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Page No 761:

Question 1:

Two cubes each of volume 27 cm3 are joined end to end to form a solid. Find the surface area of the resulting cuboid.     [CBSE 2011, 14]

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 761:

Question 2:

The volume of a hemisphere is 242512 cm3. Find its curved surface area.                                                   [CBSE 2012]

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 761:

Question 3:

If the total surface area of a solid hemisphere is 462 cm2, then find its volume.                                                   [CBSE 2014]

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 761:

Question 4:

(i) A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of 25 per metre.         [CBSE 2014]
(ii) The radius and height of a solid right-circular cone are in the ratio of 5 : 12. If its volume is 314 cm3, find the total surface area. [Take π = 3.14.]         [CBSE 2017]

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 762:

Question 5:

If the volumes of two cones are in the ratio of 1:4 and their diameters are in the ratio of 4:5, then find the ratio of their heights.

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 762:

Question 6:

The slant height of a conical mountain is 2.5 km and the area of its base is 1.54 km2. Find the height of the mountain.

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 762:

Question 7:

The sum of the radius of the base and the height of a solid cylinder is 37 metres. If the total surface area of the cylinder be 1628 sq metres, then find its volume.

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 762:

Question 8:

The surface area of a sphere is 2464 cm2. If its radius be doubled, then what will be the surface area of the new sphere?

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 762:

Question 9:

A military tent of height 8.25 m is in the form of a right circular cylinder of base diameter 30 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of canvas used in making the tent, if the breadth of the canvas is 1.5 m.   [CBSE 2012]

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 762:

Question 10:

A tent is in the shape of a right circular cylinder up to a height of 3 m and conical above it. The total height of the tent is 13.5 m and the radius of its base is 14 m. Find the cost of cloth required to make the tent at the rate of Rs 80 per square metre.

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 762:

Question 11:

A circus tent is cylindrical to a height of 3 m and conical above it. If its base radius is 52.5 m and the slant height of the conical portion is 53 m, find the area of canvas needed to make the tent.

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 762:

Question 12:

A rocket is in the form of a circular cylinder closed at the lower end and a cone of the same radius is attached to the top. The radius of the cylinder is 2.5 m, its height is 21 m and the slant height of the cone is 8 m. Calculate the total surface area of the rocket.

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 762:

Question 13:

A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them being 3.5 cm and the total height of the solid is
9.5 cm. Find the volume of the solid.                                                                                                                                         [CBSE 2012]

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 762:

Question 14:

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius on its circular face. The total height of the toy is 15.5 cm. Find the total surface area of the toy.       [CBSE 2017]

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 762:

Question 15:

A toy is in the shape of a cone mounted on a hemisphere of same base radius. If the volume of the toy is 231 cm3 and its diameter is 7 cm, then find the height of the toy.                                                                                                                                                       [CBSE 2012]

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 762:

Question 16:

A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is 4 times the radius of its base, then find the radius of the ice-cream cone.

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 762:

Question 17:

A vessel is in the form of a hemispherical bowl surmounted by a hollow cylinder. The diameter of the hemisphere is 21 cm and the total height of the vessel is 14.5 cm. Find its capacity.

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 763:

Question 18:

A toy is in the form of a cylinder with hemispherical ends. If the whole length of the toy is 90 cm and its diameter is 42 cm, then find the cost of painting the toy at the rate of 70 paise per sq cm.                                                                                                                           [CBSE 2014]

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 763:

Question 19:

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

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 763:

Question 20:

A wooden article was made by scooping out a hemisphere from each end of a cylinder, as shown in the figure. If the height of the cylinder is 20 cm and its base is of diameter 7 cm, find the total surface area of the article when it is ready.         [CBSE 2008C]

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 763:

Question 21:

A solid is in the form of a right circular cone mounted on a hemisphere. The radius of the hemisphere is 2.1 cm and the height of the cone is 4 cm. The solid is placed in a cylindrical tub full of water in such a way that the whole solid is submerged in water. If the radius of the cylinder is 5 cm and its height is 9.8 cm, find the volume of the water left in the tub.

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 763:

Question 22:

From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and base radius 6 cm is hollowed out. Find the volume of the remaining solid. Also, find the total surface area of the remaining solid.

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 763:

Question 23:

From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid.                                                                                                                               [CBSE 2014]

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 763:

Question 24:

From a solid cylinder of height 14 cm and base diameter 7 cm, two equal conical holes each of radius 2.1 cm and height 4 cm are cut off. Find the volume of the remaining solid.                                                                                                                                       [CBSE 2011]

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 763:

Question 25:

A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weight, a conical hole is drilled in the cylinder. The conical hole has a radius of 32 cm and its depth is 89 cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.                                                                                                                                                                                 [CBSE 2015]

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 763:

Question 26:

A spherical glass vessel has a cylindrical neck that is 7 cm long and 4 cm in diameter. The diameter of the spherical part is 21 cm. Find the quantity of water the vessel can hold.

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 763:

Question 27:

The adjoining figure represents a solid consisting of a cylinder surmounted by a cone at one end and a hemisphere at the other. Find the volume of the solid.

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 764:

Question 28:

From a cubical piece of wood of side 21 cm, a hemisphere is carved out in such a way that the diameter of the hemisphere is equal to the side of the cubical piece. Find the surface area and volume of the remaining piece.                                                                                 [CBSE 2014]

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 764:

Question 29:

(i) A hemisphere of maximum possible diameter is placed over a cuboidal block of side 7 cm. Find the surface area of the solid so formed.
  [CBSE 2017]

(ii) A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of 5 per 100 sq cm. [Use π = 3.14]                              [CBSE 2015]

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 764:

Question 30:

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm, respectively. The radii of the hemispherical and the conical parts are the same as that of the cylindrical part. Find the surface area of the toy, if the total height of the toy is 30 cm.

