r: h = 2 : 3

Now, volume = 1617 cm3

h = 10.5 cm

#### Page No 607:

(b)

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:

(a) 1078 cm3

We have:

#### Page No 607:

(c) halved

Suppose that the new radius is $\frac{1}{2}$r and the height is 2h.

#### Page No 607:

(b) 450

Number of coins =

#### Page No 607:

(d) 9 times

Let the new radius be $\frac{1}{3}$r.
Suppose that the new height is H.
The volume remains the same.

#### Page No 607:

(b)  1320  m2

Area covered by the roller in 1 revolution = 2$\mathrm{\pi }$rh

#### Page No 607:

(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:

#### Page No 607:

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

(b) 550 cm2

#### Page No 607:

(d) (144π) cm3

Volume of the cone =

#### Page No 608:

(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

#### Page No 608:

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

(b)  7546 cm3

#### Page No 608:

(c) 550 cm2
Let r cm be the radius of the cone.
Volume of the right circular cone  = 1232 cm3
Then we have:

#### Page No 608:

(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:

#### Page No 608:

(a) 100 %

Suppose that height of the cone becomes 2h and let its radius be r.
Then new volume of the cone =
Increase in volume = $\frac{2}{3}\mathrm{\pi }{r}^{\mathit{2}}h-\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h=\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h$
=$\frac{\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h}{\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h}×100%$=100%

Hence, the volume increases by 100%.

#### Page No 608:

(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:

$\mathrm{\pi }rl=2×\left(\mathrm{\pi }R×2l\right)\phantom{\rule{0ex}{0ex}}⇒r=4R\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{4}{1}$

#### Page No 608:

(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 = $\frac{\mathrm{\pi }{r}^{\mathit{2}}h}{\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}h}=\frac{3}{1}$

#### Page No 608:

(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:
$\mathrm{\pi }{r}^{\mathit{2}}h=\frac{1}{3}\mathrm{\pi }{r}^{\mathit{2}}H\phantom{\rule{0ex}{0ex}}⇒\frac{\mathit{h}}{\mathit{H}}=\frac{1}{3}$

#### Page No 608:

(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.

#### Page No 608:

(d) 8 times

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

#### Page No 608:

(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 =
$=\frac{\mathrm{\pi }×{3}^{2}×5}{\frac{1}{3}\mathrm{\pi }×0.{1}^{2}×1}\phantom{\rule{0ex}{0ex}}=\frac{3×9×5}{0.01}\phantom{\rule{0ex}{0ex}}=13500$

#### Page No 609:

(b) 15 m

Suppose that the height of the cone is h m.
Area of the ground =

Hence, the height of the cone is 15 m.

#### Page No 609:

(a) $\frac{32{\mathrm{\pi r}}^{3}}{3}$

#### Page No 609:

(b)  4851 cm3
Volume of the sphere =

#### Page No 609:

(d)  5544 cm2

Surface area of sphere =
=

#### Page No 609:

(c)  4851 cm3

Surface area = 1386 cm2
Let r cm be the radius of the sphere.
Then we have:

#### Page No 609:

(a) (288π) m3

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

#### Page No 609:

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

#### Page No 609:

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

$\frac{\frac{4}{3}\mathrm{\pi }{r}^{3}}{\frac{4}{3}\mathrm{\pi }{R}^{3}}=\frac{1}{8}\phantom{\rule{0ex}{0ex}}⇒{\left(\frac{r}{R}\right)}^{3}=\frac{1}{8}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{1}{2}$
∴ Ratio of surface area of spheres =$\frac{4\mathrm{\pi }{r}^{2}}{4\mathrm{\pi }{R}^{2}}={\left(\frac{r}{R}\right)}^{2}={\left(\frac{1}{2}\right)}^{2}=\frac{1}{4}$

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(d) 64

Number of balls =
= $\frac{\frac{4}{3}\mathrm{\pi }×{8}^{3}}{\frac{4}{3}\mathrm{\pi }×{2}^{3}}=\frac{512}{8}=64$

#### Page No 609:

(b) 2.1 cm

Let r cm be the radius of the sphere.
Volume of the cone = volume of the sphere

#### Page No 609:

(b)  288 m

Let h m be the length of wire.
Volume of the lead ball = volume of the wire

#### Page No 609:

(c)  126

Number of cones =

(d) 84000

#### Page No 610:

(c)  36 m
Let h m be the length of the wire.
Volume of the sphere = volume of the wire

#### Page No 610:

(a)  6.3 cm
Let r cm be the radius of base of the cone.
Volume of the sphere = volume of the cone

#### Page No 610:

(c) 2.5 cm

Let r cm be the radius of the third ball.
Volume of the original ball = volume of the three balls

#### Page No 610:

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

#### Page No 610:

(d) 88 cm2

Suppose that the radii of the spheres are r cm and (7 − r) cm. Then we have:

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

#### Page No 610:

(c) 54

Number of bottles  =

#### Page No 610:

(b) 2 : 1

Let the radii of the cone and the hemisphere be r and their respective heights be h and H.

#### Page No 610:

(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: