a. summative assessment a n - testlabz vessel is in the form of a hollow hemisphere mounted by a...
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SURFACE AREAS AND VOLUMES
A. SUMMATIVE ASSESSMENT
13
13.1 SURFACE AREA OF ACOMBINATION OF SOLIDS
1. Cuboid : For a cuboid of dimensions l,b and h, we have :
(a) Lateral surface area = 2h(l + b)
(b) Total surface area = 2(lb + bh + lh)
(c) Length of diagonal = 2 2 2l b h
2. Cube : For a cube of edge l, we have :
(a) Lateral surface area = 4l2
(b) Total surface area = 6l2
(c) Length of diagonal = 3 l
3. Cylinder : For a cylinderof radius r and height h, wehave :
(a) Area of curved surface= 2πrh
(b) Total surface area = 2πr2 + 2πrh= 2πr(r + h)
(c) Curved surface area of hollow cylinder= 2πh(R – r),where R and r are outer and inner radii
(d) Total surface area of hollow cylinder= 2πh(R + r) + 2π(R2 – r2)
4. Cone : For a cone ofheight h, radius r and slantheight l, we have :
(a) Curved surface area
= πrl = 2 2r h r
(b) Total surface area= πr2 + πrl= πr (r + l)
5. Sphere : For asphere of radius r, wehave :
Surface area = 4πr2
6. Hemisphere (solid) : For a hemisphere ofradius r we have :
(a) Curved surface area
= 2πr2
(b) Total surface area
= 3πr2
TEXTBOOK’S EXERCISE 13.1
Unless stated otherwise, take =22
7.
Q.1. 2 cubes each of volume 64 cm3 arejoined end to end. Find the surface area of theresulting cuboid. [2011 (T-II)]
Sol. Let the side of cube = y cm
Volume of cube = 64 cm3
Then, volume of cube = side3 = y3
As per condition y3 = 64 y3 = 43
Question Bank In Mathematics Class X (Term–II)
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y = 4 cmHence, side of cube is 4 cm.For the resulting cuboid
length (l ) = 4 + 4 = 8 cmbreadth (b) = 4 cm
height (h) = 4 cm Surface area of the resulting cuboid= 2(lb + bh + hl )= 2(8 × 4 + 4 × 4 + 4 × 8) cm2
= 2(32 + 16 + 32) cm2 = 2(80) cm2
= 160 cm2.
Q.2. A vessel is in the form of a hollowhemisphere mounted by a hollow cylinder. Thediameter of the hemisphere is 14 cm and thetotal height of the vessel is 13 cm. Find the innersurface area of the vessel. [2011 (T-II)]
Sol. Diameter of the hollow hemisphere= Diameter of the hollow cylinder = 14 cm
Radius of the hollow hemisphere =14
2cm = 7 cm
Radius of the base of thehollow cylinder = 7 cm
Total height of the vessel
= 13 cm
Height of the hollowcylinder = (13 – 7) cm
= 6 cm
Inner surface area of the vessel
= Inner surface area of thehemisphere + Inner surface area of the hollow cylinder
= 2 (7)2 cm2 + 2 (7) (6) cm2
= 98cm2 + 84cm2 = (98 + 84) cm2
= 182 cm2 = 182 ×22
7cm2 = 26 × 22 cm2
= 572 cm2.
Q.3. A toy is in the form of a cone of radius3.5 cm mounted on a hemisphere of same radius.The total height of the toy is 15.5 cm. Find thetotal surface area of the toy. [2011 (T-II)]
Sol. Radius of the cone = Radius of hemisphere
= 3.5 cm
Total height of the toy = 15.5 cm
Height of the cone = (15.5 – 3.5) cm = 12 cm
Slant height of the cone (l ) = 2 2r h
= 2 2(3.5) (12) cm
= 156.25 cm = 12.5 cm
Total surface area of the toy= Curved surface area of the hemisphere
+ Curved surface area of the cone
= 2 (3.5)2 cm2 + (3.5) (12.5) cm2
= 24.5 cm2 + 43.65cm2
= 68.25 cm2 =68.25 22
7
cm2 = 214.5 cm2.
Q.4. A cubical block of side 7 cm is
surmounted by a hemisphere. What is the
greatest diameter the hemisphere can have? Find
the surface area of the solid. [2011 (T-II)]
Sol. Side of cubical block = 7 cm
Side of cubical block = Diameter of hemisphere
= 7 cm 2R = 7
R =7
2cm
Surface area of solid = Surface area of the cube– Area of base of hemisphere
+ C.S.A. of hemisphere
= 6 × side2 – R2 + 2R2
= 6 (7)2 cm2 + R2
= 6 × 7 × 7 cm2 +22
7×
7
2×
7
2cm2
=7
6 49 112
cm2
=77
2942
cm2 =
588 77
2
cm2.
=665
2cm2 = 332.50 cm2
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Q.5. A hemispherical depression is cut out
from one face of a cubical wooden block such
that the diameter l of the hemisphere is equal to
the edge of the cube. Determine the surface area
of the remaining solid.
Sol. Diameter of the hemisphere = l = Side of the
cube
Radius of the hemisphere =2
l
Surface area of the remaining solid
= Surface area of hemisphere
+ Surface area of cube
– Area of base of hemisphere
=
2 222 6
2 2
l ll
=
226
2
ll
=2
264
ll
= 2
244
l .
Q.6. A medicine capsule is in the shape of a
cylinder with two hemispheres stuck to each of
its ends (see figure below). The length of the
entire capsule is 14 mm and the diameter of the
capsule is 5 mm. Find its surface area.
Sol. Diameter of capsule
= Diameter of hemisphere
= Diameter of cylinder = 5 mm
Radius of the hemisphere = r =5
2mm
Height of the cylinder
= [14 – (2.5 + 2.5)] mm = 9 mm
Surface area of the capsule = Surface area of
cylinder + 2 Surface area of hemisphere
=
22 25 5
2 (9) mm 2 2 mm2 2
π + π
= 45 mm2 + 25 mm2 = (45 + 25) mm2
= 70 mm2
= 70 ×22
7mm2 = 220 mm2.
Hence, surface area of capsule = 220 mm2
Q.7. A tent is in the shape of a cylindersurmounted by a conical top. If the height anddiameter of the cylindrical part are2.1 m and 4 m respectively and the slant heightof the top is 2.8 m, find the area of the canvasused for making the tent. Also, find the cost ofthe canvas of the tent at the rate of Rs 500 perm2 (note that the base of the tent will not becovered with canvas.)
Sol. Radius of the cone = 2 m
Radius of the cylinder = 2 m
Total surface area of the tent
= Curved surface area of the cylinder
+ Curved surface area of the cone
= 2 (2) (2.1) m2 + (2) (2.8) m2 = (8.4 + 5.6)m2
= 14 m2 = 14 ×22
7m2 = 44 m2
Cost the canvas of the tent at the rate of Rs 500
per m2 = Rs 44 × 500 = Rs 22000
Hence, cost of the canvas is Rs 22000.
Q.8. From a solid cylinder whose height is2.4 cm and diameter 1.4 cm, a conical cavity ofthe same height and same diameter is hollowedout. Find the total surface area of the remainingsolid to the nearest cm2.
Sol. Height of cylinder = 2.4 cm
Height of cone = 2.4 cm
Radius of cylinder = r =1.4
2cm = 0.7 cm
Radius of cone = 0.7 cm
Slant height, of the cone
l = 2 2(0.7) (2.4) cm = 2.5 cm
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Total surface area of the remaining solid
= C.S.A. of cylinder
+ C.S.A. of cone + Area of base
= 2rh + rl + r2
= r (2 h + l + r)
=22
7× 0.7 × (2 × 2.4 + 2.5 + 0.7) cm2
=22
7×
7
10(4.8 + 3.2) cm2
=22
7×
7
10× 8.0 cm2
=176
10cm2 = 17.6 cm2
Hence, total remaining surface area = 17.6 cm2
= 18 cm2.
Q.9. A wooden article was made by scoopingout a hemisphere from each end of a solidcylinder, as shown in figure. If the height of thecylinder is 10 cm, and its base is of radius3.5 cm, find the total surface area of the article.
Sol. Height of cylinder = 10 cm
Radius of cylinder = 3.5 cm
Total surface area of the article
= C.S.A of cylinder + 2 C.S.A. of hemisphere
= 2 (3.5 (10) cm2 + 2 [2 (3.5)2] cm2
= 70cm2 + 49 cm2 = (70 + 49) cm2
= 119 cm2 = 119 ×22
7cm2
= 17 × 22 cm2 = 374 cm2.
OTHER IMPORTANT QUESTIONS
Q.1. A cylindrical pencil sharpened at oneedge is the combination of :
(a) a cone and a cylinder(b) frustum of a cone and a cylinder(c) a hemisphere and a cylinder(d) two cylindersSol. (a) The given shape is a combination of a
cone and a cylinder.
Q.2. If each edge of a cube is increased by50%, the percentage increase in the surface areais :
(a) 25% (b) 50% (c) 75% (d) 125%
Sol. (d) Let the edge of the cube be a.
Then, its surface area = 6a2
New edge =150
100
a=
3
2
a.
Its surface area = 6 ×29
4
a=
227
2
a
Increase in surface area =227
2
a– 6a2 =
215
2
a
Per cent increase =215
2
a× 2
100
6a= 125%
Q.3. The total surface area of a hemisphereof radius 7 cm is : [2011(T-II)]
(a) 447 cm2 (b) 239 cm2
(c) 147 cm2 (c) 174 cm2
Sol. (c) Total surface area of the hemisphere= 3r2
= 3 × × 49 cm2
= 147 cm2
Q.4. If two solid hemispheres of same baseradius r are joined together along their bases,
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then curved surface area of this new solid is :
(a) 4r2 (b) 6r2
(c) 3r2 (d) 8r2
Sol. (a) The resulting solid will be a sphere of
radius r.
Its curved surface area = 4r2.
Q.5. Volumes of two cubes are in the ratio64 : 27. The ratio of their surface areas is :
(a) 3 : 4 (b) 4 : 3
(c) 9 : 16 (d) 16 : 9
Sol. (d) We have3
13
2
a
a=
64
27
1
2
a
a=
4
3
2122
6
6
a
a=
2
2
4
3=
16
9= 16 : 9
Q.6. The diameter of a solid hemisphericaltoy is 7 cm. Find its curved surface area andtotal surface area.
Sol. Diameter of the hemispherical toy = 7 cm.
... Radius of the hemispherical toy, r = 3.5 cm
Curved surface area of the toy = 2πr2
= 2 ×22
7× (3.5)2 cm2 = 77 cm2
Total surface area of the toy = 3πr2
= 3 ×22
7× (3.5)2 cm2 = 115.50 cm2.
Q.7. If a solid cone of base radius r andheight h is placed over a solid cylinder havingsame base radius and height as that of the cone,then the curved surface area of the shape is
2 2 2r r h rh . Is it true?
Sol. True. Since the curved surface area takentogether is same as the sum of curved surface areasmeasured separately.
Q.8. Two identical solid cubes of side a arejoined end to end. Then find the total surfacearea of the resulting cuboid.
Sol. The resulting solid is a cuboid of dimensions2a × a × a.
Total surface area of the cuboid= 2 (lb + bh + hl)= 2 (2a × a + a × a + a × 2a)= 10a2.
Q.9. The total surface area of atop (lattu) as shown in the figure is thesum of total surface area ofhemisphere and the total surface areaof cone. Is it true?
Sol. No, the statement is false.
Total surface area of the top (lattu) is the sum ofthe curved surface area of the hemisphere and thecurved surface area of the cone.
Q.10. Two cones with the same base radius8 cm and height 15 cm are joined together alongtheir bases. Find the surface area of the shape soformed.
Sol. Slant height of each cone = 2 28 15 cm
= 64 225 cm
= 17 cm.
Surface area of the resulting shape
= 2 (rl) = 2 ×22
7× 8 × 17 cm2
= 854.85 cm2 = 855 cm2 (approx.)
Q.11. A tent of height 8.25 m is in the formof a right circular cylinder with diameter of base30 m and height 5.5 m, surmounted by a rightcircular cone of the same base. Find the cost ofthe canvas of the tent at the rate of Rs 45 per m2.
Sol.
Height of the tent = 8.25 m.Height of the cylindrical part = 5.5 m
... Height of the conical part = (8.25 – 5.5) m
= 2.75 m.
Base radius of the tent =30
2m = 15 m.
... Slant height of the conical part
= 2 2(15) (2.75) m 225 7.5625 m+ = += 15.25 m.
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Curved surface area of the tent = curved surfacearea of the cylindrical part + curved surface area ofthe conical part
= 2πrh + πrl = πr (2h + l)
=22
7× 15 (2 × 5.5 + 15.25) m2
= 22215 26.25 m
7
× ×
= 1237.50 m2.
Rate of the canvas = Rs 45 per m2
... Cost of the canvas = Rs (1237.50 × 45)
= Rs 55687.50.
Q.12. A cone of maximum size is carved outfrom a cube of edge 14 cm. Find the surface areaof the cone and of the remaining solid left outafter the cone carved out.
Sol. Diameter of the cone = 14 cm
and height of the cone = 14 cm
Slant height of the cone = 2 2r h
= 2 27 14 cm = 245 cm = 7 5 cm.
Total surface area of the cone = rl + r2
= r (l + r)
=22
7× 7 ( 7 5 + 7) cm2
= 154 ( 5 + 1) cm2
Surface area of the cube = 6 × 142 cm2
= 1176 cm2
Surface area of the remaining solid left outafter the cone is carved out
= surface area of the cube– area of base of the cone
+ curved surface area of the cone
=222
1176 7 154 57
cm2
= 1022 154 5 cm2.
Q.13. A toy is in the form of a cone mountedon a hemisphere of common base radius7 cm. The total height of the toy is 31 cm. Findthe total surface area of the toy.
[2007, 2011 (T-II)]
Sol.
Height of the toy = 31 cmBase radius of the cone
= radius of the hemisphere= 7 cm
Height of the cone = (31 – 7) cm = 24 cm
Slant height of the cone = 2 2r h
= 2 27 24 cm
= 49 576 cm
= 625 cm = 25 cm Total surface area of the toy
= Curved surface area of the hemisphere+ Curved surface area of the cone
= 2r2 + rl = r (2r + l)
=22
7× 7 (14 + 25) cm = 858 cm2.
Q.14. A solid is in the form of a rightcircular cylinder with hemispherical ends. Thetotal height of the solid is 58 cm and thediameter of the cylinder is 28 cm. Find the totalsurface area of the solid. [2006]
Sol.
Radius of the each hemisphere = base
radius of the cylinder = 14 cm
Total height of the toy = 58 cm
Height of the cylinder
= [58 – (14 + 14)] cm = 30 cm
Total surface area of the solid
= 2r2 + 2rh + 2r2 = 2r (2r + h)
= 2 ×22
7× 14 (2 × 14 + 30) cm2
= 88 × 58 cm2 = 5104 cm2.
Q.15. A toy is in the shape of a right circularcylinder with a hemisphere on one end and acone on the other. The radius and height of the
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cylindrical part are 5 cm and 13 cm respectively.The radii of the hemisphercial and conical partsare the same as that of the cylindrical part. Findthe surface area of the toy if the total height ofthe toy is 30 cm. [2002]
Sol.
Radius of the cone = Radius of the cylinder= radius of the hemisphere= 5 cm.
Total height of the toy = 30 cmHeight of the cylinder h = 13 cm Height of the cone = [30 – (13 + 5)] cm
= 12 cm.
Slant height of the cone = 2 212 5 cm
= 144 25 cm = 13 cm.Total surface area of the toy
= curved surface area of the hemisphere+ curved surface area of the cylinder
+ curved surface area of the cone
= 2r2 + 2rh + rl = r (2r + 2h + l )
=22
7× 5 (2 × 5 + 2 × 13 + 13) cm2
=22
7× 5 × 49 cm2 = 770 cm2.
Q.16. The internal and external diameters ofa hollow hemispherical vessel are 24 cm and25 cm respectively. If the cost of painting 1 cm2
of the surface area is Rs 5.25, find the total costof painting the vessel all over. [2001]
Sol. External radius (R) of the vessel = 12.5 cm
Internal radius (r) of the vessel = 12 cm
Total surface area of the vessel
= 2R2 + 2r2 + (R2 – r2)
= [2 × (12.5)2 + 2 × 122 + (12.52 – 122)] cm2
= [312.5 + 288 + 12.25] cm2
=22
7× 612.75 cm2 = 1925.78 cm2.
Required cost of painting
= Rs 5.25 × 1925.78 = Rs 1010.38.
Choose the correct option (Q 1 – 7) :1. A funnel is the combination of :
(a) a cone and a cylinder
(b) frustrum of a cone and a cylinder
(c) a hemisphere and a cylinder
(d) a hemisphere and a cone.
2. A plumbline (shahul) is the combination of :
(a) a cone and a cylinder(b) a hemisphere and a cone(c) frustrum of a cone and a cylinder(d) a sphere and a cylinder
3. A shuttle cock used for playing badmintonhas the shape of the combination of : [2011 (T-II)]
(a) a cylinder and a cone(b) a cylinder and a hemisphere(c) a sphere and a cone(d) frustrum of a cone and a hemisphere
4. The height of a conical tent is 14 m and itsfloor area is 346.5 m2. The length of 1.1 m wide
canvas required to built the tent is :
(a) 490 m (b) 525 m (c) 665 m (d) 860 m
5. The ratio of the total surface area to thelateral surface area of a cylinder with base diameter160 cm and height 20 cm is :
(a) 1 : 2 (b) 2 : 1 (c) 3 : 1 (d) 5 : 1
6. The radius of the base of a cone is 5 cm andits height is 12 cm. Its curved surface area is :
(a) 30π cm2 (b) 65π cm2
(c) 80π cm2 (d) none of these
7. A right circular cylinder of radius r cm andheight h cm (h > 2r) just encloses a sphere ofdiameter :
(a) r cm (b) 2r cm (c) h cm (d) 2h cm
8. Two identical solid hemispheres of equalbase radius r cm are stuck together along theirbases. The total surface area of the combination is6r2. Is it true?
PRACTICE EXERCISE 13.1 A
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Unless stated otherwise, take =22
7.
Q.1. A solid is in the shape of a conestanding on a hemisphere with both their radiibeing equal to 1 cm and the height of the coneis equal to its radius. Find the volume of thesolid in terms of .
Sol.
TEXTBOOK’S EXERCISE 13.2
9. A solid cylinder of radius r and height h is
placed over other cylinder of same height and
radius. The total surface area of the shape formed
is 4rh + 4r2. Is it true?
10. A solid ball is exactly fitted inside the
cubical box of side a. Surface area of the ball is
4a2. Is it true?
11. From a solid cylinder whose height is 2.4
cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out.
Find the total surface area of the remaining solid
to the nearest cm2.12. A decorative block shown below, is made
of two solids – a cube and a hemisphere. Thebase of the block is a cube with edge 5 cm, andthe hemisphere fixed on the top has a diameter
4.2 cm. Find the total surface area of the block.
22
7
π = . [2011 (T-II)]
13. A tent of height 3.3 m is in the form of
a right circular cylinder of diameter 12 m and
height 2.2 m, surmounted by a right circular cone
of the same diameter. Find the cost of canvas of
the tent at the rate of Rs 500 per m2.
14. Three cubes each of side 5 cm are joined
end to end. Find the surface area of the resulting
cuboid.
15. A solid is composed of a cylinder withhemispherical ends. If the whole length of thesolid is 108 cm and the diameter ofhemispherical ends is 36 cm, find the cost ofpolishing the surface at the rate of7 paise per cm2.
16. A rocket is in the form of a cone of height28 cm, surmounted over a right circular cylinderof height 112 cm. The radius of the bases of coneand cylinder are equal, each being 21 cm. Find
the total surface area of the rocket.22
7
π =
13.2 VOLUME OF A COMBINATION OFSOLIDS
1. Volume of a cuboid of dimensions l, b andh = l × b × h.
2. Volume of a cube of edge l = l3.3. Volume of a cylinder of base radius r and
height h = r2h.
4. Volume of a cone of base radius r and height
h =1
3r2h.
5. Volume of a sphere of radius r =4
3r3.
6. Volume of a hemisphere of radius r =2
3r3.
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Radius of the hemisphere
= Radius of cone = 1 cmHeight of cone = h = 1 cm
Volume of hemisphere =2
3r3 =
2
3 (1)3 cm3
=2
3 cm3 ..(i)
Volume of cone =1
3r2h =
1
3 (1)2 (1) cm3
=1
3 cm3 ..(ii)
Volume of the solid= Volume of the hemisphere + Volume of cone
=2 1
3 3
π + π cm3 = cm3.
Q.2. Rachel, an engineering student, wasasked to make a model shaped like a cylinderwith two cones attached at its two ends by usinga thin aluminium sheet. The diameter of themodel is 3 cm and its length is 12 cm. If eachcone has a height of 2 cm, find the volume of aircontained in the model that Rachel made.(Assume the outer and inner dimensions of themodel to be nearly the same.)
Sol.
For conical portion :
Radius of the base (r) =3
2cm = 1.5 cm
Height of cone (h1) = 2 cm
We know that, volume of cone =1
3r2h
Volume of cone OAB =1
3r2h1
=1
3(1.5)2 (2) cm3 = 1.5 cm3 ...(i)
Volume of cone ABO =1
3r2h1
=1
3(1.5)2 × (2) cm3 = 1.5 cm3 ...(ii)
For cylindrical portion :
Radius of the base (r) = 1.5 cm
Height of cylinder h2= 12 cm – (2 + 2) cm = 8 cm
Volume of cylinder = r2h2
= (1.5)2 (8) cm3 = 18 cm3 ..(iii)
Adding equations (i), (ii) and (iii), we have
Total volume of the model = volume of the twocones + volume of the cylinder.
= 1.5 cm3 + 1.5 cm3 + 18 cm3 = 21 cm3
= 21 ×22
7cm3 = 66 cm3
Hence, the volume of the air contained in the modelthat Rachel made is 66 cm3.
Q.3. A gulab jamun,contained sugar syrup up toabout 30% of its volume. Findapproximately how much syrupwould be found in 45 gulabjamuns, each shaped like acylinder with two hemisphericalends with length 5 cm anddiameter 2.8 cm (see figure).
[2011 (T-II)]
Sol. Gulab jamun is in the shape of cylinder withtwo hemispherical ends.
Diameter of cylinder = 2.8 cm Radius of cylinder = 1.4 cmHeight of cylindrical part
= (5 – 1.4 – 1.4) cm = (5 – 2.8) cm = 2.2 cm
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Volume of a gulab jamun
=2
3 (1.4)3 cm3 + (1.4)2 (2.2) cm3 +
2
3 (1.4)3cm3
=4
3 (1.4)3 cm3 + (1.4)2 (2.2)cm3
= (1.4)24 1.4
2.23
cm3
= (1.96)5.6 6.6
3
cm3
=(1.96) (12.2)
3
cm3
Volume of 45 gulab jamuns
= 45 ×(1.96) (12.2)
3
cm3
= 15 (1.96) (12.2) cm3
= 15 ×22
7× 1.96 × 12.2 cm3
= 15 × 22 × 0.28 × 12.2 = 1127.28 cm3
Volume of syrup = 1127.28 ×30
100cm3
= 338.184 = 338 cm3 (approximately)
Q.4. A pen stand made of wood is in the
shape of a cuboid with four conical depressions
to hold pens. The dimensions of the cuboid are
15 cm by 10 cm by 3.5 cm. The radius of each of
the depressions is 0.5 cm and the depth is 1.4
cm. Find the volume in the entire stands. (See
figure).
Sol. Length of cuboid, l = 15 cm
Width of cuboid, b = 10 cm
Height of cuboid, h = 3.5 cm
Volume of the cuboid
= 15 × 10 × 3.5 cm3 = 525 cm3
Volume of a conical depression
=1
3 (0.5)2 (1.4) cm3
=1
3×
22
7× 0.25 ×
14
10cm3
=11
30cm3
Volume of four conical depressions
= 4 ×11
30cm3 =
22
15cm3 = 1.47 cm3
Volume of the wood in the pen stand
= (525 – 1.47) cm3 = 523.53 cm3.
Q.5. A vessel is in the form of an inverted
cone. Its height is 8 cm and the radius of its top,
which is open, is 5 cm. It is filled with water up
to the brim. When lead shots, each of which is a
sphere of radius 0.5 cm are dropped into the
vessel, one-fourth of the water flows out. Find
the number of lead shots dropped in the vessel.
Sol. Radius of cone = 5 cm
Height of cone = 8 cm
Volume of cone =1
3r2h =
1
3 (5)2 8 cm3
=200
3 cm3
Radius of spherical lead shot, r1 = 0.5 cm
Volume of a spherical lead shot
=4
3 3
1r =4
3 (0.5)3 cm3 =
6
cm3
Volume of water that flows out
=1
4× volume of the cone
=1 200
4 3
=50
3
cm3
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Let the number of lead shots dropped in thevessel be n.
Volume of n lead shots =6
ncm3
As per condition,
6
n=
50
3
n =50 6
3
n = 100
Hence, the number of lead shots dropped in thevessel is 100.
Q.6. A solid iron pole consist of a cylinder of
height 220 cm and base diameter 24 cm, which
is surmounted by another cylinder of height 60
cm and radius 8 cm. Find the mass of the pole,
given that 1 cm3 of iron has approximately 8 g
mass. (Use = 3.14)
Sol. Diameter of cylinder ABCD = 24 cm
Base radius of cylinder ABCD, r =24
2cm3
= 12 cm
Height of cylinder ABCD (h) = 220 cm
Volume of cylinderABCD= r2h = (12)2 (220)cm3 = 31680 cm3
Base radius of cylinder ABCD, R = 8 cm
Height of cylinder ABCD (H) = 60 cm
Volume of cylinder ABCD
= R2h = (8)2 (60) cm3 = 3840 cm3
Volume of solid iron pole
= Volume of the cylinder ABCD
+ Volume of the cylinder ABCD
= 31680 cm3 + 3840 cm3
= 35520 cm3
= 35520 × 3.14 cm3 = 111532.8 cm3
Mass of the pole = 111532.8 × 8 g
= 892262.4 g = 892.26 kg
Hence, the mass of the pole is 892.26 kg(approximately).
Q.7. A solid consisting of a right circular coneof height 120 cm and radius 60 cmstanding on a hemisphere of radius 60 cmis placed upright in a right circularcylinder full of water such that it touchesthe bottom. Find the volume of water left inthe cylinder, if the radius of the cylinder is60 cm and its height is 180 cm.
Sol.
Radius of the cone OAB (r) = 60 cm
Height of cone OAB (h1) = 120 cm
Volume of cone OAB
=1
3r2h1 =
1
3 (60)2 (120) cm3
= 144000 cm3
Radius of the hemisphere (r) = 60 cm
Volume of hemisphere =2
3r3
=2
3 (60)3 cm3
= 144000 cm3
Radius of the cylinder (r) = 60 cm
Height of cylinder (h2) = 180 cm
Volume of cylinder = r2h2
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Radius of cylindrical neck = 1 cm
Height of cylindrical neck = 8 cm
Amount of water it holds
= ( )3
3 2 34 8.5cm 1 8 cm
3 2
π + π
=4
3× 3.14 × 4.25 × 4.25 × 4.25 cm3 + 8 × 3.14 cm3
= 321.39 cm3 + 25.12 cm3 = 346.51 cm3
Hence, she is correct. The correct volume is346.51 cm3.
= (60)2 (180) cm3
= 648000 cm3
Volume of water left in the cylinder= Volume of the cylinder – [Volume of the cone
+ Volume of the hemisphere]= 648000 cm3 – [144000 + 144000] cm3
= 648000 cm3 – 288000 cm3
= 360000 cm3
=360000
100 100 100
π× ×
m3 = 0.36 m3
= 0.36 ×22
7m3 = 1.131 m3 (approx.)
Q.8. A spherical glass vessel has a
cylindrical neck 8 cm long, 2 cm in diameter; the
diameter of the spherical part is 8.5 cm. By
measuring the amount of water it holds, a child
finds its volume to be 345 cm3. Check whether
she is correct, taking the above as the inside
measurements, and = 3.14.
Sol. Diameter of sphere = 8.5 cm
So, r =8.5
2cm
Q.1. Volume of the largest right circular conethat can be cut out from a cube of edge4.2 cm is :
(a) 9.7 cm3 (b) 77.6 cm3
(c) 58.2 cm3 (d) 19.4 cm3
Sol. (d) Radius of the cone =4.2
2cm = 2.1 cm.
Height of the cone = 4.2 cm.
Volume of the cone=1
3r2h
=1
3×
22
7× 2.1 × 2.1 × 4.2 cm3 = 19.404 cm3.
Q.2. A hollow cube of internal edge 22 cm isfilled with spherical marbles of diameter 0.5 cm
and it is assumed that1
8space of the cube
OTHER IMPORTANT QUESTIONS
remains unfilled. Then the number of marblesthat the cube can accommodate is :
(a) 142296 (b) 142396
(c) 142496 (d) 142596
Sol. (a) Volume of the cube
= 223 cm3 = 10648 cm3
Space which remains unfilled
=10648
8cm3 = 1331 cm3
Remaining space = (10648 – 1331) cm3
= 9317 cm3
Volume of 1 marble =4
3 (0.25)3 cm3
Let n marbles can be accommodated.
Then, n ×4
3×
22
7× (0.25)3 = 9317
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n = 3
9317 3 7
4 22 (0.25)
= 142296.
Q.3. A medicine capsule is in the shape of acylinder of diameter 0.5 cm with twohemispheres stuck to each of its ends. The lengthof entire capsule is 2 cm. The capacity of thecapsule is :
(a) 0.36 cm3 (b) 0.35 cm3
(c) 0.34 cm3 (d) 0.33 cm3
Sol. (a)
Height of the cylindrical part= (2 – 0.5) cm = 1.5 cm
Radius of each hemispherical part= Radius of the cylindrical part= 0.25 cm.
Capacity of the capsule
=4
3r3 + r2h = r2
4
3r h
=22
7× (0.25)2
40.25 1.5
3
cm3
=22
7× (0.25)2
5.5
3
cm3 = 0.36 cm3
Q.4. A solid piece of iron in the form of acuboid of dimensions 49 cm × 33 cm × 24 cm ismoulded to form a solid sphere. The radius of thesphere is : [2011 (T-II)]
(a) 25 cm (b) 21 cm(c) 19 cm (d) 23 cm
Sol. (b) Volume of sphere = Volume of cuboid
4
3r3 = (49 × 33 × 24) cm3 = 38808 cm3
r3 =3 338808 3 7
cm = 9261 cm4 22
× ××
r = 21 cm
Q.5. The volume of a sphere (in cu.cm) isequal to its surface area (in sq. cm). Thediameter of the sphere (in cm) is : [2011 (T-II)]
(a) 3 (b) 6(c) 2 (d) 4
Sol. (b)4
3r3 = 4r2
r = 3 d = 2r = 2 × 3 = 6 cm
Q.6. The ratio of the volumes of two spheresis 8 : 27. The ratio between their surface areasis : [2011 (T-II)]
(a) 2 : 3 (b) 4 : 27(c) 8 : 9 (d) 4 : 9
Sol. (d )
31
1
3 22
48 23
4 27 3
3
rr
rr
π= ⇒ =
π
Ratio between surface areas = 4 : 9
Q.7. The ratio between the radius of the baseand the height of the cylinder is 2 : 3. If itsvolume is 1617 cm3, the total surface area of thecylinder is : [2011 (T-II)]
(a) 208 cm2 (b) 77 cm2
(c) 707 cm2 (d) 770 cm2
Sol. (d) Let the radius and height of the cylinder
be 2x and 3x respectively.Then, volume of the cylinder = r2h
1617 =22
7× (2x)2 × 3x
x3 =1617 7
22 4 3
=
343
8
x = 3.5 cm. Total surface area of the cylinder
= 2r (h + r)
= 2 ×22
7× 7 (10.5 + 7) cm2
= 44 × 17.5 cm2 = 770 cm2.
Q.8. On increasing each of the radius of thebase and the height of a cone by 20%, its volumewill be increased by :
(a) 25% (b) 40%(c) 50% (d) 72.8%
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Sol. (d) Volume of the original cone =1
3r2h
New radius =120
100
r=
6
5
r
New height =120
100
h=
6
5
h
New volume =1
3
26
5
r
×6
5
h
=72
125r2h.
Increase in volume =72
125r2h –
1
3r2h
=216 125
375
r2h
=91
375r2h.
Per cent increase in volume
= 2
2
91 100
1375
3
r h
r h
=91 100 3
375
= 72.8%.
Q.9. A sphere and a cube have the samesurface. Show that the ratio of the volume of
sphere to that of the cube is 6 : π[2011 (T-II)]
Sol. Let the radius of the sphere be r and the edge
of the cube be x. Whole surface area of sphere = 4r2
and whole surface area of cube = 6x2.
According to question, 4r2 = 6x2.
2
2
6 3 3
4 2 2= = ⇒ =
π π πr r
xx
Now,
3
3
4Volume of sphere 3Volume of cube
r
x
π=
=
3 24 4
3 3
r r r
x x x
π = π ×
=4 3 3 2 3 2 3 6
3 2 2 2
×π × × = = =
π π π π πHence, ratio of the volume of sphere to that of
cube = 6 : π
Q.10. The internal and external radii of ahollow spherical shell are 3 cm and 5 cmrespectively. If it is melted to form a solid
cylinder of height 102
3cm, find the diameter of
the cylinder. [2011 (T-II)]Sol. Let the radius of the base of the cylinder be
r cm. Then, volume of the metallic solid cylinder of
height2
103
cm.
= Volume of the metal in the spherical shell
( )2 3 332 45 3
3 3rπ × × = π −
( )232 4125 27
3 3r = −
r2 =3 4
9832 3
× ×
r2 =49 7
cm4 2
r⇒ =
Hence, the diameter of the base of the cylinder
= 7 cm.
Q.11. A solid ball is exactly fitted inside thecubical box of side a. The volume of the ball is
4
3a3. Is it true?
Sol. Diameter of the ball = side of the cube
Radius of the ball =2
a
Volume of the ball =4
3 ×
3
8
a=
3
6
a
Hence, the statement is false.
Q.12. From a solid cube of side 7 cm, aconical cavity of height 7 cm and radius 3 cm ishollowed out. Find the volume of the remainingsolid.
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Sol. Volume of the cube = 73 cm3
= 343 cm3
Volume of the cone =1
3 × 32 × 7 cm3
= 66 cm3
Volume of the remaining solid= (343 – 66) cm3
= 277 cm3.
Q.13. The difference between the outer andinner curved surface areas of a hollow rightcircular cylinder 14 cm long is88 cm2. If the volume of metal used in makingcylinder is 176 cm3, find the outer and innerdiameters of the cylinder. [2010]
Sol. Let the inner and outer radii of the cylinder
be r cm and R cm respectively.Then, the height of the cylinder = 14 cm.
Inner surface of the cylinder = 2πr × 14 cm2
= 28πr cm2
Outer surface of the cylinder = 2πR × 14 cm2
= 28R cm2
Difference of the two surfaces = (28πR – 28πr)⇒ 88 = 28π (R – r)
⇒ (R – r) =88 7
28 22
××
= 1
⇒ R – r = 1 ...(i)
Volume of the metal used in making the cylinder= π(R2 – r2) × 14 cm3
... 176 = π (R + r) (R – r) × 14
⇒ (R + r) =176 7
22 1 14
×× ×
= 4
⇒ R + r = 4 ...(ii)
Solving (i) and (ii), we have
R = 2.5 cm and r = 1.5 cm
Hence, inner and outer diameters of the cylinder
are 3 cm and 5 cm respectively.
Q.14. An ice cream cone, full of ice cream ishaving radius 5 cm and height 10 cm as shown.Calculate the volume of ice cream provided that
its1
6part is left unfilled with ice cream.
Sol.
Radius of the hemispherical portion= 5 cm
= radius of the cone.
Height of the conical portion
= (10 – 5) cm = 5 cm.
Capacity of the shape =2
3r3 +
1
3r2h
=1
3r2 (2r + h)
=1
3×
22
7× 5 × 5 (2 × 5 + 5) cm3
=22 25
21
× 15 cm3 =
2750
7cm3.
Space which remains unfilled =1
6×
2750
7cm3.
Required volume of the ice cream
=2750 2750
7 6 7
cm3
=2750 5
7 6 cm3 = 327.4 cm3.
Q.15. A solid toy is in the form of ahemisphere surmounted by a right-circular cone.The height of the cone is 4 cm and the diameterof the base is 8 cm. Determine the volume of thetoy. If a cube circumscribes the toy, then find thedifference of the volumes of cube and the toy.Also, find the total surface area of the toy.
Sol. Volume of the toy = Volume of the cone +
Volume of the hemisphere
=1
3r2h +
2
3r3 =
1
3r2 (h + 2r)
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=1
3×
22
7× 4 × 4 (4 + 8) cm3 =
1408
7cm3.
A cube circumscribes this toy, hence edge of thecube = 8 cm.
Volume of the cube = 83 cm3 = 512 cm3
Required difference in the volumes of the toy
and the cube
=1408
5127
cm3
=2176
7cm3 = 310.86 cm3.
Total surface area of the toy
= curved curface area of the cone
+ curved surface area of the hemisphere
=2 2 22r h r r
=2 2 2r h r r π + +
=22
4 16 16 2 47
cm2
=22
4 4 2 87
cm2
= 88 42 2
7
cm2
=88 4
7
× 3.41 cm2 = 171.47 cm2.
Q.16. 16 glass spheres each of radius 2 cmare packed into a cubical box of internaldimensions 16 cm × 8 cm × 8 cm and then thebox is filled with water. Find the volume of waterfilled in the box.
Sol. Capacity of the box
= 16 × 8 × 8 cm3 = 1024 cm3
Volume of the 16 glass spheres
= 16 ×4
3r3
= 16 ×4
3×
22
7× 2 × 2 × 2 cm3
=11264
21cm3
Volume of water filled in the box
=11264
102421
cm3 =
10240
21cm3
= 487.61 cm3.
Q.17. A building is in the form of a cylindersurmounted by a hemispherical valuted dome
and contains19
4121
m3 of air. If the internal
diameter of the dome is equal to its total heightabove the floor, find the height of the building.
[2001]
Sol. Let the internal height of the
cylindrical part be h and the internal
radius be r.Then, total height of the building
= h + r
Also, 2r = h + r h = r.Now, volume of the building
= Volume of the cylindrical part +
Volume of the hemispherical part
19
4121
= r2h +2
3r3
880
21= r3 +
2
3r3 [ r = h]
880
21=
35
3
r
r3 =880 3 7
21 5 22
= 8
r = 2
Hence, height of the building = h + r
= (2 + 2) m = 4 m.
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Q.18. A godown building is in the form asshown in the figure. The vertical cross sectionparallel to the width side of the building is arectangle 7 m × 3 m, mounted by a semicircle ofradius 3.5 m. The inner measurements of thecuboidal portion of the building are 10 m × 7 m× 3 m. Find the volume of the godown and thetotal interior surface area excluding the floor
(base).2 2
7
π = .
Sol. The godown building consists of cuboid at
the bottom and the top of the building is in the formof half of the cylinder.
Length of the cuboid = 10 m,
Breadth of the cuboid = 7 m
Height of the cuboid = 3 m
Volume of the cuboid = lbh = 10 × 7 × 3 m3
= 210 m3.
Radius of the cylinder = 3.5 mLength of the cylinder = 10 m
Volume of the half of the cylinder =1
2πr2h
=1 22
2 7× × (3.5)2 × 10 m3
= 192.5 m3
Volume of the godown = volume of the cuboid
+ volume of the half cylinder
= (210 + 192.5) m3
= 402.5 m3
Interior surface area of the cuboid= Area of four walls
= 2 (l + b) h
= 2(10 + 7) 3 m2 = 102 m2
Interior curved surface area of half of the cylinder
= πrh =22
7× 3.5 × 10 m2 = 110 m2
Interior area of two semicircles
= 221
2r
π = πr2
=22
7× (3.5)2 m2 = 38.5 m2
Total interior surface area excluding the base floor= area of the four walls
+1
2(curved surface area of the cylinder)
+ 2 (area of the semicircle)
= (102 + 110 + 38.5) m2
= 250.5 m2.
Q.19. A tent is in the shape of a cylindersurmounted by a conical top. If the height anddiameter of the cylindrical part are 2.1 m and 4m respectively and the slant height of the top is2.8 m, find the area of canvas used for makingthe tent. Find the cost of the canvas of the tent atthe rate of Rs 550 per m2. Also, find the volumeof air enclosed in the tent. [2008C]
Sol.
Height of the cone, H
= 2 22.8 2 m
= 7.84 4 m = 1.95 m
Area of canvas required for making the tent
= Curved surface area of the tent
= Curved surface area of the cylindrical part
+ curved surface area of the conical part
= 2rh + rl = r (2h + l )
=22
7× 2 (2 × 2.1 + 2.8) m2
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=44
7× 7 m2 = 44 m2.
Cost of canvas = Rs 500 × 44 = Rs 22000.
Volume of the air enclosed in the tent
= Volume of the cylindrical part
+ Volume of the conical part
= r2h +1
3r2H = r2 ×
H
3h
=22
7× 22
1.952.1
3
m3
=88 8.25
7 3 m3 = 34.57 m3.
Q.20. From a solid cylinder whose height is8 cm and radius 6 cm, a conical cavity of height8 cm and of base radius 6 cm, is hollowed out.Find the volume of the remaining solid correct totwo places of decimal. Also find the total surfacearea of the remaining solid. (Take = 3.14)
[2008, 2011 (T-II)]
Sol.
Radius of the cylinder
= radius of the cone = 6 cm.
Height of the cylinder
= height of the cone = 8 cm.
Volume of the remaining solid
= r2h –1
3r2h =
2
3r2h
=2
3× 3.1416 × 36 × 8 cm3
= 603.19 cm3
Slant height of the cone, l
= 2 2r h
= 36 64 cm
= 10 cm
Total surface area of the remaining solid= curved surface area of the cylinder
+ area of top + curved surface area of the cone= 2rh + r2 + rl= r (2h + r + l)
= 3.14 × 6 (16 + 6 + 10) cm2
= 18.84 × 32 cm2
= 602.88 cm2.
Q.21. A juice seller serves hiscustomers using a glass as shownin the figure. The inner diamaterof the cylindrical glass is 5 cm,but the bottom of the glass has ahemispherical portion raisedwhich reduces the capacity of theglass. If the height of the glass is10 cm, find the apparent capacity of the glassand its actual capacity.(Use = 3.14) [2009]
Sol. Radius of the cylindrical glass r = 2.5 cm
Height of the glass = 10 cm
Apparent capacity of the glass = r2h= 3.14 × 2.5 × 2.5 × 10 cm3
= 196.25 cm3
Volume of the hemispherical portion
=2
3r3 =
2
3× 3.14 × 2.5 × 2.5 × 2.5 cm3
= 32.71 cm3
Actual capacity of the glass
= (196.25 – 32.71) cm3
= 163.54 cm3.
Q.22. A cylindrical vessel with internaldiamater 10 cm and height 10.5 cm is full ofwater. A solid cone of the diameter7 cm and height of 6 cm is completely immersedin water. Find the volume of
(i) water displaced out of the cylindricalvessel.
(ii) water left in the cylindrical vessel.
[Take =22
7] [2009]
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Sol. Radius of the cylinder, r = 5 cm
Height of the cylinder, h = 10.5 cm
Capacity of the vessel = r2h
=22
7× 5 × 5 × 10.5 cm3 = 825 cm3
Volume of the cone =1
3r2h
=1
3×
22
7× 3.5 × 3.5 × 6 cm3 = 77 cm3.
(i) Water displaced out of the cylinder
= Volume of the cone = 77 cm3
(ii) Water left in the cylindrical vessel= Capacity of the vessel
– Volume of the cone
= (825 – 77) cm3 = 748 cm3.
Q.23. A pen stand made of wood is in theshape of a cuboid with four conical depressionsand a cubical depression to hold pens and pinsrespectively. The dimensions of the cuboid are
10 cm, 5 cm and 4 cm. The radius of each of theconical depressions is 0.5 cm and depth is 2.1cm. The edge of the cubical depression is 3 cm.Find the volume of the wood in the entire stand.
Sol. Volume of a cuboid = 10 × 5 × 4 cm3
= 200 cm3.
Volume of the conical depression
=1
3r2h =
1
3×
22
7× (0.5)2 × 2.1 cm3
Volume of 4 conical depressions
=4
3×
22
7× (0.5)2 × 2.1 cm3
= 2.2 cm3
Volume of cubical depression= 33 cm3 = 27 cm3.
Volume of wood in the entire stand
= [200 – (2.2 + 27)] cm3
= 170.8 cm3.
PRACTICE EXERCISE 13.2A
Choose the correct option (Q 1 – 5) :
1. The surface area of a sphere is 154 cm2. Thevolume of the sphere is :
(a) 1792
3cm3 (b) 359
1
2cm3
(c) 12152
3cm3 (d) 1374
1
3cm3
2. The ratio of the volumes of two spheres is8 : 27. The ratio between their surface areas is :
(a) 2 : 3 (b) 4 : 27
(c) 8 : 9 (d) 4 : 9
3. The curved surface area of a cylinder is264 m2 and its volume is 924 m3. The height of thecylinder is :
(a) 3 m (b) 4 m
(c) 6 m (d) 8 m
4. The radii of the base of a cylinder and a cone
of same height are in the ratio 3 : 4. The ratio be-
tween their volumes is :
(a) 9 : 8 (b) 9 : 4
(c) 3 : 1 (d) 27 : 16
5. The capacity of a cylindrical vessel with a
hemispherical portion raised upward at the bottom
as shown in the figure is :
(a) r2h
(b) 2
3 23
rh r
(c) 2
3 23
rh r
(d) ( )3
3 43
rh r
π+
6. Two solid cones A and B are placed in a
cylindrical tube as shown in the figure. The ratio
of their capacities is 2 : 1. Find the heights and
capacities of the cones. Also, find the volume of
the remaining portion of the cylinder.
7. Marbles of diameter 1.4 cm are dropped intoa cylindrical beaker of diameter 7 cm containing
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some water. Find the number of marbles that shouldbe dropped into the beaker so that the water levelrises by 5.6 cm.
8. A solid is in the form of a right circular conemounted on a hemisphere. The radius of thehemisphere is 3.5 cm and the height of the cone is4 cm. The solid is placed in a cylindrical tub, fullof water, in such a way that the whole solid issubmerged in water. If the radius of the cylinder is5 cm and height 10.5 cm, find the volume of waterleft in the cylindrical tub.
9. The largest possible sphere is carved out
from a solid cube of side 7 cm. Find the volume
of the sphere.
10. A cylindrical boiler, 2 m high, is 3.5 m in
diameter. It has a hemispherical lid. Find the
volume of its interior, including the part covered
by the lid.2 2
7
π =
11. An ice cream cone consists of a rightcircular cone of height 14 cm and the diameter ofthe circular top is 5 cm. It has a hemisphericalscoop of ice cream on the top with the samediameter as of the circular top of the cone. Findthe volume of ice cream in the cone.
12. A solid toy is in the form of a hemispheresurmounted by a right circular cone. Height ofthe cone is 2 cm and the diameter of the base is4 cm. If a right circular cylinder circumscribesthe toy, find how much more space it will cover.
[2011 (T-II)]
13. A cylindrical tub of radius 12 cm
contains water to a depth of 20 cm. A spherical
iron ball is dropped into the tub and thus the
level of water is raised by 6.75 cm. What is the
radius of the ball?
14. From a solid cylinder of height 12 cm
and base diameter 10 cm, a conical cavity with
the same height and diameter is carved out. Find
the volume of the remaining solid.
15. A building is in the form of a cylindersurmounted by a hemispherical dome as shown inthe figure. The base diameter of the dome is equal
to2
3of the total height of the building. Find the
height of the building, if it contains 167
27m3 of
air. [2011 (T-II)]
16. A heap of rice is in the form of a cone ofdiameter 9 m and height 3.5 m. Find the volume ofthe rice. How much canvas cloth is required to justcover the heap?
17. 500 persons are taking a dip into a cuboidalpond which is 80 m long and 50 m broad. What isthe rise of water level in the pond, if the averagedisplacement of the water by a person is 0.04 m3.
18. A rocket is in the form of a right circularcylinder closed at the lower end and surmountedby a cone with the same radius as that of thecylinder. The diameter and height of the cylinderare 6 cm and 12 cm respectively. If the slant heightof the conical portion is 5 cm, find the total surfacearea and volume of the rocket. (Take = 3.14)
13.3 CONVERSION OF SOLID FROM ONE SHAPE TO ANOTHER
TEXTBOOK’S EXERCISE 13.3
Take π =22
7, unless stated otherwise.
Q.1. A metallic sphere of radius 4.2 cm ismelted and recast into the shape of a cylinder of
radius 6 cm. Find the height of the cylinder.Sol. Radius of sphere = 4.2 cm
Volume of sphere =4
3r3 =
4
3 (4.2)3 cm3
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Volume of cylinder = R2H
= (6)2H cm3
As per condition,Volume of the sphere = Volume of the cylinder
4
3 (4.2)3 = (6)2H
H =
3
2
4 4.2
3 6
H = 2.74Hence, the height of the cylinder is 2.74 cm.
Q.2. Metallic spheres of radii 6 cm, 8 cmand 10 cm, respectively, are melted to form asingle solid sphere. Find the radius of theresulting sphere.
Sol. We know that, volume of the sphere =4
3r3
Volume of sphere of radius 6 cm
=4
3 (6)3 cm3 ...(i)
Volume of sphere of radius 8 cm
=4
3 (8)3 cm3 ...(ii)
Volume of sphere of radius 10 cm
=4
3 (10)3 cm3 ...(iii)
Let the radius of the resulting sphere be R cm.Then volume of the resulting sphere
=4
3R3 cm3 ...(iv)
As per condition,
4
3R3 =
4
3 (6)3 +
4
3 (8)3 +
4
3 (10)3
R3 = (6)3 + (8)3 + (10)3
R3 = 1728
R = 3 1728 R = 12
Hence, the radius of the resulting sphere is 12 cm.
Q.3. A 20 m deep well with diameter 7 m isdug and the earth from digging is evenly spreadout to form a platform 22 m by 14 m. Find theheight of the platform. [2011 (T-II)]
Sol. Diameter of well = 7 m
Radius (r) =7
2m
Depth (h) = 20 m
Volume of well = r2h =
27
2
20 m3
= 245 m3
Length of platform (L) = 22 m
Width of platform (B) = 14 m
Let the height of the platform be H.
Then, volume of the platform = LBH
= 22 × 14 × H m3 = 308H m3
As per condition,
Volume of platform = Volume of well
308 H = 245
H =245
308
H =
245 22
308 7
H= 2.5
Hence, the height of the platform is 2.5 m.
Q.4. A well of diameter 3 m dug 14 m deep.The earth taken out of it has been spread evenlyall around it in the shape of a circular ring ofwidth 4 m to form an embankment. Find theheight of the embankment. [2011 (T-II)]
Sol. For well :
Diameter = 3 m
Radius of well (r) =3
2m
Depth of well (h) = 14 m Volume of earth taken out = r2h
=
23
2
(14) m3 =63
2 m3
Width of the embankment = 4 m
Let the height of the embankment be H m.
Radius of the well with embankment, R
=3
4 cm2
+ =11
2m
Volume of earth= Volume of the embankment
R2H – r2H = (R2 – r2) H
=2 2
311 3H m
2 2
π − =
3121 9H m
4 4
π −
= 28H m3
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As per condition,
28H =63
2 H =
63
2 28 H =
9
8OR, H = 1.125
Hence, the height of the embankment is 1.125 m.
Q.5. A container shaped like a right dircularcylinder having diameter 12 cm and height 15cm is full of ice cream. The ice cream is filledinto cones of height 12 cm and diameter 6 cm,having a hemispherical shape on the top. Findthe number of such cones which can be filledwith ice cream. [2008, 2011(T-II)]
Sol. Diameter of cylinder = 12 cm
Radius of cylinder (r) =12
2cm = 6 cm
Height of cylinder (h) = 15 cmVolume of cylinder = r2h = (6)2 (15) cm3
= 540 cm3
Diameter of cone = 6 cm
Radius of cone (R) =6
2cm = 3 cm
Height of cone (H) = 12 cm
Volume of cone =1
3R2H
=1
3(3)2 (12) cm3 = cm3
Radius of hemisphere = 3 cm
Volume of hemisphere =2
3(3)3 cm3
= 18 cm3
Volume of ice cream in one cone-volume ofcone + volume of hemisphere
= (36 + 18) cm3 = 54 cm3
Let n cones be filled with ice cream.Then, volume of n cones = n (54) cm3
As per conditionn (54) = 540
n =540
54
ππ
n = 10
Hence, 10 cones can be filled with ice cream.
Q.6. How many silver coins, 1.75 cm indiameter and of thickness 2 mm, must be meltedto form a cuboid of dimension 5.5 cm × 10 cm× 3.5 cm?
Sol. Diameter of silver coin = 1.75 cm
Radius of silver coin (r) =1.75
2cm =
7
8cm
Thickness of silver coin (h) = 2 mm =2
10cm
=1
5cm
Volume of a silver coin = r2h =
27 1
8 5
cm3
=49
320 cm3
Let n coins be melted.
Then, volume of coins = n ×49
320 cm3
Length of cuboid (L)= 5.5 cmBreadth of cuboid (B)= 10 cmHeight of cuboid (H)= 3.5 cmVolume of the cuboid
= Length × Breadth × Height= (5.5 × 10 × 3.5) cm3 = 192.5 cm3
=1925
10cm3 =
385
2cm3
As per condition,
n ×49
320 =
385
2
n =385 320
2 49
n =385 320 7
2 49 22
n = 400Hence, 400 coins must be melted to form the
required cuboid.
Q.7. A cylindrical bucket, 32 cm high andwith radius of base 18 cm, is filled with sand.This bucket is emptied on the ground and aconical heap of sand is formed. If the height ofthe conical heap is 24 cm, find the radius andslant height of the heap.
Sol. For cylindrical bucket :
Radius of cylindrical bucket (r)= 18 cmHeight of cylindrical bucket (h) = 32 cmVolume of cylindrical bucket = r2h= (18)2 (32) cm3 = 10368 cm3
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Height of cone (H) = 24 cmLet the radius of conical heap be R cm.
Then,volume of conical heap =1
3R2H
=1
3R2 (24) cm3= 8R2 cm3
As per condition,
8R2 = 10368
8R2 = 10368
R2 =10368
8
R2 = 1296
R = 1296 R = 36Hence, the radius of the conical heap is 36 cm.
Slant height (l ) = 2 2R H+
= 2 236 24 cm = 1296 576 cm
= 1872 cm = 12 12 13 cm = 12 13 cmHence, the slant height of the conical heap is
12 13 cm.
Q.8. Water in a canal 6 m wide and 1.5 mdeep is flowing with a speed of 10 km/h. Howmuch area will it irrigate in 30 minutes, if 8 cmof standing water is needed? [2011 (T-II)]
Sol. Width of canal = 6 m
Depth of canal = 1.5 m =15
10m =
3
2m
Speed of flowing water
= 10 km/h = 10 × 1000 m/h
= 10000 m/h
=10000
60m per min
=500
3m/min
=500 30
3
m per 30 minutes
= 5000 m per 30 minutes
Volume of water that flows in 30 minutes
= 6 ×3
2× 5000 m3 = 45000 m3
The area it will irrigate in 30 minutes with 8 cmof standing water
=45000
8
100
m2 =4500000
8m2
= 562500 m2 =562500
10000hectare
[1 hectare = 10000 m2, 1 m2 =1
10000hectare]
= 56.25 hectare.