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 764:

Question 31:

The inner diameter of a glass is 7 cm and it has a raised portion in the bottom in the shape of a hemisphere, as shown in the
figure. If the height of the glass is 16 cm, find the apparent capacity and the actual capacity of the glass.

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 764:

Question 32:

A wooden toy is in the shape of a cone mounted on a cylinder, as shown in the figure. The total height of the toy is 26 cm, while the height of the conical part is 6 cm. The diameter of the base of the conical part is 5 cm and that of the cylindrical part is 4 cm. The conical part and the cylindrical part are respectively painted red and white. Find the area to be painted by each of these colours. [Take π=227]

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 784:

Question 1:

A solid metallic cuboid of dimensions 9 m × 8 m × 2 m is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed.       [CBSE 2017]

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 784:

Question 2:

A cone of height 20 cm and radius of base 5 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the diameter of the sphere.                                                                                                                                                                                    [CBSE 2011]

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 784:

Question 3:

Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively are melted to form a single solid sphere. Find the radius of the resulting sphere.
                                                                                                                                                                                                           [CBSE 2012]

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 784:

Question 4:

A solid metal cone with base radius of 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls thus formed.

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 784:

Question 5:

The radii of internal and external surfaces of a hollow spherical shell are 3 cm and 5 cm, respectively. It is melted and recast into a solid cylinder of diameter 14 cm. Find the height of the cylinder.                                                                                                     [CBSE 2012]

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 785:

Question 6:

The internal and external diameters of a hollow hemispherical shell are 6 cm and 10 cm, respectively. It is melted and recast into a solid cone of base diameter 14 cm. Find the height of the cone so formed.

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 785:

Question 7:

A copper rod of diameter 2 cm and length 10 cm is drawn into a wire of uniform thickness and length 10 m. Find the thickness of the wire.
                                                                                                                                                                                                        [CBSE 2012]

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 785:

Question 8:

A hemispherical bowl of internal diameter 30 cm contains some liquid. This liquid is to be poured into cylindrical bottles of diameter 5 cm and height 6 cm each. Find the number of bottles required.

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 785:

Question 9:

A solid metallic sphere of diameter 21 cm is melted and recast into a number of smaller cones, each of diameter 3.5 cm and height 3 cm. Find the number of cones so formed.

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 785:

Question 10:

A spherical cannon ball, 28 cm in diameter, is melted and recast into a right circular conical mould with base diameter of 35 cm. Find the height of the cone.

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 785:

Question 11:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.

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 785:

Question 12:

A spherical shell of lead, whose external and internal diameters are 24 cm and 18 cm, is melted and recast into a right circular cylinder 37 cm high. Find the diameter of the base of the cylinder.

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 785:

Question 13:

A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm. Find the radius of the base of the cone.

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 785:

Question 14:

A spherical ball of diameter 21 cm is melted and recast into cubes, each of side 1 cm. Find the number of cubes so formed.

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 785:

Question 15:

How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?

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 785:

Question 16:

A solid sphere of radius 3 cm is melted and then cast into small spherical balls, each of diameter 0.6 cm. Find the number of balls obtained.

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 785:

Question 17:

The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire.

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 785:

Question 18:

The diameter of a copper sphere is 18 cm. It is melted and drawn into a long wire of uniform cross-section. If the length of the wire is 108 m, find its diameter.

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 786:

Question 19:

A hemispherical bowl of internal radius 9 cm is full of water. Its contents are emptied into a cylindrical vessel of internal radius 6 cm. Find the height of water in the cylindrical vessel.                                                                                                                                   [CBSE 2012]

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 786:

Question 20:

A hemispherical tank, full of water, is emptied by a pipe at the rate of 257 litres per second. How much time will it take to empty half the tank if the diameter of the base of the tank is 3 m?                                                                                                                       [CBSE 2012]

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 786:

Question 21:

The rain water from a roof of 44 m × 20 m drains into a cylindrical tank having diameter of base 4 m and height 3.5 m. If the tank is just full, then find the rainfall in cm.                                                                                                                                                                  [CBSE 2014]

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 786:

Question 22:

The rain water from a 22 m × 20 m roof drains into a cylindrical vessel of diameter 2 m and height 3.5 m. If the rain water collected from the roof fills 45th of the cylindrical vessel, then find the rainfall in centimetre.                                                                                      [CBSE 2015]

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 786:

Question 23:

A solid right circular cone of height 60 cm and radius 30 cm is dropped in a right circular cylinder full of water, of height 180 cm and radius 60 cm. Find the volume of water left in the cylinder, in cubic metres.                                                                        [CBSE 2015]

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 786:

Question 24:

Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m per second. Determine the rise in level of water in the tank in half an hour.                                                                         [CBSE 2013]

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 786:

Question 25:

Water is flowing at the rate of 6 km/hr through a pipe of diameter 14 cm into a rectangular tank which is 60 m long and 22 m wide. Determine the time in which the level of water in the tank will rise by 7 cm.                                                                                   [CBSE 2011]

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 786:

Question 26:

Water in a canal, 5.4 m wide and 1.8 m deep, is flowing with a speed of 25 km/hr. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?    [CBSE 2017]

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 786:

Question 27:

A farmer connects a pipe of internal diameter 25 cm from a canal into a cylindrical tank in his field, which is 12 m in diameter and 2.5 m deep. If water flows through the pipe at the rate of 3.6 km/hr, then in how much time will the tank be filled? Also, find the cost of water if the canal department charges at the rate of 0.07 per m3.                                                                                                                      [CBSE 2009C]

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 786:

Question 28:

Water running in a cylindrical pipe of inner diameter 7 cm, is collected in a container at the rate of 192.5 litres per minute. Find the rate of flow of water in the pipe in km/hr.                                                                                                                                                [CBSE 2013]

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 786:

Question 29:

150 spherical marbles, each of diameter 1.4 cm, are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel.                                                        [CBSE 2014]

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 787:

Question 30:

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm, containing some water. Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm.

