Volume of the cylindrical tank = Volume of the rectangle tank
πr²h = 28 x 16 x 11 m³

\(h=\frac { 16\times 11 }{ 88 } =2m\)

Clearly bucket forms frustum of a cone such that the radii of its circular ends are r₁ = 28 cm, r₂ = 7 cm, h = 45 cm
Capacity of the bucket = volume of the frustum
\(\Rightarrow \frac { 1 }{ 3 } \times \pi h[{ r }_{ 1 }^{ 2 }+{ r }_{ 2 }^{ 2 }+{ r }_{ 1 }{ r }_{ 2 }]\)
\(\Rightarrow \frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 45[{ 28 }^{ 2 }+{ 7 }^{ 2 }+28\times 7)]\)
= 22 x 15 x (28 x 4 + 7 + 28)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)

Let r be the radius of the hemisphere.
Given that, base area = \(\pi\)r² = 1386 sq. m
T.S.A. = 3 \(\pi\)r² sq.m
= 3 x 1386 = 4158
Therefore, T.S.A. of the hemispherical solid is 4158 m².

Let r₁ and r₂ be the radii of the balloons.
Given that, \(\frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 12 }{ 16 } =\frac { 3 }{ 4 } \)
Now, ratio of C.S.A. of balloons \(=\frac { 4\pi { r }_{ 1 }^{ 2 } }{ 4\pi { r }_{ 2 }^{ 2 } } =\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } ={ \left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) }^{ 2 }={ \left( \frac { 3 }{ 4 } \right) }^{ 2 }=\frac { 9 }{ 16 } \)
Therefore, ratio of C.S.A. of balloons is 9:16.

Given that, radius r = 7 cm
Now, total surface area of the cone = \(\pi\)r(l + r)sq. units
T.S.A = 704 cm²
704 = \(\frac{22}{7}\times7(l+7)\)
32 = l + 7 implies l = 25 cm
Therefore, slant height of the cone is 25 cm.

Given that, diameter d = 2.8 m and height = 3 m
radius r = 1.4 m
Area covered in one revolution = curved surface area of the cylinder
= 2\(\pi\)rh sq. units
\(2\times \frac { 22 }{ 7 } \times 1.4\times 3=26.4\)
Area covered in 1 revolution = 26.4 m²
Area covered in 8 revolutions = 8 x 26.4 = 211.2
Therefore, area covered is 211.2 m²

Given that, C.S.A. of the cylinder = 88 sq. cm
2\(\pi\)rh = 88
\(2\times \frac { 22 }{ 7 } \times 14=88\) (given h = 14cm)
2r = \(\frac { 88\times 7 }{ 22\times 14 } =2\)
Therefore, diameter = 2 cm

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.

Let h and r be the height and radius of the cone and cylinder.
Let l be the slant height of the cone.
Given that, h = 2.4 cm and d = 1.4 cm ; r = 0.7 cm
Here, total surface area of the remaining solid} C.S.A. of the cylinder + C.S.A. of the cone + area of the bottom
= 2\(\pi\)rh + \(\pi\)rl + \(\pi\)r² sq.units
Now, \(\\ \\ l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { 0.49+5.76 } =\sqrt { 6.25 } =2.5cm\)
Area of the remaining solid = 2\(\pi\)rh + \(\pi\)rl + \(\pi\)r² sq.units
= \(\pi\)r(2h + l + r)
\(\frac { 22 }{ 7 } \times 0.7\times [(2\times 2.4)+2.5+0.7]\)
Therefore, total surface area of the remaining solid is 17.6 m²