Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 17 Volumes And Surface Areas Of Solids are provided here with simple step-by-step explanations. These solutions for Volumes And Surface Areas Of Solids are extremely popular among Class 10 students for Maths Volumes And Surface Areas 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 10 Maths Chapter 17 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 10 Maths are prepared by experts and are 100% accurate.

Page No 786:

Answer:



As, volume of a cube=27 cm3edge3=27edge=273edge=3 cmThe length of the resulting cuboid, l=3+3=6 cm,its breadth, b=3 cm and its height, h=3 cmNow, the surface area of the resulting cuboid=2lb+bh+hl=26×3+3×3+3×6=218+9+18=2×45=90 cm2

Page No 786:

Answer:

As, volume of hemisphere=242512 cm323πr3=24251223×227×r3=48512r3=4851×3×72×2×22r3=441×3×72×2×2r3=21323r=212 cm

So, the curved surface area of the hemisphere=2πr2=2×227×212×212=693 cm2

Page No 786:

Answer:

As, the total surface area of the solid hemisphere=462 cm23πr2=4623×227×r2=462r2=462×73×22r2=49r2=49r=7 cm

Now, the volume of the solid hemisphere=23πr3=23×227×7×7×7=21563 cm3=71823 cm3
= 718.67 cm3

Page No 786:

Answer:

(i)
We have,the height of the cone, h=24 m, the base diameter of the cone, d=14 mAlso, the base radius of the cone, r=d2=142=7 mThe slant height of the cone, l=h2+r2=242+72=576+49=625=25 mThe curved surface area of the tent=πrl=227×7×25=550 m2The area of cloth required to make the tent=550 m2The length of the cloth=5505=110 mSo, the cost of cloth used=110×25=2750

(ii)
The ratio of radius and height of a solid right-circular cone is 5:12.
Let radius, r = 5x and height, h=12x.
Volume = 314 cm3.
13πr2h=31413×3.14×5x2×12x=314x3=314×33.14×5×5×12x3=1x=1
So, radius r = 5 cm and height h = 12 cm.
Using Pythagoras Theorem, slant height is given by l=r2+h2=25+144=169=13 cm
Total Surface Area of Cone =πrr+l=3.14×5×5+13=3.14×5×18=282.6 cm2.



Page No 787:

Answer:

Let r and R be the base radii, h and H be the heights, v and V be the volumes of the two given cones.We have,2r2R=45 or rR=45         .....iandvV=1413πr2h13πR2H=14r2hR2H=14rR2×hH=14452×hH=14          Using i1625×hH=14hH=1×254×16hH=2564 h:H=25:64

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

Page No 787:

Answer:

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

As, the area of the base=1.54 km2πr2=1.54227×r2=1.54r2=1.54×722r2=0.49r=0.49r=0.7 km

Now,h=l2-r2=2.52-0.72=6.25-0.49=5.76=2.4 km

So, the height of the mountain is 2.4 km.

Page No 787:

Answer:

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

We have,(r+h)=37 mAs, the total surface area of the cylinder=1628 m22πrr+h=16282×227×r×37=1628r=1628×72×22×37r=7 mSo, h=37-7=30 m

Now, the volume of the solid cylinder=πr2h=227×7×7×30=4620 m3

Page No 787:

Answer:

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

We have,R=2r and the surface area of the given sphere=2464 cm2i.e. 4πr2=2464 cm2The surface area of the new sphere=4πR2=4π2r2=4π4r2=44πr2=4×2464=9856 cm2

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

Page No 787:

Answer:


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

The slant height of the cone, l=r2+H2=152+2.752=225+7.5625=232.5625=15.25 m

The area of the canvas used in making the tent=CSA of cylinder+CSA of cone=2πrh+πrl=πr2h+l=227×152×5.5+15.25=227×1511+15.25=227×15×26.25=1237.5 m2

As, the width of the canvas = 1.5 m

So, the length of the canvas=237.51.5=825 m

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

Page No 787:

Answer:



Radius of the cylinder = 14 m
Radius of the base of the cone = 14 m
Height of the cylinder (h) = 3 m
Total height of the tent = 13.5 m
Surface area of the cylinder=2πrh=2×227×14×3 m2=264 m2

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

Surface area of the cone = πrr2+h2
 
=πrr2+h2=227×14×142+10.52 m2=44×196+110.25 m2 =44×306.25 m2=44×17.5 m2=770 m2

Total surface area =(264+770) m2=1034 m2

∴ Cost of cloth=Rs 1034×80=Rs 82720 

Page No 787:

Answer:



Radius of the cylinder = 52.5 m
Radius of the base of the cone = 52.5 m
Slant height (l) of the cone = 53 m
Height of the cylinder (h) = 3 m
Curved surface area of the cylindrical portion=2πrh=2×227×52.5×3 m2=990 m2
Curved surface area of the conical portion=πrl=227×52.5×53 m2=8745 m2

Thus, the total area of canvas required for making the tent=8745+990 m2=9735 m2

Page No 787:

Answer:



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

Curved surface area of the cylinder =2πrh=2×227×2.5×21=330 m2
Radius of the cone = 2.5 m
Slant height of the cone = 8 m
Curved surface area of the cone=πrl=227×2.5×8=62.86 m2

Area of circular base=πr2=227×2.5×2.5=19.643 m2
∴ Total surface area of rocket=330+62.86+19.643=412.503 m2

Page No 787:

Answer:


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

Now,Volume of solid=Volume of cone+Volume of hemisphere=13πr2h+23πr3=13πr2h+2r=13×227×3.5×3.5×6+2×3.5=13×227×3.5×3.5×6+7=13×227×3.5×3.5×13=500.53166.83 cm3

So, the volume of the solid is 166.83 cm3.

Page No 787:

Answer:

The radius of hemisphere as well as that of base of cone is r = 3.5 cm.
The height of toy = 15.5 cm.
The height of hemisphere, h = 3.5 cm and height of cone = 15.5-3.5 = 12 cm.
Using Pythagoras Theorem, slant height of cone is l=3.52+122=12.25+144=156.25=12.5 cm.
Therefore, the Total surface area of toy = Curved surface of cone + Curved surface of hemisphere
=πrl+2πr2=227×3.5×12.5+2×227×3.52=2752+77=275+1542=4292=214.5 cm2.

Page No 787:

Answer:



We have,Base radius of cone=Base radius of hemisphere=r=72=3.5 cm,As, the volume of the toy=231 cm3Volume of cone+Volume of hemisphere=23113πr2h+23πr3=23113πr2h+2r=23113×227×3.5×3.5×h+2×3.5=23138.53×h+7=231h+7=231×338.5h+7=18h=18-7h=11 cmSo, the height of the toy=h+r=11+3.5=14.5 cm

Page No 787:

Answer:


We have,the base radius of the cylindrical container, R=6 cm,the height of the container, H=15 cm,Let the base radius and the height of the ice-cream cone be r and h, respectively.Also, h=4rNow, the volume of the cylindrical container=πR2H=227×6×6×15=118807 cm3the volume of the ice-cream distributed to 10 children=118807 cm310×Volume of a ice-cream cone=11880710×Volume of the cone+Volume of the hemisphere=11880710×13πr2h+23πr3=11880710×13πr2×4r+23πr3=11880710×43πr3+23πr3=11880710×63πr3=11880710×2×227×r3=118807r3=11880×77×10×2×22r3=27r=273 r=3 cm

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

Page No 787:

Answer:

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

Volume of the hemisphere =23πr3=23×227×10.5×10.5×10.5=2425.5 cm3

Height of the cylinder =Total height of the vessel-Radius of the hemisphere=14.5-10.5=4 cm
Volume of the cylinder =πr2h=227×10.5×10.5×4=1386 cm3

Total volume =Volume of hemispherical part+Volume of cylinder=1386+2425.5=3811.5 cm3



Page No 788:

Answer:


We have,the total height of the toy=90 cm andthe radius of the toy, r=422=21 cmAlso, the height of the cylinder, h=90-42=48 cmNow, the total surface area of the toy=CSA of cylinder+2×CSA of a hemisphere=2πrh+2×2πr2=2πrh+2r=2×227×21×48+2×21=44×3×48+42=44×3×90=11880 cm2So, the cost of painting the toy=11880×70100=Rs 83,16

Page No 788:

Answer:


We have,the total height of the capsule=14 mm andthe radius of the capsule, r=52 mmAlso, the height of the cylinder, h=14-2×52=14-5=9 mmNow, the surface area of the capsule=CSA of the cylinder+2×CSA of a hemisphere=2πrh+2×2πr2=2πrh+2r=2×227×52×9+2×52=227×5×9+5=227×5×14=220 mm2

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

Page No 788:

Answer:

The height h of cylinder = 20 cm and diameter of its base = 7 cm.
The radius r of its base = 3.5 cm.
Curved surface area of cylinder = 2πrh=2×227×3.5×20=440 cm2.

Now, curved surface area of one hemisphere =2πr2=2×227×3.52=77 cm2.

Total surface area of the article = Curved surface area of cylinder + 2(curved surface area of hemisphere)
=440+277=440+154=594 cm2.

Page No 788:

Answer:

The object is shown in the figure below.

Radius of hemisphere = 2.1 cm
Volume of hemisphere=23πr3=23×227×2.1×2.1×2.1=19.404 cm3

Height of cone = 4 cm
Volume of cone =13πr2h=13×227×2.1×2.1×4=18.48 cm3
Volume of the object=18.48+19.404=37.884 cm3

Volume of cylindrical tub=πr2h=227×5×5×9.8=770 cm3

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

Volume of water left in the tub = 770-37.884=732.116 cm3

Page No 788:

Answer:




Volume of the solid left = Volume of cylinder - Volume of cone=πr2h-13πr2h=23×227×8×6×6=603.428 cm3

The slant length of the cone, l=r2+h2=36+64=10cm

Total surface area of final solid = Area of base circle + Curved surface area of cylinder + Curved surface area of cone
=πr2+2πrh+πrl=πr(r+2h+l)=227×6×6+16+10=603.42 cm2

Page No 788:

Answer:


We have,the height of the cone=the height of the cylinder=h=2.8 cm andthe radius of the base, r=4.22=2.1 cmThe slant height of the cone, l=r2+h2=2.12+2.82=4.41+7.84=12.25=3.5 cmNow, the total surface area of the remaining solid=CSA of cylinder+CSA of cone+Area of a base=2πrh+πrl+πr2=πr2h+l+r=227×2.1×2×2.8+3.5+2.1=22×0.3×5.6+5.6=6.6×11.2=73.92 cm2

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

Page No 788:

Answer:

We have,the height of the cylinder, H=14 cm,the base radius of cylinder, R=72 cm,the base radius of each conical holes, r=2.1 cm andthe height of each conical holes, h=4 cmVolume of the remaining solid=Volume of the cylinder-Volume of 2 conical holes=πR2H-2×13πr2h=227×72×72×14-23×227×2.1×2.1×4=539-36.96=502.04 cm3

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

Page No 788:

Answer:

We have,the base radius of the cylinder, R=3 cm,the height of the cylinder, H=5 cm,the base radius of the conical hole, r=32 cm andthe height of the conical hole, h=89 cmNow,Volume of the cylinder, V=πR2H=π×32×5=45π cm3Also,Volume of the cone removed from the cylinder, v=13πr2h=π3×322×89=2π3 cm3So, the volume of metal left in the cylinder, V'=V-v=45π-2π3=133π3 cm3 The required ratio=V'v=133π32π3=1332=133:2

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

Page No 788:

Answer:


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

Volume of spherical part of the vessel =43πr3=43×227×10.5×10.5×10.5=4851 cm3
Diameter of cylinder = 4 cm
Radius of cylinder = 2 cm
Height of cylinder = 7 cm

Volume of the cylindrical part of the vessel =πr2h=227×2×2×7=88 cm3

Total volume of the vessel=4851+88=4939 cm3

Page No 788:

Answer:

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

Volume of hemisphere=23πr3=23×227×3.5×3.5×3.5=89.83 cm3

Volume of the cylinder=πr2h=227×3.5×3.5×6.5=250.25 cm3

Height of cone =12.8-6.5=6.3 cm

Volume of the cone=13πr2h=13×227×3.5×3.5×6.3=80.85 cm3

Total volume=89.83+250.25+80.85=420.93 cm3



Page No 789:

Answer:


We have,the edge of the cubical piece, a=21 cm andthe radius of the hemisphere, r=a2=212 cmThe surface area of the remaining piece=TSA of cube+CSA of hemisphere-Area of circle=6a2+2πr2-πr2=6a2+πr2=6×21×21+227×212×212=21×216+227×4=21×216+1114=21×2184+1114=21×3952=2992.5 cm2

Also,Volume of the remaining piece=Volume of the cube-Volume of the hemisphere=a3-23πr3=21×21×21-23×227×212×212×212=21×21×21×1-23×227×12×12×12=21×21×21×11-1142=21×21×21×42-1142=21×21×312=6835.5 cm3

Page No 789:

Answer:

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

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

Radius (r) of hemispherical part =  = 3.5 cm

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

− Area of base of hemispherical part

= 6 (Edge)2  = 6 (Edge)2 +



(ii)

We have,the edge of the cubical block, a=10 cmThe largest diameter of the hemisphere=a=10 cmAlso, the radius of the hemisphere, r=102=5 cmNow,Total surface area of the solid=TSA of cube+CSA of hemisphere-Area of circle=6a2+2πr2-πr2=6a2+πr2=6×10×10+3.14×5×5=600+78.5=678.5 cm2As, the rate of painting the solid=5 per 100 cm2So, the cost of painting the solid=678.5×510033.92

hence, the cost of painting the total surface area of the solid is ₹33.92.

Page No 789:

Answer:


We have,the base radius of cone=the base radius of cylinder=the base radius of hemisphere=r=5 cm,the height of the cylinder, H=13 cm,the total height of the toy=30 cmAlso, the height of the cone, h=30-13+5=12 cmThe slant heigt of the cone, l=r2+h2=52+122=25+144=169=13 cmNow, the surface area of the toy=CSA of cone+CSA of cylinder+CSA of hemisphere=πrl+2πrH+2πr2=πrl+2H+2r=227×5×13+2×13+2×5=227×5×13+26+10=227×5×49=770 cm2

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

Page No 789:

Answer:


We have,the height of the glass, h=16 cm andthe base radius of cylinder=the base radius of hemisphere, r=72 cmNow,The apparent capacity of the glass=Volume of the cylinder=πr2h=227×72×72×16=616 cm3Also,The actual capacity of the glass=Volume of cylinder-Volume of hemisphere=616-23πr3=616-23×227×72×72×72=616-5396=31576 cm3526.17 cm3

Page No 789:

Answer:


We have,the base radius of the conical part, r=52=2.5 cm,the base radius of the cylindrical part, R=42=2 cm,the total height of the toy=26 cm,the height of the conical part, h=6 cmAlso, the height of the cylindrical part, H=26-6=20 cmAnd, the slant height of the conical part, l=r2+h2=2.52+62=6.25+36=42.25=6.5 cmNow,The area to be painted by red colour=CSA of cone+Area of base of conical part-Area of base of cylindrical part=πrl+πr2-πR2=227×2.5×6.5+227×2.5×2.5-227×2×2=227×16.25+227×6.25-227×4=227×16.25+6.25-4=227×18.558.14 cm2Also,The area to be painted by white colour=CSA of cylinder+Area of base of cylinder=2πRH+πR2=πR2H+R=227×2×2×20+2=227×2×42=264 cm2



Page No 809:

Answer:

The volume of solid metallic cuboid is 9×8×2=144 m3.
This cuboid has been recasted into solid cubes of edge 2 m whose volume is given by 23=8 m3.
Therefore, the total number of cubes so formed =the volume of solid metallic cuboidthe volume of solid cubes=1448=18.

Page No 809:

Answer:

We have,the base radius of the cone, r=5 cm andthe height of the cone, h=20 cmLet the radius of the sphere be R.As,Volume of sphere=Volume of cone43πR3=13πr2hR3=πr2h×33×4πR3=r2h4R3=5×5×204R3=125R=1253R=5 cmDiameter of the sphere=2R=2×5=10 cm

So, the diameter of the sphere is 10 cm.

Page No 809:

Answer:

We have,the radii r1=6 cm, r2=8 cm and r3=10 cmLet the radius of the resulting sphere be R.As,Volume of resulting sphere=Volume of three metallic spheres43πR3=43πr13+43πr23+43πr3343πR3=43πr13+r23+r33R3=r13+r23+r33R3=63+83+103R3=216+512+1000R3=1728R=17283R=12 cm

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

Page No 809:

Answer:

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

Volume=13πr2h=13π×12×12×24=48×24×π cm3

Radius of each ball = 3 cm
Volume of each ball=43πr3=43π×3×3×3=36π cm3
Total number of balls formed by melting the cone=Volume of coneVolume of a ball=48×24π36π=32

Page No 809:

Answer:

We have,the internal base radius of spherical shell, r1=3 cm,the external base radius of spherical shell, r2=5 cm andthe base radius of solid cylinder, r=142=7 cmLet the height of the cylinder be h.As,Volume of solid cylinder=Volume of spherical shellπr2h=43πr23-43πr13πr2h=43πr23-r13r2h=43r23-r137×7×h=4353-3349×h=43125-27h=43×9849 h=83 cm

So, the height of the cylinder is 83 cm.



Page No 810:

Answer:

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

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

Volume of hemispherical shell=23π53-33=1963×227=6163 cm3

Radius of cone = 7 cm
Let the height of the cone be h cm.
Volume of cone =13πr2h=13×227×7×7h=154h 3cm3

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

154h3=6163h=616154=4 cm

Page No 810:

Answer:

We have,the radius of the copper rod, R=22=1 cm,the height of the copper rod, H=10 cm andthe height of the wire, h=10 m=1000 cmLet the radius of the wire be r.As,Volume of the wire=Volume of the rodπr2h=πR2Hr2h=R2Hr2×1000=1×10r2=101000r2=1100r=1100r=110r=0.1 cmThe diameter of the wire=2r=2×0.1=0.2 cm The thickness of the wire=0.2 cm

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

Page No 810:

Answer:

Inner diameter of the bowl = 30 cm
Inner radius of the bowl=30 cm2=15 cm
Inner volume of the bowl = Volume of liquid =23πr3=23×π×153 cm3

Radius of each bottle = 2.5 cm
Height = 6 cm

Volume of each bottle=πr2h=π×52×52×6=75π2cm3

Total number of bottles required =23π×15×15×1575π2=2π×15×15×15×23×75π=15×4=60 

Page No 810:

Answer:

Diameter of sphere = 21 cm
Radius of sphere =212cm

Volume of sphere=43πr3=4×21×21×21π3×2×2×2=21×21×21π3×2 cm3

Diameter of the cone = 3.5 cm
Radius of the cone =3.52=74cm
Height = 3 cm

Volume of each cone=13πr2h=13π×3×742=742π cm3

Total number of cones=Volume of sphereVolume of a cone=21×21×21π3×2742π=21×21×21×π×4×43×2×π×7×7=504 

Page No 810:

Answer:

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

Volume of ball =43πr3=43π×143 cm3

Diameter of base of cone = 35 cm

Radius of base of cone =352cm
Let the height of the cone be h cm.

Volume of cone =13πr2h=13π×3522×h cm3

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

Or, 43π×143=13π×3522×h h=43π×14313π×3522=4×14×14×14×2×235×35=35.84 cm

Page No 810:

Answer:

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

=43π×33-43π×323+43π×23=43π×3×3×3-43π×32×32×32+43π×2×2×2=4π×3×3-π×3×32+43π×2×2×2=36π-π92+323π=36×6-9×3-32×26π=216-27-646π=125π6

Therefore,

 43πr3=125π6Or, r=125×34×63=12583=52=2.5 cm

Page No 810:

Answer:

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

Volume of the shell =43π123-93=43π1728-729=43π×999=4π×333 cm3

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

Volume of cylinder =πr2h=37πr2 cm3

Volume of the shell = Volume of cylinder 

Or, 4π×333=37πr2 r2=4×33337=4×9 r=4×9=36=6 cm

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

Page No 810:

Answer:

Radius of hemisphere = 9 cm

Volume of hemisphere =23πr3=23π×9×9×9 cm3

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

Volume of the cone =13πr2h=13πr2×72 cm3

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

13πr2×72=23π×9×9×9r2=2×9×9×972=814r=814=92=4.5 cm

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

Page No 810:

Answer:

Diameter of the spherical ball= 21 cm

Radius of the ball =212 cm
Volume of spherical ball  =43πr3=43×227×212×212×212=11×21×21=4851 cm3
Volume of each cube =13=1 cm3

Number of cubes = Volume of spherical ballVolume of each cube=48511=4851

Page No 810:

Answer:

Radius of the sphere = R = 8 cm
Volume of the sphere =43πR3=43π×8×8×8=43π×512 cm3

Radius of each new ball = r = 1 cm
Volume of each ball =43πr3=43π×1×1×1=43π×1 cm3

Total number of new balls that can be made =Volume of sphereVolume of each ball=43π×51243π×1=512

Page No 810:

Answer:

Radius of solid sphere = 3 cm
Volume of the sphere=43πr3=43π×3×3×3 cm3

Radius of each new ball = 0.3 cm
Volume of each new ball =43πr3=43π×310×310×310 cm3

Total number of balls =43π×3×3×343π×310×310×310=1000

Page No 810:

Answer:

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

Volume of sphere=43πr3=43π×21×21×21 cm3

Diameter of wire = 2.8 cm
Radius of wire = 1.4 cm
Let the length of the wire be cm.
Volume of the wire=πr2l=π×1.4×1.4×l

The volume of the sphere is equal to the volume of the wire. 
​Therefore,
π×1.4×1.4×l=43π×21×21×21l=4×21×21×213×1.4×1.4=6300 cm=63 m

So, the wire is 63 m long.

Page No 810:

Answer:

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

Volume of sphere=43πr3=43π×9×9×9 cm3

Length of wire = 108 m = 10800 cm
Let radius of the wire be r cm.
Volume of the wire=πr2l=π×r2×10800 cm3

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

π×r2×10800=43π×9×9×9r2=4×9×9×93×10800=4×9×93×4×3=0.09r=0.09=0.3 cmThus, d=2r=2×0.3 cm=0.6 cm

The diameter of the wire is 0.6 cm.



Page No 811:

Answer:

We have,the radius of the hemispherical bowl, R=9 cm andthe internal base radius of the cylindrical vessel, r=6 cmLet the height of the water in the cylindrical vessel be h.As,Volume of water in the cylindrical vessel=Volume of hemispherical bowlπr2h=23πR3r2h=23R36×6×h=23×9×9×9h=23×9×9×96×6h=272 h=13.5 cm

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

Page No 811:

Answer:

We have,the radius of the hemispherical tank, r=32 mVolume of the hemispherical tank=23πr3=23×227×32×32×32=9914 m3Now,Volume of half tank=12×9914=9928 m3=9928 kL=9900028 LAs, the rate of water emptied by the pipe=257 L/sSo, the time taken to empty half the tank=9900028257=9900025×4=990 s=99060 min=16.5 min=16 min 30 s

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

Page No 811:

Answer:

We have,the length of the roof, l=44 m,the width of the roof, b=20 m,the height of the cylindrical tank, H=3.5 m andthe base radius of the cylindrical tank, R=42=2 mLet the height of the rainfall be h.Now,Volume of rainfall=Volume of cylindrical tanklbh=πR2H44×20×h=227×2×2×3.5h=227×2×2×3.544×20h=120 mh=10020 cm h=5 cm

So, the height of the rainfall is 5 cm.

Page No 811:

Answer:

We have,the length of the roof, l=22 m,the width of the roof, b=20 m,the base radius of the cylindrical vessel, R=22=1 m andthe height of the cylindrical vessel, H=3.5 mLet the height of the rainfall be h.Now,Volume of rainfall=Volume of rain water collected in the cylindrical vessellbh=45×Volume of cylindrical vessel22×20×h=45×πR2H440h=45×227×1×1×3.5h=45×227×3.5440h=0.02 m h=2 cm

So, the height of the rainfall is 2 cm.

Page No 811:

Answer:

We have,height of cone, h=60 cm,the base radius of cone, r=30 cm,the height of cylinder, H=180 cm andthe base radius of the cylinder, R=60 cmNow,Volume of water left in the cylinder=Volume of cylinder-Volume of cone=πR2H-13πr2h=227×60×60×180-13×227×30×30×60=227×30×30×602×2×3-13=227×5400012-13=227×54000×353=1980000 cm3=19800001000000 m3=1.98 m3

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

Page No 811:

Answer:

We have,the internal radius of the cylindrical pipe, r=22=1 cm andthe base radius of cylindrical tank, R=40 cm,Also, the rate of water flow, h=0.4 m/s=40 cm/sLet the rise in level of water be H.Now,The volume of water flowing out of the cylindrical pipe in 1 sec=πr2h =π×1×1×40 =40π cm3So, the volume of water flowing out of the cylindrical pipe in half an hour (30 min)=40π×60×30=72000π cm3As,Volume of water in the cylindrical tank=Volume of standing water in cylindrical pipe for half an hourπR2H=72000πR2H=7200040×40×H=72000H=7200040×40 H=45 cm

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

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

Page No 811:

Answer:

We have,Speed of the water flowing through the pipe, H=6 km/h=600000 cm3600 s=5003 cm/s,Radius of the pipe, R=142=7 cm,Length of the rectangular tank, l=60 m=6000 cm,Breadth of the rectangular tank, b=22 m=2200 cm andRise in the level of water in the tank, h=7 cmNow,Volume of the water in the rectangular tank=lbh=6000×2200×7=92400000 cm3Also,Volume of the water flowing through the pipe in 1 s=πR2H=227×7×7×5003=770003 cm3So,The time taken=Volume of the water in the rectangular tankVolume of the water flowing through the pipe in 1 s=92400000770003=92400×377=3600 s=1 hr

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

Page No 811:

Answer:

Width of the canal = 5.4 m
Depth of the canal = 1.8 m
Height of the standing water needed for irrigation = 10 cm = 0.1 m
Speed of the flowing water = 25 km/h = 2500060=12503 m/min
Volume of water flowing out of the canal in 1 min
= Area of opening of canal × 12503
5.4×1.8×12503
= 4050 m3
∴ Volume of water flowing out of the canal in 40 min = 40 × 4050 m3 = 162000 m3
Now,
Area of irrigation=Volume of water flowing out from canal in 40 minHeight of the standing water needed for irrigation=1620000.1=1620000 m2=162 hectare        1 hectare=10000 m2
Thus, the area irrigated in 40 minutes is 162 hectare.

Page No 811:

Answer:

We have,the radius of the cylindrical tank, R=122=6 m=600 cm,the depth of the tank, H=2.5 m=250 cm,the radius of the cylindrical pipe, r=252=12.5 cm,speed of the flowing water, h=3.6 km/h=360000 m3600 s=100 cm/sNow,Volume of water flowing out from the pipe in a hour=πr2h=227×12.5×12.5×100 cm3Also,Volume of the tank=πR2H=227×600×600×250 cm3So, the time taken to fill the tank=Volume of the tankVolume of water flowing out from the pipe in a hour=227×600×600×250227×12.5×12.5×100=5760 s=57603600=1.6 hAlso, the cost of water=0.07×227×6×6×2.5=19.80

 

Page No 811:

Answer:

We have,Radius of cylindrical pipe, r=72 cm andThe rate of flow of water=192.5 L/min=192.5 L1 min=192.5×1000 cm31 min           As, 1 L=1000 cm3=192500 cm3/minThe volume of water flowing out from the cylindrical pipe in 1 min=192500 cm3Now, the rate of flow of water in the pipe=The volume of water flowing out from the cylindrical pipe in 1 minπr2=192500227×72×72=192500×277=5000 cm/min=5000×601×100000 km/hr=3 km/hr

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

Page No 811:

Answer:

We have,the radius of spherical marble, r=1.42=0.7 cm andthe radius of the cylindrical vessel, R=72 cm=3.5 cmLet the rise in the level of water in the vessel be H.Now,Volume of water rised in the cylindrical vessel=Volume of 150 spherical marblesπR2H=150×43πr3R2H=200r33.5×3.5×H=200×0.7×0.7×0.7H=200×0.7×0.7×0.73.5×3.5 H=5.6 cm

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

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



Page No 812:

Answer:

Diameter of each marble = 1.4 cm
Radius of each marble = 0.7 cm
Volume of each marble =43πr3=43π×0.73 cm3

The water rises as a cylindrical column.
Volume of cylindrical column filled with water=πr2h=π×722×5.6 cm3

Total number of marbles

=Volume of cylindrical water columnVolume of marble=π×722×5.643π×0.73=7×7×5.6×32×2×4×0.7×0.7×0.7=150

Page No 812:

Answer:



We have,Radius of well, R=102=5 m,Depth of the well, H=14 m andWidth of the embankment=5 m,Also, the outer radius of the embankment, r=R+5=5+5=10 mAnd, the inner radius of the embankment=R=5 mLet the height of the embankment be h.Now,Volume of the embankment=Volume of the earth taken outVolume of the embankment=Volume of the wellπr2-πR2h=πR2Hπr2-R2h=πR2Hr2-R2h=R2H102-52h=5×5×14100-25h=25×1475h=25×14h=25×1475 h=143 m

So, the height of the embankment is 143 m.

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

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

Page No 812:

Answer:

We have,Length of the fiel, l=35 m,Width of the field, b=22 m,Depth of the well, H=8 m andRadius of the well, R=142=7 m,Let the rise in the level of the field be h.Now,Volume of the earth on remaining part of the field=Volume of earth dug outArea of the remaining field×h=Volume of the wellArea of the field-Area of base of the well×h=πR2Hlb-πR2×h=πR2H35×22-227×7×7×h=227×7×7×8770-154×h=1232616×h=1232h=1232616 h=2 m

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

Page No 812:

Answer:

We have,Diameter of the coppe wire, d=6 mm=0.6 cm,Radius of the copper wire, r=0.62=0.3 cm,Length of the cylinder, H=18 cm,Radius of the cylinder, R=492 cmThe number of rotations of the wire on the cylinder=Length of the cylinder, HDiameter of the copper wire, d=180.6=30The circumference of the base of the cylinder=2πR=2×227×492=154 cmSo, the length of the wire, h=30×154=4620 cm=46.2 mNow, the volume of the wire=πr2h=227×0.3×0.3×4620=1306.8 cm3Also, the weight of the wire=Volume of the wire×Density of the wire=1306.8×8.8=11499.84 g11.5 kg

Page No 812:

Answer:


We have,In ABC, B=90°, AB=l1=15 cm and BC=l2=20 cmLet OD=OB=r, AO=h1 and CO=h2Using Pythagoras theorem,AC=AB2+BC2=152+202=225+400=625h=25 cmAs, arABC=12×AC×BO=12×AB×BCAC×BO=AB×BC25r=15×20r=15×2025r=12 cmNow,Volume of the double cone so formed=Volume of cone 1+Volume of cone 2=13πr2h1+13πr2h2=13πr2h1+h2=13πr2h=13×3.14×12×12×25=3768 cm3Also,Surace area of the solid so formed=CAS of cone 1+CSA of cone 2=πrl1+πrl2=πrl1+l2=227×12×15+20=227×12×35=1320 cm2

Page No 812:

Answer:

Diameter of cylinder (d) = 2 m
Radius of cylinder (r) = 1 m
Height of cylinder (H) = 5 m
Volume of cylindrical tank, Vc = πr2H = π×12×5=5π m

Length of the park (l) = 25 m
Breadth of park (b) = 20 m
height of standing water in the park = h​
Volume of water in the park = lbh = 25×20×h

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

5π=25×20×h5π25×20=hh=π100 mh=0.0314 m

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



Page No 822:

Answer:

We have,Height of the frustum, h=14 cm,Base radii, R=162=8 cm and r=122=6 cmThe capacity of the glass=Volume of the frustum=13πhR2+r2+rR=13×227×14×82+62+8×6=13×22×2×64+36+48=443×148=65123 cm32170.67 cm3

So, the capacity of the glass is 2170.67 cm3.

Page No 822:

Answer:

We have,Height, h=8 cm,Base radii, R=18 cm and r=12 cmAlso, the slant height, l=R-r2+h2=18-122+82=62+82=36+64=100=10 cmNow,Total surface area of the solid frustum=πR+rl+πR2+πr2=3.14×18+12×10+3.14×182+3.14×122=3.14×30×10+3.14×324+3.14×144=3.14×300+324+144=3.14×768=2411.52 cm2

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

Page No 822:

Answer:

We have,Height, h=24 cm,Upper base radius, R=14 cm and lower base radius, r=7 cmAlso, the slant height, l=R-r2+h2=14-72+242=72+242=49+576=625=25 cmi Volume of the bucket=13πhR2+r2+Rr=13×227×24×142+72+14×7=227×8×196+49+98=1767×343=8624 cm3So, the volume of water which can completely fill the bucket is 8624 cm3.ii Surface area of the bucket=πR+rl+πr2=227×14+7×25+227×7×7=227×21×25+22×7=22×3×25+22×7=1650+154=1804 cm2So, the area of the metal sheet used to make the bucket is 1804 cm2.



Page No 823:

Answer:

We have,Height, h=24 cm,Upper radius, R=20 cm andLower radius, r=8 cmNow,Volume of the container=13πhR2+r2+Rr=13×227×24×202+82+20×8=1767×400+64+160=1767×624=1098247 cm3=109.8247 L         As, 1000 cm3=1 LSo, the cost of the milk in the container=109.8247×21=329.472329.47

Hence, the cost of milk which can completely fill the container is ₹329.47.

Page No 823:

Answer:

Let r=8 cm, R=20 cm, h=16 cm.
l=R-r2+h2=20-82+162=144+256=400l=20 cm.
(i)
Total metal sheet required to make the container = Curved surface area of frustum + Area of the base
=πr+Rl+πr2=π8+2020+π82=560π+64π=624π=1961.14 cm2.
The cost for 100 cm2 of sheet = Rs. 10.
The cost of 1 cm2 of sheet = 10100=Rs. 0.1.
The cost of 1961.14 cm2 of sheet = 1961.14×0.1=Rs. 196.11
DISCLAIMER: The answer calculated above is not matching with the answer provided in the textbook.

(ii)
The volume of frustum =
 πh3r2+R2+rR=227×16382+202+160=227×16364+400+160=227×163×624=732167 cm3.
we know that 1 l=1000 cm3 or 1 cm3 = 0.001 l.
Volume=732167×0.001=732167×11000=73.2167 l.
Cost of milk is Rs. 35 per litre.
Hence, the cost at which this frustum can be filled = 73.2167×35=Rs 366.08.

Page No 823:

Answer:

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

Using the formula for height of a frustum:

Height =  h =
=l2-R-r2=102-33-272=100-62=100-36=64=8 cm

Capacity of the frustum
 =13πhR2+r2+Rr=13×227×8332+272+33×27=22×83×7×2709=22704 cm3

Surface area of the frustum

=πR2+πr2+πlR+r=πR2+r2+lR+r=227332+272+1033+27=2271089+729+1060=22×24187=7599.43 cm2

Page No 823:

Answer:

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

Capacity of the frustum
=13πhR2+r2+Rr=13×227×15282+212+28×21=22×57×1813=28490 cm3=28.49 litres

Page No 823:

Answer:

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

Slant height, l, of the frustum
=h2+R2-r2=162+20-82=256+122=256+144=400=20 cm

Surface area of the frustum
 =πr2+πlR+r=πr2+lR+r=22782+2020+8=22764+2028=22×6247=1961.14 cm2

100 cm2 of metal sheet costs Rs 15.
So, total cost of the metal sheet used for fabrication
=1961.14100×15=Rs 294.171 

Page No 823:

Answer:

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

Slant height, l
=h2+R-r2=242+15-52=576+102=576+100=676=26 cm
  
Capacity of the frustum

=13πhR2+r2+Rr=13×3.14×24152+52+15×5=3.14×8×325=8164 cm3=8.164 litres

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

Surface area of the bucket
=πr2+πlR+r=π52+2615+5=3.1425+2620=3.1425+520=1711.3 cm2

Cost of 100 cm2 of metal sheet is Rs 10.
So, cost of metal used for making the bucket =1711.3100×10=Rs 171.13 

Page No 823:

Answer:

We have,Height, h=14 cm,Radius of upper end, R=352=17.5 cm andRadius of lower end, r=302=15 cmNow,Volume of the container=13πhR2+r2+Rr=13×227×14×17.52+152+17.5×15=443×306.25+225+262.5=443×793.75=349253 cm3So, the mass of the oil that is completely filled in the container=349253×1.2=13970 kg=13.97 kg The cost of the oil in the container=40×13.97=558.80

Page No 823:

Answer:

We have,Radius of upper end, R=28 cm andRadius of lower end, r=21 cmLet the height of the bucket be h.Now,Volume of water the bucket can hold=28.49 LVolume of bucket=28490 cm3                 As, 1L=1000 cm313πhR2+r2+Rr=2849013×227×h×282+212+28×21=2849022h21×49×16+9+12=2849022h3×7×37=28490h=28490×322×7×37 h=15 cm

So, the height of the bucket is 15 cm.

Page No 823:

Answer:

We have,Height, h=15 cm,Radius of the upper end, R=14 cm,Radius of lower end=r,As,Volume of the bucket=5390 cm313πhR2+r2+Rr=539013×227×15×142+r2+14r=53901107×196+r2+14r=5390196+r2+14r=5390×7110196+r2+14r=343r2+14r-147=0r2+21r-7r-147=0rr+21-7r+21=0r+21r-7=0r+21=0 or r-7=0r=-21 or r=7As, r cannot be negative r=7 cm

So, the value of r is 7 cm.

Page No 823:

Answer:

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

Surface area of the frustum

=πR2+πr2+πlR+r=πR2+r2+lR+r=332+272+1033+27π=1089+729+1060π=2418×3.14=7592.52 cm2



Page No 824:

Answer:

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

Height of the frustum = h = 24 m

Slant height = l
=h2+R-r2=242+10-32=576+49=625=25 m

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

Slant height, L, of the cone = 32+42=9+16=25=5 m

Total quantity of canvas = Curved surface area of the frustum + curved surface area of the conical top
=πl(R+r)+πLr=πl(R+r)+Lr=22725×13+5×3=227325+15=1068.57 m2 

Page No 824:

Answer:

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

Height of the frustum= h = 8 m
Slant height l =
h2+R-r2=82+13-72=64+36=100=10 m

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

Total surface area of the tent = Curved area of frustum + Curved area of the cone
=πlR+r+πrL=227×1013+7+227×7×12=227200+84=892.57 m2

Page No 824:

Answer:

We have,Perimeter of upper end, C=48 cm,Perimeter of lower end, c=36 cm andHeight, h=11 cmLet the radius of upper end be R and the radius of lower end be r.As, C=48 cm2πR=48R=482πR=24π cmSimilarly, c=36 cm r=362πr=18π cmAnd, l=R-r2+h2=24π-18π2+112=6π2+112=6×7222+112=21112+112=441+121×121121=441+14641121=1508211 cmNow,Volume of the frustum=13πhR2+r2+Rr=13×π×11×24π2+18π2+24π×18π=11π3×576π2+324π2+432π2=11π3×1332π2=113×1332π=113×1332×722=1554 cm3Also,Curved surface area of the frustum=πR+rl=227×24π+18π×1508211=227×42π×1508211=227×42×722×150821142×11.164436468.91 cm2

Page No 824:

Answer:


We have,Radius of solid cone, R=CP=10 cm,Let the height of the solid cone be, AP=H,the radius of the smaller cone, QD=r andthe height of the smaller cone be, AQ=h.Also, AQ=AP2 i.e. h=H2 or H=2h          .....iNow, in AQD and APC,QAD=PAC        Common angleAQD=APC=90°So, by AA criteriaAQD~APCAQAP=QDPChH=rRh2h=rR                 Using i12=rRR=2r                .....iiAs,Volume of smaller cone=13πr2hAnd,Volume of solid cone=13πR2H=13π2r2×2h           Using i and ii=83πr2hSo,Volume of frustum=Volume of solid cone-Volume of smaller cone=83πr2h-13πr2h=73πr2hNow, the ratio of the volumes of the two parts=Volume of the smaller coneVolume of the frustum=13πr2h73πr2h=17=1:7

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

Page No 824:

Answer:



We have,Height of the given cone, H=20 cmLet the radius of the given cone be R,the height of the smaller cone be h andthe radius of the smaller cone be r.Now, in AQD and APC,QAD=PAC        Common angleAQD=APC=90°So, by AA criteriaAQD~APCAQAP=QDPChH=rR               .....iVolume of smaller cone=18×Volume of the given coneVolume of smaller coneVolume of the given cone=1813πr2h13πR2H=18rR2×hH=18hH2×hH=18                Using ihH3=18hH=183h20=12h=202h=10 cm PQ=H-h=20-10=10 cm

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

Page No 824:

Answer:



We have,Height of the solid metallic cone, H=20 cm,Height of the frustum, h=202=10 cm andRadius of the wire=124 cmLet the length of the wire be l, EG=r and BD=R.In AEG,tan30°=EGAG13=rH-h13=r20-10r=103 cmAlso, in ABD,tan30°=BDAD13=RH13=R20R=203 cmNow,Volume of the wire=Volume of the frustumπ1242l=13πhR2+r2+Rrl576=13×10×2032+1032+203103l=5763×10×4003+1003+2003l=5763×10×7003l=448000 cm l=4480 m

So, the length of the wire is 4480 m.

Page No 824:

Answer:



We have,Radius of open side, R=10 cm,Radius of upper base, r=4 cm andSlant height, l=15 cmNow,The area of material used=πR+rl+πr2=227×10+4×15+227×4×4=227×14×15+227×4×4=227×14×15+4×4=227×210+16=227×226=49727 cm2710.28 cm2

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

Page No 824:

Answer:



We have,Height of the cylindrical portion, h=10 cm,Height of the frustum of cone portion, H=22-10=12 cm,Radius of the cylindical portion=Radius of smaller end of frustum portion, r=82=4 cm andRadius of larger end of frustum portion, R=182=9 cmAlso, the slant height of the frustum, l=R-r2+H2=9-42+122=52+122=25+144=169=13 cmNow,The area of the tin sheet required=CSA of frustum of cone+CSA of cylinder=πR+rl+2πrh=227×9+4×13+2×227×4×10=227×13×13+227×80=227×169+80=227×249782.57 cm2

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



Page No 825:

Answer:

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



Since ABO~AEP   (AA Similarity)
AOAP=BOEP3h2h=rr1r1=2r3        .....1
Also, ABO~AGQ     (AA Similarity)
AOAQ=BOGQ3hh=rr2r2=r3         .....2
Now,
Volume of cone AGF,
V1=13πr22h     =13πr32h      From 2     =127πr2h
Voulme of the frustum GFDE,
V2=13πr12+r22+r1r2h     =13π4r29+r29+2r29h       From 1 and 2     =727πr2h
Voulme of the frustum EDCB,
V3=13πr2+r12+r1rh     =13πr2+4r29+2r23h       From 1 and 2     =1927πr2h
∴ Required ratio = V1:V2:V3=127πr2h:727πr2h:1927πr2h=1:7:19



Page No 826:

Answer:

We have,Depth of the river, h=1.5 m,Width of the river, b=36 m,Speed of the flowing water, l=3.5 km/hr=3.5×1000 m60 min=1753 m/minNow,The amount of water that runs into the sea per minute=lbh=1753×36×1.5=3150 m3/min

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

Page No 826:

Answer:

Let the edge of the cube be a.As,Volume of the cube=729 cm3a3=729a=7293a=9 cmNow,Surface area of the cube=6a2=6×9×9=486 cm2

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

Page No 826:

Answer:

We have,Edge of the cube, a=10 cm andEdge of the cubical box, l=1 m=100 cmNow,The number of cubes that can be put in the box=Volume of the cubical boxVolume of the cube=l3a3=1003103=103=1000

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

Page No 826:

Answer:

We have,Edges of the cubes: a1=6 cm, a2=8 cm and a3=10 cmLet the edge of the new cube so formed be a.As,Volume of the new cube so formed=a13+a23+a33a3=63+83+103a3=216+512+1000a3=1728a=17283 a=12 cm

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

Page No 826:

Answer:

We have,Length of the resulting cuboid, l=5×5=25 cm,Breadth of the resulting cuboid, b=5 cm andHeight of the resulting cuboid, h=5 cmNow,Volume of the resulting cuboid=lbh=25×5×5=625 cm3

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

Page No 826:

Answer:

Let the edges of the cubes be a and b.As,Volume of the first cubeVolume of the second cube=827a3b3=827ab=8273ab=23                    .....iNow,The ratio of the surface areas of the cubes=Surface area of the first cubeSurface area of the second cube=6a26b2=ab2=232                 Using i=49=4:9

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

Page No 826:

Answer:

We have,Height=Base radius i.e. h=rAs,Volume of the cylinder=2517 cm3πr2h=1767227×h2×h=1767h3=176×77×22h3=8h=83 h=2 cm

So, the height of the cylinder is 2 cm.

Page No 826:

Answer:

Let the radius of the base and the height of the cylinder be r and h, respectively.We have,r:h=2:3 i.e. rh=23or h=3r2                    .....iAs,Volume of the cylinder=12936 cm3πr2h=12936227×r2×3r2=12936                    Using i337×r3=12936r3=12936×733r3=2744r=27443 r=14 cm

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

Page No 826:

Answer:

Let the radii of the cylinders be r1 and r2; and their heights be h1 and h2.We have,r1:r2=2:3 or r1r2=23                     .....iand h1:h2=5:3 or h1h2=53          .....iiNow,The ratio of the volumes of the cylinders=Volume of the first cylinderVolume of the second cylinder=πr12h1πr22h2=r1r22×h1h2=232×53                Using i and ii=2027=20:27

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

Page No 826:

Answer:

We have,Radius of wire, r=12=0.5 mm=0.05 cmLet the length of the wire be l.As,Volume of the wire=66 cm3πr2l=66227×0.05×0.05×l=66l=66×722×0.05×0.05 l=8400 cm=84 m

So, the length of the wire is 84 m.



Page No 827:

Answer:

We have,Height=84 cmLet the radius and the slant height of the cone be r and l, respectively.As,Area of the base of the cone=3850 cm2πr2=3850227×r2=3850r2=3850×722r2=1225r=1225 r=35 cmNow,l=h2+r2=842+352=7056+1225=8281=91 cm

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

Page No 827:

Answer:

We have,Base radius of the cylinder, r=8 cm,Height of the cylinder, h=2 cm andHeight of the cone, H=6 cmLet the base radius of the cone be R.Now,Volume of the cone=Volume of the cylinder13πR2H=πr2hR2=3r2hHR2=3×8×8×26R2=64R=64 R=8 cm

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

Page No 827:

Answer:

Let the radius and height of the cone be r and h, respectively. Then,Radius of the cylindrical vessel=r andHeight of the cylindrical vessel=hNow,The number of cones=Volume of the cylindrical vesselVolume of a cone=πr2h13πr2h=3

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

Page No 827:

Answer:

Let the radius of the sphere be r.As,Volume of the sphere=4851 cm343πr3=485143×227×r3=4851r3=4851×3×74×22r3=92618r=926183r=212 cmNow,Curved surface area of the sphere=4πr2=4×227×212×212=1386 cm2

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

Page No 827:

Answer:

Let the radius of the sphere be r.As,Curved surface area of the sphere=5544 cm24πr2=55444×227×r2=5544r2=5544×74×22r2=441r=441r=21 cmNow,Volume of the sphere=43πr3=43×227×21×21×21=38808 cm3

So, the volume of the sphere is 38808 cm3.

Page No 827:

Answer:

Let the radii of the two spheres be r and R.As,Surface area of the first sphereSurface area of the second sphere=4254πr24πR2=425rR2=425rR=425rR=25                 .....iNow,The ratio of the volumes of the two spheres=Volume of the first sphereVolume of the second sphere=43πr343πR3=rR3=253                    Using i=8125=8:125

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

Page No 827:

Answer:

We have,Radius of the solid metallic sphere, R=8 cm andRadius of the spherical ball, r=2 cmNow,The number spherical balls obtained=Volume of the solid metallic sphereVolume of a spherical ball=43πR343πr3=Rr3=823=43=64

So, the number of spherical balls obtained is 64.

Page No 827:

Answer:

We have,Radius of a lead shot, r=32=1.5 mm=0.15 cm andDimensions of the cuboid are 9 cm×11 cm×12 cmNow,The number of the lead shots=Volume of the cuboidVolume of a lead shot=9×11×1243πr3=9×11×1243×227×0.15×0.15×0.15=84000

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

Page No 827:

Answer:

We have,Radius of the metallic cone, r=12 cm,Height of the metallic cone, h=24 cm andRadius of the sphere, R=2 cmNow,The number of spheres so formed=Volume of the metallic coneVolume of a sphere=13πr2h43πR3=r2h4R3=12×12×244×2×2×2=108

So, the number of spheres so formed is 108.

Page No 827:

Answer:

We have,Radius of the hemisphere, R=6 cm andHeight of the cone, h=75 cmLet the radius of the base of the cone be r.Now,Volume of the cone=Volume of the hemisphere13πr2h=23πR3r2=2R3hr2=2×6×6×675r2=5.76r=5.76 r=2.4 cm

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

Page No 827:

Answer:

We have,Radius of the sphere, R=182=9 cm andRadius of the wire, r=42=2 mm=0.2 cmLet the length of the wire be l.Now,Volume of the wire=Volume of the copper sphereπr2l=43πR3l=4R33r2l=4×9×9×93×0.2×0.2 l=24300 cm=243 m

So, the length of the wire is 243 m.

Page No 827:

Answer:

We have,Height of the frustum, h=6 cm,Radii of the circular ends, R=14 cm and r=6 cmLet the slant height of the frustum be l.Now,l=R-r2+h2=14-62+62=82+62=64+36=100=10 cm

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

Page No 827:

Answer:

Let the radius of the shere be R and the edge of the cube be a.As, the sphere is fit inside the cube.So, diameter of the sphere=edge of the cube2R=a                 .....iNow,The ratio of the volume of the cube to that of the sphere=Volume of the cubeVolume of the sphere=a343πR3=2R343πR3              Using i=3×8R34πR3=6π=6:π

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

Page No 827:

Answer:

Let the radius of the sphere be r.We have,The radius of the cone=The radius of the cylinder=The radius of the sphere=r andThe height of the cylinder=The height of the cone=The height of the sphere=2rNow,Volume of the cylinder=πr22r=2πr3,Volume of the cone=13πr22r=23πr3 andVolume of the sphere=43πr3So,The ratio of the volumes of the cylinder, the cone and the sphere=2πr3:23πr3:43πr3=1:13:23=3:1:2

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

Page No 827:

Answer:

Let the edge of the cube be a.As,Volume of the cube=125 cm3a3=125a=1253a=5 cmSo,Length of the resulting cuboid, l=2×5=10 cm,Breadth of the resulting cuboid, b=5 cm andHeight of the resulting cuboid, h=5 cmNow,Surface area of the resulting cuboid=2lb+bh+hl=2×10×5+5×5+5×10=2×50+25+50=2×125=250 cm2

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

Page No 827:

Answer:

We have,Edges of the cubes: a1=3 cm, a2=4 cm and a3=5 cmLet the edge of the new cube be a.Now,Volume of the new cube=a13+a23+a33a3=33+43+53a3=27+64+125a3=216a=2163 a=6 cm

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

Page No 827:

Answer:

We have,Radius of the metallic sphere, R=82=4 cm andHeight of the cylindrical wire, h=12 m=1200 cmLet the radius of the base be r.Now,Volume of the cylindrical wire=Volume of the metallic sphereπr2h=43πR3r2=4R33hr2=4×4×4×43×1200r2=16225r=16225r=415 cm The width of the wire=2r=2×415=815 cm

So, the width of the wire is 815 cm.



Page No 828:

Answer:

We have,Width of the cloth, B=5 m,Radius of the conical tent, r=142=7 m andHeight of the conical tent, h=24 mLet the length of the cloth used for making the tent be L.Also,The slant height of the conical tent, l=r2+h2=72+242=49+576=625=25 mNow,The curved surface of the conical tent=πrl=227×7×25The area of the cloth used for making the tent=550 m2LB=550L=550BL=5505L=110 mSo, the cost of the cloth used=25×110=2750

So, the cost of the cloth used for making the tent is ₹2750.

Page No 828:

Answer:



We have,Radius of the cylinder=Radius of the hemispher=r=3.5 cm andHeight of the cylinder, h=10 cmNow,Volume of the toy=Volume of the cylinder-Volume of the two hemispheres=πr2h-2×23πr3=πr2h-4r3=227×3.5×3.5×10-4×3.53=38.5×10-143=38.5×163=6163 cm3205.33 cm3

So, the volume of wood in the toy is 6163 cm3 or 205.33 cm3.

Page No 828:

Answer:

Let the edge of the metal cubes be 3x, 4x and 5x.Let the edge of the single cube be a.As,Diagonal of the single cube=123 cma3=123a=12 cmNow,Volume of the single cube=Sum of the volumes of the metallic cubesa3=3x3+4x3+5x3123=27x3+64x3+125x31728=216x3x3=1728216x3=8x=83x=2So, the egdes of the cubes are 3×2=6 cm, 4×2=8 cm and 5×2=10 cm.

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

Page No 828:

Answer:

We have,External radius of the hollow sphere, R1=82=4 cm,Internal radius of the hollow sphere, R2=42=2 cm andBase radius of the cone, r=82=4 cmLet the height of the cone be h.Now,Volume of the cone=Volume of the hollow sphere13πr2h=43πR13-43πR2313πr2h=43πR13-R23h=4r2R13-R23h=44×443-23h=1464-8h=14×56 h=14 cm

So, the height of the cone is 14 cm.

Page No 828:

Answer:

We have,Height of the frustum, h=24 cm,Radius of the open end, R=422=21 cm andRadius of the close end, r=282=14 cmNow,Volume of the bucket=13πhR2+r2+Rr=13×227×24×212+142+21×14=1767×441+196+294=1767×931=23408 cm3=23.408 L         As, 1000 cm3=1 L The cost of the milk=30×23.408=702.24

So, the cost of the milk which the bucket can hold is ₹702.24.

Page No 828:

Answer:



We have,Radius of the cylinder=Radius of the cone=r=4.22=2.1 m,Height of the cylinder, H=4 m andHeight of the cone, h=2.8 mAlso,The slant height of the cone, l=r2+h2=2.12+2.82=4.41+7.84=12.25=3.5 mNow,The outer surface area of the building=CSA of the cylinder+CSA of the cone=2πrH+πrl=πr2H+l=227×2.1×2×4+3.5=6.6×11.5=75.9 m2

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

Page No 828:

Answer:

We have,Height of the cone, h=84 cm andBase radius of the cone, r=21 cmLet the radius of the solid sphere be R.Now,Volume of the solid sphere=Volume of the solid cone43πR3=13πr2hR3=r2h4R3=21×21×844R3=21×21×21R=21 cm Diameter=2R=2×21=42 cm

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

Page No 828:

Answer:



We have,Radius of the hemisphere=Radius of the cone=r=3.5 cm andHeight of the cone=15.5-3.5=12 cmAlso,The slant height of the cone, l=h2+r2=122+3.52=144+12.25=156.25=12.5 cmNow,Total surface area of the toy=CSA of cone+CSA of hemisphere=πrl+2πr2=πrl+2r=227×3.5×12.5+2×3.5=11×12.5+7=11×19.5=214.5 cm2

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

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

Page No 828:

Answer:

We have,Height, h=28 cm,Radius of the upper end, R=28 cm andRadius of the lower end, r=7 cmAlso,The slant height, l=R-r2+h2=28-72+282=212+282=441+784=1225=35 cmNow,Capacity of the bucket=13πhR2+r2+Rr=13×227×28×282+72+28×7=883×784+49+196=883×1029=30184 cm3Also,Total surface area of the bucket=πlR+r+πr2=227×35×28+7+227×7×7=110×35+154=3850+154=4004 cm2

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

Page No 828:

Answer:

We have,Radius of the upper end, R=20 cm andRadius of the lower end, r=12 cmLet the height of the bucket be h.As,Volume of the bucket=12308.8 cm313πhR2+r2+Rr=12308.813×3.14×h×202+122+20×12=12308.83.14h3×400+144+240=12308.83.14h3×784=12308.8h=12308.8×33.14×784 h=15 cm

So, the height of the bucket is 15 cm.

Page No 828:

Answer:

We have,Radius of the upper end, R=20 cm andRadius of the lower end, r=8 cmLet the height of the container be h.As,Volume of the container=1045937 cm313πhR2+r2+Rr=73216713×227×h×202+82+20×8=73216722h21×400+64+160=73216722h21×624=732167h=73216×217×22×624h=16 cmAlso,The slant height of the container, l=R-r2+h2=20-82+162=122+162=144+256=400=20 cmNow,Total surface area of the container=πlR+r+πr2=227×20×20+8+227×8×8=227×20×28+227×64=227×560+64=227×624=137287 cm2So, the cost of metal sheet=1.4×137287=2745.60

Hence, the cost of the metal sheet used for making the milk container is ₹2745.60.

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



Page No 829:

Answer:

We have,Radius of the metallic sphere, R=282=14 cm,Radius of the smaller cone, r=12×423=12×143=73 cm andHeight of the smaller cone, h=3 cmNow,The number of cones so formed=Volume of the metallic sphereVolume of a smaller cone=43πR313πr2h=4R3r2h=4×14×14×1473×73×3=672

So, the number of cones so formed is 672.

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

Page No 829:

Answer:

We have,Internal radius of the cylindrical vessel, R=102=5 cm,Height of the cylindrical vessel, H=10.5 cm,Radius of the solid cone, r=72=3.5 cm andHeight of the solid cone, h=6 cmiVolume of water displaced out of the cylinder=Volume of the solid cone=13πr2h=13×227×3.5×3.5×6=77 cm3ii As,Volume of the cylindrical vessel=πR2H=227×5×5×10.5=825 cm3So, the volume of water left in the cylindrical vessel=Volume of the cylindrical vessel-Volume of the solid cone=825-77=748 cm3

Page No 829:

Answer:

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

Page No 829:

Answer:

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



Page No 830:

Answer:

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

Page No 830:

Answer:

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

Page No 830:

Answer:

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

Page No 830:

Answer:

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

Page No 830:

Answer:

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

Page No 830:

Answer:

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

Page No 830:

Answer:

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



Page No 831:

Answer:

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

Page No 831:

Answer:

(b) 21 cm
Volume of the cuboid =l×b×h = 49×33×24 cm3
Let the radius of the sphere be r cm.
Volume of the sphere=43πr3

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

    43πr3=49×33×2443×227×r3=49×33×24
r3=49×33×24×2188r3=21×21×21r3=213r=21 cm

Hence, the radius of the sphere is 21 cm.

