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10th Standard
Maths
A shuttle cock used for playing badminton has the shape of a frustum of a cone is mounted on a hemisphere. The diameters of the frustum are 5 cm and 2 cm. The height of the entire shuttle cock is 7 cm. Find its external surface area.
What is the ratio of the volume of a cylinder, a cone, and a sphere. If each has the same diameter and same height?
Find the number of coins, 1.5 cm is diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.
A spherical ball of iron has been melted and made into small balls. If the radius of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made?
A wooden article was made by scooping out a hemisphere from each end of a cylinder as shown in figure. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm find the total surface area of the article.
The diameter of a roller is 84 cm and its length is 120 cm, It takes 500 complete revolutions to move once over to level a playground. Find the area of the play ground (in sq. m).
The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white washing its curved surface area at the rate of Rs 210 per 100 m2?
The diameter of the moon is approximately one fourth of the diameter of the Earth. Find the ratio of their surface areas.
The radii of circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Find its total surface area.
The sum of the radius of the base and the height of a solid cyrinder is 37 m If the total surface area of the solid cylinder is 1628 m2, find the circumference its base and volume of the cylinder.
A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained
The diameter of a metallic ball is 4.2 cm. what is the mass of ball if the density of the metal is 8.9 g per cm3?
A bucket is in the form of a cone. Its depth is 24 cm and the diameters of the top and bottom ends are 30 cm and 10 cm respectively. Find the capacity of the bucket.
Find the number of spherical lead shots, each of diameter 6 cm that can be made from a solid cuboids of lead having dimensions 24 cm x 22 cm x 12 cm
A cylindrical bucket 32 cm high and with radius of base 18 cm, is filled with sand completely This bucket is emptied on the ground and a conical hehp of sand is formed. The height of the conical heap is 24 cm, find the radius and slant height of the heap.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly and around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
Answers
External surface area of the cock = Surface area of frustum + CSA of hemisphere
CSA of frustum = π(R + r)l sq. units.
Here R = \(\frac{5}{2}cm\)
\(r=\frac { 2 }{ 2 } =1cm\)
\(l=\sqrt { ({ R-r) }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { (2.5-1) }^{ 2 }+{ 6 }^{ 2 } } \)
\(=\sqrt { { 1.5 }^{ 2 }+36 } \)
\(=\sqrt { 2.25+36 } \)
\(=\sqrt { 38.25 } \)
\(\cong 6.18\)
∴ CSA of the frustum \(=\frac { 22 }{ 7 } \times 3.5\times 6.1=\frac { 469.7 }{ 7 } \)
= 67.1 cm2
CSA of hemisphere = 2π2
\(=2\times \frac { 22 }{ 7 } \times 1\times 1\)
= 6.28cm2
∴ Total external surface area
= 67.1 + 6.28
= 73.38cm2
= 73.39 cm2(approx.)
Volume of a cylinder \(=\pi { r }^{ 2 }h\)
Volume of a cone \(=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
Volume of a sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }\)
Their ratio V1 : V2 : V3
\(h:\frac { h }{ 3 } :\frac { 4r }{ 3 } \)
3h: h: 4r
3h : h : 2(2r) (where 2r = h)
∴ V1 : V2 : V3 = 3: 1 : 2
No. of coins required \(=\frac{Volume\ of\ the\ cylinder}{Volume\ of\ 1\ coin}\)
\(=\frac { \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \pi { r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { \pi \times \frac { 42 }{ 20 } \times \frac { 45 }{ 20 } \times 10 }{ \pi \times \frac { 15 }{ 20 } \times \frac { 15 }{ 20 } \times \frac { 2 }{ 10 } } \)
= 450
No. of balls required \(=\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \frac { 4 }{ 3 } \pi \times { \left( \frac { r }{ 4 } \right) }^{ 3 } } \)
\(={ r }^{ 3 }\times \frac { 4 }{ r } \times \frac { 4 }{ r } \times \frac { 4 }{ r } =64\)
Radius of the cylinder be r
Height of the cylinder be h
Total surface area of the article
= CSA of cylinder + CSA of 2 hemispheres
