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×b×h
                                  =(12×8×4.5) cm3= 432 cm3

Total Surface area = 2(lb + lh+ bh)
                              =2(12×8 + 12×4.5 +8×4.5) cm2=2(96 +54 + 36) cm2=2× 186  cm2=372  cm 2

Lateral surface area = 2l+b×h
                              =212+8×4.5 cm2=2(20)×4.5  cm2=40×4.5  cm2=180 cm2

(ii)
Here, l = 26 m; b = 14 m; h =6.5 m
Volume of the cuboid =  l×b×h
                                =26×14×6.5m3=2366 m3

Total surface area = 2(lb + lh+ bh)
                             =2(26×14+26×6.5+6.5×14) m2=2(364+169+91) m2=2×624 m2=1248 m2

Lateral surface area = 2l+b×h 
                              =226+14×6.5 m2=2×40×6.5 m2=520 m2

(iii)
Here, l = 15 m; b = 6 m; h = 5 dm = 0.5 m
Volume of the cuboid =  l×b×h
                                =(15×6×0.5) m3=45 m3


Total surface area = 2(lb + lh+ bh
                          =2(15×6+15×0.5+6×0.5) m2=2(90+7.5 +3) m2=2×100.5 m2= 201m2

Lateral surface area = 2l+b×h 
                               =215+6×0.5 m2=2×21×0.5 m2=21 m2

(iv)
Here, l = 24 m; b = 25 cm = 0.25 m; h =6 m
Volume of the cuboid =  l×b×h 
                                =24×0.25×6 m3=36 m3

Total Surface area = 2(lb + lh+ bh)
                            =2(24×0.25 +24×6 +0.25×6) m2=2(6+144 + 1.5) m2=2×151.5 m2=303 m2

Lateral surface area = 2l+b×h 
                             =224+0.25×6 m2=2×24.25×6 m2=291 m2

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 = 500001000 = 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 = Volume of the tankLength × Height=5010×2.5=5025 = 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

=Volume of the godownVolume of each wooden crate=150000.9375=16000

Page No 555:

Answer:

Number of planks = volume of the pit in cm3volume of 1 plank in cm3

Volume of one plank = (l×b×h) cm3
                                  =500×25×10  cm3= 125000 cm3
Volume of the pit=(l×b×h) cm3Here, l=20 m=2000 cm; b=6 m=600 cm; h=80 cmi.e., volume of the pit =2000×600×80 cm3                                  =96000000 cm3

∴ Number of planks = 96000000125000 = 96000125=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=800×22.5×600 cm3=10800000 cm3

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 =25×11.25×6 =1687.5 cm3

∴ Number of bricks = volume of the wallvolume of one brick = 108000001687.5=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(lbbhhl)

= 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(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:

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 =  =1500×30×400 cm3= 18000000 cm3

Now, volume of each brick=22×12.5×7.5 cm3                                         =2062.5 cm3

Also, volume of the mortar = 112×volume of the wall
                                        =1800000012=1500000 cm3
Total volume of the bricks in the wall = volume of the wall − volume of the mortar
                                                           = (18000000 − 1500000) cm3= 16500000 cm3

∴ Number of bricks = volume of bricksvolume of one brick=165000002062.5=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 = 594001000 = 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 = total costcost per metre square 

Let h m be the height of the box.
Then area of the sheet = total surface area of the box    
                                 =2(lb+lh +bh) m2=2(5×3+5×h+3×h) m2=2(15+8h) =( 30+16h )m2

Now, 30+16h=648012030+16h=5416h = 24h= 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=16×3x×2x1536=16×6x2x2=153696=16x=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 = (20×16×4.5) m3=1440 m3

Volume of air required by each person = 5 m3

∴ Capacity of the dining hall = volume of dining hallvolume of air required by each person=14405=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 × breadth = 10 × 6.4 m2
No. of students=area of the floorarea given to one student on the floor =10×6.41.6=64016=40 students

Now, volume of the classroom=10×6.4×5 m3

 Air required by each student=volume of the roomnumber of students=10×6.4×540 m3=8 m3

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(14×11+14×x +11×x)758 = 2(154 +14x +11x)758  =2(154+25x) 758  = 308 +50x50x = 758-308 = 450x=45050=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×depth
                                                                 =20000×0.05   m3=1000 m3

Page No 556:

Answer:

Here, a = 9 m
Volume of the cube = a3=93 m3=729 m3
Lateral surface area of the cube = 4a2=4×92m2=4×81 m2=324 m2
Total surface area of the cube = 6a2= 6×92m2=6×81 m2=486 m2
∴ Diagonal of the cube = 3a=3×9=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 =6x2x2=11766=196x =196=14 
i.e., the side of the cube is 14 cm.

∴ Volume of the cube = x3 =143 cm3=2744 cm3

Page No 556:

Answer:

Suppose that the side of cube is x cm.
Lateral surface area of the cube = 900 cm2
Then 900= 4x2x2=9004=225x=225=15
i.e., the side of the cube is 15 cm.
∴ Volume of the given cube = x3 cm3=153 cm3=3375 cm3

Page No 556:

Answer:

Suppose that the side of the given cube is x cm.
Volume of the cube = 512 cm3
Then 512 =x3x =5123 =8i.e., the side of the cube is 8 cm.

∴ Surface area of the cube = 6x2 cm2=6×82 cm2=384 cm2

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
                                        =33+43+53cm3=(27+64+125) cm3=216 cm3

Suppose the edge of the new cube = x cm
Then we have:
Then 216=x3x=2163 =6 i.e., the edge of the new cube is 6 cm.

∴ Lateral surface area of the new cube = 4x2 cm2=4×62 cm2=144 cm2

Page No 556:

Answer:

Length of the longest pole = length of the diagonal of the room
                                      =l2+b2+h2 m=102+102+52 m=100+100+25=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.

∴ lbh = 19           .....(1)

Also,

Length of the diagonal = 11 cm

l2+b2+h2=11l2+b2+h2=121                  .....2

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:

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 aa+50100a = 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

=Increase in surface area of the cubeInitial surface area of the cube×100%=7.5a26a2×100%=125%

Page No 556:

Answer:


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,

SV=2ab+bc+caabcSV=2ababc+bcabc+caabcSV=21c+1a+1b1V=2S1a+1b+1c

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

=Volume of water flowing out of the canalHeight of standing water on the field=720000.09=800000 m2
=80000010000                1 hectare = 10000 m2=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 = Volume of the solid metallic cuboidVolume of each solid cube=1448 = 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=2πrh                                                             =2×227×14×40  cm2                                                            =2×22×2×40 cm2                                                            =3520 cm2


Total surface area of the cylinder=2πrh + 2πr2                                                        = 3520 +2×227×142 cm2                                                       =(3520+1232 )=4752 cm2

 Volume of the cylinder=πr2h                                          =227×142×40 cm3                                         =22×14×2×40 cm3                                         =24640 cm3

Page No 573:

Answer:


Radius of the bowl, r72 cm

Height of soup in the bowl, h = 4 cm

Volume of soup in one bowl = πr2h=227×722×4=154 cm3

∴ 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 = πr2h=227×0.22×10 m3

∴ Amount of concrete mixture required to build 14 such pillars

= Volume of concrete mixture used in each pillar × 14

= 227×0.22×10 ×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, r72 cm

Height of plastic cylinder, h = 10 cm

∴ Volume of soft drink in plastic cylinder = πr2h=227×722×10 = 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 = 502 = 25 cm = 0.25 m        (1 m = 100 cm)

Height of each pillar, h = 4 m

∴ Surface area of each pillar = 2πrh=2×227×0.25×4 cm2

Surface area of 20 pillars = Surface area of each pillar × 20 = 2×227×0.25×4 ×20 cm2

Rate of cleaning = ₹ 14 per m2

∴ Total cost of cleaning the 20 pillars

= Surface area of 20 pillars × Rate of cleaning 

2×227×0.25×4 ×20 ×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.

2πrh=4.42×227×0.7×h=4.4h=4.4×72×22×0.7=1 m

Thus, the height of the cylinder is 1 m.

(ii) Volume of the cylinder = πr2h=227×0.72×1 = 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.

2πrh=94.2 cm22×3.14×r×5=94.2r=94.22×3.14×5=3 cm
Thus, the radius of the cylinder is 3 cm.

(ii) Volume of the cylinder = πr2h=3.14×32×5 = 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

πr2h=0.0154 m3227×r2×1=0.0154r=0.0154×722=0.07 m
∴ Area of the metal sheet needed to make the cylinder 

= Total surface area of the cylinder

=2πrhr+h=2×227×0.07×1×0.07+1=2×227×0.07×1×1.07=0.4708 m2
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, r242 = 12 cm

Outer radius of the wooden pipe, R282 = 14 cm

Length of the wooden pipe, h = 35 cm

∴ Volume of wood in the pipe = πR2-r2h=227×142-122×35=5720 cm3

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 = 34321000 = 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, r52 = 2.5 cm = 0.025 m         (1 m = 100 cm)

∴ Total radiating surface in the system

= Curved surface area of the cylindrical pipe

=2πrh=2×227×0.025×28=4.4 m2
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=πr2h                                       =227×(10.5)2×60 cm3                                      =22×10.5×1.5×60 cm3=20790 cm3

∴ Weight of cylinder = volume of cylinder ×weight of cylinder per gram
                              =20790×5  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 =2πrh1210 =2×227×10×h h =1210×72×22×10=11×72×2=19.25 cm

 Volume of the cylinder=πr2h                                            =227×102×19.25 cm3                                           = 2200×2.75 cm3 =6050 cm3

Page No 573:

Answer:

Let r be the radius and h be the height of the cylinder.
Circumference of its base(circle) = 110 cm.
2πr=110 r=1102πr=1102×227r=110×72×22r=352cm
Curved surface area of a cylinder = 4400 cm2.
2πrh=4400h=44002πrh=44002×227×352h=4400×7×22×22×35h=40 cm
Also, Volume of the cylinder = πr2h=227×3522×40=22×35×35×407×2×2=38500 cm3.

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=πr2h =227×(2x)2×3x                   = 227×12x3x3=1617×722×12=42.875x=42.8753 = 3.5 cm

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

 Total surface area of the cylinder=2πrh+2πr2                                                         =2×227(7×10.5+7×7) cm2                                                       =447×(73.5+49) cm2 =770  cm2

Page No 573:

Answer:

Total surface area = 462 cm2

Given: Curved surface area =13×total surface area = 13×462=154 cm2

Now, total surface area − curved surface area = 2πrh +2πr2 -2πrh
462 -154 =2πr2308 =2×227×r2r2=308×744=49r = 7 cm

Now, curved surface area = 154 cm2

2πrh = 1542×227×7×h=154h =15444=3.5 cm

  Volume of the cylinder=πr2h                                              =227×72×3.5                                           =539 cm3

Page No 573:

Answer:

Curved surface area = 23×total surface area = 23×231 = 2×77 =154 cm2

Now, total surface area − curved surface area = 2πrh +2πr2-2πrh

Then 231-154=2πr22×227×r2=77r2=77×744=12.25r=3.5 cmAlso, curved surface area=154 cm22πrh = 1542×227×3.5×h =154h=154×744×3.5=7 cm

Volume of the cylinder=πr2h                                           =227×(3.5)2×7=269.5 cm3



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 2πrh +2πr2 =616 cm2  and 2πrh = 308 cm2 2πrh +2πr2 - 2πrh =616-3082πr2=3082×227×r2=308r2=308×744=49r= 7 cmNow, 2πrh = 308 cm22×227×7×h =308h =30844=7 cm

 Volume of the cylinder=πr2h  cubic cm                                          =227×72×7                                           =1078 cm3

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
πr2h1 =l×b×h227×142×72=66×28×hh =22×14×2×7266×28=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 = πr2h 
                               =227×0.25×0.25×7 =22×0.25×0.25=1.375  cm3

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., 15×1000 cm3 = 330×11.375×15×1000=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 cm3 of gold is drawn into a wire of diameter 0.1 mm.
Here, r 0.1 mm=0.120cm=1200cm
 πr2h=1227×1200×1200×h =1h=40000×722=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×(2.5)2×100-π×1.52×100=227×100×6.25-2.25=22007×4=88007cm3
1 cm3 of cast iron weighs 21 g.

∴ Weight of 88007cm3cast iron = 88007×211000 kg=88×0.3=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                                         = πR2h -πr2h       (where R and r are the outer and inner radii, respectively)                                        =227×25×62-5.22                                         =227×25×36-27.04                                           =227×25×8.96=704  cm3

Page No 574:

Answer:


Height of the cylindrical tank, h = 1 m

Radius of the cylindrical tank, r1402 = 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

=2πrhr+h=2×227×0.7×1×0.7+1=2×227×0.7×1×1.7=7.48 m2
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 = πR2H=π×152×32 cm3

Radius of the glass, r = 3 cm

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

∴ Volume of orange juice in each glass = πr2h=π×32×8 cm3

Number of glasses of orange juice sold by the juiceseller 

=Volume of orange juice in the vasselVolume of orange juice in each glass=π×152×32π×32×8=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, r102 = 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

πR2-r2H=πr2hH=52×8.412.52-7.52H=210100H=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

90001000              (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, R1.42 = 0.7 m

Height of the water tank, H = 2.1 m

∴ Capacity of the water tank = πR2H=π0.72×2.1 m3

Speed of the water flow = 2 m/s

Radius of the pipe, r3.52 = 1.75 cm = 0.0175 m 

Area of the cross section of the pipe = πr2=π0.01752 m2

Volume of the water flowing out of the pipe in one second = Area of the cross section of the pipe × 2 m = π0.01752×2 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

π0.01752×2×t m3

Now,

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

π0.01752×2×t=π0.72×2.1t=0.72×2.10.01752×2t=1680 s
t=168060

⇒ 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 = 562 = 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 = πr2h=π×282×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.

227×282×h=32×22×14h=32×22×14×722×282h=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, r32 m

∴ Volume of the tube-well = πr2h=227×322×280 = 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 = 2πrh=2×227×32×280 = 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 = 132008.4 cm3

Let the length of the copper wire be l cm.

Radius of the copper wire, r42 = 2 mm = 0.2 cm            (1 cm = 10 mm)

∴ Volume of the copper wire = πr2l=227×0.22×l cm3

227×0.22×l=132008.4l=13200×722×0.22×8.4l=12500 cm


l=12500100          1 m = 100 cml=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

=Total cost of painting the inner curved surface of the cylindrical vasselRate of painting=330030=110 m2
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 = 2πrh=2×227×r×10

2×227×r×10=110r=110×72×22×10r=1.75 m
Thus, the inner radius of the base is 1.75 m.

(iii) Capacity of the vassel = πr2h=227×1.752×10=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)

2πRh-2πrh=882×227×14×R-r=88R-r=88×72×22×14R-r=1 cm         .....1
Volume of the tube = 176 cm3                (Given)

πR2-r2h=176227×R-r×R+r×14=176R+r=176×722×1×14            Using  1R+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 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.

2πr1=30r1=302π cm

h1 = 18 cm

V1=πr12h1=π302π2×18 cm3

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. 

2πr2=18r2=182π cm

h2 = 30 cm

V2=πr22h2=π182π2×30 cm3

Now,

V1V2=π302π2×18π182π2×30=53

⇒ 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 = πrl=227×5.25×10 = 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, r242 = 12 m

∴ Total surface area of the cone = πrr+l=227×12×12+21=227×12×33=87127 cm2

Thus, the total surface area of the cone is 87127 cm2.

Page No 588:

Answer:


Radius of the cone, r = 7 cm

Height of the cone, h = 24 cm

∴ Slant height of the cone, l=r2+h2=72+242=49+576=625 = 25 cm

Area of the sheet required to make one cap = Curved surface area of the cone = πrl=227×7×25 = 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:

Answer:


Slant height, l = 14 cm

Let the radius of the base be r cm.

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

πrl=308227×r×14=308r=308×722×14r=7 cm
∴ Total surface area of the cone = πrr+l=227×7×7+14=227×7×21 = 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:

Answer:


Slant height of the conical tomb, l = 25 m

Radius of the conical tomb, r = 142 = 7 m

∴ Curved surface area of the conical tomb = πrl=227×7×25 = 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:

Answer:


Radius of the conical tent, r = 24 m

Height of the conical tent, h = 10 m
 
∴ Slant height of the conical tent, l=r2+h2=242+102=576+100=676 = 26 m

Curved surface area of the conical tent = πrl=227×24×26 m2

The cost of 1 m2 canvas is ₹ 70.

∴ Cost of canvas required to make the tent

= Curved surface area of the conical tent × ₹ 70

=227×24×26×70

= ₹ 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 = 402 = 20 cm = 0.2 m         (1 m = 100 cm)

Height of each cone, h = 1 m
 
∴ Slant height of each cone, l=r2+h2=0.22+12=0.04+1=1.04 = 1.02 m

Curved surface area of each cone = πrl=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:

Answer:


Radius of the cone, r = 35 cm

Height of the cone, h = 12 cm
 
∴ Slant height of the cone, l=r2+h2=352+122=1225+144=1369 = 37 cm

(i) Volume of the cone = 13πr2h=13×227×352×12 = 15400 cm3

(ii) Curved surface area of the cone = πrl=227×35×37 = 4070 cm2

(iii) Total surface area of the cone = πrr+l=227×35×35+37=227×35×72 = 7920 cm2

Page No 588:

Answer:

Height of the cone, h = 6 cm
Slant height of the cone, l = 10 cm
Radius, r = l2-h2=100-36 =64 =8 cm

Volume of the cone = πr2h 
                              =13×3.14×82×6=401.92 cm3

Curved surface area of the cone = πrl 
                                                =3.14×8×10=251.2 cm2

∴ Total surface area = πrl+πr2 
                               =251.2 + 3.14×82=452.16 cm2

Page No 588:

Answer:

Radius of the conical pit, r = 3.52 m 

Depth of the conical pit, h = 12 m
 
∴ Capacity of the conical pit = 13πr2h=13×227×3.522×12 = 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 = 92 m = 4.5 m

Height of the heap, h = 3.5 m
 
​∴ Volume of the heap of wheat = 13πr2h=13×3.14×4.52×3.5 = 74.1825 m3

Now, 

Slant height of the heap, l=r2+h2=4.52+3.52=20.25+12.25=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

=πrl3.14×4.5×5.780.541 m2
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

πrl=550227×7×l=550l=55022=25 m

Now,

Height of the tent, h=l2-r2=252-72=625-49=576 = 24 m

∴ Volume of the tent = 13πr2h=13×227×72×24 = 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=r2+h2 =49 +576  =625   =25 m

 Curved surface area of the cone =πrl                                                       =227×7×25                                                 = 550 m2Here, area of the cloth =curved surface area of the cone = 550 m2

Width of the cloth = 2.5 m
∴ Length of the cloth = area of the clothwidth of the cloth=5502.5=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 = volume of the first conevolume of the second cone
                            