Page No 831:

Answer:

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

So, the radius of the base of the largest cone = 4.22 = 2.1 cm

Hence, the correct answer is option (a).

Page No 831:

Answer:

We have,Radius of the solid sphere, R=9 cm andRadius of the solid cylinder, r=9 cmLet the height of the cylinder be h.Now,Volume of the cylinder=Volume of the sphereπr2h=43πR3h=4R33r2h=4×9×9×93×9×9 h=12 cm

Hence, the correct answer is option (a).

Page No 831:

Answer:

We have,Length of the rectangular sheet, l=40 cm,Width of the rectangular sheet, b=22 cm andHeight of the hollow cylinder, h=40 cmLet the radius of the cylinder be r.As, l=hSo, the circumference of base of the cylinder=b2πr=222×227×r=22r=22×72×22 r=3.5 cm

Hence, the correct answer is option (a).

Page No 831:

Answer:

We have,Radius of the spher, r=62=3 cm,Height of the cylinder, H=45 cm andRadius of the cylinder, R=42=2 cmNow,The number of sphere that can be made=Volume of the cylinderVolume of the sphere=πR2H43πr3=3R2H4r3=3×2×2×454×3×3×3=5

Hence, the correct answer is option (c).

Page No 831:

Answer:

Let the radius of the two spheres be r and R.As,Surface area of the first sphereSurface area of the second sphere=1694πR24πr2=169Rr2=169Rr=169Rr=43              .....iNow,The ratio of their volumes=Volume of the first sphereVolume of the second sphere=43πR343πr3=Rr3=433             Using i=6427=64:27

Hence, the correct answer is option (a).

Page No 831:

Answer:

Let the radius of the sphere be r.As,Surface area of the sphere=616 cm24πr2=6164×227×r2=616r2=616×74×22r2=49r=49r=7 cm Diameter of the sphere=2r=2×7=14 cm

Hence, the correct answer is option (b).

Page No 831:

Answer:

Let the radius of the given sphere and that of the new sphere be r and R, respectively;Also, the volume of the given sphere and that of the new sphere be v and V, respectively.We have,R=3r        .....iNow,Volume of the new sphereVolume of the given sphere=43πR343πr3=Rr3=3rr3             Using i=33=27

Hence, the correct answer is option (d).

Page No 831:

Answer:

We have,Height of the frustum, h=16 cm,Radii of the circular ends, R=402=20 cm and r=162=8 cmNow,The slant height of the frustum, l=R-r2+h2=20-82+162=122+162=144+256=400=20 cm

Hence, the correct answer is option (a).

Page No 831:

Answer:

We have,Radius of the sphere, r=182=9 cm andRadius of the cylindrical vessel, R=362=18 cmLet the rise of water level be H.Now,Volume of the water rised=Volume of the sphereπR2H=43πr3H=4r33R2H=4×9×9×93×18×18 H=3 cm

Hence, the correct answer is option (a).



Page No 832:

Answer:


Let the radii of the smaller and given cones be r and R, respectively; and their heights be h and H, respectively.We have,H=2h         .....iIn AQD and APC,QAD=PAC       Common angleAQD=APC=90°So, by AA criteriaAQD~APCAQAP=QDPChH=rRh2h=rR           Using i12=rRR=2r            .....iiNow,The ratio of the volume of the smaller cone to the whole cone=Volume of the smaller coneVolume of the whole cone=13πr2h13πR2H=rR2×hH=r2r2×h2h               Using i and ii=122×12=18=1:8

Hence, the correct answer is option (d).

Page No 832:

Answer:

We have,Height of the bucket, h=40 cm,Radius of the upper end, R=24 cm andRadius of the lower end, r=15 cmNow,The slant height, l=R-r2+h2=24-152+402=92+402=81+1600=1681=41 cm

Hence, the correct answer is option (a).

Page No 832:

Answer:

Let the radius of the hemisphere or the radius of the cone be r and the slant height of the cone be l.Now,Surface area of the hemisphere=Surface area of the cone2πr2=πrlπr2πrl=12rl=12 r:l=1:2

Hence, the correct answer is option (a).

Page No 832:

Answer:

Let the base radius and height of the original cylinder be r and h, respectively.Also,The radius of the new cylinder, R=r2 and its height, H=h.Now,The ratio of the volume of the new cylinder to that of the original cylinder=Volume of the new cylinderVoilume of the original cylinder=πr2hπR2H=πr2hπr22h=4πr2hπr2h=41=4:1

Hence, the correct answer is option (d).

Page No 832:

Answer:

(c) 363
The edge of the cubical ice-cream brick = a = 22 cm
Volume of the cubical ice-cream brick=a3
                                               =22×22×22 cm3

Radius of an ice-cream cone = 2 cm
Height of an ice-cream cone = 7 cm
Volume of each ice-cream cone=13πr2h
                                              =13×227×2×2×7 cm3

Number of ice-cream cones=Volume of the cubical ice cream brickVolume of each ice cream cone

                                        =22×22×22×3×722×2×2×7=363
Hence, the number of ice-cream cones is 363.

Page No 832:

Answer:

(c) 11200
Volume of wall = 270×300×350 cm3
18th of the wall is covered with mortar.
So,
Volume of the wall filled with bricks=78×270×300×350 cm3
Volume of each brick=22510×1125100×875100 cm3
                                =9×225×3532 cm3

Number of bricks used to construct the wall=Volume of the wall composed of bricksVolulme of each brick
                                                              
                                                              =7×270×300×350×328×9×225×35
                                                               = 11200

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

Page No 832:

Answer:

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


As R=Diameter2, R=22 cm=1 cm h=16 cm

Therefore,
    12×43πr3=πR2h12×43r3=R2h

12×43d23 = 12×1616×d38 = 16
d3 = 8d = ±2Since d cannot be negative, thus, d = 2


Hence, the diameter of each sphere is 2 cm.

Page No 832:

Answer:

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

Then, R=442 cm=22 cm, r=242 cm=12 cm, h=35 cm

Capacity of the bucket = Volume of the frustum of the cone
                                 =13πhR2+r2+Rr cm3
                                 =13×227×35×222+122+22×12 cm3=1103×892 cm3=110×8923×1000 litres=32.7 litres

Hence, the capacity of the bucket is 32.7 litres.

Page No 832:

Answer:

(d) 4950 cm2
Let the radii of the top and bottom of the bucket be R and r and let its slant height be l.
Then, R=28 cm, r=7 cm, l=45 cm
Curved surface area of the bucket=πlR+r
                                                 =227×45×28+7 cm2=4950 cm2

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



Page No 833:

Answer:

(b) 16 : 9
Let the radii of the spheres be R and r.
Then, ratio of their volumes
=43πR343πr3

Therefore,

    43πR343πr3=6427R3r3=6427

Rr3=433Rr=43
Hence, the ratio of their surface areas=4πR24πr2
                                                       
                                                     =R2r2=Rr2=432=169=16:9

Page No 833:

Answer:

(a) 142296
Since 18th of the cube remains unfulfilled,
volume of the cube =  22×22×22 cm3

Space filled in the cube=78×22×22×22 cm3
                                 =7×1331 cm3

Radius of each marble=0.52 cm
                                =520 cm=14 cm

Volume of each marble=43πr3
                                 =43×227×14×14×14 cm3=1124×7 cm3

Therefore, number of marbles required=7×1331×24×711
                                                        = 142296

Page No 833:

Answer:

(b) 14 cm
Let the internal and external radii of the spherical shell be r and R, respectively.
Volume of the spherical shell=43πR3-r3
                                         =43π43-23 cm3   Since r42=2 cm, R=82=4 cm=43π×56 cm3
Radius of the cone = 82= 4 cm
Volume of the cone=13πr2h
                             =13π×4×4×h cm3
Therefore,
13π×4×4×h=43π×56163×h=43π×56h=4×56×33×16h=14 cm

Page No 833:

Answer:

(d) 0.36 cm3
Radius of the capsule
=0.52 cm
 = 0.25 cm

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

Then,

0.25+x+0.25=20.5+x=2x=1.5

Hence, the capacity of the capsule=2×Volume of the hemisphere+Volume of the cylinder
                                                 =2×23πr3+πr2h
                                                 =43×227×14×14×14+227×14×14×1.5    Since 0.25=14
                                                 =11168+33112 cm3
                                                 =121336 cm3 = 0.36 cm3

Page No 833:

Answer:

(d) 17 m
Length of the longest pole that can be kept in a room =Length of the diagonal of the room
                                                                            =l2+b2+h2 m=122+92+82 m=289 m=17 m 

Page No 833:

Answer:

(b) 216 cm2

Let the edge of the cube be a cm.
Then, length of the diagonal = 3a
Or,
  3a=63a=6 cm
Therefore, the total surface area of the cube = 6a2
                                                               =6×6×6 cm3=216 cm3

Page No 833:

Answer:

(b) 1176 cm2
Let the edge of the cube be a cm.
Then, volume of the cube = a3
Or,
a3=2744a3=23×73a=2×7a=14 cm

Therefore, surface area of the cube=6a2
                                                        =6×14×14 cm2=1176 cm2

Page No 833:

Answer:

(c) 1728 cm3
Let the edge of the cube be a.
Total surface area of the cube = 6a2
Therefore,
6a2=864a2=144a=12 cm
Therefore, volume of the cube = a3
                                               =12×12×12 cm3=1728 cm3

Page No 833:

Answer:

(b) 6400

Volume of the wall=800×600×22.5 cm3
Volume of each brick = 25×11.25×6 cm3

Number of bricks=Volume of the wallVolume of each brick
                         
                        =800×600×22.525×11.25×6=6400

Page No 833:

Answer:

(b) 4 m
Area of the base of a rectangular tank =6500 cm2

                                                          =6500100×100 m2=1320 m2

Let the depth of the water be d metres.
Then,

  1320×d=2.6d=2610×2013 md=4 m

Page No 833:

Answer:

Note : It should be 128 m3 instead of 12.8 m3

(b) 40 cm
Let the breadth of the wall be x cm.
Then, its height = 5x cm
and its length=8×5x cm
                   = 40x cm
Hence, the volume of the wall=40x×x×5x cm3
It is given that the volume of the wall = 128 m3.
Therefore,

  40x×x×5x=128×100×100×100200x3=128000000x3=640000

x3=40×40×40x=40 cm



Page No 834:

Answer:

(c) xyz
Let the length of the cuboid = l
breadth of the cuboid = b
and height of the cuboid = h
Since, the areas of the three adjacent faces are x, y and z, we have:

lb=xbh=ylh=z

Therefore,

  lb×bh×lh=xyzl2b2h2=xyzlbh=xyz

Hence, the volume of the cuboid = lbh=xyz.

Page No 834:

Answer:

(c) 236 cm2
Let l, b and h be the length, breadth and height of the cuboid.
Then,

  l+b+h=19l+b+h2=192

Therefore,

  l2+b2+h2+2lb+bh+lh=361552+2lb+bh+lh=3612lb+bh+lh=361-1252lb+bh+lh=236 cm2

Hence, the surface area of the cuboid is 236 cm2.

Page No 834:

Answer:

(d) 125%
Let the original edge of the cube be a units.
Then, the original surface area of the cube = 6a2 units

New edge of the cube = 150% of a
                                =150a100=3a2
Hence, new surface area=6×3a22
                                    =27a22
Increase in area=27a22-6a2
                       =15a22
% increase in surface area=15a22×16a2×100%
                                    = 125 %

Page No 834:

Answer:

(d) 225
Volume of the cuboidal granary=8 m×6 m×3 m
Volume of each bag=0.64 m3

Number of bags that can be stored in the cuboidal granary =Volume of the cuboidal granaryVolume of each bag
                                                                                    =8×6×30.64
                                                                                    = 225

Page No 834:

Answer:

(d) 27
Volume of the given cube=6×6×6 cm3
Volume of each small cube=2×2×2 cm3

Number of cubes formed =Volume of the given cubeVolume of each small cube
                                     =6×6×62×2×2
                                      = 27

Page No 834:

Answer:

(c) 1000 m3
Volume of water that falls on 2 hectares of land=Area of the ground×Amount of rain in m
                                                          =2×1000×5100 m3         ∵5 cm =5100 m , 2 heactares = 2×1000 m2=1000 m3

Page No 834:

Answer:

(c) 1 : 9
Let the edges of the two cubes be a and b.
Then, ratio of their volumes=a3b3
Therefore,
  a3b3=127ab3=133ab=13
The ratio of their surface areas=6a26b2
Therefore,
6a26b2=a2b2       =19   Since ab=13        =1:9
Hence, the ratio of their surface areas is 1:9.