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 787:

Question 31:

In a village, a well with 10 m inside diameter, is dug 14 m deep. Earth taken out of it is spread all around to a width 5 m to form an embankment. Find the height of the embankment. What value of the villagers is reflected here?                                      [CBSE 2014]

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 787:

Question 32:

In a corner of a rectangular field with dimensions 35 m×22 m, a well with 14 m inside diameter is dug 8 m deep. The earth dug out is spread evenly over the remaining part of the field. Find the rise in the level of the field.                                                                                    [HOTS]

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 787:

Question 33:

A copper wire of diameter 6 mm is evenly wrapped on a cylinder of length 18 cm and diameter 49 cm to cover its whole surface. Find the length and the volume of the wire. If the density of the copper be 8.8 g per cm3, then find the weight of the wire.                           [HOTS]

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 787:

Question 34:

A right triangle whose sides are 15 cm and 20 cm (other than hypotenuse), is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate)                                                                       [HOTS]

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 787:

Question 35:

In a hospital, used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of hospital whose length is 25 m and breadth is 20 m. If the tank is filled completely then what will be the height of standing water used for irrigating the park? Write your views on recycling of water.              [CBSE 2017]

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 797:

Question 1:

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 16 cm and 12 cm. Find the capacity of the glass.                                                                                                                                                               [CBSE 2012)

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 797:

Question 2:

The radii of the circular ends of a solid frustum of a cone are 18 cm and 12 cm and its height is 8 cm. Find its total surface area.
[Use π = 3.14]                                                                                                                                                               [CBSE 2011]

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 797:

Question 3:

A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm, respectively. Find
(i) the volume of water which can completely fill the bucket;
(ii) the area of the metal sheet used to make the bucket.                                                                                                                    [CBSE 2014]

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 798:

Question 4:

A container, open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm, respectively. Find the cost of milk which can completely fill the container at the rate of 21 per litre.                         [CBSE 2014]

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 798:

Question 5:

A container made of a metal sheet open at the top is of the form of frustum of cone, whose height is 16 cm and the radii of its lower and upper circular edges are 8 cm and 20cm respectively. Find
(i) the cost of metal sheet used to make the container if it costs ₹ 10 per 100 cm2            [CBSE 2013]
(ii) the cost of milk at the rate of ₹ 35 per litre which can fill it completely.                       [CBSE 2017]

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 798:

Question 6:

The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm, and its slant height is 10 cm. Find its capacity and total surface area.

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 798:

Question 7:

A bucket is in the form of a frustum of a cone. Its depth is 15 cm and the diameters of the top and the bottom are 56 cm and 42 cm, respectively. Find how many litres of water can the bucket hold.

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 798:

Question 8:

A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm and radii of its lower and upper ends are 8 cm and 20 cm, respectively. Find the cost of the bucket if the cost of metal sheet used is Rs 15 per 100 cm2.

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 798:

Question 9:

A bucket made up of a metal sheet is in the form of frustum of a cone. Its depth is 24 cm and the diameters of the top and bottom are 30 cm and 10 cm, respectively. Find the cost of completely filling the bucket with milk at the rate of Rs 20 per litre and the cost of metal sheet used if it costs Rs 10 per 100 cm2.

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 798:

Question 10:

A container in the shape of a frustum of a cone having diameters of its two circular faces as 35 cm and 30 cm and vertical height 14 cm,
is completely filled with oil. If each cm3 of oil has mass 1.2 g, then find the cost of oil in the container if it costs 40 per kg.   [CBSE 2009C]

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 798:

Question 11:

A bucket is in the form of a frustum of a cone and it can hold 28.49 litres of water. If the radii of its circular ends are 28 cm and 21 cm, then find the height of the bucket.                                                                                                                                                              [CBSE 2012]

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 798:

Question 12:

The radii of the circular ends of a bucket of height 15 cm are 14 cm and r cm (r < 14). If the volume of bucket is 5390 cm3, then find the value of r.                                                                                                                                                                                            [CBSE 2011]

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 798:

Question 13:

The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Find its total surface area. [Use π=3.14.]

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 799:

Question 14:

A tent is made in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20 m and 6 m, respectively, and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required.

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 799:

Question 15:

A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper and lower circular ends of the frustum be 14 m and 26 m, respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be 12 m, find the area of the canvas required to make the tent. (Assume that the radii of the upper circular end of the frustum and the base of the surmounted conical portion are equal.)

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 799:

Question 16:

The perimeters of the two circular ends of a frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm, then find its volume and curved surface area.                                                                                                                                                                     [CBSE 2014]

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

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Question 17:

A solid cone of base radius 10 cm is cut into two parts through the midpoint of its height, by a plane parallel to its base. Find the ratio of the volumes of the two parts of the cone.                                                                                                                                                [CBSE 2013]

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.

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Question 18:

The height of a right circular cone is 20 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be 18 of the volume of the given cone, then at what height above the base is the section made?                                                           [HOTS] [CBSE 2014]

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.

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Question 19:

A solid metallic right circular cone 20 cm high and whose vertical angle is 60°, is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 112 cm, then find the length of the wire.
                                                                                                                                                                                             [HOTS] [CBSE 2014]

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.

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Question 20:

A fez, the cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, radius at the upper base is
4 cm and its slant height is 15 cm, then find the area of material used for making it. Use π=227

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.

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Question 21:

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm,
diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, then find the area of the tin sheet required to make the funnel.

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 800:

Question 22:

A right circular cone is divided into three parts by trisecting its height by two planes drawn parallel to the base. Show that the volume of the three portions starting from the top are in the ratio 1 : 7 : 19.            [CBSE 2017]

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 801:

Question 1:

A river 1.5 m deep and 36 m wide is flowing at the rate of 3.5 km/hr. Find the amount of water (in cubic metres) that runs into the sea per minute.

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 801:

Question 2:

The volume of a cube is 729 cm3. Find its surface area.

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 801:

Question 3:

How many cubes of 10 cm edge can be put in a cubical box of 1 m edge?

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 801:

Question 4:

Three cubes of iron whose edges are 6 cm, 8 cm and 10 cm, respectively are melted and formed into a single cube. Find the edge of the new cube formed.

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 801:

Question 5:

Five identical cubes, each of edge 5 cm, are placed adjacent to each other. Find the volume of the resulting cuboid.

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 801:

Question 6:

The volumes of two cubes are in the ratio 8 : 27. Find the ratio of their surface areas.

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 801:

Question 7:

The volume of a right circular cylinder with its height equal to the radius is 2517 cm3. Find the height of the cylinder.

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 801:

Question 8:

The ratio between the radius of the base and the height of a cylinder is 2 : 3. If the volume of the cylinder is 12936 cm3, then find the radius of the base of the cylinder.

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 801:

Question 9:

The radii of two cylinders are in the ratio of 2 : 3 and their heights are in the ratio of 5 : 3. Find the ratio of their volumes.

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 801:

Question 10:

66 cubic cm of silver is drawn into a wire 1 mm in diameter. Calculate the length of the wire in metres.

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 802:

Question 11:

If the area of the base of a right circular cone is 3850 cm2 and its height is 84 cm, then find the slant height of the cone.

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 802:

Question 12:

A cylinder with base radius 8 cm and height 2 cm is melted to form a cone of height 6 cm. Calculate the radius of the base of the cone.

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 802:

Question 13:

A right cylindrical vessel is full of water. How many right cones having the same radius and height as those of the right cylinder will be needed to store that water?

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 802:

Question 14:

The volume of a sphere is 4851 cm3. Find its curved surface area.

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 802:

Question 15:

The curved surface area of a sphere is 5544 cm2. Find its volume.

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 802:

Question 16:

The surface areas of two spheres are in the ratio of 4 : 25. Find the ratio of their volumes.

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 802:

Question 17:

A solid metallic sphere of radius 8 cm is melted and recast into spherical balls each of radius 2 cm. Find the number of spherical balls obtained.

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 802:

Question 18:

How many lead shots each 3 mm in diameter can be made from a cuboid of dimensions 9 cm × 11 cm × 12 cm?

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 802:

Question 19:

A metallic cone of radius 12 cm and height 24 cm is melted and made into spheres of radius 2 cm each. How many spheres are formed?

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 802:

Question 20:

A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. Find the radius of the base of the cone.

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 802:

Question 21:

A copper sphere of diameter 18 cm is drawn into a wire of diameter 4 mm. Find the length of the wire.

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 802:

Question 22:

The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6 cm, respectively. Find the slant height of the frustum.

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 802:

Question 23:

Find the ratio of the volume of a cube to that of a sphere which will fit inside it.

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 802:

Question 24:

Find the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?

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 802:

Question 25:

Two cubes each of volume 125 cm3 are joined end to end to form a solid. Find the surface area of the resulting cuboid.            [CBSE 2013C]

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 802:

Question 26:

Three metallic cubes whose edges are 3 cm, 4 cm and 5 cm, are melted and recast into a single large cube. Find the edge of the new cube formed.                                                                                                                                                                                       [CBSE 2011]

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 802:

Question 27:

A solid metallic sphere of diameter 8 cm is melted and drawn into a cylindrical wire of uniform width. If the length of the wire is 12 m, then find its width.                                                                                                                                                                                      [CBSE 2013]

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 803:

Question 28:

A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used, at the rate of 25 per metre.                                                                                                                                                                                                  [CBSE 2014]

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.

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Question 29:

A wooden toy was made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm
and its base is of radius 3.5 cm, then find the volume of wood in the toy.                                                                                         [CBSE 2013]

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.

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Question 30:

Three cubes of a metal whose edges are in the ratio 3 : 4 : 5 are melted and converted into a single cube whose diagonal is 123 cm. Find the
edges of the three cubes.                                                                                                                                                                  [CBSE 2013C]

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.

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Question 31:

A hollow sphere of external and internal diameters 8 cm and 4 cm, respectively is melted into a solid cone of base diameter 8 cm. Find the height of the cone.                                                                                                                                                                      [CBSE 2013C]

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.

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Question 32:

A bucket of height 24 cm is in the form of frustum of a cone whose circular ends are of diameter 28 cm and 42 cm. Find the cost of milk at the rate of 30 per litre, which the bucket can hold.                                                                                                                    [CBSE 2013C]

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.

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Question 33:

The interior of a building is in the form of a right circular cylinder of diameter 4.2 m and height 4 m surmounted by a cone of same diameter.
The height of the cone is 2.8 m. Find the outer surface area of the building.                                                                                    [CBSE 2014]

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.

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Question 34:

A metallic solid right circular cone is of height 84 cm and the radius of its base is 21 cm. It is melted and recast into a solid sphere. Find the
diameter of the sphere.                                                                                                                                                                        [CBSE 2014]

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.

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Question 35:

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total
surface area of the toy.                                                                                                                                                                        [CBSE 2012]

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.

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Question 36:

If the radii of the circular ends of a bucket 28 cm high, are 28 cm and 7 cm, then find its capacity and total surface area.             [CBSE 2011]

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.

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Question 37:

A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm3 of water. The radii of the top and bottom circular ends are
20 cm and 12 cm, respectively. Find the height of the bucket. [Use π = 3.14]                                                                              [CBSE 2006C]

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.

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Question 38:

A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 1045937 cm3. The radii of its lower and upper circular ends are 8 cm and 20 cm, respectively. Find the cost of metal sheet used in making the container at the rate of 1.40 per cm2.
                                                                                                                                                                                                           [CBSE 2010]

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.



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Question 39:

A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 423 cm and height 3 cm. Find the number of cones so formed.

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.

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Question 40:

A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the volume of water
(i) displaced out of the cylinder
(ii) left in the cylinder.                                                                                                                                                                       [CBSE 2009]

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 804:

Question 1:

A cylindrical pencil sharpened at one end is a combination of
(a) a cylinder and a cone
(b) a cylinder and frustum of a cone
(c) a cylinder and a hemisphere
(d) two cylinders

Answer:

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

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Question 2:

A shuttlecock used for playing badminton is a combination of


(a) cylinder and a hemisphere
(b) frustum of a cone and a hemisphere
(c) a cone and a hemisphere
(d) a cylinder and a sphere

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.



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Question 3:

A funnel is a combination of


(a) a cylinder and a cone
(b) a cylinder and a hemisphere
(c) a cylinder and frustum of a cone
(d) a cone and hemisphere

Answer:

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

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Question 4:

A surahi is a combination of


(a) a sphere and a cylinder
(b) a hemisphere and a cylinder
(c) a cylinder and a cone
(d) two hemispheres

Answer:

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

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Question 5:

The shape of a glass (tumbler) is usually in the form of


(a) a cylinder
(b) frustum of a cone
(c) a cone
(d) a sphere

Answer:

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

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Question 6:

The shape of the gilli used in a gilli-danda game is a combination of


(a) a cone and a cylinder
(b) two cylinders
(c) two cones and a cylinder
(d) two cylinders and a cone

Answer:

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

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Question 7:

A plumbline (sahul) is a combination of


(a) a hemisphere and a cone
(b) a cylinder and a cone
(c) a cylinder and frustum of a cone
(d) a cylinder and a sphere

Answer:

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

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Question 8:

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


(a) a cone
(b) a sphere
(c) a cylinder
(d) frustum of a cone

Answer:

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

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Question 9:

During conversion of a solid from one shape to another, the volume of the new shape will
(a) decrease
(b) increase
(c) remain unaltered
(d) be doubled

Answer:

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



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Question 10:

In a right circular cone, the cross-section made by a plane parallel to the base is a
(a) sphere
(b) hemisphere
(c) circle
(d) a semicircle

Answer:

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

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Question 11:

A solid piece of iron in the form a cuboid of dimensions (49 cm × 33 cm × 24 cm) is moulded into a solid sphere. The radius of the sphere is
(a) 19 cm
(b) 21 cm
(c) 23 cm
(d) 25 cm

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.

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Question 12:

Choose the correct answer of the following:

The radius (in cm) of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is

(a) 2.1                         (b) 4.2                         (c) 8.4                         (d) 1.05                                                                                     [CBSE 2011]

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).

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Question 13:

Choose the correct answer of the following question:

A metallic solid sphere of radius 9 cm is melted to form a solid cylinder of radius 9 cm. The height of the cylinder is

(a) 12 cm                         (b) 18 cm                         (c) 36 cm                        (d) 96 cm                                                                    [CBSE 2014]

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).

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Question 14:

Choose the correct answer of the following question:

A rectangular sheet of paper 40 cm × 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is

(a) 3.5                        (b) 7                        (c) 80                        (d) 5                                                                                                [CBSE 2014]

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).

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Question 15:

Choose the correct answer of the following question:

The number of solid spheres, each of diameter 6 cm, that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm, is

(a) 2                       (b) 4                       (c) 5                       (d) 6                                                                                                         [CBSE 2014]

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).

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Question 16:

Choose the correct answer of the following question:

The surface areas of two spheres are in the ratio 16 : 9. The ratio of their volumes is

(a) 64 : 27                          (b) 16 : 9                          (c) 4 : 3                          (d) 163 : 93                                                             [CBSE 2013C]

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).

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Question 17:

Choose the correct answer of the following question:

If the surface area of a sphere is 616 cm2, its diameter (in cm) is

(a) 7                         (b) 14                         (c) 28                         (d) 56                                                                                           [CBSE 2012]

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).

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Question 18:

Choose the correct answer of the following question:

If the radius of a sphere becomes 3 times, then its volume will become

(a) 3 times                         (b) 6 times                         (c) 9 times                         (d) 27 times                                                         [CBSE 2011]

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).

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Question 19:

Choose the correct answer of the following question:

If the height of a bucket in the shape of frustum of a cone is 16 cm and the diameters of its two circular ends are 40 cm and 16 cm, then its slant height is

(a) 20 cm                         (b) 125 cm                         (c) 813 cm                         (d) 16 cm                                                     [CBSE 2013C]

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).

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Question 20:

Choose the correct answer of the following question:

A sphere of diameter 18 cm is dropped into a cylindrical vessel of diameter 36 cm, partly filled with water. If the sphere is completely submerged, then the water level rises by

(a) 3 cm                         (b) 4 cm                         (c) 5 cm                         (d) 6 cm                                                                          [CBSE 2011]
 

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).



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Question 21:

Choose the correct answer of the following question:

A solid right circular cone is cut into two parts at the middle of its height by a plane parallel to its base. The ratio of the volume of the smaller cone to the whole cone is

(a) 1 : 2                            (b) 1 : 4                            (c) 1 : 6                            (d) 1 : 8                                                                     [CBSE 2012]

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).

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Question 22:

Choose the correct answer of the following question:

The radii of the circular ends of a bucket of height 40 cm are 24 cm and 15 cm. The slant height (in cm) of the bucket is

(a) 41                      (b) 43                      (c) 49                      (d) 51                                                                                                  [CBSE 2012]

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).

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Question 23:

Choose the correct answer of the following question:

A solid is hemispherical at the bottom and conical (of same radius) above it. If the surface areas of the two parts are equal, then the ratio of its radius and the slant height of the conical part is

(a) 1 : 2                        (b) 2 : 1                        (c) 1 : 4                        (d) 4 : 1                                                                                  [CBSE 2011]

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).

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Question 24:

Choose the correct answer of the following question:

If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is

(a) 1 : 2                         (b) 2 : 1                         (c) 1 : 4                         (d) 4 : 1                                                                             [CBSE 2012]

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).

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Question 25:

A cubical ice-cream brick of edge 22 cm is to be distributed among some children by filling ice-cream cones of radius 2 cm and height 7 cm up to the brim. How many children will get the ice-cream cones?
(a) 163
(b) 263
(c) 363
(d) 463

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.

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Question 26:

A mason constructs a wall of dimensions (270 cm × 300 cm × 350 cm) with bricks, each of size (22.5 cm × 11.25 cm × 8.75 cm) and it is assumed that 18 space is covered by the mortar. Number of bricks used to construct the wall is
(a) 11000
(b) 11100
(c) 11200
(d) 11300

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.

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Question 27:

Twelve solid spheres of the same size are made by melting a solid metallic cylinder of base diameter 2 cm and height 16 cm. The diameter of each sphere is
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 6 cm

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.

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Question 28:

The diameters of two circular ends of a bucket are 44 cm and 24 cm, and the height of the bucket is 35 cm. The capacity of the bucket is
(a) 31.7 litres
(b) 32.7 litres
(c) 33.7 litres
(d) 34.7 litres

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.

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Question 29:

The slant height of a bucket is 45 cm and the radii of its top and bottom are 28 cm and 7 cm, respectively. The curved surface area of the bucket is
(a) 4953 cm2
(b)  4952 cm2
(c) 4951 cm2
(d) 4950 cm2

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.



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Question 30:

The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is
(a) 9 : 16
(b) 16 : 9
(c) 3 : 4
(d) 4 : 3

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 808:

Question 31:

A hollow cube of internal edge 22 cm is filled with spherical marbles of diameter 0.5 cm and 18 space of the cube remains unfilled. Number of marbles required is
(a) 142296
(b) 142396
(c) 142496
(d) 142596

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

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Question 32:

A metallic spherical shell of internal and external diameters 4 cm and 8 cm, respectively, is melted and recast in the form of a cone of base diameter 8 cm. The height of the cone is
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 8 cm

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

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Question 33:

A medicine capsule is in the shape of a cylinder of diameter 0.5 cm with a hemisphere tucked at each end. The length of the entire capsule is 2 cm. The capacity of the capsule is
(a) 0.33 cm2
(b) 0.34 cm2
(c) 0.35 cm2
(d) 0.36 cm2

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 808:

Question 34:

The length of the longest pole that can be kept in a room (12 m × 9 m ×8 m) is
(a) 29 m
(b) 21 m
(c) 19 m
(d) 17 m

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 

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Question 35:

The length of the diagonal of a cube is 63 cm. Its total surface area is
(a) 144 cm2
(b) 216 cm2
(c) 180 cm2
(d) 108 cm2

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 808:

Question 36:

The volume of a cube is 2744 cm2. Its surface area is
(a) 196 cm2
(b) 1176 cm2
(c) 784 cm2
(d) 588 cm2

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

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Question 37:

The total surface area of a cube is 864 cm2. Its volume is
(a) 3456 cm3
(b) 432 cm3
(c) 1728 cm3
(d) 3456 cm3

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 808:

Question 38:

How many bricks, each measuring (25 cm × 11.25 cm × 6 cm), will be required to construct a wall (8 m × 6 m × 22.5 cm)?
(a) 8000
(b) 6400
(c) 4800
(d) 7200

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 808:

Question 39:

The area of the base of a rectangular tank is 6500 cm2 and the volume of water contained in it is 2.6 m3. The depth of water in the tank is
(a) 3.5 m
(b) 4 m
(c) 5 m
(d) 8 m

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

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Question 40:

The volume of a wall, 5 times as high as it is broad and 8 times as long as it is high, is 128 m3. The breadth of the wall is
(a) 30 cm
(b) 40 cm
(c) 22.5 cm
(d) 25 cm

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 809:

Question 41:

If the areas of three adjacent faces of a cuboid are x, y and z, respectively, the volume of the cuboid is
(a) xyz
(b) 2xyz
(c) xyz
(d) 3xyz

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.

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Question 42:

The sum of length, breadth and height of a cuboid is 19 cm and its diagonal is 55 cm. Its surface area is
(a) 361 cm2
(b) 125 cm2
(c) 236 cm2
(d) 486 cm2

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 809:

Question 43:

If each edge of a cube is increased by 50%, the percentage increase in the surface area is
(a) 50%
(b) 75%
(c) 100%
(d) 125%

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 809:

Question 44:

How many bags of grain can be stored in a cuboidal granary (8 m × 6 m × 3 m), if each bag occupies a space of 0.64 m3?
(a) 8256
(b) 90
(c) 212
(d) 225

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

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Question 45:

A cube of side 6 cm is cut into a number of cubes, each of side 2 cm. The number of cubes formed is
(a) 6
(b) 9
(c) 12
(d) 27

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

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Question 46:

Rainfall in an area is 5 cm. The volume of the water that falls on 2 hectares of land is
(a) 100 m3
(b) 10 m3
(c) 1000 m3
(d) 10000 m3

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 809:

Question 47:

The volumes of two cubes are in the ratio 1 : 27. The ratio of their surface area is
(a) 1 : 3
(b) 1 : 8
(c) 1 : 9
(d) 1 : 18

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 809:

Question 48:

The diameter of the base of a cylinder is 4 cm and its height is 14 cm. The volume of the cylinder is
(a) 176 cm3
(b) 196 cm3
(c) 276 cm3
(d) 352 cm3

Answer:

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

Page No 809:

Question 49:

The diameter of a cylinder is 28 cm and its height is 20 cm. The total surface area of the cylinder is
(a) 2993 cm2
(b) 2992 cm2
(c) 2292 cm2
(d) 2229 cm2

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 809:

Question 50:

The height of a cylinder is 14 cm and its curved surface area is 264 cm2. The volume of the cylinder is
(a) 308 cm3
(b) 396 cm3
(c) 1232 cm3
(d) 1848 cm3

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 809:

Question 51:

The curved surface are of a cylinder is 1760 cm2 and its base radius is 14 cm. The height of the cylinder is
(a) 10 cm
(b) 15 cm
(c) 20 cm
(d) 40 cm

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 809:

Question 52:

The ratio of the total surface area to the lateral surface area of a cylinder with base radius 80 cm and height 20 cm is
(a) 2 : 1
(b) 3 : 1
(c) 4 : 1
(d) 5 : 1

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 810:

Question 53:

The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. The height of the pillar is
(a) 4 m
(b) 5 m
(c) 6 m
(d) 7 m

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 810:

Question 54:

The ratio between the radius of the base and the height of a cylinder is 2 : 3. If its volume is 1617 cm3, the total surface area of the cylinder is
(a) 308 cm2
(b) 462 cm2
(c) 540 cm2
(d) 770 cm2

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 810:

Question 55:

The radii of two cylinders are in the ratio 2 : 3 and their heights are in the ratio 5 : 3. The ratio of their volumes is
(a) 27 : 20
(b) 20 : 27
(c) 4 : 9
(d) 9 : 4

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 810:

Question 56:

The heights of two circular cylinders of equal volume are in the ratio 1 : 2. The ratio of their radii is
(a) 1:2
(b) 2:1
(c) 1 : 2
(d) 1 : 4

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 810:

Question 57:

The radius of the base of a cone is 5 cm and its height is 12 cm. Its curved surface area is
(a) 60π cm2
(b) 65π cm2
(c) 30π cm2
(d) None of these

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 810:

Question 58:

The diameter of the base of a cone is 42 cm and its volume is 12936 cm3. Its height is
(a) 28 cm
(b) 21 cm
(c) 35 cm
(d) 14 cm

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 810:

Question 59:

The area of the base of a right circular cone is 154 cm2 and its height is 14 cm. Its curved surface area is
(a) 1545cm2
(b) 1547cm2
(c) 777cm2
(d) 775cm2

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 810:

Question 60:

On increasing the radii of the base and the height of a cone by 20%, its volume will increase by
(a) 20%
(b) 40%
(c) 60%
(d) 72.8%

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 810:

Question 61:

The radii of the base of a cylinder and a cone are in the ratio 3 : 4. If their heights are in the ratio 2 : 3, the ratio between their volumes is
(a) 9 : 8
(b) 3 : 4
(c) 8 : 9
(d) 4 : 3

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 810:

Question 62:

A metallic cylinder of radius 8 cm and height 2 cm is melted and converted into a right circular cone of height 6 cm. The radius of the base of this cone is
(a) 4 cm
(b) 5 cm
(c) 6 cm
(d) 8 cm

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 810:

Question 63:

The height of a conical tent is 14 m and its floor area is 346.5 m2. How much canvas, 1.1 wide, will be required for it?
(a) 490 m
(b) 525 m
(c) 665 m
(d) 860 m

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 811:

Question 64:

The diameter of a sphere is 14 cm. Its volume is
(a) 1428 cm3
(b) 1439 cm3
(c) 143713cm3
(d) 1440 cm3

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 811:

Question 65:

The ratio between the volume of two spheres is 8 : 27. What is the ratio between their surface areas?
(a) 2 : 3
(b) 4 : 5
(c) 5 : 6
(d) 4 : 9

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 811:

Question 66:

A hollow metallic sphere with external diameter 8 cm and internal diameter 4 cm is melted and moulded into a cone of base radius 8 cm. The height of the cone is
(a) 12 cm
(b) 14 cm
(c) 15 cm
(d) 18 cm

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 811:

Question 67:

A metallic cone of base radius 2.1 cm and height 8.4 cm is melted and moulded into a sphere. The radius of the sphere is
(a) 2.1 cm
(b) 1.05 cm
(c) 1.5 cm
(d) 2 cm

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 811:

Question 68:

The volume of a hemisphere is 19404 cm3. The total surface area of the hemisphere is
(a) 4158 cm2
(b) 16632 cm2
(c) 8316 cm2
(d) 3696 cm2

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 811:

Question 69:

The surface area of a sphere is 154 cm2. The volume of the sphere is
(a) 17923cm3
(b) 35913cm3
(c) 143713cm3
(d) None of these

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 811:

Question 70:

The total surface area of a hemisphere of radius 7 cm is
(a) (588π) cm2
(b) (392π) cm2
(c) (147π) cm2
(d) (98π) cm2

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 811:

Question 71:

The circular ends of a bucket are of radii 35 cm and 14 cm and the height of the bucket is 40 cm. Its volume is
(a) 60060 cm3
(b) 80080 cm3
(c) 70040 cm3
(d) 80160 cm3

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 811:

Question 72:

If the radii of the ends of a bucket are 5 cm and 15 cm and it is 24 cm high, then its surface area is
(a) 1815.3 cm2
(b) 1711.3 cm2
(c) 2025.3cm2
(d) 2360 cm2

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 811:

Question 73:

A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, the total area of canvas required is
(a) 1760 m2
(b) 2640 m2
(c) 3960 m2
(d) 7920 m2

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 812:

Question 74:

Match the following columns:

Column I Column II
(a) A solid metallic sphere of
radius 8 cm is melted and the
material is used to make solid
right cones with height 4 cm
and base radius of 8 cm.
How many cones are formed?
(p) 18
(b) A 20-m-deep well with
diameter 14 m is dug up
and the earth from digging
is evenly spread out to form
a platform 44 m by 14 m.
The height of the platform
is ...........m.
(q) 8
(c) A sphere of radius 6 cm is
melted and recast in the
shape of a cylinder of radius
4 cm. Then, the height of the
cylinder is ......... cm.
(r) 16 : 9
(d) The volumes of two spheres
are in the ratio 64 : 27. The
ratio of their surface areas is ....... .
(s) 5

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 812:

Question 75:

Match the following columns:

Column I Column II
(a) The radii of the circular ends of
a bucket, in the form of the frustum of a cone of height 30 cm, are 20 cm
and 10 cm respectively. The
capacity of the bucket is ........cm3.
(p) 2418π
(b) The radii of the circular ends
 of a conical bucket of height
15 cm are 20 and 12 cm
respectively. The slant height
of the bucket is ........ cm.
(q) 22000
(c) The radii of the circular ends of
a solid frustum of a cone are 33 cm
and 27 cm and its slant height is
10 cm. The total surface area of
the bucket is .........cm2.
(r) 12
(d) Three solid metallic spheres of
radii 3 cm, 4 cm and 5 cm are
melted to form a single solid
sphere. The diameter of the
resulting sphere is ........ cm.
(s) 17

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 813:

Question 76:

Assertion (A)
If the radii of the circular ends of a bucket 24 cm high are 15 cm and 5 cm, respectively, then the surface area of the bucket is 545π cm2.

Reason(R)
If the radii of the circular ends of the frustum of a cone are R and r, respectively, and its height is h, then its surface area is π{R2+r2+l(R-r)}, where l2=h2+(R-r)2.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

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 814:

Question 77:

Assertion (A)
The outer surface of a hemisphere of radius 7 cm is to be painted. The total cost of painting at Rs 5 per cm2 is Rs 2300.

Reason (R)
The total surface area of a hemisphere is 3π r2.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

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 814:

Question 78:

Assertion (A)
The number of coins of diameter 1.75 cm and 2 mm thickness that can be formed from a melted cuboid (10 cm × 5.5 cm × 3.5 cm) is 400.

Reason (R)
Volume of a cylinder of base radius r and height h =(πr2h) cubic units. And area of a cuboid = (l × b × h) cubic units.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

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 814:

Question 79:

Assertion (A)
If the volumes of two spheres are in the ratio 27 : 8, then their surface areas are in the ratio 3 : 2.

Reason (R)
Volume of a sphere = 43πR3
Surface area of a sphere = 4πR2

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

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 814:

Question 80:

Assertion (A)
The curved surface area of a cone of base radius 3 cm and height 4 cm is 15π cm2.

Reason (R)
Volume of a cone = πr2h

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

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 824:

Question 1:

Find the number of solid spheres, each of diameter 6 cm, that could be moulded to form a solid metallic cylinder of height 45 cm and diameter 4 cm.

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.

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Question 2:

Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii?

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.

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Question 3:

A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, then find the total area of the canvas required.

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.

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Question 4:

The radii of the top and bottom of a bucket of slant height 45 cm are 28 cm and 7 cm, respectively. Find the curved surface area of the bucket.

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

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Question 5:

A solid metal cone with base radius 12 cm and height 24 cm is melted to form solid spherical balls of diameter 6 cm each. Find the number of balls formed.

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

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Question 6:

A hemispherical bowl of internal diameter 30 cm is full of a liquid. This liquid is poured into cylindrical bottles of diameter 5 cm and height 6 cm each. How many bottles are required?

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

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Question 7:

A solid metallic sphere of diameter 21 cm is melted and recast into small cones of diameter 3.5 cm and height 3 cm each. Find the number of cones so formed.

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

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Question 8:

The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of diameter 2.8 cm. Find the length of the wire.

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.

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Question 9:

A drinking glass is in the shape of the frustum of a cone of height 21 cm with 6 cm and 4 cm as the diameters of its two circular ends. Find the capacity of the glass.

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

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Question 10:

Two cubes, each of volume 64 cm3, are joined end to end. Find the total surface area of the resulting cuboid.

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

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Question 11:

The radius of the base and the height of a solid right circular cylinder are in the ratio 2 : 3 and its volume is 1617 cm3. Find the total surface area of the cylinder.

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                                                              

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Question 12:

A toy is in the form of a cone mounted on a hemisphere of common base radius 7 cm. The total height of the toy is 31 cm. Find the total surface area of the toy.

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



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Question 13:

A hemispherical bowl of internal radius 9 cm is full of water. This water is to be filled in cylindrical bottles of diameter 3 cm and height 4 cm. Find the number of bottles needed in which the water can be filled.

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

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Question 14:

The surface areas of a sphere and a cube are equal. Find the ratio of their volumes.

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.

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Question 15:

The slant height of the frustum of a cone is 4 cm and the perimeters (i.e. circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

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

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Question 16:

A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each hemispherical end is 7 cm, find the surface area of the solid.

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

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Question 17:

From a solid cylinder whose height is 15 cm and diameter 16 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. (Use π = 3.14)

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.

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Question 18:

A solid rectangular block of dimensions 4.4 m, 2.6 m and 1 m is cast into a hollow cylindrical pipe of internal radius 30 cm and thickness
​5 cm. Find the length of the pipe.

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.

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Question 19:

An open metal bucket is in the shape of a frustum of a cone, mounted on a hollow cylindrical base made of the same metallic sheet. The diameters of the two circular ends of the bucket are 45 cm and 25 cm, the total vertical height of the bucket is 40 cm and that of the cylindrical base is 6 cm. Find the area of the metallic sheet used to make the bucket. Also, find the volume of water the bucket can hold, in litres.

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

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Question 20:

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank which is 10 m in diameter and 2 m deep. If the water flows through the pipe at the rate of 4 km/hr, then in how much time will the tank be filled completely?

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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