                                               =13π×3r2×h13π×r2×3h=9r2h3r2h=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., 2πrh = 8x and πrl = 5x πrh2+r2=5xHence 4π2r2h2=64x2 and π2r2(h2+r2)=25x2 Ratio of curved surface areas=4π2r2h2π2r2(h2+r2) =6425  4π2r2h2π2r2(h2+r2) =64254h2(h2+r2)=6425     25h2 =16h2 +16r2   9h2=16r2   r2h2 =916    rh=34                                             

∴ 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., 13π×1.62×3.6 =13π×1.22×hh  = 1.6×1.6×3.61.2×1.2=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
                                   = πrl +2πrH  
                                  =227×105253 +2×3 =227×1052×53+6=9735 m2

But area of the canvas = length × breadth

∴ Length of the canvas = areabreadth=97355=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
                                       = πr2h +13πr2H   
                                      =227×102280+13×42=227×280+14=227×100×294=92400 cm3

∴ Weight of the pillar = volume of the pillar × weight per cubic cm
                                 =92400×7.51000=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
                                                 =πr2h -13πr2h  =23πr2h=23×3.14×62×10=753.6 cm3

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)21000 cm3 = 196.4 cm3
Radius of the the conical vessel = 40 cm
Depth of the vessel = 24 cm
Volume of the vessel = 13π(20)224=10057.1 cm3
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
x=10057.1196.4=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 = πr2=227×52=5507 m2

Average area occupied by a student on the ground = 57 m2

∴ Number of students who can sit in the tent

=Area of the base of the conical tentAverage area occupied by a student on the ground=550757=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

πrl=165227×5×l=165l=165×722×5l=10.5 m
Let the height of the tent be h m.

h=l2-r2=10.52-52=110.25-25=85.25 ≈ 9.23 m

∴Volume of the tent = 13πr2h=13×227×52×9.23 ≈ 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 = 43πr3 
                     =43×227×3.5×3.5×3.5=179.67 cm3

∴ Surface area = 4πr2
                       =4×227×3.5×3.5=154 cm2

(ii) Radius of the sphere=4.2 cm
Now, volume = 43πr3 
                      =43×227×4.2×4.2×4.2=310.46 cm3

∴ Surface area = 4πr2 
                         =4×227×4.2×4.2=221.76 cm2

(iii) Radius of sphere=5 m
Now, volume = 43πr3 
                      =43×227×53=523.81 cm3

∴ Surface area = 4πr2 
                       =4×227×52= 314.29 cm2

Page No 600:

Answer:

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

 43πr 3=38808r3=38808×3×74×22=9261r =92613 =21 cm

∴ Surface area of the sphere =4πr2 
                                            =4×227×21×21=5544 cm2



Page No 601:

Answer:

Volume of the sphere = 606.375 m3

Then 43πr3=606.375r3=606.375×3×74×22=144.703r =5.25 m

∴ Surface area = 4πr2 
                      =4×227×5.25×5.25=346.5 m2

Page No 601:

Answer:


Let the radius of the sphere be r cm.

Surface area of the sphere = 154 cm2

4πr2=1544×227×r2=154r=154×74×22=12.25r=3.5 cm
∴ Volume of the sphere = 43πr3=43×227×3.53179.67 m3

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 4πr2=576πr2=5764=144r=12 cm

  Volume of the sphere=43×π×12×12×12 cm3                                           =2304π cm3

Page No 601:

Answer:

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

Volume of the cuboid =l×b×h                                         =12×11×9                                        = 1188 cm3

Radius of one lead shot = 3 mm= 0.32cm

 Volume of one lead shot =43×227×0.323                                                         =11×97000                                          =0.014 cm3

 Number of lead shots =volume of the cuboidvolume of one lead shot                                           =11880.014                                            =84857.1484857 

Page No 601:

Answer:

Radius of the sphere = 8 cm

Volume of the sphere = 43πr3 =43×227×83=2145.52 cm3
Radius of one lead ball = 1 cm
Volume of one lead ball = 43×227×13=4.19 cm3

∴ Number of lead balls = volume of the spherevolume of one lead ball=2145.524.19=512.05512

Page No 601:

Answer:

Radius of the solid sphere = 3 cm

Volume of the solid sphere = 43πr3 
                                         = 43×227×33 cm3

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

Volume of the spherical ball = 43×227×0.33 cm3

Now, number of small spherical balls = volume of the spherevolume of the spherical ball
                                                        =43π×2743π×0.33 =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 = 43πr3=43π(10.5)3 cm3
Radius of each smaller cone, r2 = 3.05 cm
Height of each smaller cone = 3 cm
Volume of each smaller cone =13πr22h=13π(3.05)2×3 cm3
Number of cones obtained = volume of the spherevolume of each smaller cone
                                        =43πr313πr22h=4×10.5×10.5×10.53.5×3.5×3=126.006126