Page No 834:

Answer:

(a) 176 cm3
Volume of the cylinder=πr2h
                                 =227×2×2×14   d=4 cmr=42cm=2 cm=176 cm3

Page No 834:

Answer:

(b) 2992 cm2
The total surface area of the cylinder=2πr+2r2
                                                     =2πrh+r=2×227×14×20+14 cm2     d=28 cmr=282cm=14 cm=209447 cm2=2992 cm2

Page No 834:

Answer:

(b) 396 cm3

Curved surface area of the cylinder =2πrh=2×227×r×14

Therefore,
   2×227×r×14=264r=26488r=3 cm

Hence, the volume of the cylinder=πr2h
                                                 =227×3×3×14 cm3=27727 cm3=396 cm3

Page No 834:

Answer:

(c) 20 cm
Curved surface area of the cylinder=2πrh
                                                   =2×227×14×h
Therefore,
   2×227×14×h=1760h=176088 cm h=20 cm
Hence, the height of the cylinder is 20 cm.

Page No 834:

Answer:

(d) 5 : 1
Ratio of the total surface area to the lateral surface area=Total surface areaLateral surface area
                                                                                        =2πrh+r2πrh=h+rh=20+8020=10020=51=5:1
Hence, the required ratio is 5:1.



Page No 835:

Answer:

(c) 6 m
The curved surface area of a cylindrical pillar
=2πrh
Therefore, 2πrh=264
Volume of a cylinder=πr2h
Therefore, πr2h=924

Hence,

   πr2h2πrh=924264r2=924264r=924×2264r=7 m

Therefore,

2πrh=2642×227×7×h=264h=26444h=6 m

Hence, the height of the pillar is 6 m.

Page No 835:

Answer:

(d) 770 cm2
Let the common multiple be x.
Let the radius of the cylinder be 2x cm and its height be 3x cm.
Then, volume of the cylinder=πr2h
                                          =227×2x2×3x
Therefore,
   227×2x2×3x=1617227×4x2×3x=1617227×12x3=1617x3=1617×722×112

x3=72×72×72x3=723x=72
Now, r=7 cm and h=212 cm
Hence, the total surface area of the cylinder=2πrh+2πr2
                                                               =2πrh+r=2×227×7×212+7 cm2=2×227×7×352 cm2=770 cm2

Page No 835:

Answer:

(b) 20 : 27
Let the radii of the two cylinders be 2r and 3r and their heights be 5h and 3h, respectively.

Then, ratio of their volumes=π×2r2×5hπ×3r2×3h
                                       =4r2×59r2×3=2027=20:27

Page No 835:

Answer:

(b) 2:1
Let the radii of the two cylinders be r and R and their heights be h and 2h, respectively.
Since the volumes of the cylinders are equal, therefore:

    π×r2×h=π×R2×2h
r2R2=21rR2=21rR=21  r:R =2:1       

Hence, the ratio of their radii is 2:1.

Page No 835:

Answer:

(b) 65π cm2
Given: r = 5 cm, h = 12 cm

Slant height of the cone, l=r2+h2
                                     =52+122=25+144=169=13 cm

Hence, the curved surface area of the cone=πrl
                                                              =π×5×13 cm2=65π cm2

Page No 835:

Answer:

(a) 28 cm
Let h be the height of the cone.
Diameter of the cone = 42 cm
Radius of the cone = 21 cm

Then, volume of the cone=13πr2h
                              =13×227×21×21×h =22×21×h

Therefore,

22×21×h=12936h=1293622×21h=28 cm
Hence, the height of the cone is 28 cm.

Page No 835:

Answer:

(a) 1545cm2
Area of the base of the of a right circular cone=πr2
Therefore,
   πr2=154227×r2=154r2=154×722r2=49r=7 cm
Now, r = 7 cm and h = 14 cm
Then, slant height of the cone, l=r2+h2
                                              =72+142=49+196=245=75 cm
Hence, the curved surface area of the cone=πrl
                                                             =227×7×75 cm2=1545 cm2

Page No 835:

Answer:

(d) 72.8%
Let the original radius of the cone be r and height be h.
Then, original volume=13πr2h

Let 13πr2h=V
New radius = 120% of r
                =120r100=6r5

New height = 120% of h
                 =120h100=6h5


Hence, the new volume = 13π×6r52×6h5
                                   =21612513πr2h=216125V

Increase in volume=216125V-V

                           =91V125

Increase in % of the volume=91V125×1V×100%

                                         = 72.8%

Page No 835:

Answer:

(a) 9 : 8
Let the radii of the base of the cylinder and cone be 3r and 4r and their heights be 2h and 3h, respectively.
Then, ratio of their volumes=π3r2×2h13π4r2×3h
                                         =9r2×2×316r2×3=98=9:8

Page No 835:

Answer:

(d) 8 cm
Radius of the cylinder = 8 cm
Height of the cylinder = 2 cm
Height of the cone = 6 cm

Volume of the cylinder  = Volume of the cone
Therefore,
   π×8×8×2=13π×r2×6128=r2×2r2=1282r2=64r=8 cm
Hence, the radius of the base of the cone is 8 cm.

Page No 835:

Answer:

(b) 525 m
Area of the floor of a conical tent=πr2
Therefore,

   πr2=346.5227×r2=346.5r2=346510×722r2=4414r2=2122r=212 m

Height of the cone = 14 m

Slant height of the cone, l=r2+h2

                                    =2122+142=4414+196=12244=354 m

Area of the canvas = Curved surface area of the conical tent
                           =πrl=227×212×352 m2=577.5 m2

Length of the canvas=Area of the canvasWidth of the canvas

                              =577.51.1m=525 m



Page No 836:

Answer:

(c) 143713cm3
Volume of the sphere=43πr3

                              =43×227×7×7×7 cm3 d=14 cm⇒r=142cm=7 cm=43123 cm3=143713 cm3

Page No 836:

Answer:


(d) 4 : 9
Let the radii of the spheres be R and r, respectively.
Then, ratio of their volumes=43πR343πr3
Therefore,
43πR343πr3=827R3r3=827Rr3=233Rr=23
Hence, the ratio between their surface areas=4πR24πr2
                                                                      =R2r2=Rr2=232=49=4:9

Page No 836:

Answer:

DISCLAIMER : The answer to the question does not match the options given.

External diameter = 8 cm
Internal diameter = 4 cm
Let the external and internal radii of the hollow metallic sphere be R and r, respectively.
Then,
External radius = 82= 4 cm
Internal Radius = 42= 2 cm
Then, volume of the hollow sphere:
43πR3-r3
                                                  
=43π43-23

Therefore,
Volume of the hollow sphere =  Volume of the cone formed
  43π43-23=13π×82×h464-8=64×h224=64×hh=22464h=3.5 cm

Page No 836:

Answer:

(a) 2.1 cm
Radius of cone = 2.1 cm
Height of cone = 8.4 cm

Volume of cone = 13πr2h = 13π2.12×8.4

Volume of the sphere=43πr3
Therefore,

Volume of cone = Volume of sphere

   13π×2.12×8.4=43πr34.41×8.4=4r337.044=4r3r3=37.0444r3=2.13r=2.1 cm

Hence, the radius of the sphere is 2.1 cm.

Page No 836:

Answer:

(a) 4158 cm2
Volume of hemisphere=23πr3
Therefore,

   23πr3=1940423×227×r3=19404r3=19404×2144r3=213r=21 cm

Hence, the total surface area of the hemisphere=3πr2
                                                                    =3×227×21×21 cm2=4158 cm2

Page No 836:

Answer:

(a) 17923cm3
Surface area of a sphere=4πr2
Therefore,
  4πr2=1544×227×r2=154r2=154×788r2=494r2=722r=72 cm
Volume of the sphere=43πr3
                               =43×227×72×72×72cm3=5393cm3=17923 cm3

Page No 836:

Answer:

(c) (147π) cm2

Radius of the hemisphere = 7 cm
Total surface area of the hemisphere = Curved surface area of hemisphere + Area of the circle=3πr2
                                                     =3π×7×7 cm2=147π cm2

Page No 836:

Answer:

(b) 80080 cm3
Let R and r be the radii of the top and base of the bucket, respectively, and let h be its height.

Then,

 R=35 cm, r=14 cm, h=40 cm
Volume of the bucket = Volume of the frustum of the cone
                                 =13πhR2+r2+Rr cm3
                                 =13×227×40×352+142+35×14 cm3=88021×1911 cm3=80080 cm3
Hence, the volume of the bucket is 80080 cm3.

Page No 836:

Answer:

(b) 1711.3 cm2
Let R and r be the radii of the top and base of the bucket, respectively, and let h and l be its height and slant height.

Then, 

R=15 cm, r=5 cm, h=24 cm

l=h2+R-r2
 =242+15-52=576+100=676=26 cm

Surface area of the bucket=πr2+lR+r
                                      =3.14×52+2615+5=3.14×26×20+25cm2=1711.3 cm2

Page No 836:

Answer:

(d) 7920 m2
Total area of the canvas required = (Curved surface area of the cylinder) + (Curved surface area of the cone)
                                                =2πrh+πrl=2×227×1052×4+227×1052×40 d=105cmr=1052cm=1320+6600 m2=7920 m2



Page No 837:

Answer:

(a)
Volume of the sphere=43πr3
                               =43π×83cm3

Volume of each cone=13πr2h
                               =13π×82×4 cm3

Number of cones formed=Volume of the sphereVolume of each cone

                                     =4π×8×8×8×33×π×8×8×4=8
Hence, aq

(b)
Volume of the earth dug out = Volume of the cylinder
                                         =πr2h=227×7×7×20
Let the height of the platform be h.
Then, volume of the platform = volume of the cuboid
                                           =44×14×h m3
Therefore,

  227×7×7×20=44×14×h3080=616×hh=3080616h=5 m

Hence, bs

(c)
Volume of the sphere
=43πr3
                             
 =43π×6×6×6

Let h be the height of the cylinder.
Then, volume of the cylinder=πr2h
                                           =π×4×4×h
Therefore,

   43π×6×6×6=π×4×4×h43×6×6×6=4×4×h228=16×hh=22816h=18 cm

Hence, cp

(d)
Let the radii of the spheres be R and r respectively.
Then, ratio of their volumes=43πR343πr3
Therefore,
43πR343πr3=6427R3r3=6427Rr3=433Rr=43
Hence, the ratio of their surface areas=4πR24πr2
                                                               =R2r2=Rr2=432=169=16:9
Hence, dr

Page No 837:

Answer:

(a)
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 20 cm, r = 10 cm and h = 30 cm.
Capacity of the bucket = Volume of the frustum of the cone
                                 =πh3R2+r2+Rr
                                 =227×13×30×202+102+20×10 cm3=2207×400+100+200 cm3=2207×700 cm3=22000 cm3
Hence, aq

(b)
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 20 cm, r = 12 cm and h = 15 cm.
Slant height of the bucket, l=h2+R-r2
                                       =152+20-122=225+64=289=17 cm
Hence, bs

(c)
Let R and r be the top and base of the bucket and let l be its slant height.
Then, R = 33 cm, r = 27 cm and h = 10 cm
Total surface area of the bucket=πR2+r2+lR+r
                                              =π×332+272+10×33+27=π×1089+729+600=2418π cm2
Hence, cp

(d)
Let the diameter of the required sphere be d.
Then, volume of the sphere=43πr3
                                         =43πd23
Therefore,
43πd23=43π33+43π43+43π5343πd38=43π×33+43+53d38=216
d3
= 1728
d3
= 123
d  = 12 cm

Hence, dr



Page No 838:

Answer:

Assertion (A):
Let R and r be the top and base of the bucket and let h be its height.
Then, R = 15 cm, r = 5 cm and h = 24 cm
Slant height, l=h2+R-r2
                  =242+15-52=576+100=676=26 cm
Surface area of the bucket=πR2+r2+lR+r
                                      =π×152+52+26×15+5=π×225+25+520=770π cm2
Thus, the area and the formula are wrong.