Q.9. A farmer connects a pipe of internaldiameter 20 cm from a canal into a cylindricaltank in her field, which is 10 m in diameter and2 m deep. If water flows through the pipe at therate of 3 km/h, in how much time will the tank befilled? [2011(T-II)]
Sol. Diameter of cylindrical tank = 10 m
Radius of cylindrical tank (r) =10
2m = 5 m
Depth of cylindrical tank (h) = 2 m
Volume of cylindrical tank
= r2h = (5)2 (2) cm3 = 50 m3
Rate of flow of water = 3 km/h = 3000 m/h
=3000
60m/min = 50 m/min
Internal diameter of pipe = 20 cm
Internal radius of pipe (R) =20
2cm = 10 cm
=10
100m = 0.1 m
Volume of water that flow per minute in pipe
= r2 (50) m3 = (0.1)2 (50) m3
= 0.5 m3 =5
10
m3 =
2
m3
Required time =50
2
= 100 minutes
Hence, the tank will filled in 100 minutes.
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Q.1. A cubical ice cream brick of edge 22 cmis to be distributed among some children byfilling ice cream cones of radius 2 cm and height7 cm up to its brim. How many children will getthe ice cream cones?
(a) 163 (b) 263(c) 363 (d) 463Sol. (c) Required number of children
=capacity of cubical brick
capacity of an ice cream cone
=3
2
22
1 222 7
3 7
=22 22 3
4
= 363
Q.2. A metallic spherical shell of internaland external diameters 4 cm and 8 cmrespectively is melted and recast into the form ofa cone of base diameter 8 cm. The height of thecone is :
(a) 12 cm (b) 14 cm(c) 15 cm (d) 18 cmSol. (b) Internal and external radii of the ball are
2 cm and 4 cm respectively.Volume of metal used in the spherical ball
=4
3 (43 – 23) cm3 =
224
3 cm3
Let the height of the cone be h. Then,
1
3 × 42 × h =
224
3
h =224
16= 14 cm.
Q.3. Twelve solid spheres of the same sizeare made by melting a solid metallic cylinder ofbase diameter 2 cm and height 16 cm. Thediameter of each sphere is :
(a) 2 cm (b) 3 cm
(c) 4 cm (d) 6 cmSol. (a) Let the radius of each sphere be r.
Then, 12 ×4
3r3 = × 12 × 16
r3 =16 3
12 4
= 1 r = 1
Hence, diameter of each sphere = 2 cm.
Q.4. A spherical iron ball is melted andrecast into 8 identical balls. The radius of each
new ball is1
8th the radius of the original ball. Is
it true?
Sol. We have, 8 ×4
3r3 =
4
3R3
r3 =3
8
R
r =2
R
Hence, radius of each smaller ball is half of theoriginal ball.
Thus, the statement is false.
Q.5. A solid metallic sphere of radius10.5 cm is melted and recast into a number ofsmaller cones, each of radius 3.5 cm and height3 cm. Find the number of cones so formed.
Sol. Volume of the solid metallic sphere
=4
3× (10.5)3 cm3
Volume of a cone =1
3 × (3.5)2 × 3 cm3
Number of cones so formed
=Volume of the sphere
Volume of a cone
=
3
2
410.5
31
3.5 33
= 126.
Q.6. The circumference of the circular end ofa hemispherical bowl is 132 cm. Find thecapacity of the bowl. [2011 (T-II)]
Sol. Let r be radius of the edge of a
hemispherical bowl. Its circumference = 2r
2r = 132 r =132 7
2 22
×× = 21 cm
OTHER IMPORTANT QUESTIONS
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Now, capacity of the bowl
32 2 2221 21 21
3 3 7r= π = × × × × = 44 × 441
= 19404 cm3
Q.7. Three cubes of a metal whose edges arein the ratio 3 : 4 : 5 are melted and convertedinto a single cube whose diagonal is
12 3 cm. Find the edges of the three cubes.
Sol. Let the edges of the three cubes be 3x, 4x
and 5x. Then,Total volume of these three cubes
= 27x3 + 64x3 + 125x3 = 216x3.Let the edge of the new cube formed be a. Then,a3 = 216x3 a = 6x
Diagonal of the new cube = 3a = 6 3x
6 3x = 12 3 x = 2
Edge of the three cubes are 3 × 2 cm, 4 × 2 cm
and 5 × 2 cm or 6 cm, 8 cm and 10 cm.
Q.8. If the diameter of the cross-section of awire is decreased by 5%, how much per cent willthe length be increased so that the volumeremains the same?
Sol. Let the diameter and length of the cross-
section of the wire by 2r and h respectively.Then, volume of the wire = r2h
New diameter = 2r –2 × 5
100
r
= 2r –19
10 10
r r
New radius =19
20
r
To keep the volume same, let the new length be H.
...
219
20
r
× H = r2h 361
400r2 × H = r2h
H =400
361
h
... Increase in length =
400
361
h– h =
39
361
h
% increase =
39100
39 100361361
h
h
10.80%.
Q.9. A spherical ball of lead 3 cm indiameter is melted and recast into three sphericalballs. The diameters of two of these balls are1 cm and 1.5 cm. Find the diameter of the thirdball.
Sol. Radius of the bigger ball =3
2cm = 1.5 cm.
Volume of the bigger ball
=4
3 (1.5)3 cm3 ...(i)
Radii of two smaller balls are1
2cm and
1.5
2cm
= 0.5 cm and 0.75 cm.
... Total volume of the two smaller balls
= 3 34 4(0.5) (0.75)
3 3
π + π cm3
=4
3 [(0.5)3 + (0.75)3] cm3 ...(ii)
Let the radius of the third ball be r cm.
... Volume of the third ball =
4
3r3 cm3 ...(iii)
From (i), (ii) and (iii), we have
4
3 (1.5)3 =
4
3 [(0.5)3 + (0.75)3] +
4
3 r3
(1.5)3 = (0.5)3 + (0.75)3 + r3
r3 = (1.5)3 – (0.5)3 – (0.75)3
r3 = 3.375 – 0.125 – 0.421875
r3 = 2.828125 r = 1.41
... Diameter of the third ball
= 2 × 1.41 cm = 2.82 cm.
Q.10. Find the number of metallic discs with1.5 cm base diameter and of height 0.2 cm, to bemelted to form a right circular cylinder of height10 cm and diamater 4.5 cm.
Sol. Volume of the cylinder = ×
24.5
2
× 10
cm3
Volume of 1 circular disc
= ×
21.5
2
× 0.2 cm3
Let n circular discs are needed.
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Then,
n × ×
21.5
2
× 0.2 = ×
24.5
2
× 10
n =4.5 4.5 10
0.2 1.5 1.5
= 450.
Hence, 450 circular discs are needed.
Q.11. Water is flowing at the rate of 5 km/hrthrough a pipe of diameter 14 cm into arectangular tank, which is 50 m long and 44 mwide. Determine the time in which the level ofwater in the tank will rise by 7 cm. [2011 (T-II)]
Sol. Suppose the level of the water in the tank
will rise by 7 cm in x hours. Since the water is flowingat the rate of 5 km/hr. Therefore, length of the watercolumn in x hours = 5x km = 5000x metres.
Clearly, the water column forms a cylinder whoseradius
14 7cm = m
2 100r =
Length = h = 5000x metres Volume of the water flowing through the
cylindrical pipe in x hours
2 3 322 7 75000 m 77 m
7 100 100r h x x= π = × × × =
Also,Volume of the water that falls into the tank in
x hours = 50 × 44 ×7
100m3 = 154 m3
But, volume of the water flowing through thecylindrical pipe in x hours = Volume of the water thatfalls in the tank in x hours
77x = 154 x =154
77= 2
Hence, the level of the water in the tank willrise by 7 cm in 2 hours.
Q.12. Water is flowing at the rate of 3 km/hrthrough a circular pipe of 20 cm internaldiameter into a cylindrical cistern of diameter10 m and depth 2 m. In how much time will the
cistern be filled?22
Take7
π = [2011 (T-II)]
Sol. Suppose the cistern is filled in x hours. Since
water is flowing at the rate of 3 km/hr. Therefore,length of the water column in x hours = 3x km= 3000x metres.
Clearly, the water column forms a cylinder ofradius.
r =20
2cm = 10 cm =
1
10m and
h = height (length) = 3000x metres Volume of the water that flows in the cistern
in x hours 2 322 1 13000 m
7 10 10r h x
= π = × × × Also,
Volume of the cistern =22
5 5 27
× × × m3
[ ]5 , 2r m h m= =
Since the cistern is filled in x hours.
Volume of the water that flows in the cistern in
x hours = Volume of the cistern.
22 1 1 22
3000 5 5 27 10 10 7
x× × × = × × ×
5 5 2 10 10 5
hrs = hrs3000 3
x× × × × =
= 1 hour 40 minutes
Q.13. A rectangular water tank of base 11 m× 6 m contains water up to a height of 5 m. If thewater in the tank is transferred to a cylindricaltank of radius 3.5 m, find the height of the waterlevel in the tank.
Sol. Volume of water in the tank = 11 × 6 × 5 m3.
Let the required height of the water level be h.Then, 11 × 6 × 5 = × (3.5)2 × h
h =11 6 5 7
22 3.5 3.5
= 8.6 m.
Hence, height of water level in the tank = 8.6 m.
Q.14. How many coins 1.75 cm in diameterand 2 mm thick must be melted to form a cuboidof dimensions 11 cm × 10 cm × 7 cm?
22Take
7
π = [2011 (T-II)]
Sol. Here each coin is a cylinder with
r =1.75
2cm = 8.75 mm and h = 2 mm
Volume of each coin= r2h = (8.75)2 × 2 cu mm
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Volume of the cuboid = 11 × 10 × 7 cu cm= 770 cu cm = 770000 cu mm
Required number of coins
7700001600
8.75 8.75 2= =
π × × ×Q.15. Water flows at the rate of 10 m/min
through a cylindrical pipe 5 mm in diameter.How long would it take to fill a conical vesselwhose diamater at the base is 40 cm and depth24 cm?
Sol.Volume of water which flows out through the
pipe in 1 minute = ×0.5
2× 1000 cm3.
Capacity of the conical vessel
=1
3× × (20)2 × 24 cm3
Time taken by the pipe to fill the conical vessel
=
2
2
20 243
0.51000
2
min
= 51.2 min = 51 min 12 sec.
Q.16. A solid iron cuboidal block ofdimensions 4.4 m × 2.6 m × 1 m is recast into ahollow cylindrical pipe of internal radius 30 cmand thickness 5 cm. Find the length of the pipe.
Sol. Volume of the cuboid = 4.4 × 2.6 × 1 m3.
Internal radius of the pipe = 30 cm = 0.3 m.Thickness of pipe = 5 cm. External radius of the pipe
= 35 cm = 0.35 m.Let the length of the pipe be h m.Then, 4.4 × 2.6 × 1 = × (0.352 – 0.32) × h
h =4.4 2.6 1 7
22 0.65 0.05
× × ×× ×
m = 112 m.
Hence, length of the pipe = 112 m.
Q.17. A hemispherical bowl of internalradius 9 cm is full of a liquid. The liquid is to befilled into cylindrical shaped bottles each ofradius 1.5 cm and height 4 cm. How many suchbottles are needed to empty the bowl? [2005]
Sol. Capacity of the hemispherical bowl
=2
3× × 93 cm3 =
1458
3 cm3
Capacity of 1 cylindrical bottle= × (1.5)2 × 4 cm3
Let n cylindrical bottles are required.
n × × (1.5)2 × 4 =1458
3
n =1458
1.5 1.5 4 3 = 54
Hence, 54 cylindrical bottles are required.
Q.18. A solid right circular cone of diameter14 cm and height 8 cm is melted to form a hollowsphere. If the external diameter of the sphere is10 cm, find the internal diameter of the sphere.
[2007]
Sol.Volume of the cone =1
3 × 72 × 8 cm3
Let the internal radius of the sphere be r.
Then,1
3 × 72 × 8 =
4
3× × (53 – r3)
53 – r3 =49 8
4
= 98
r3 = 125 – 98 = 27 r = 3
Hence, internal diameter of the sphere = 6 cm.
Q.19. An iron spherical ball has been meltedand recast into smaller balls of equal size. If the
radius of each of the smaller balls is1
4of the
radius of the original ball, how many such ballsare made? Compare the surface area of all thesmaller balls combined together with that of theoriginal ball? [2007]
Sol. Let the radius of the bigger ball be r.
Then, the radius of each smaller ball =4
r.
Let n smaller balls are formed.
Then,4
3r3 = n ×
4
3
3
4
r
n = 4 × 4 × 4 = 64Hence, 64 smaller balls are formed.Also, surface area of the bigger ball = 4r2.
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PRACTICE EXERCISE 13.3A
Choose the correct option (Q 1 – 5) :1. If a marble of radius 2.1 cm is put into a
cylindrical cup full of water of radius5 cm and height 6 cm, then how much water
flows out of the cylindrical cup?
(a) 38.8 cm3 (b) 55.4 cm3
(c) 19.4 cm3 (d) 471.4 cm3
And, surface area of the 64 smaller balls
= 64 × 4
2
4
r
= 16r2
Surface area of the 64 smaller balls together
Surface area of the original ball
=2
2
16
4
r
r
= 4 : 1.
Q.20. The rain water collected on the roof ofa building of dimensions 22 m × 20 m, is drainedinto a cylindrical vessel having base diameter2 m and height 3.5 m. If the vessel is full up tothe brim, find the height of rain water on the
roof.22
7
π = [2010, 2011 (T-II)]
Sol. Let the height of rain water on the roof be
h m. Then,Volume of water collected on the roof of the
building = (20 × 20 × h) m3.Radius of the cylindrical vessel = 1 mHeight of the cylindrical vessel = 3.5 m. Capacity of the cylindrical vessel
= × 12 × 3.5 m3.But, according to the question20 × 20 × h = × 12 × 3.5
h =22
7× 3.5 ×
1
20×
1
20=
11
400m
=11
400× 100 cm = 2.75 cm.
Q.21. The surface area of a solid metallicsphere is 616 cm2. It is melted and recast into acone of height 28 cm. Find the diameter of thebase of the cone so formed. [2010, 2011 (T-II)]
Sol. Let r be the radius of the sphere.Then, 4r2 = 616
r2 =616 7
4 22
××
= 49 r = 7
Volume of the sphere =4
3r3
=4
3×
22
7× 7 × 7 × 7 cm3
Let R be the radius and h be the height of thecone.
Then, volume of sphere =1
3r2h
=1
3×
22
7× r2 × 28 cm3
But, according to the question,
4
3×
22
7× 7 × 7 × 7 =
1
3×
22
7× r2 × 28
r2 =4 7 7 7
28
× × ×= 7 × 7
r = 7Hence, base diameter of the cone
= 2 × 7 cm = 14 cm.
Q.22. A hemispherical bowl of internaldiameter 36 cm is full of liquid. This liquid is tobe filled in cylindrical bottles of radius 3 cm andheight 6 cm. How many such bottles are requiredto empty the bowl? [2011 (T-II)]
Sol. Diameter of hemispherical bowl = 36 cm
Radius of hemispherical bowl = 18 cmVolume of hemispherical bowl
=32 2 22
18 18 18 cu cm3 3 7
rπ = × × × ×
Height of bottle = 6 cm, radius of bottle = 3 cm
Volume of bottle = r2h =22
3 3 6 cu cm7
× × ×
No. of bottles required
Volume of liquid in the bottle
Volume of 1 bottle=
2 2218 18 18
3 7 72 bottles22
3 3 67
× × × ×= =
× × ×
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2. A solid piece of aluminium in the form ofa cuboid of dimensions 49 cm × 33 cm× 24 cm is moulded to form a solid sphere. Theradius of the sphere is :
(a) 21 cm (b) 23 cm (c) 25 cm (d) 19 cm
3. During conversion of a solid from oneshape to another, the volume of the new shapewill :
(a) increase (b) decrease(c) remains unaltered (d) be doubled
4. A solid sphere of radius 3 cm is melted andthen cast into small spherical balls each of diam-eter 0.6 cm. The number of small balls thus ob-tained is :
(a) 50 (b) 100 (c) 500 (d) 1000
5. Acubical block having edge 44 cm is meltedand then recast into spherical bullets, each of ra-dius 2 cm. Number bullets thus formed is :
(a) 2000 (b) 2541 (c) 2451 (d) 2145
6. Three metallic solid cubes whose edges are3 cm, 4 cm and 5 cm are melted and formed into asingle cube. Find the edge of the cube so formed.
7. How many spherical bullets each havingdiameter 3 cm can be made from a cuboidal solidlead of dimensions 9 cm × 11 cm × 12 cm?
8. The radii of the internal and external surfacesof a hollow spherical shell are 3 cm and 5 cmrespectively. If it is melted and recast into a solid
cylinder of height8
3cm, find the diameter of the
cylinder. [2011 (T-II)]
9. A right circular cone is 3.6 cm high andhas base radius 1.6 cm. It is melted and recastinto a right circular cone with radius of its baseas 1.2 cm, find its height.
10. A piece of metal in the form of a cone ofradius 3 cm and height 7 cm is melted and recastinto a cube. Find the side of the cube.
11. The diameter of a copper sphere is 6 cm.The sphere is melted and drawn into a long wireof uniform cross section. Find the radius of thewire if its length is 36 cm.
12. A vessel in the form of a hemisphericalbowl is full of water. The contents are emptiedinto a cylinder. The internal radii of the bowl andcylinder are 6 cm and 4 cm respectively. Find theheight of water level in the cylinder.
13. How many lead balls each of radius1 cm, can be made from a sphere of radius8 cm?
14. A metallic toy in the form of a cone ofradius 11 cm and height 62 cm mounted on ahemisphere of the same radius is melted andrecast into a solid cube. Find the surface area ofthe cube thus formed.
15. A canal is 3 m wide and 1.2 m deep. Thewater in the canal is flowing with a speed of20 km/h. How much area will it irrigate in20 minutes if 8 cm of standing water is required?
16. How many spherical lead shots each ofdiameter 4.2 cm can be obtained from a solidrectangular lead piece with dimensions 66 cm,42 cm and 21 cm?
17. A solid metallic hemisphere of radius 8 cmis melted and recast into a right circular cone ofbase radius 6 cm. Determine the height of the cone.
18. Water flows through a cylindrical pipe,whose inner radius is 1 cm, at the rate of80 cm/sec in an empty cylindrical tank, the radiusof whose base is 40 cm. What is the rise of waterlevel in the tank in half an hour?
[2007, 2011 (T-II)]
13.4 FRUSTUM OF A CONE
1. When a cone is cut by a plane parallel tothe base of the cone, then the portion betweenthe plane and the base is called the frustum ofthe cone.(a) Slant height of the frustum,
l = 22 Rh r
(b) Volume of thefrustum of the cone
=3
h[R2 + r2 + Rr]
(c) Lateral surface areaof the frustum of the cone= l (R + r)
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TEXTBOOK’S EXERCISE 13.4
(d) Total surface area of the frustum of thecone = (area of the base) + (area of the top)
+ (lateral surface area)
= [R2 + r2 + l (R + r)]
= [R2 + r2 + l (R + r)].
Q.1. A drinking glass is in the shape of afrustum of a cone of height 14 cm. The diametersof its two circular ends are 4 cm and 2 cm. Findthe capacity of the glass.
Sol. Let radius of lower end (r1) =2
2cm = 1 cm
Radius of upper end (r2) =4
2cm = 2 cm
Height of the glass (h) = 14 cm
Capacity of the glass = Volume of the frustum
=1
3h (r1
2 + r22 + r1r2)
=1
3×
22
7× 14 [(2)2 + (1)2 + (2) (1] cm3
=1
3×
22
7× 14 × 7 cm3 =
308
3cm3
= 1022
3cm2
Hence, capacity of the glass is 1022
3cm2.
Q.2. The slant height of a frustum of a coneis 4 cm and the perimeters (circumference) of itscircular ends are 18 cm and 6 cm. Find thecurved surface area of the frustum.
Sol. Let r1 be the radius of upper end and r2 be the
radius of lower end.Then, slant height l = 4 cmCircumference of upper end, 2r1 = 18 cm r1 = 9 cmCircumference of lower end, 2r2 = 6 cm r2 = 3 cmCurved surface area of the frustum
= (r1 + r2) l = (r1 + r2) l
= (9 + 3) 4 = 48 cm2.
Q.3. A fez, the cap used by the Turks, isshaped like the frustum of a cone (see figure). Ifits radius on the open side is 10 cm, radius at theupper base is 4 cm and its slant height is 15 cm,find the area of material used for making it.
Sol.
Radius of upper end, r1 = 10 cm
Radius of lower end, r2 = 4 cm
Slant height of frustum, l = 15 cm
Surface area of the frustum
= (r1 + r2) l + r22
=22
7(10 + 4) (15) cm2 + (4)2 cm2
= 660 cm2 + 16 cm2 = 600 cm2 + 16 ×22
7cm2
= 600 cm2 +352
7cm2 =
4620 352
7
cm2
=4972
7cm2 = 710
2
7cm2
Hence, the area of material used for making it is
7102
7cm2.
Q.4. A container, opened from the top ismade up of a metal sheet, is in the form of afrustum of a cone of height 16 cm with radii ofits lower and upper ends as 8 cm and 20 cm,respectively. Find the cost of the milk which cancompletely fill the container, at the rate of Rs 20per litre. Also, find the cost of metal sheet usedto make the container, if it costs Rs 8 per100 cm2. (Take = 3.14).
Sol. Height of container (h) = 16 cm
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Radius of upper end, r1 = 20 cm
Radius of lower end, r2 = 8 cm
Volume of the container= Capacity of the container
=1
3h (r1
2 + r22 + r1r2)
=1
3(3.14) (16) {(20)2 + (8)2 + (20) (8)} cm3
=1
3(3.14) (16) (400 + 64 + 160) cm3
=1
3(3.14) (16) (624) cm3
= (3.14) (16) (208) cm3 = 10449.92 cm3
= 10.44992 litres Cost of the milk = Rs 10.44992 × 20
= Rs 208.9984 Rs 209Total surface area
= (r1 + r2) 2 21 2( )h r r + r2
2
= [(3.14) (20 + 8) 2 216 20 8 + (3.14) (8)2] cm2
= [(3.14) (28) 256 144 + (3.14) (64)] cm2
= [(3.14) (28) (20) + 200.96] cm2 = [1758.4 + 200.96] cm2
= 1959.36 cm2
Area of the metal sheet used = 1959.36 cm2
Cost of metal sheet = Rs 1959.36 ×8
100= Rs 156.7488 = Rs 156.75
Q.5. A metallic right circular cone 20 cmhigh and whose vertical angle is 60° is cut intotwo parts at the middle of its height by a planeparallel to its base. If the frustum so obtained be
drawn into a wire of diameter1
16cm, find the
length of the wire.Sol.
In ODB, tan 30° =2
10
r
1
3= 2
10
r r2 =
10
3cm
In EOF, tan 30° =1
20
r
1
3=
1
20
r r1 =
20
3cm
Height of the frustum = h = 10 cm
Volume of the frustum =1
3h (r1
2 + r22 + r1r2)
=1
3×
22
7× 10 ×
2 220 10 20 10
3 3 3 3
cm3
=1
3×
22
7× 10 ×
400 100 200
3 3 3
cm3
=1
3×
22
7× 10 ×
700
3cm3 =
22000
9cm3
Diameter of the wire =1
16cm
Radius of the wire r
=1
2×
1
16cm =
1
32cm
Let length of the wire be x cm.Then, volume of the wire = r2 x
=22
7×
21
32
x =11
3584
xcm3
According to the question,
11
3584
x=
22000
9
x =22000 3584
11 9
x =
2000 3584
9
x =7168000
9
x = 796444.44 cm x = 7964.4 m
Hence, the length of the wire is 7964.4 m.
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OTHER IMPORTANT QUESTIONS
Q.1. The shape of a glass (tumbler) is usu-ally in the form of :
(a) cone (b) frustum of a cone(c) a cylinder (d) a sphereSol. (b) The shape of a glass (tumbler) is usually
in the form of frustum of a cone.
Q.2. A cone is cut through a plane parallel toits base and then the cone that is formed on oneside of that plane is removed. The new part thatis left over on the other side of the plane iscalled :
(a) a frustum of the cone (b) cone(c) cylinder (d) sphereSol. (a) The desired shape is called a frustum of
the cone.
Q.3. In a right circular cone, the cross-sec-tion made by a plane parallel to the base is a:
(a) circle (b) frustum of a cone(c) sphere (d) hemisphereSol. (a) The desired cross-section is a circle.
Q.4. A frustum of a right circular cone ofheight 16 cm with radii of its circular ends as 8cm and 20 cm has its slant height equal to :
[2011 (T-II)](a) 18 cm (b) 16 cm(c) 20 cm (d) 24 cm
Sol. (c) Slant height = 2 2(R )h r+ −
= 2 216 (20 8)+ − cm
= 256 144+ cm
= 400 cm = 20 cm
Q.5. The radii of the ends of the frustum ofa cone of height h are R and r. The volume of thefrustum of the cone is :
(a)3
h(R2 + r2 + Rr) (b)
3
(R2 + r2 – Rr)
(c)3
h(R2 + r2 – Rr) (d)
3
(R2 + r2 + Rrhh)
Sol. (a) Required volume =3
h(R2 + r2 + Rr)
Q.6. The radii of the top and bottom of abucket of slant height 45 cm are 28 cm and 7 cmrespectively. The curved surface area of thebucket is :
(a) 4950 cm2 (b) 4951 cm2
(c) 4952 cm2 (d) 4953 cm2
Sol. (a) Curved surface area of the bucket
= l (R + r)
=22
7× 45 (28 + 7) cm2 = 4950 cm2.
Q.7. The diameters of two circular ends of abucket are 44 cm and 24 cm. If the height of thebucket is 35 cm, the capacity of the bucket is :
(a) 32.7 litres (b) 33.7 litres(c) 34.7 litres (d) 31.7 litres
Sol. (a) Capacity of the bucket
=3
h[R2 + r2 + Rr]
=22
7×
35
3[222 + 122 + 22 × 12] cm3
= 32706.6 cm3 = 32.7 litres.
Q.8. The total surface area of a frustum of acone is [R2 + r2 + l (R + r)],
where l = 22h R r , R, and r are the
radii of the two ends of the frustum and h is theheight. Is it true?
Sol. Yes, the statement is true.
Q.9. The slant height of the frustion of a coneis 5 cm. If the difference between the radii of itstwo circular ends is 4 cm, write the height of thefrustum. [2010]
Sol. We have, l = 2 21 2( )h r r+ −
5 = 2 24h + h2 = 25 – 16 = 9
h = 3 cm.Q.10. A cone of radius 8 cm and height 12
cm is divided into two parts by a plane throughthe mid-point of its axis parallel to its base. Findthe ratio of the volumes of two parts.
[2011 (T-II)]
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Sol.
Let h be the height of the given cone. On dividing
the cone through the mid-point of its axis and parallelto its base into two parts, we get the shapes as shownin the figure.
OAB ~ OCD
OB
OD=
AB
CD
2
h
h=
8
r
2 =8
r r = 4 cm.
Volume of the smaller cone
Volume of the frustum of the cone
=
2
2 2
14
3 2
8 4 8 43 2
h
h
=16
112=
1
7
Hence, required ratio = 1 : 7.
Q.11. A bucket is in the form of a frustum ofa cone and holds 28.490 litres of water. The radiiof the top and bottom are 28 cm and 21 cm,respectively. Find the height of the bucket.
[2008C]Sol. Here, R = 28 cm, r = 21 cm
Volume = 28.490 litres = 28490 cm3. Volume of the bucket
=1
3h (R2 + r2 + Rr)
28490 =1
3×
22
7× h × (282 + 212 + 28 × 21)
28490 21
22
= h × 1813
h =28490 21
22 1813
= 15
Hence, height of the bucket = 15 cm.
Q.12. A milk container of height 16 cm ismade of metal sheet in the form of a frustum ofa cone with radii of its lower and upper ends as8 cm and 20 cm respectively. Find the cost ofmilk at the rate of Rs 22 per litre which the con-tainer can hold.
