Length of arc of a sector = \(\frac{\theta}{360^{\circ}} \times 2 \pi r_{1}\)
\(=\frac{216^{\circ}}{360^{\circ}} \times 2 \pi \times 21 \)
\(=\frac{3}{5}(2 \pi)(21)=\frac{126 \pi}{5} \mathrm{~cm}\)
Since, sector is made into a cone by bringing the bounding radii together.
\(\left.\begin{array}{l}\text { Circumference of base } \\ \text { of the cone }\end{array}\right\}=\left\{\begin{array}{l}\text { length of are } \\ \text { of a sector }\end{array}\right.\)
\(\therefore 2 \pi r=\frac{126 \pi}{5}\)
radius of base of a cone, \(r=\frac{63}{5} \mathrm{~cm}\)
and radius of sector = slant height of cone
21 = l
Height of cone, h \(=\sqrt{l^{2}-r^{2}}=\sqrt{(21)^{2}-\left(\frac{63}{5}\right)^{2}} \)
\(=\sqrt{441-158.76} \)
\(=\sqrt{282.24}=16.8 \mathrm{~cm}\)
Volume of cone \(=\frac{1}{3} \pi r^{2} h cu. units \)
\(=\frac{1}{3} \times \frac{22}{7} \times 12.6 \times 12.6 \times 16.8 \)
\(=\frac{58677.696}{21}\)
= 2794.176 = 2794.18 cm³

Volume of a cone = 1005 \(\frac{5}{7}\) cu.cm
\(\text { i.e., } \frac{1}{3} \pi r^{2} h=1005 \frac{5}{7}\)
area of base = area of circle
\( =201 \frac{1}{7} \text { sq. units } \)
\(i.e \ \pi r^{2}=201 \frac{1}{7} \Rightarrow r^{2}=64 \)
Substituting in (1), r = 8 cm
\(\frac{1}{3}\left(201 \frac{1}{7}\right) \mathrm{h}=1005 \frac{5}{7} \)
\(\frac{1}{3}\left(\frac{1408}{7}\right) \mathrm{h}=\frac{7040}{7} \)
\(h=\frac{7040}{7} \times \frac{7}{1408} \times 3=15 \mathrm{~cm}\)
Slant height of cone \(l =\sqrt{h^{2}+r^{2}} \)
\(=\sqrt{15^{2}+8^{2}}=\sqrt{225+64} \)
\(=\sqrt{289}=17 \mathrm{~cm} \)

External diameter of hollow hemisphere
= 2R = 14 cm
External radius R = 7 cm
Given volume = \(\frac{436 \pi}{3} \mathrm{~cm}^{3}\)
\(\frac{2}{3} \pi\left(R^{3}-r^{3}\right) =\frac{436 \pi}{3} \)
\(R^{3}-r^{3} =218 \)
(7)³ - r³ = 218
r³ = 343 - 218 = 125 = 5 cm
Thickness = R - r = 7 - 5 = 2 cm

Slant height l = 4 m
Perimeter of larger circle = 2nR = 18 cm
\(R=\frac{9}{\pi}=\frac{63}{22} \mathrm{~m}\)
Perimeter of smaller circle = \(2 \pi r\) = 16 m
\(r=\frac{8}{\pi}=\frac{56}{22} \mathrm{~m}\)
C.S.A of Frustum of cone = \(\pi l(\mathrm{R}+r) \text { sq. units }\)
\(=\frac{22}{7} \times 4 \times\left(\frac{63}{22}+\frac{56}{22}\right) \)
\(=\frac{4}{7}(119)=68 \mathrm{~m}^{2} \)
Cost of painting per sq. m = Rs. 100
Cost of painting for 68 sq. m = 68 x 100
= Rs.6800

Hollow metallic cylinder
External radius R = 4.3 cm
Internal radius r = 1.1 cm
Length = height = h = 4 cm
Volume \(=\pi\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right) \mathrm{h} \text { cu. units } \)
\(=\pi\left((4.3)^{2}-(1.1)^{2}\right)(4) \)
\(=\pi(18.49-1.21) 4 \)
\(=69.12 \pi \mathrm{cm} 3 \)
Solid cylinder
height h = 12 cm
radius r = ?
volume \(=\pi r^{2} h\ sq. units \)
\(=\pi r^{2}(12) \)
Given, Hollow cylinder is melted to form solid cylinder
Volume of cylinder = Volume of hollow cylinder
\(\pi r^{2}(12) =69.12 \pi \)
\(r^{2} =\frac{69.12}{12}=5.76 \)
r = 2.4 cm
Diameter of cylinder  = 2r = 4.8 cm

Coin is in the form of a cylinder
Diameter of the coin = 1.5 cm
Radius of the coin = \(\frac{1.5}{2}\)
Thickness = height = 2 mm = \(\frac{2}{10}=0.2 \mathrm{~cm}\)
Volume of coin (cylinder) = \(\pi r^{2} h\)
\(=\pi\left(\frac{1.5}{2}\right)^{2}(0.2) \)
\(=0.1125 \pi \mathrm{cm}^{3} \)
Diameter of cylinder = 4.5 cm
radius = \(\frac{4.5}{2}=2.25 \mathrm{~cm}\)
height = 10 cm
volume = \(\pi r^{2} h\ sq. units \)
= \(\pi(2.25)^{2}(10) \)
= \(50.625 \pi \)
No.of coins \(=\frac{\text { Volume of cylinder }}{\text { Volume of Coin }} \)
\(=\frac{50.625 \pi}{0.1125 \pi}=450 \text { coins. } \)

Area of tin sheet required
= C.S.A of cylinder + C.S.A of frustum
Cylinder:
Radius = 4 cm
Height = 10 cm
C.S.A  \(=2 \pi r h\ sq. units \)
\(=2 \times \frac{22}{7} \times 4 \times 10=\frac{1760}{7}\ sq. units\)
Frustum of a cone
r₁ = radius of top = 9 cm
r₂ = radius of bottom = 4 cm
Height = 22 - 10 = 12 cm
Slant height \(l =\sqrt{h^{2}+\left(r_{1}-r_{2}\right)^{2}} \)
\(=\sqrt{12^{2}+(9-4)^{2}} \)
\(=\sqrt{144+25}=\sqrt{169}=13 \mathrm{~cm} \)
C.S.A  \(=\pi\left(r_{1}+r_{2}\right) l \text { sq, units } \)
\(=\frac{22}{7}(9+4)(13) \)
\(=\frac{3718}{7} \text { sq. units } \)
Area of tin sheet \(=\frac{1760}{7}+\frac{3718}{7}\)
\(=\frac{5478}{7}=782.57 \mathrm{~cm}^{2}\)

Let the radius of hemispherical bowl = r
Volume of hemispherical bowl
\(=\frac{2}{3} \pi r^{3} \text { cu. units }\)
Let the height of cylindrical vessel = h
Given \(r=h+h \frac{50}{100} \Rightarrow \mathrm{r}=\mathrm{h}\left(1+\frac{50}{100}\right) \)
\(\mathrm{h}= \frac{2}{3} r \)
Now, Volume of cylindrical vessel
\(=\pi r^{2}\left(\frac{2 r}{3}\right)=\frac{2}{3} \pi r^{3}\)
Hence, Volume of juice in the cylindrical vessel
\(=\frac{\frac{2}{3} \pi r^{3}}{\frac{2}{3} \pi r^{3}} \times 100 \%=100 \%\)

Solid sphere
radius = 6 cm
Volume \(=\frac{4}{3} \pi r^{3} \text { cu. units } \)
\(=\frac{4}{3} \pi(6)^{3} \)
\(=\frac{4}{3} \pi(216)=288 \pi \mathrm{cm}^{3} \)
Hollow cylinder
Internal radius = 'r'
External radius = 'R' = 5 cm
Height h = 32 cm
Volume of Hollow Cylinder
\(=\pi h\left(\mathrm{R}^{2}-r^{2}\right)\ cu. units \)
\(=\pi(32)\left(25-r^{2}\right) \mathrm{cm}^{3} \)
Given that solid sphere is melted to form a hollow cylinder.
Volume of Hollow Cylinder = Volume of Sphere
\(32 \pi\left(25-r^{2}\right) =288 \pi \)
\(25-r^{2} =\frac{288}{32}=9 \)
r² = 25 - 9 = 16
Internal radius r = 4 cm
Thickness = External radius - Internal radius
= R - r = 5 - 4 = 1 cm.