Note:
Question seems to be incorrect.



Page No 839:

Answer:

(d) Assertion (A) is false and Reason (R) is true.
Assertion (A):
Total surface area of the hemisphere=3πr2
                                                     =3×227×7×7 cm2=462 cm2

Cost of painting at Rs 5 per cm2=Rs 462×5

                                                = Rs 2310
Hence, Assertion (A) is false.

Reason (R): The given statement is true.

Page No 839:

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
Assertion (A):
Volume of the cuboid =l×b×h
                                =10×5.5×3.5 cm3
Volume of each coin=πr2h
                              =227×175200×175200×15   d=1.75cmr=1.752cm, h=2mmh=210cm

Number of coins=Volume of the cuboidVolume of each coin

                        =10×55×35×7×200×200×510×10×22×175×175=400
Hence, Assertion (A) is true.

Reason (R): The given statement is true.

Page No 839:

Answer:

(d) Assertion (A) is false and Reason (R) is true.
Assertion (A):
Let R and r be the radii of the two spheres.
Then, ratio of their volumes=43πR343πr3
Therefore,
43πR343πr3=278R3r3=278Rr3=323Rr=32
Hence, the ratio of their surface areas=4πR24πr2
                                                       =R2r2=Rr2=322=94=9:4
Hence, Assertion (A) is false.

Reason (R): The given statement is true.

Page No 839:

Answer:

(c) Assertion (A) is true and Reason (R) is false.
Assertion (A):
Curved surface area of a cone=πrr2+h2
                                              =π×3×32+42=π×3×9+16=π×3×25=15π cm2
Hence, Assertion (A) is true.

Reason (R): The given statement is false.



Page No 849:

Answer:

We have,Radius of the sphere, R=62=3 cm,Radius of the cylinder, r=42=2 cm andHeight of the cylinder, h=45 cmNow,The number of solid spheres=Volume of the CylinderVolume of the sphere=πr2h43πR3=3r2h4R3=3×2×2×454×3×3×3=5

So, the number of solid spheres so moulded is 5.

Page No 849:

Answer:

Let the radii of the two cylinders be r and R; and the heights be h and H.We have,hH=12         .....iNow,Volume of the first cylinder=Volume of the second sphereπr2h=πR2HhH=R2r212=R2r2r2R2=12rR2=21rR=21 r:R=2:1

So, the ratio of their radii is 2 : 1.

Page No 849:

Answer:

We have,Height of the cylindrical part, H=4 m,Radius of the base, r=1052 m andSlant height of the conical part, l=40mNow,The total area of canvas required=CSA of conical part+CSA of cylindrical part=πrl+2πrH=πrl+2H=227×1052×40+2×4=11×15×48=7920 m2

So, the area of the canvas required to make the tent is 7920 m2.

Page No 849:

Answer:

Let R and r be the radii of the top and base of the bucket, respectively, and let l be its slant height.
Then, curved surface area of the bucket
= Curved surface area of the frustum of the cone
=πlR+r=227×45×28+7 cm2=227×45×35 cm2=4950 cm2

Page No 849:

Answer:

Radius of cone = 12 cm
Height of cone = 24 cm
Volume of the metallic cone=13πr2h
                                         =13π×122×24

Radius of spherical ball = 62 cm = 3 cm

Volume of each spherical ball=43πr3
                                           =43π×33

Number of balls formed=Volume of the metallic coneVolume of each spherical ball

                                   =π×12×12×24×33×4×π×3×3×3=32

Page No 849:

Answer:

Radius of hemispherical ball = 302= 15 cm
Volume of hemispherical bowl=23πr3
                                                 =23π×15×15×15 cm3

Radius of each bottle = 52cm

Height of each bottle = 6 cm

Volume of each bottle=πr2h
                                =π×52×52×6 cm3

Number of bottles required=Volume of the hemispherical bowlVolume of each bottle

                                        =2×π×15×15×15×2×23×π×5×5×6=60

Page No 849:

Answer:

Radius of sphere = 212 cm
Volume of the metallic sphere=43πr3
                                           =43π×212×212×212 cm3

Radius of cone = 3.52 cm
Height of cone =  3 cm

Volume of each small cone=13πr2h
                               =13π×3520×3520×3 cm3

Number of cones=Volume of the metallic sphereVolume of each cone
                         
                        =4×π×21×21×3×20×203×2×2×2×π×35×3×3×3=504

Page No 849:

Answer:

Radius of the sphere = 422 = 21 cm
Volume of the sphere=43πr3
                               =43π×21×21×21 cm3

Radius of the wire = 2.82 = 1.4 cm
Let the length of the wire be h cm. Then,
Volume of the wire=πr2h
                            =π×1410×1410×h cm3

Therefore,

  43π×21×21×21=π×1410×1410×h12348=4925×hh=12348×2549h=6300 cmh=6300100 mh=63 m

Hence, the length of the wire is 63 m.

Page No 849:

Answer:

Let R and r be the radii of the top and base, respectively, of the drinking glass and let its height be h.

Then, R62cm=3 cm, r42 cm=2 cm, h=21 cm

Capacity of the glass = Capacity of the frustum of the cone
                              =πh3R2+r2+Rr=227×13×21×32+22+3×2 cm3=22×19 cm3=418 cm3

Page No 849:

Answer:

Volume of the cube = a3

Therefore,
  a3=64a3=43a=4 cm

Each side of the cube = 4 cm

Then,
Length of the cuboid 2×4 cm=8 cm
Breadth of the cuboid = 4 cm
Height of the cuboid = 4 cm

Total surface area of the cuboid =2lb+bh+lh
                                             =28×4+4×4+8×4 cm2=2×80 cm2=160 cm2

Page No 849:

Answer:

Let the radius of the cylinder be 2x cm and its height be 3x cm.

Then, volume of the cylinder=πr2h
                                          =227×2x2×3x
Therefore,
   227×2x2×3x=1617227×4x2×3x=1617227×12x3=1617x3=1617×722×12

x3=72×72×72x3=723x=72
Now, r=7 cm and h=212 cm
Hence, the total surface area of the cylinder:

2πrh+2πr2
​ =2πrh+r=2×227×7×212+7 cm2=2×227×7×352 cm2=770 cm2                                                              

Page No 849:

Answer:

Radius of the hemisphere = Radius of the cone = 7 cm

Height of the cone=31-7cm = 24 cm

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

Total surface area of the toy = (Curved surface area of the hemisphere) + (Curved surface area of the cone)
                                        =2πr2+πrl=π×r×2r+l=227×7×14+25 cm2=858 cm2



Page No 850:

Answer:

Radius of hemisphere = 9 cm
Volume of hemisphere=23πr3
                                      =23π×9×9×9cm3
Radius of each bottle = 32 cm
Height of each bottle = 4 cm
Volume of each bottle=πr2h
                                =π×32×32×4 cm3

Number of bottles=Volume of the hemisphereVolume of each bottle
=2π×9×9×9×2×23×π×3×3×4=54

Page No 850:

Answer:

Surface area of the sphere=4πr2
Surface area of the cube=6a2
Therefore,

   4πr2=6a22πr2=3a2r2=3a22πr=32πa

Ratio of their volumes=43πr3a3=4πr33a3

                                =4π3a3×33a32π2π  Since r=32π a=232π=2×32×227=2×3×72×2×11=2111

Thus, the ratio of their volumes is 21 : 11.

Page No 850:

Answer:

Let R and r be the radii of the top and base of the frustum of the cone, respectively, and its slant height be l.

Then,
        2πR=18πR=9   ...(i)2πr=6πr=3      ....(ii) 

Curved surface area of the frustum=πlR+r
                                                   =l×πR+πr=4×9+3   Since l=4 cm    (from (i) and (ii))=48 cm2

Page No 850:

Answer:

Radius of the hemispherical end = 7 cm
Height of the hemispherical end = 7 cm
Height of the cylindrical part=104-2×7cm=90 cm
Surface area of the solid = 2(curved surface area of the hemisphere) + (curved surface area of the cylinder)
                                   =22πr2+2πrh
                                  =2πr2r+h=2×227×7×2×7+90 cm2=44×104 cm2=4576 cm2

Page No 850:

Answer:



We have,Height of the cylinder=Height of the cone=h=15 cm andRadius of the cylinder=Radius of the cone=r=162=8 cmAlso, the slant height of the cone, l=h2+r2=152+82=225+64=289=17 cmNow,The total surface area of the remaining solid=CSA of the cone+CSA of the cylinder+Area of the base=πrl+2πrh+πr2=πrl+2h+r=3.14×8×17+2×15+8=3.14×8×55=1381.6 cm2

So, the total surface area of the remaining solid is 1381.6 cm2.

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

Page No 850:

Answer:

We have,Length of the rectangular block, l=4.4 m,Breadth of the rectangular block, b=2.6 m,Height of the rectangular block, h=1 m,Internal radius of the cylindrical pipe, r=30 cm=0.3 m andThickness of the pipe=5 cm=0.05 mAlso, the external radius of the pipe=0.3+0.05=0.35 mLet the length of the pipe be H.Now,Volume of the pipe=Volume of the blockπR2H-πr2H=lbhπR2-r2H=lbh227×0.352-0.32H=4.4×2.6×1227×0.1225-0.09H=4.4×2.6227×0.0325×H=4.4×2.6H=4.4×2.6×722×0.0325 H=112 m

So, the length of the pipe is 112 m.

Page No 850:

Answer:


We have,Radius of the upper end of the frustum, R=452 cm,Radius of the lower end of the frustum=Radius of the cylinder=r=252 cm,Height of the cylinder, h=6 cm andTotal height of the bucket=40 cmAnd, the height of the frustum, H=40-6=34 cmAlso, the slant height of frustum, l=R-r2+H2=452-2522+342=102+342=100+1156=125635.44 cmNow,The area of the metallic sheet used to make the bucket=CSA of the frustum+CSA of the cylinder+Area of the base of the cylinder=πR+rl+2πrh+πr2=227×452+252×35.44+2×227×252×6+227×252×252=227×35×35.44+227×150+227×6254=227×1240.4+150+156.25=227×1546.65=4860.9 cm2Also,The volume of the water that the bucket can hold=Volume of the frustum=13πHR2+r2+Rr=13×227×34×4522+2522+452252=13×227×34×4522+2522+452252=74821×20254+6254+11254=74821×37754=33615.48 cm3=33.61548 L     As, 1000 cm3=1 L33.61 L

Page No 850:

Answer:

We have,Internal radius of the pipe, r=202=10 cm=0.1 m,Radius of the cylindrical tank, R=102=5 m andHeight of the cylindrical tank, H=2 mAlso, the speed of the water flow in the pipe, h=4 km/hr=4×1000 m1 hr=4000 m/hrNow,The volume of the water flowing out of the pipe in a hour=πr2h=227×0.1×0.1×4000=8807 m3And,The volume of the cylindrical tank=πR2H=227×5×5×2=11007 m3So,The time taken to fill the tank=Volume of the cylindrical tankVolume of water flowing out of the pipe in a hour=110078807=1100880=54 hr=114 hr=1 hr and 14×60 min=1 hr 15 min

So, the tank will be completely filled in 1 hour 15 minutes.



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