= 2ㅠrh + 2πr2 = 2πr(h + 2r)
\(=2\times \frac { 22 }{ 7 } \times 3.5\times (10+2\times 3.5)\)
= 22 x 17 = 374 cm2
Diameter of roller = 84 cm
\(\text { radius }=\frac{84}{2}=42 \mathrm{~cm}=0.42 \mathrm{~m}\)
Iength = h = 120cm = 1.2 m
Area covered by roller in one revolution = C.S.A of the roller
\(=2 \pi r h \)
\(=2 \times \frac{22}{7} \times 0.42 \times 1.2 \)
\(=3.168 \mathrm{~m}^{2} \)
Area covered in 500 revolutions = Area of the playground
= 500 x 3.168 = 1584 sq.m
diameter = 14 m ; radius = 7 cm;
Slant height l = 25 m
\(\text { C.S.A }=\pi r l=\frac{22}{7} \times 7 \times 25\)
= 500 m2
Given, Cost of white washing per 100 m2 is Rs 210
Cost of white washing the conical tomb
\(=550 \times \frac{210}{100}=\text { Rs } 1155\)
Let the diameter of the Earth be ' R'
\(\text { Radius of the Earth }=\frac{R}{2}\)
\(\text { Diameter of the Moon }=\frac{1}{4} R\)
\(\text { Radius of the Moon }=\frac{R}{8}\)
\(\therefore \text { Ratio of surface areas of moon and earth }\)
\(=\frac{4 \pi\left(\frac{R}{8}\right)^{2}}{4 \pi\left(\frac{R}{2}\right)^{2}}\)
\(\therefore \text { Ratio }=\frac{1}{16}=1: 16\)
Given R = 33cm, r = 27 cm and I = 10cm
\(\therefore \text { T.S.A of frustum }=\pi\left(R^{2}+r^{2}+l(R+r)\right)\)
\(=\frac{22}{7}\left((33)^{2}+(27)^{2}+10(33+27)\right) \)
\(=\frac{22}{7}(1089+729+600) \)
\(=\frac{53196}{7}=7599.43 \mathrm{~cm}^{2} \)
\(\text { radius } =\mathrm{r}, \text { height }=\mathrm{h} \mid \)
\(\text { T.S. } \mathrm{A} =1628 \)
\(2 \pi r(h+r) =1628 \)
\(2 \pi r(37) =1628 \)
\(\Rightarrow \mathrm{r} =7 \mathrm{~m} \)
\(\text { and } 7+\mathrm{h} =37 \)
\(\Rightarrow \mathrm{h} =37-7=30 \mathrm{~m}
\)
\(\text { Circumference } =2 \pi r \)
\(=2 \times \frac{22}{7} \times 7=44 \mathrm{~m} \)
\(\text { Volume } =\pi r^{2} h \)
\(=\frac{22}{7} \times 7 \times 7 \times 30=4620 \mathrm{~m}^{3}
\)
Triangle ABC is revolved about the side AB
= 12 cm
\(\text { Volume of cone }=\frac{1}{3} \pi r^{2} h \mathrm{cu} \text {. units }\)
\(=\frac{1}{3} \pi(5)^{2}(12) \)
\(=100 \times \frac{22}{7} \)
\(=314 \mathrm{~cm}^{3}(\mathrm{app}) \)
Diameter = 4.2 cm;
Radius = 2.1 cm
\(\text { Volume }=\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7} \times(2.1)^{3}\)
\(=38.808 \mathrm{~cm}^{3}\)
Mass of the ball = Volume x Density
\(=38.808 \times 8.9=345.3912 \mathrm{~g}\)
\(\text { Given: } R=\frac{30}{2}=15 \mathrm{~cm}, \mathrm{r}=\frac{10}{2}=5 \mathrm{~cm} \text { and }\)
h = 24 cm
\(\text { Volume } =\frac{1}{3} \pi h\left(R^{2}+R r+r^{2}\right) \text { cubic units. } \)
\(=\frac{1}{3} \times \frac{22}{7} \times 24 \times\left(15^{2}+15 \times 5+5^{2}\right)
\)
\(=\frac{22}{7} \times 8 \times(225+75+25) \)
\(=\frac{57200}{7}=8171.42 \mathrm{~cm}^{3}
\)
Diameter of spherical lead shots = 6 cm
Radius = 3 cm
Volume of cuboid = l x b x h
\(\text { Volume of sphere }=\frac{4}{3} \pi r^{3}\)
\(\text { No. of lead shots }=\frac{24 \times 22 \times 12}{\frac{4}{3} \times \frac{22}{7} \times 3 \times 3 \times 3}\)
\(=\frac{24 \times 22 \times 12 \times 3 \times 7}{4 \times 22 \times 3 \times 3 \times 3}\)
No. of lead shots = 56
Cylinder:
Height,H = 32 cm;
Radius,R =18 cm;
Cone:
Height,h = 24 cm;
Radius = r
Volume of cylinder = Volume of cone
\(\pi \mathrm{R}^{2} \mathrm{H} =\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h} \)
\(18 \times 18 \times 32 =\frac{1}{3} \mathrm{r}^{2} \times 24 \)
\(\mathrm{r}^{2} =18 \times 18 \times 4 \)
\(\mathrm{r} =18 \times 2=36 \mathrm{~cm} \)
\(l =\sqrt{r^{2}+h^{2}} \)
\(=\sqrt{36^{2}+24^{2}} \)
\(=\sqrt{1296+576} \)
\(=\sqrt{1872} \)
\(l=43.27 \mathrm{~cm} \)
Cylindrical Well
D = 3m
\(r=\frac{3}{2} m\)
h = 14m
Hollow Cylindrical embankment
w = 4
\(\therefore \ \mathrm{R} =\frac{3}{2}+4 \)
\(=\frac{11}{2} \mathrm{~m} \)
h = 2
Volume of cylindrical well = Volume of hollow cylindrical embankment
\(\pi r^{2} h=\pi\left(R^{2}-r^{2}\right) h\)
\(\frac{3}{2} \times \frac{3}{2} \times 14=\left(\frac{121}{4}-\frac{9}{4}\right) h\)
\(\frac{3 \times 3 \times 7}{2}=\frac{112}{4} \times h\)
\(\frac{3 \times 3 \times 7}{2} \times \frac{4}{112}=\mathrm{h}\)
\(\mathrm{h}=\frac{9}{8} \mathrm{~m}\)
\(\mathrm{h}=1.12 \mathrm{~m}\)
