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
\(\pi r^{2} h=\pi\left(R^{2}-r^{2}\right) h\)
\(\frac{3}{2} \times \frac{3}{2} \times 14=\left(\frac{121}{4}-\frac{9}{4}\right) h\)
\(\frac{3 \times 3 \times 7}{2}=\frac{112}{4} \times h\)
\(\frac{3 \times 3 \times 7}{2} \times \frac{4}{112}=\mathrm{h}\)
\(\mathrm{h}=\frac{9}{8} \mathrm{~m}\)
\(\mathrm{h}=1.12 \mathrm{~m}\)

Cylinder:
Height, H = 32 cm;
Radius, R =18 cm;
Cone:
Height, h = 24 cm;
Radius = r
Volume of cylinder = Volume of cone
\(\pi \mathrm{R}^{2} \mathrm{H} =\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h} \)
\(18 \times 18 \times 32 =\frac{1}{3} \mathrm{r}^{2} \times 24 \)
\(\mathrm{r}^{2} =18 \times 18 \times 4 \)
\(\mathrm{r} =18 \times 2=36 \mathrm{~cm} \)
\(l =\sqrt{r^{2}+h^{2}} \)
\(=\sqrt{36^{2}+24^{2}} \)
\(=\sqrt{1296+576} \)
\(=\sqrt{1872} \)
\(l=43.27 \mathrm{~cm} \)

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

\(\text { Given: } R=\frac{30}{2}=15 \mathrm{~cm}, \mathrm{r}=\frac{10}{2}=5 \mathrm{~cm} \text { and }\)
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 = 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
\(=38.808 \times 8.9=345.3912 \mathrm{~g}\)

Triangle ABC is revolved about the side AB
= 12 cm
\(\text { Volume of cone }=\frac{1}{3} \pi r^{2} h \mathrm{cu} \text {. units }\)
\(=\frac{1}{3} \pi(5)^{2}(12) \)
\(=100 \times \frac{22}{7} \)
\(=314 \mathrm{~cm}^{3}(\mathrm{app}) \)

\(\text { radius } =\mathrm{r}, \text { height }=\mathrm{h} \mid \)
\(\text { T.S. } \mathrm{A} =1628 \)
\(2 \pi r(h+r) =1628 \)
\(2 \pi r(37) =1628 \)
\(\Rightarrow \mathrm{r} =7 \mathrm{~m} \)
\(\text { and } 7+\mathrm{h} =37 \)
\(\Rightarrow \mathrm{h} =37-7=30 \mathrm{~m} \)
\(\text { Circumference } =2 \pi r \)
\(=2 \times \frac{22}{7} \times 7=44 \mathrm{~m} \)
\(\text { Volume } =\pi r^{2} h \)
\(=\frac{22}{7} \times 7 \times 7 \times 30=4620 \mathrm{~m}^{3} \)

Given R = 33cm, r = 27 cm and I = 10cm
\(\therefore \text { T.S.A of frustum }=\pi\left(R^{2}+r^{2}+l(R+r)\right)\)
\(=\frac{22}{7}\left((33)^{2}+(27)^{2}+10(33+27)\right) \)
\(=\frac{22}{7}(1089+729+600) \)
\(=\frac{53196}{7}=7599.43 \mathrm{~cm}^{2} \)

Let the diameter of the Earth be ' R'
\(\text { Radius of the Earth }=\frac{R}{2}\)
\(\text { Diameter of the Moon }=\frac{1}{4} R\)
\(\text { Radius of the Moon }=\frac{R}{8}\)
\(\therefore \text { Ratio of surface areas of moon and earth }\)
\(=\frac{4 \pi\left(\frac{R}{8}\right)^{2}}{4 \pi\left(\frac{R}{2}\right)^{2}}\)
\(\therefore \text { Ratio }=\frac{1}{16}=1: 16\)

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
\(=550 \times \frac{210}{100}=\text { Rs } 1155\)

Diameter of roller = 84 cm
\(\text { radius }=\frac{84}{2}=42 \mathrm{~cm}=0.42 \mathrm{~m}\)
length = 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

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

No. of balls required \(=\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \frac { 4 }{ 3 } \pi \times { \left( \frac { r }{ 4 } \right) }^{ 3 } } \)
\(={ r }^{ 3 }\times \frac { 4 }{ r } \times \frac { 4 }{ r } \times \frac { 4 }{ r } =64\)

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

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

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.)