Sol. Here, h = 16 cm, r = 8 cm, R = 20 cm
Capacity of the container
=3
h(R2 + r2 + Rr)
=22
7×
16
3(202 + 82 + 20 × 8) cm3
=352
21× 624 cm3 =
73216
7cm3
= 10.46 litresCost of milk which this container can hold
= Rs 22 × 10.46 = Rs 230.12.
Q.13. A bucket is in the form of a frustum ofa cone of height 30 cm with radii of its lower andupper ends as 10 cm and 20 cm respectively.Find the capacity and the surface area of thebucket. [2011 (T-II)]
Sol. Here, h = 30 cm, R = 20 cm, r = 10 cm
Capacity of the bucket =3
h(R2 + r2 + Rr)
=22
7×
30
3(202 + 102 + 20 × 10) cm3
=220
7× 700 cm3 = 22000 cm3 = 22 litres.
Now, l = 2 2(R )h r+ −
= 2 230 (20 10) cm
= 900 100 cm
= 1000 cm = 10 10 cmTotal surface area of the bucket
= Curved surface area of the bucket+ surface area of the bottom
[l (R + r) + r2]
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=22
7[10 10 (20 + 10) + 100] cm2
=22
7(300 10 + 100) cm2
=22
7× 100 (9.49 + 1) cm2
=22
7× 100 × 10.49 cm2
= 3295.86 cm2.
Q.14. The height of a cone is 30 cm. A smallcone is cut off at the top by a plane parallel to
the base. If its volume be1
27th of the volume of
the given cone, at which height above the base is
the section made? [2011 (T-II)]
Sol. Height of the cone = 30 cm.
Let its radius be R cm
Let the height of the small cone= h and its radius = r
Volume of the given cone
=2 2R H R
3 3
π π = (30) cm3 ...(i)
Volume of the smaller cone =3
πr2h ...(ii)
In similar OAB and OCD, we have
R 30
r h= ... (iii)
As per the question,
Volume of smaller cone
Volume of the given cone=
1
27
⇒
2
2
1327
R (30)3
r h
[From (i) and (ii)]
⇒2
2
1
2730 R
r h
⇒ h =
2
2
30 R
27 r
⇒ h =
230 30
27 h
[From (iii)]
⇒3 3
1 1
30 27 3
h = =
⇒1 30
30 3 3
hh= ⇔ = = 10 cm.
... Height of the frustum
= 30 cm – 10 cm = 20 cm.
Q.15. A tent is made in the form of a frustumof a cone surmounted by another cone. The di-ameters of the base and the top of the frustum are20 m and 6 m respectively and the height is 24 m.If the height of the tent is 28 m and the radius ofthe conical part is equal to the radius of the topof the frustum, find the quantity of canvas re-quired. [2011 (T-II)]
Sol. Diameter of the base of the frustum = 20 m
Diameter of the top of the frustum = 6 mHeight of the frustum = 24 mLet l be the slant height of the frustum. Then
l 2 = (r1 – r2)2 + h2
⇒ l 2 = (10 – 3)2 + (24)2
= (7)2 + (24)2 = 49 + 576 = 625
⇒ l = 25 m
Curved surface of the frustum = π (r1 + r2) l
=22
7× (10 + 3) × 25 cm2
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=22
7× 13 × 25 cm2
= 1021.43 m2
Diameter of the base of the cone = 6 m.
Height of the cone = 4 m
Let l be the slant height of the cone.
l2 = (3)2 + (4)2 = 9 + 16 = 25
⇒ l = 5
Curved surface of the cone
= πrl =22
7× 3 × 5 m2 = 47.14 m2
Area of the canvas required
= curved surface area of frustum
+ curved surface area of cone
= (1021.43 + 47.14) m2
= 1068.57 m2.
Q.16. A shuttle cock used for playing bad-minton has the shape of a frustum of a conemounted on a hemisphere (see figure). The exter-nal diameters of the frustum are 5 cm and 2 cmand the height of the entire shuttle cock is 7 cm.Find its external surface area. [2011 (T-II)]
Sol. For the upper portion of the shuttle cock :
r1 = 1 cm, r2 =5
2cm = 2.5 cm
and h = (7 – 1) cm = 6 cm.
... l = 2 2
2 1( )h r r+ −
2 2(6) (2.5 1) cm
36 2.25 cm
= 38.25 cm = 6.2 cm (approx).
... CSA of frustum part = πl (r1 + r2)
= π × 6.2 × (2.5 + 1) cm2
= π × 6.2 × 3.5 cm2 = 21.7 π cm2
CSA of hemispherical part with radius 1 cm
= 2π 21r = 2π(1)2 cm2 = 2π cm2
... Total surface area = (21.7π + 2π) cm2
= 23.7π cm2
= 23.7 ×22
7cm2 = 74.49 cm2
≅ 74.5 cm2 (approx).
Q.17. A hollow cone is cut by a plane paral-lel to the base and the upper portion is removed.If the curved surface of the remainder is 8/9 ofthe curved surface of the whole cone, find theratio of line segments into which the altitude ofthe cone is divided by the plane. [2004]
Sol. Let L be the slant height of the given cone
and R be its base radius.Then, the smaller cone as shown in the figure has
been removed.
Let l and r be the slant height and radius respec-tively of the smaller cone.
Then,OAB ~ OCD
AB
CD=
OB
OD=
OA
OC...(i)
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PRACTICE EXERCISE 13.4A
Choose the correct option (Q 1 – 5) :1. The shape of a bucket is usually in the
form of :(a) a cone (b) a cylinder(c) sphere (d) frustum of a cone2. Slant height of frustum of cone of a height
h and radii of its ends as R and r is :
(a) 2 2R r+ (b) ( )2 2 2Rh r+ +
(c) ( )22 Rh r+ − (d) ( )22 Rh r− +
3. Curved surface area of a frustum of acone of height h and radii of its ends as R1 andR2 is :
(a) h (R1 + R2)
(b) (R1 + R2) {h2 + (R1 – R2)2}
(c) (R1 + R2) ( )221 2R Rh + −
(d) (R1 + R2)
R
r=
L
l r =
R
L
l...(ii)
Now, curved surface area of the remainder
=8
9of the curved surface area of the whole cone
Curved surface area of the smaller cone
=1
9of the curved surface area of the whole cone
rl =1
9RL
R
L
l l =
1
9RL [From (ii)]
9l2 = L2
L
l=
1
3.... (iii)
Now, from (i)OB
OD=
OA
OC
L
l=
OA
OC=
3
1[From (iii)]
OC =OA
3=
OC + AC
3
2OC = AC OC
AC=
1
2
Hence, required ratio = 1 : 2.
Q.18. The radius of base of a right circularcone is r. It is cut by a plane parallel to the baseat a height h from the base. The distance of theboundary of the upper surface from the centre of
the base of the frustum is2
2
9
rh . Show that
the volume of the frustum is213
27
r h.
Sol. We have OA = r, OO = h, OA =2
2
9
rh
Now, in OOA, we haveOA2 = OO2 + OA2
OA = 2 2OA OO
=2
2 2
9
rh h =
3
r
Volume of the frustum
=3
h(OA2 + OA2 + OA × OA)
=
22
3 9 3
h r rr r
= 2 2 29 3
3 9
r r rh =
213
27
r hProved.
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4. If the radii of the ends of a frustum of a cone,which is 45 cm high are 28 cm and 7 cm, then thevolume of the frustum of the cone is :
(a) 48510 cm3 (b) 45810 cm3
(c) 41510 cm3 (d) 48501 cm3
5. The volume of a frustum of a cone is616 cm3. If the radii of its ends are 10 cm and 6 cm,then the height of frustum of the cone is :
(a) 3 cm (b) 4 cm(c) 5 cm (d) 7 cm6. An open metallic bucket is in the shape of
a frustum of a cone, mounted on a hollow cylindri-cal base made of the same metallic sheet. The sur-face area of the metallic sheet used is equal tocurved surface area of frustum of the cone + areaof circular base + curved surface area of the cylin-der. Is this statement true?
7. A cone of radius 4 cm is divided into twoparts by drawing a plane through the mid-point ofits axis and parallel to its base. Compare the vol-umes of the two parts.
8. A bucket of height 8 cm and made up ofcopper sheet is in the form of a frustum of a rightcircular cone with radii of its lower and upper endsas 3 cm and 9 cm respectively. Calculate :
(i) the height of the cone of which the bucketis a part.
(ii) the volume of water which can be filledin the bucket.
(iii) the area of copper sheet required tomake the bucket. [2003]
9. If the radii of the ends of a bucket 24 cmhigh are 5 cm and 15 cm, find the surface area ofthe bucket.
10. The radii of the top and bottom of a 12 cmdeep tub are 20 cm and 10 cm. Find its volume andcost of tin sheet used for making the tub at the rateof Rs 1.20 per dm2.
11. An inverted cone of vertical height 12 cmand radius of base 9 cm contains water to a depth
of 4 cm. Find the area of interior surface of conenot in contact with water. (Use π = 3.14)
12. The radii of the circular ends of a conicalcurd container, 6 cm high are 16 cm and24 cm. Find the volume of the container in litres.(Take π = 3.14)
13. An open metallic bucket is in the shapeof a frustum of a cone mounted on a hollowcylindrical base made of metallic sheet. If thediameters of the two ends of the bucket are 45cm and 25 cm, the total vertical height of thebucket is 30 cm and that of the cylindricalportion is 6 cm, find the area of the metallic sheetused to make the bucket. Also find the volume of
water it can hold22
Take7
. [2011 (T-II)]
14. The height of a cone is 42 cm. A smallcone is cut off at the top by a plane parallel to the
base. If its volume is1
27of the volume of the
given cone, at what height above the base, the
section has been made?
15. A bucket made up of metal sheet is in theform of a frustum of a cone. Its depth is24 cm and the diameters of the top and bottom are30 cm and 10 cm respectively. Find the cost of milkwhich can completely fill the bucket at the rate ofRs 20 per litre and the cost of the metal sheet used,if it costs Rs 10 per 100 cm2 (use π = 3.14).
[2006, 2011 (T-II)]
TEXTBOOK’S EXERCISE 13.5 (OPTIONAL)
Q.1. A copper wire, 3 mm in diameter, iswound about a cylinder whose length is 12 cm,and diameter 10 cm, so as to cover the curved
surface of the cylinder. Find the length and massof the wire, assuming the density of copper to be8.88 g per cm3.
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Sol. Diameter of copper wire = 3 mm
Radius of copper wire =Diameter
2=
3
20cm
Height of the cylinder = 12 cmRadius of cylinder = 5 cmLength of wire used in one turn
= Circumference of cylinder.Length of wire used in one turn
= 2r = 2 × 3.14 × 5 =157
5cm
Number of turns used
=Height of the cylinder
Diameter of wire
=12
3
10
=120
3= 40
Total length of wire used
= Number of turns
× Length of wire used in one turn
= 40 ×157
5cm = 1256 cm = 12.56 m
Volume of wire used
= r2h =22
7×
3
20×
3
20× 1256 cm3
= 88.81 cm3.
Mass of 1 cm3 = 8.88 g [Given]
Mass of 88.81 cm3 = 88.81 × 8.88 g
= 789 g (approx.)
Q.2. A right triangle, whose side are 3 cmand 4 cm (other than hypotenuse is made torevolve about it a hypotenuse. Find the volumeand surface area of the double cone so formed.(Choose value of as found appropriate).
Sol. Let ABC be the triangle, right angled at A.
In ABC (By Pythagorus Theorem)
(BC)2 = (AB)2 + (AC)2
BC2 = (4)2 + (3)2 = 16 + 9 = 25
BC= 5 cm
Now, area of ABC =1
2× 3 × 4 cm2 = 6 cm2
Also, area of ABC =1
2× BC × AO = 6 cm2
1
2× AO × 5 = 6 AO =
12
5
Volume of double cone= (Volume of cone ABA)
+ (Volume of cone ACA)
=1
3OA2OB +
1
3OA2OC
=1
3OA2 (OB + OC)
=1
3×
22
7×
12
5×
12
5×
9 16
5 5
=1
3×
22
7×
12
5×
12
5×
25
5
=1056
35= 30
6
35cm3
= 30.1714 cm3.Surface area of double cone
= (Surface area of cone ABA)+ (Surface area of cone ACA)
= × OA × AB + × AO × AC
=22 12
37 5
cm2 +
22 124
7 5
× × cm2
=22
7×
12
5× 7 cm2 =
264
5cm2 = 52.8 cm2
Q.3. A cistern, internally measuring 15 cm ×120 cm × 110 cm, has 129600 cm3 of water in it.Porous bricks are placed in the water until thecistern is full to the brim. Each brick absorbsone-seventeenth of its own volume of water. Howmany bricks can be put in without overflowingthe water, each brick being 22.5 cm × 7.5 cm× 6.5 cm?
Sol. Volume of 1 brick = 22.5 × 7.5 × 6.5 cm3
= 1096.875 cm3
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Volume of cistern = 150 × 120 × 10 cm3
= 1980000 cm3
Volume of water = 129600 cm3
Volume of cistern to be filled
= (1980000 – 129600) cm3
= 1850400 cm3
Let n bricks be needed.
Then,water absorbed by n bricks
= n ×1096.875
17cm3
1850400 + n ×1096.875
17= n (1096.875)
16
17
n× 1096.875 = 1850400
n =1850400 17
16 1096.875
n = 1792.4102
= 1792 (approx.)
Hence, number of bricks used = 1792.
Q.4. In one fortnight of a given month, therewas a rainfall of 10 cm in a river valley. If thearea of the valley is 97280 km2, check whetherthe total rainfall was approximately equivalentto the addition to the normal water of threerivers each 1072 km long, 75 m wide and 3 mdeep.
Sol. Volume of total rainfall in the valley
= 97280 ×10
1000 100
= 9.728 km3
Volume of water of three rivers
= 3 × 1072 ×75
1000×
3
1000= 0.7236 km3
Hence, the two are not approximately equivalent.
Q.5. An oil funnel made of tin sheet consistsof a 10 cm long cylindrical portion attached to afrustum of a cone. If the total height is 22 cm,diameter of the cylindrical portion is 8 cm andthe diameter of the top of the funnel is18 cm, find the area of the tin sheet required tomake the funnel (see figure).
