Radius of hemisphere = r units
Radius of cone = Radius of hemisphere
Height of cone = Radius of hemisphere
Maximum volume of cone = \(\frac{1}{3} \pi r^{2} h \text { cu. units }\)
\(=\frac{1}{3} \pi\left(r^{2}\right) r=\frac{1}{3} \pi r^{3} \text { cu. units }\)

Radius of conical flask = 'r' units
Height of conical flask = 'h' units
Volume of conical flask = Volume of water
\(=\frac{1}{3} \pi r^{2} h \text { cu. units }\)
Since, water is poured into the cylindrical flask
Volume of cylinder = Volume of water
\(\pi(\mathrm{xr})^{2} H=\frac{1}{3} \pi r^{2} h\)
[xr - radius of cylinder, H - height]
\(\mathrm{X}^{2} \mathrm{r}^{2} \mathrm{H}=\frac{r^{2}}{3} h\)
Height of the water in cylinder flask
\(\mathrm{H}=\frac{h}{3 x^{2}}\)

Diameter of cylindrical pipe = 14 cm
Radius = 7 cm
Length of the pipe = Speed of the water
= 15 km = 15000 m
Length of the water tank = 50 m
Width of the water tank = 44 m
Height of the water tank = Water level
= 21 cm
= 0.21 cm
volume of water tank = l x b x h cu. units
= 50 x 44 x 0.21 = 462 m³
Volume of cylindrical Pipe = Volume of Rectangular tank
\(\frac{\pi r^{2} h}{} h =462 \)
\(\frac{22}{7} \times 0.07 \times 0.07 \times h =462 \)
\(\mathrm{h} =\frac{462 \times 7}{22 \times 0.07 \times 0.07} \)
\(=\frac{3234}{0.1078}=30000 \)  
Time required \(=\frac{30000}{15000}=2 \text { hrs. } \)

Radius of well = 5 m
Depth of well = 14 m
Volume of earth taken out \(=\pi r^{2} h \)
\(=\frac{22}{7} \times(5)^{2} \times 14 \)
= 1100 m³
Now, it is spread to form an embankment, which is in the form of hollow cylinder
Inner radius = 5m
Width of embankment = 5 m
Outer radius = 5 + 5 = 10 m
height = h
Volume of hollow cylinder = \(\pi h\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)\)
\(\therefore \pi h\left(\mathrm{R}^{2}-\mathrm{r}^{2}\right)=1100 \)
\(\frac{22}{7} \times h\left(10^{2}-5^{2}\right)=1100 \)
height of the embankment
\(h=\frac{1100 \times 7}{22 \times 75}=4.67 \mathrm{~m}\)

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 : \(\sqrt3\)

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

Let the radii of two cones be r₁ and r₂ and heights be h₁ and h₂
Given ratio of their radii = \(\frac{r_{1}}{r_{2}}=\frac{1}{3}\)
\(r_{1}=\frac{r_{2}}{3} \)
\(h_{1}=3 r_{1}, h_{2}=3 r_{1} \)
[ r₁ is the radius of smaller cone]
Slant heights \(l_{1} =\sqrt{h_{1}^{2}+r_{1}^{2}} \)
\(=\sqrt{9 r_{1}^{2}+r_{1}^{2}}=\sqrt{10} r_{1} \)
\(l_{2} =\sqrt{h_{2}^{2}+r_{2}^{2}} \)
\(=\sqrt{9 r_{1}^{2}+9 r_{1}^{2}}=\sqrt{18 r_{1}^{2}}=3 \sqrt{2} r_{1}\)
Ratio of curved surface areas
\(=\frac{\text { CSA of I cone }}{\text { CSA of II cone }} \)
\(=\frac{\pi r_{1} l_{1}}{\pi r_{2} l_{2}} =\frac{r_{1}\left(\sqrt{10} r_{1}\right)}{\left(3 r_{1}\right)\left(3 \sqrt{2} r_{1}\right)} \)
\(=\frac{\sqrt{10}}{9 \sqrt{2}}= \frac{\sqrt{5} \sqrt{2}}{9 \sqrt{2}}=\frac{\sqrt{5}}{9} \)
Ratio of C.S.A = \(\sqrt{5}: 9\)

Each person requires 22 m² of floor area.
Required base area = 22 x 4 = 88 m²
\(\pi r^{2} =88 \)
\(r^{2} =\frac{88 \times 7}{22}=4 \times 7 \)
\(r =2 \sqrt{7} \mathrm{~m} \)
slant height = 19 m
height of the tent, h \(=\sqrt{l^{2}-r^{2}}\)
\(=\sqrt{(191)^{2}-(2 \sqrt{7})^{2}} \)
\(=\sqrt{361-28} =\sqrt{330}=18.25 \mathrm{~m} \)
Height of the tent = 18.25 m

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.