∴ 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=43πr3=43π(6)3 cm3
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=πR2h=π(4)2×90 cm3

Number of spheres= volume of the cylindervolume of the sphere
                              =πR2h43πr3=42×90×34×63=12×90216 =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
43πR3=πr2h4×32=0.12×hh=4×90.1×0.1=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 43πR3=πr2h43×93=r2×10800r2=4×7293×10800=4×813×1200=9100r=310=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
43πr3=13πR2h4×7.83=R2×31.2R2=4×7.8×7.8×7.831.2=60.84R = 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
43πR3=13πr2h4×143=17.52×hh=4×14×14×1417.5×17.5=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
43π×33=43π×1.53+43π×23+43π×r327 =3.375+8+r3r3=27-11.375 = 15.625r = 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  = 4πr24π2r2=r24r2=14
                                                     = 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:
4πr24πR2=14rR=14=12

Now, ratio of the volumes = 43πr343πR3=rR3=123 =18

∴ 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
43πr3=π×122×6.75r3=144×6.75×34=36×6.5×3=729r = 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
43π×93=π×152×hh=4×7293×15×15=4×2725=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 =43×22763-43                                                            =8821×216-64                                                             =8821×152                                                             =636.95 cm3


 Outer surface area =4×227×6×6                                 =452.57 cm2

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 =  43πR3-r3 
                            =43π93-83=43×227×729-512=4×22×21721=88×313=27283cm3

Weight of the shell = volume of the shell × density per cubic cm
                             = 27283×4.54092 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

23π×93=13π×r2×72r2=2×9×9×972=814r =92=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  = volume of the hemispherical bowlvolume of a cylindrical shaped bottle
                                                                        =23π×93π×1.52×4=2×9×9×93×1.5×1.5×4=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
                                                                       =23π4.53-43 =23×227×91.125-64=23×227×27.125= 56.83 cm3

∴ 56.83 cm3 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 = 2πr2=2×227×5.252 = 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 = 10.52 = 5.25 cm

∴ Inner curved surface area of the bowl = 2πr2=2×227×5.252 = 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

=173.25×32100

= ₹ 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.

∴ 2r14 × 2R        (Given)

⇒ r=R4            .....(1)

Volume of the moonVolume of the earth=43πr343πR3=R43R3=164             [Using (1)]

Thus, the volume of the moon is 164 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 = 3πr2

Numercial value of volume of the solid hemisphere = 23πr3

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

23πr3=3πr22r=9 units
Thus, the diameter of the hemisphere is 9 units.



Page No 604:

Answer:

(c) 810 cm3

 Volume of the cuboid=l×b×h                                          = 15×12×4.5  cm3                                     =810 cm3

Page No 604:

Answer:

(b) 552 cm2

Total surface area of the cuboid =2lb+bh+lh cm2                                                 =2(12×9+8×9+12×8) cm2                                                 =2108+72+96 cm2                                                 =552 cm2



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(lb) = 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 ×weight of iron per cubic metre
                              =9×0.4×0.2×50=36 kg

Page No 605:

Answer:

(a) 15 m

Length of longest rod = diagonal of the room
                                 = diagonal of a cuboid
                                 =l2+b2+h2 =100+100+25 m=225 m= 15 m

Page No 605:

Answer:

(d) 11.2 cm
Maximum length of the pencil =  diagonal of the box
                                             =l2+b2+h2 =82+62+52 cm=64+36+25 cm=125 cm=11.2 cm

Page No 605:

Answer:

(b)  192

Number of planks =  volume of the pitvolume of 1 plank
                             =40×12×164×5×2=768040=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 = volume of the pitvolume of 1 plank

                                  =20×6×0.55×0.25×0.1=60×1000125=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 = volume of the wallvolume of 1 brick

                                             =800×600×22.525×11.25×6=108000001687.5=6400 

Page No 605:

Answer:

(b) 270

Number of persons  = volume of the hallvolume of air required by 1 person

                               =20×15×4.55=20×3×4.5=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 =1.5×30×3000 m3                                                                           =135000 m3

∴ Volume of water that runs into the sea per minute = 13500060=2250 m3

Page No 605:

Answer:

(d) 512  m3

Suppose that a m be the edge of the cube.
We have:
4a2=256a2=2564= 64a =8 m

∴ Volume of the cube = a3  m3  =83 m3 = 512  m3 

Page No 605:

Answer:

(c)  64 cm3

Let a cm be the edge of the cube.
We have:
6a2=96a2=16a=4 cm

∴ Volume of the cube = a3 cm3 =43 cm3 = 64 cm3 

Page No 605:

Answer:

(b) 384 cm2

Suppose that a cm is the edge of the cube.
We have:
a3 = 512a =5123 = 8 cm

Total surface area of cube =6a2 cm2=6×8×8 cm2= 384 cm2

Page No 605:

Answer:

(d)  103 cm103 cm
Length of the longest rod = body diagonal of the vessel
                                 =3a =3×10 =103 cm



Page No 606:

Answer:

(b)  384 cm2
We have:
3a=83a =8 cm

∴ Surface area of the cube = 6a2=6×8×8= 384 cm2

Page No 606:

Answer:

Let a be the edge of the cube.
Then the surface area is 6a2=S (say)

Now, increased edge = a+50100a = 150100a = 32a

Then, new surface area=632a2 =6×94a2 =94SIncrease in surface area =94S - S  = 54S Percentage increase in surface area =54SS = 54×100% = 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

a3=33+43+53     =27+64+125      =216     = 6 cm

∴ Lateral surface area of the new cube = 4a2 = 4×6×6 = 144 cm2

Page No 606:

Answer:

(d) 1000 m3

 Area of the land=2 sq hec=2000 sq mAmount of rainfall=5 cm=0.05 m Volume of the water=area of the land ×amount of rainfall                              =2000×0.05                             =1000 m3

Page No 606:

Answer:

(c) 1 : 9
Suppose that the edges of the cubes are a and b.
We have:
a3b3=127ab3=127ab=13

∴ Ratio of the surface areas = 6a26b2=ab2=132 =19

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 = 2a3=8a3  

Hence, the volume becomes 8 times the original volume.

Page No 606:

Answer:

(b)  396 cm3

 Volume of the cylinder=πr2h                                      =227×32×14                                      =22×9×2                                       = 396 cm3

Page No 606:

Answer:

(b)  1760 cm2

Curved surface area of the cylinder=2πrh                                                             =2×227×14×20                                                            =44×40                                                             =1760 cm2

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:
2πrh = 17602×227×14×h = 1760h=1760×744×14=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:

2πrh =2642×227×r×14 =264r=264×744×14=3 cm

  Volume of the cylinder=227×32×14                                =22×9×2                               = 396 cm3

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:
2πrh = 264  and  πr2h =924rh =2642πh =2642r×πNow, πr2h =π×r2×2642r×π=924r=924×2264r=7 m h =264×72×7×22=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 = 2π(2r)(5h)2π(3r)(3h)=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=π2r2×5hπ3r2×3h=4×59×3=2027

Page No 607:

Answer:

(d) 770 cm2
We have:
  r: h = 2 : 3
 rh=23h= 32r
Now, volume = 1617 cm3
πr2h = 1617227×r2×32r =1617r3=1617×1466=343r= 7 cm

h = 10.5 cm

Hence, total surface area=2πrh+2πr2                                           =227(2×7×10.5 +2×72)                                          =227147+2×49                                          =227×245                                           =770 cm2

Page No 607:

Answer:

(b) 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:

πr2h=πR2×2hr2R2=21rR=21r:R = 2 :1

Page No 607:

Answer:

(a) 1078 cm3

We have:

2πrh2πrh+2πr2=124πrh = 2πrh +2πr22πrh =2πr2rh=11

Also, 2πrh +2πr2=6162πr2+2πr2=6164πr2=616πr2=154r2=154πcm2

 Volume of the cylinder=πr2h  =π×154π×r=154×154×722=154×7=1078 cm3

Page No 607:

Answer:

(c) halved

Suppose that the new radius is 12r and the height is 2h.
 Volume =π×12r2×2h             =π×r24×2h             =12πr2h 

Page No 607:

Answer:

(b) 450

Number of coins = volume of cylinder to be formedvolume of a coin
 
                          =π×2.25×2.25×10π×0.75×0.75×0.2=225×225×575×75=450  

Page No 607:

Answer:

(d) 9 times

Let the new radius be 13r.
 Suppose that the new height is H.
The volume remains the same.
i.e., πr2h =π×13r2×Hh = 19H H =9h  The new height becomes nine times the original height.

Page No 607:

Answer:

(b)  1320  m2

Area covered by the roller in 1 revolution = 2πrh
 =2×227×42×100=44×600=26400 cm2  Area covered in 50 revolutions =26400 ×500 cm2                                                           =1320 m2

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:
π0.252×h =2.2×10×10×10h =2200×722×0.0625    =7000.0625    =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 = 2πrh  

Page No 607:

Answer:

(b) 550 cm2

   l =r2+h2  =72+242  =49+576   =625   =25 cm Curved surface area of the cone =πrl                                                  = 227×7×25                                                   =550 cm2

Page No 607:

Answer:

(d) (144π) cm3

Volume of the cone = 13πr2h 
                       =13π×62×12 =π×36×4=144π  cm3



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
L×B=πrlL×2.5=227×7×72+2422.5L=22×49 + 576L=22×252.5=220 m

Page No 608:

Answer:

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

Then 13×3.14×r2×15 = 1570r2 =15703.14×5=100r = 10 cm

Page No 608:

Answer:

(b)  7546 cm3

Radius of the cone, r=l2-h2                                 =282-212                                 =784-441                                 =343 cm

 Volume of the cone =13πr2h                            =13×227×343×21                             =22×343                             = 7546 cm3

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:
13×227×r2×24=1232r2=1232×2122×24r2=49r = 7 cm

 Curved surface area of the cone =πrl                                                        =227×7×72+242                                                         =22×49+576                                                         =22×25                                                         =550 cm2

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:
13π4r2h13π5r2H=1416r2h25r2H=14hH=2564 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 = 13πr2(2h) = 23πr2h = 2×volume of the cone
Increase in volume = 23πr2h-13πr2h=13πr2h
 Percentage increase=increase in volume initial volume ×100 %=13πr2h13πr2h×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:

πrl=2×(πR×2l)r=4RrR=41

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 = πr2h13πr2h=31

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:
πr2h=13πr2HhH=13

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=π3r2×2h13π4r2×3h=3×9×216×3=98

Page No 608:

Answer:

(d) 8 times

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

New volume of the cone=13π2r2×2h=83πr2h =8×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 = volume of the cylindervolume of 1 cone
                             =π×32×513π×0.12×1=3×9×50.01=13500 



Page No 609:

Answer:

(b) 15 m

Suppose that the height of the cone is h m.
Area of the ground = 11×4 = 44 m2

 πr2=44r2=44×722=14Also, 13πr2h =22013×227×14h = 220h = 220×2122×14=15 m
Hence, the height of the cone is 15 m.

Page No 609:

Answer:

(a) 32πr33

Volume of the sphere of radius 2r=43π2r3 =323πr3

Page No 609:

Answer:

(b)  4851 cm3
Volume of the sphere = 43πr3  
                                 =43×227×10.5×10.5×10.5=88×0.5×10.5×10.5 = 4851 cm3

Page No 609:

Answer:


(d)  5544 cm2

Surface area of sphere = 4πr2 
                                     = 4×227×21×21=5544 cm2

Page No 609:

Answer:

(c)  4851 cm3

Surface area = 1386 cm2
Let r cm be the radius of the sphere.
Then we have:
4πr2=1386r2=1386×74×22=110.25r = 10.5 cm Volume of the sphere=43πr3                                          =43×227×10.5×10.5×10.5                                             =4851 cm3

Page No 609:

Answer:

(a) (288π) m3

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

4πr2=144πr2=1444=36r =6 m Volume of the sphere = 43πr3                                                  =43π×6×6×6                                                   =288π  m3

Page No 609:

Answer:

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

43πr3=38808r3=38808×3×74×22=9261r=21 cm Curved surface area=4πr2 = 4×227×21×21                                                         = 5544 cm3

Page No 609:

Answer:

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

43πr343πR3=18rR3=18rR=12
∴ Ratio of surface area of spheres =4πr24πR2=rR2=122=14

Page No 609:

Answer:

(d) 64

Number of balls = volume of solid metal ballvolumeof 1 small ball
                        = 43π×8343π×23=5128=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
13π×2.1×2.1×8.4 =43πr3r3 =2.1×2.1×2.1r = 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
43π×63=π×0.12hh = 4×2163×0.01=28800 cm =288 m

Page No 609:

Answer:

(c)  126

Number of cones =volume of the spherevolume of 1 cone
                          =43π×10.5×10.5×10.513π×3.5×3.5×3=4×3×3×3.5 =126

Page No 609:

Answer:

(d) 84000

Number of lead shots =volume of cuboidvolume of 1 lead shot                                     =9×11×1243×227×0.15×0.15×0.15                                     =9×11×3×3×722×0.15×0.15×0.15                                     =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
43π×33 =π×0.12×hh= 4×90.001   =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
43π×6.33=13π×r2×25.2r2=4×6.3×6.3×6.325.2=39.69r = 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
43π×33 =43π×1.53+43π×23+43πr327 =3.375+8 +r3r3 = 27- 11.375 = 15.625r = 2.5 cm

Page No 610:

Answer:

(a)  1 : 4
Ratio of the surface areas of balloon = 2π×622π×122=36144=14

Page No 610:

Answer:

(d) 88 cm2

Suppose that the radii of the spheres are r cm and (7 − r) cm. Then we have:
43π7-r343πr3=6427(7-r)r=4321-3r=4r21=7rr=3 cm

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

Required difference=4×227×42  -4×227×32=4×227×16-9 = 4×227×7=88 cm2

Page No 610:

Answer:

(c) 54

Number of bottles  = volume of bowlvolume of 1 bottle
                         
                           =23×π×93π×1.52×4=2×7292.25×12=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 13π×r2×h = 23π×r2×HhH=21

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:

13πr2h23πr3πr2h =13πr323πr3πr3  =123  =   1 : 2 : 3

Page No 610:

Answer:

(c) 3 units

We have:

43πr3= 4πr213r =1r = 3 units



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