Sol. Slant height of the frustum of the cone, l
= 221 2h r r
= 2
2 18 822 10
2 2
cm
= 2 212 5 cm
= 144 25 cm = 169 cm = 13 cm
Area of metal sheet
= C.S.A. of cylinder + C.S.A. of frustum
= 2 (4) (10) cm2 + (4 + 9) 13 cm2
= (80 + 169) cm2 = 249 cm2
= 249 ×22
7cm2
=5478
7cm2 = 782
4
7cm2.
Q.6. Derive the formula for the curvedsurface area and total surface area of thefrustum of a cone.
Sol. Let the frustum be ABCD. The frustum can be
viewed as a difference of two right circular cones OABand OCD.
Let OA = OB = l1, OP = h1,
PB = r1, OC = OD = l2,
OQ = h2 and QD = r2
Let height (PQ) of the frustum be h and slant height(BD or AC) of the frustum be l.
Then l2 = l1 – l ....(i)
In DBL, DB2 = DL2 + BL2
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l= 221 2h r r ...(ii)
Curved surface area of the frustum
= r1l1 – r2l2 ..(iii)
Now, in similar triangles OPB and OQD, we have
OB
OD=
PB
QD
1
2
l
l=
1
2
r
r l1 =
1 2
2
r l
r=
1 1
2
r l l
r
l1r2 = r1l1 – r1l l1(r1 – r2) = r1l
l1 = 1
1 2
r l
r r...(iv)
Also, from (i) and (ii), we have,
l2 = 1
1 2
r l
r r– l = 1 1 2
1 2
r l r l r l
r r
l2 = 2
1 2
r l
r r....(v)
Substituting the values of l1 and l2 from (iv) and (v)in (iii), we have curved surface area of the frustum ofthe cone
2 21 2
1 2 1 2
.r l r l
r r r r
=
1 2 1 2
1 2
r r r r l
r r
= l (r1 + r2) Proved.
Now, total surface area of the frustum= curved surface area + area of the circular base
+ area of the circular top
= l (r1 + r2) + r12 + r2
2 Proved.
Q.7. Derive the formula for volume of thefrustum of a cone.
Sol. Let the frustum be ABCD. The frustum can be
seen as a difference of two right circular cones OAB andOCD.
Let OA = OB = l1, OP = h1, PB = r1
OC = OD = l2, OQ = h2 and QD = r2
Let height (PQ) of the frustum be h.
Then h = h1 – h2 ....(i)
In similar triangles OPB and OQD, we have
OP
OQ=
PB
QD 1
2
h
h=
1
2
r
r
h1 = 1 2
2
r h
r=
( )1 1
2
r h h
r
−
r2h1 = r1h1 – r1h h1(r1 – r2) = r1h
h1 = 1
1 2
r h
r r...(ii)
Also, from (i) and (ii), we have
h2 = 1
1 2
r h
r r– h = 1 1 2
1 2
r h r h r h
r r
h2 = 2
1 2
r h
r r....(iii)
Now, volume of the frustum of the cone= Volume of the cone OAB
– Volume of the cone OCD.
=1
3r1
2h1 –1
3r2
2h2
=1
3
3 31 2
1 2 1 2
r h r h
r r r r
[From (i) and (iii)]
=1
3h
3 31 2
1 2
r r
r r
=1
3h
2 21 2 1 2 1 2
1 2
r r r r r r
r r
=1
3h (r1
2 + r22 + r1r2) Proved.
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B. FORMATIVE ASSESSMENT
Activity 1
Objective : To compare the curved surface areas and the total surface areas of two right circularcylinders which are formed from rectangular sheets of paper having same dimensions.
Materials Required : Rectangular sheets of paper, a pair of scissors, sellotape, geometrybox, etc.
Procedure :
1. Take two rectangular sheets ABCD of dimensions 22 cm × 11cm.
Figure 1
2. Fold one of the sheets ABCD along AB such that AD and BC coincide. Use sellotape to geta cylinder as shown below.
Figure 2
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3. Put the cylinder on a white sheet of paper and draw the outline of its base. Using a pair of scissorscut out the circular region. Make one more such circular cut out.
Figure 3
4. Using sellotape, fix these circular cut outs on the top and bottomof the hollow cylinder as shown in the figure.
5. Take another sheet ABCD and fold it along AD such that AB and DC coincide. Use sellotape toget a cylinder as shown.
Figure 5
6. Repeat steps 3 and 4 for the cylinder obtained above.
Figure 6
Figure 4
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Observations :
1. Let r1
be the base radius of the cylinder in figure 2.
Then circumference of its base = 2 r1
= 22 r1
=22 7
2 22
×× cm = 3.5 cm.
Also, height, h1
of the cylinder = 11 cm.
2. So, curved surface area of the cylinder in figure 2 = 2 r1h
1= 2 ×
22
7× 3.5 × 11 cm2
= 242 cm2.
3. Total surface area of the cylinder in figure 4 = curved surface area + 2 r1
2
= (242 + 2 ×22
7× 3.5 × 3.5) cm2 = 319 cm2
4. Let r2
be the base radius of the cylinder in figure 5.
Then circumference of its base = 2 r2
= 11 r2
=11 7
2 22
××
cm = 1.75 cm.
Also, height, h2
of the cylinder = 22 cm.
5. So, curved surface area of the cylinder in figure 5
= 2 r2h
2= 2 ×
22
7× 1.75 × 22 cm2 = 242 cm2.
6. Total surface area of the cylinder in figure 6 = curved surface area + 2 r2
2
= (242 + 2 ×22
7× 1.75 × 1.75) cm2 = 261.25 cm2.
7. From 2 and 5, we see that the curved surface area of both the cylinders is same.
8. From 3 and 6, we see that the total surface area of the two cylinders is different.
Conclusion :
1. The curved surface areas of two cylinders formed by folding the rectangular sheets havingsame dimensions is same.
2. The total surface areas of two cylinders formed by folding the rectangular sheets having same
dimensions are not same.
Activity 2
Objective : To compare the volumes of two right circular cylinders which are formed fromrectangular sheets of paper having same dimensions.
Materials Required : Rectangular sheets of paper, a pair of scissors, sellotape, geometrybox, etc.Procedure :
1. Take two rectangular sheets ABCD of dimensions 22 cm × 11 cm.
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Figure 7
2. Fold one of the sheets ABCD along AB such that AD and BC coincide. Use sellotape to geta cylinder as shown below.
Figure 8
3. Take another sheet ABCD and fold it along AD such that AB and DC coincide. Use sellotapeto get a cylinder as shown.
Figure 9
Observations :1. Let r
1be the radius of the base of the cylinder in figure 8.
Then, circumfrence of its base = 2r1
= 22 r1
=22 7
2 22
×× cm = 3.5 cm.
Also, height h1
of the cylinder = 11 cm.
2. Volume of this cylinder = r1
2h1
=22
7× 3.5 × 3.5 × 11 cm3 = 423.50 cm3
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3. Let r2
be the radius of the base of the cylinder in figure 9.
Then circumference of its base = 2r2
= 11 r2
=11 7
2 22
××
cm = 1.75 cm.
Also, height, h2
of the cylinder = 22 cm.
4. Volume of this cylinder = r2
2h2
=22
7× 1.75 × 1.75 × 22 cm3 = 211.75 cm3.
5. From 2 and 4 above, we see that the two volumes are different.
Conclusion : The volumes of two cylinders formed by folding the rectangular sheets having same
dimensions are different.
Activity 3
Objective : To make a cone of given slant height l andbase circumference.
Materials Required : White sheets of paper, a pair ofscissors, colour pencils, sellotape, geometry box, etc.
Procedure : Let us make a cone of slant height, l = 7 cmand base circumference = 11 cm
1. On a white sheet of paper, draw a circle of radius 7
cm and centre O.
Figure 11
2. Mark a sector OAB such that AOB = 90°. Cut the sector OAB and bring the radii OA and
OB together. Use sellotape to get a cone as shown above.
Observations :1. The radius of the circle in figure 11 becomes slant height of the cone.
or, l = 7 cm
2. AOB = 90°
So, length of the arc AB =180
rπ θ°°
, where = 90° =
227 90
7180
× × °
° cm = 11cm
Hence, base circumference of the cone = length of the arc AB = 11 cm
Conclusion : From the above activity, we see that a cone of given slant height and base
circumference can be made from a sector of a circle.
Figure 10
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Activity 4
Objective : To give a suggestivedemonstration of the formula for the lateralsurface area of a cone.
Materials Required : White sheets ofpaper, a pair of scissors, colour pencils,geometry box, sellotape, gluestick, etc.
Procedure :
1. On a white sheet of paper, draw a circleof any convenient radius and with centre O.Mark a sector OAB on the circle and cut itout. From the previous activity, we know thatthis sector can be folded to make a cone.
2. Fold the sector such that OA and OB coincide.Again fold it as shown below. Press it to make creases.
3. Unfold the cut out and draw lines along the creases. Now, cut thesector along the creases to get four smaller sectors as shown below.
Figure 14
4. Paste the four smaller sectors on a white sheet of paper to getan approximate parallelogram as shown alongside.
Observations :
1. From figure, slant height of the cone = l and basecircumference of the cone = 2r, where r is the base radius of thecone.
2. Figure 15 is an approximate parallelogram whose base is half of the base circumference ofthe cone and height is approximately the slant height of the cone.
i.e, base of the parallelogram =1
2× 2r = r
Height of the parallelogram = l Area of the parallelogram = base × height = r × l = rlThus, curved surface area of the cone = area of the parallelogram = rl
Conclusion : The curved (lateral) surface area of a cone of base radius r and slant height l is
equal to rl.
Figure 12
Figure 13
Figure 15
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Activity 5
Objective : To give a suggestive demonstration of the formula for the volume of a right circularcone.
Materials Required : Cones and cylinders of same base radius and height, sand, etc.
Procedure :
1. Take a set of cone and cylinder having same base radius (r) and height (h).
2. Fill the cone with sand
3. Pour the sand from the cone into the cylinder
4. Repeat step 3 until the cylinder gets completely filled with sand.
5. Repeat steps 2 to 4 for another set of cone and cylinder having same base radius and same height.
Observations :
1. We see that after filling the cone with sand and pouring the sand into the cylinder, it needsthree pourings to fill the cylinder completely.
So, volume of the cylinder = 3 times the volume of the cone
or, volume of the cone =1
3volume of the cylinder..
2. Volume of the cylinder = r2h
3. So, volume of the cone =1
3r2h
Conclusion : From the above activity, it is verified that the volume of a cone of base radius r
and height h is given by1
3r2h.
Activity 6
Objective : To give a suggestive demonstration of the formula for the volume of a sphere interms of its radius.
Materials Required : A hollow sphere and two cylinders whose base diameter and height areequal to the diameter of the sphere, sand, etc.
Procedure :
1. Take a hollow sphere and two cylinders whone base diameter and height are equal to thediameter of the sphere.
2. Fill the sphere with sand and pour it into one of the cylinders.
3. Fill the sphere with sand second time and pour it into the same cylinder as in step 2, till itgets completely filled.
4. Pour the remaining sand into the other cylinder.
5. Again fill the sphere with sand and pour it into the semi-filled cylinder to fill the cylinder withsand completely.
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Observations :
1. We see that the total sand poured in three pourings completely fill the two cylinders.
2. So, three times the volume of the sphere = two times the volume of the cylinder
= 2r2h = 2r2(2r) [ h = 2 r for each cylinder]
3 × volume of the sphere = 4r3
Volume of the sphere =4
3r3
Conclusion : From the above activity, it is verified that the volume of a sphere of radius r is
given by4
3r3.
Angles in a Cube
Find the angle between the two dotted lines drawn on the surface of a cube.
Cutting Cubes
A cube of side 4 cm painted red, yellow and green on opposite faces. It was cut into cubicalblocks each of the side 1 cm.
1. How many cubes are obtained?
2. How many cubes are there whose only one face is painted?
3. How many cubes are there whose no face is painted?
4. How many cubes are there whose two faces are painted?
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5. How many cubes are there whose three faces are painted?
6. How many cubes are there whose two faces are painted with the same colour?
7. How many cubes are there whose three faces are painted all with different colours.
8. How many cubes are there whose all faces are painted?
9. At most how many painted faces a cube can have?
ANSWERS
A. SUMMATIVE ASSESSMENT
Practice Exercise 13.1A
1. (b) 2. (b) 3. (d) 4. (b)5. (a) 6. (b) 7. (b) 8. No9. No 10. No 11. 17.6 cm2 12. 163.86 cm2
13. Rs 99,000 14. 350 cm2 15. Rs 855.36 16. 18480 cm2
Practice Exercise 13.2A1. (a) 2. (d) 3. (c) 4. (d)5. (b) 6. 14 cm, 7 cm, 132 cm3, 66 cm3, 396 cm3 7. 1508. 688.83 cm3 9. 179.66 cm3 10. 30.48 cm3 11. 124.4 cm3
12. 8 cm3 13. 3 cm 14. 628.57 cm3 15. 6 m16. 74.25 m3, 80.61 m2 17. 0.5 m 18. 301.44 cm2, 377.1 cm3
Practice Exercise 13.3A1. (a) 2. (a) 3. (c) 4. (d)5. (b) 6. 6 cm 7. 84 8. 14 cm9. 6.4 cm 10. 4.04 cm (approx.) 11. 1 cm 12. 9 cm
13. 512 14. 2904 cm2 15. 30 hectares 16. 150017. 28.44 cm 18. 90 cm
Practice Exercise 13.4A1. (d) 2. (c) 3. (c) 4. (a)5. (a) 6. Yes 7. 1 : 78. (i) 12 cm (ii) 312 cm3 (iii) 405.43 cm2 9. 2420 cm2 10. 8800 cm3, Rs 21.44
11. 376.8 cm2 12. 7.64 litres 13. 3822.5 cm2, 26.6 litres14. 28 cm 15. Rs 163.28, Rs 17.13
B. FORMATIVE ASSESSMENT
Angles in a cube
Complete the triangle. The sides of this triangle are diagonals of the faces of the cube. So, the
triangle is equilateral. So, the required angle is 60°.
Cutting cubes
1. 84 2. 24 3. 8 4. 24 5. 8 6. 0 7. 8 8. 0 9. 3