Let r and h be the radius and height of the cone respectively.
Given that, volume of the cone = 11088 cm³
\(\frac { 1 }{ 3 } { \pi r }^{ 2 }h=11088\)
\(\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times { r }^{ 2 }\times 24=11088\)
\({ r }^{ 2 }=441\)
Therefore, radius of the cone r = 21 cm.

Let r and h be the radius and height of the cylinder respectively.
Given that, height h = 2 m, base area = 250 m²
Now, volume of a cylinder = \(\pi\)r h ² cu. units
= base area x h
= 250 x 2 = 500 m³
Therefore, volume of the cylinder = 500 m³

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

Let h and r be the height and radius of the cylinder respectively.
Given that, h = 15 cm, r = 6 cm
Volume of the container V = \(\pi\)r²h cubic units.
Let, r₁ = 3 cm, h₁ = 9 cm be the radius and height of the cone.
Also, r₁ = 3 cm is the radius of the hemispherical cap.
Volume of one ice cream cone = (Volume of the cone + Volume of the hemispherical cap)
\(=\frac { 1 }{ 3 } \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 }+\frac { 2 }{ 3 } \pi { r }_{ 1 }^{ 3 }\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 3\times 3\times 9+\frac { 2 }{ 3 } \times \frac { 22 }{ 7 } \times 3\times 3\times 3\)
\(=\frac { 22 }{ 7 } \times 9(3+2)=\frac { 22 }{ 7 } \times 45\)
\(Number\ of\ cones=\frac { volume\ of\ the\ cylinder }{ volume\ of\ one\ ice\ cream\ cone } \)
Number of ice cream cones needed \(=\frac { \frac { 22 }{ 7 } \times 6\times 6\times 15 }{ \frac { 22 }{ 7 } \times 45 } =12\)
Thus 12 ice cream cones are required to empty the cylindrical container.

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

Let the number of small spheres obtained be n.
Let r be the radius of each small sphere and R be the radius of metallic sphere.
Here, R = 16 cm, r = 2 cm
Now, n\(\times\)(Volume of a small sphere) = Volume of big metallic sphere
\(n\left( \frac { 4 }{ 3 } { \pi r }^{ 2 } \right) =\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
\(n\left( \frac { 4 }{ 3 } \pi \times { 2 }^{ 3 } \right) =\frac { 4 }{ 3 } \pi \times { 16 }^{ 3 }\)
8n = 4096 gives n = 512
Therefore, there will be 512 small spheres.

Let r and R be the inner and outer radii of the hollow sphere.
Given that, inner diameter d = 14 cm; inner radius r = 7 cm; thickness = 1 mm = \(\frac{1}{10}\)cm
Outer radius R = 7 + \(\frac { 1 }{ 10 } =\frac { 71 }{ 10 } =7.1cm\)
Volume of hollow sphere \(=\frac { 4 }{ 3 } \pi \left( { R }^{ 3 }-{ r }^{ 3 } \right) cu.cm\)
\(=\frac { 4 }{ 3 } \times \frac { 22 }{ 7 } (357.91-343)=62.48cm^{ 3 }\)
But, weight of brass in 1 cm³ = 17.3 gm
Total weight = 17362 x 62.48 = 1080.90 gm
Therefore, total weight is 1080.90 grams.

Let r be the radius of the hemisphere.
Given that, volume of the hemisphere = 29106 cm³
Now, volume of new hemisphere = \(\frac{2}{3}\)(Volume of original sphere)
= \(\frac{2}{3}\) x 29106
Volume of new hemisphere = 19404 cm³
\(\frac { 2 }{ 3 } \pi { r }^{ 3 }=19404\)
\({ r }^{ 3 }=\frac { 19404\times 3\times 7 }{ 2\times 22 } =9261\)
\(r=\sqrt [ 3 ]{ 9261 } =21cm\)
Therefore, r = 21 cm

Let r and h be the radius and height of the cylinder respectively.
Given that, volume of the tank = 1.078 x 10⁶ = 1078000 litre
1078 m³ (since 1l = \(\frac{1}{1000}m^3\))
diameter = 7m gives radius = \(\frac{7}{2}\)m
volume of the tank = \(\pi\)r h 2 cu. units
1078 = \(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times h\)
Therefore, height of the tank is 28 m