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 r₁ and h₁ be the radius and height of the cone - I and let r₂ and h₂ be the radius and height of the cone-II.
Given h₂ = 2h₁ = 2 and \(\frac { Volume\ of\ the\ cone\ I }{ Volume\ of\ the\ cone\ II } =\frac { 2 }{ 3 } \)
\(\frac { \frac { 1 }{ 3 } { \pi r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \frac { 1 }{ 3 } { \pi r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } \times \frac { { h }_{ 1 } }{ 2{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } =\frac { 4 }{ 3 } \text {gives} \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 2 }{ \sqrt { 3 } } \)
Therefore, ratio of their radii = 2 : \(\sqrt3\)

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

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

Radius of sphere = 12 cm
Volume of sphere = \(\frac{4}{3} \pi r^{3} cu. units \)
= \(\frac{4}{3} \pi(12)^{3} \)
\(=2304 \pi \mathrm{cm}^{3}\)
Radius of cylinder = 8 cm
height = h cm
Volume of cylinder = \(\pi r^{2} h \) cu. units
= \(\pi(8)^{2} h \)
= \(64 \pi \mathrm{h} \mathrm{cm}^{3}\)
Given that sphere is melted and cast into a cylinder
Volume of cylinder = Volume of sphere
\(64 \pi h =2304 \pi \)
\(h =\frac{2304 \pi}{64 \pi}=36 \)
Height of the cylinder = 36 cm.

Let h₁ and h₂ be the heights of a cone and cylinder respectively.
Also, let r be the radius of the cone.
Given that, height of the cone h₁ = 24 cm; radius of the cone and cylinder r = 6 cm
Since, Volume of cylinder = Volume of cone
\({ \pi r }^{ 2 }=\frac { 1 }{ 3 } { \pi r }^{ 2 }{ h }_{ 1 }\)
\({ h }_{ 2 }=\frac { 1 }{ 3 } \times { h }_{ 1 }\quad gives\quad { h }_{ 2 }=\frac { 1 }{ 3 } \times 24=8\)
Therefore, height of cylinder is 8 cm