Hollow Hemisphere 
Internal diameter = 6 cm
Internal radius 'r' = 3 cm
External diameter = 10 cm
External radius 'R' = 5 cm
\(\left.\begin{array}{l} \text { Volume of hemisphere (or) } \\ \text {Volume of material used } \end{array}\right\}=\frac{2}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \text { cu. units }\)
\(=\frac{2}{3} \pi\left(5^{3}-3^{3}\right) \)
\(=\frac{2}{3} \pi(125-27)=\frac{196 \pi}{3} \mathrm{~cm}^{3} \)
Cylinder
Diameter = 14 cm
radius = 7 cm
height = h
Volume of cylinder \(=\pi r^{2} h\ cu. units \)
\(=\pi(7)^{2} h \)
\(=49 \pi h \mathrm{~cm}^{3} \)
Given that hollow hemisphere is melted and cast into a solid cylinder
Volume of cylinder = volume of hollow hemisphere
\(49 \pi h =\frac{196 \pi}{3} \)
\(h =\frac{196}{3 \times 49}=\frac{4}{3}=1.33 \)
Height of the cylinder = 1.33 cm.

(i) Surface area of a sphere Radius of sphere = r
Surface area = 4r² sq. units
(ii) Curved surface area of cylinder 
Radius of cylinder = r
Height of cylinder = r + r = 2r
Curved surface area \(=2 \pi r h\ sq. units \)
\(=2 \pi r(2 \mathrm{r}) \)
\(=4 \pi r^{2} \text { sq. units } \)
(iii) Ratio of the areas \(=\frac{\text { Surface area of sphere }}{\text { CSA of cylinder }} \)
\(=\frac{4 \pi r^{2}}{4 \pi r^{2}}=\frac{1}{1} \)
Ratio = 1 : 1.

Edge of cube = 7 cm
surface area of a cube = 6a² sq. units
= 6(7)²
= 294 cm²
radius of hemisphere = \(\frac{7}{2} \mathrm{~cm}\)
[Only C.S.A is considered as the hemisphere surmounted]
C.S.A of hemisphere \(=2 \pi r^{2} \text { sq. units } \)
\(=2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
= 77 Cm²
Surface area of = T.S.A of cube + C.S.A the solid of hemisphere area of circular region (bottom of hemisphere)

\(=294+77-\left(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right)\)
= 371 - 38.5
= 332.5 cm²

Radius of hemisphere = 6 cm
Volume of hemisphere \(=\frac{2}{3} \pi r^{3} \text { cu. units }\)
\(=\frac{2}{3} \pi(6)^{3} \)
\(=\frac{2}{3} \pi(216) \)
\(=144 \pi \mathrm{cm}^{3} \)
base of cone = 6 cm
Height of the cone = 12 cm
Volume of the cone \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \pi(6)^{2}(12) \)
= 144 cm³
volume of the solid = Volume of cone + Volume of hemisphere
\(=144 \pi+144 \pi=288 \pi\)
Volume of water displaced
= Volume of the solid placed in the cylinder
\(=288 \pi=288 \times \frac{22}{7}\)
= 905.14 cm³

Let r be the radius of the hemisphere.
Given that, diameter of the hemisphere = side of the cube = l
Radius of the hemisphere = \(\frac{l}{2}\)
TSA of the remaining solid = Surface area of the cubical part + C.S.A. of the hemispherical part − Area of the base of the hemispherical part
= 6 x (Edge)² + 2\(\pi\)r²−\(\pi\)r²
= 6 x (Edge)² +  \(\pi\)r²
\(=6{ \times (l) }^{ 2 }+\pi { \left( \frac { l }{ 2 } \right) }^{ 2 }=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)
Total surface area of the remaining solid \(=\frac { 1 }{ 4 } (24+\pi ){ l }^{ 2 }\)sq. units

Let R, r be the top and bottom radii of the frustum.
Let h₁, h₂ be the heights of the frustum and cylinder respectively.
Given that, R = 12 cm, r = 6 cm, h₂ = 12 cm
Now, h₁ = 20 – 12 = 8 cm
Here, Slant height of the frustum l = \(\sqrt { \left( R-r \right) ^{ 2 }+{ h }_{ 1 }^{ 2 } } units\)
\(=\sqrt { 36+64 } \)
l = 10 cm
Outer surface area = \(2\pi r{ h }_{ 2 }+\pi (R+r)\quad l\quad sq.units\)
\(=\pi [2r{ h }_{ 2 }+(R+r)l]\)
\(=\pi [(2\times 6\times 12)+(18\times 10)]\)
\(=\pi [144+180]\)
\(=\frac { 22 }{ 7 } \times 324=1018.28\)
Therefore, outer surface area of the funnel is 1018.28 cm².