Let Height h = h cm,
radius = 20 cm
Volume = 8800 cm³
\(\text { Volume } =\frac{1}{3} \pi r^{2} h \)
\(\frac{1}{3} \pi r^{2} h =8800 \)
\(\frac{1}{3} \times \frac{22}{7} \times 20 \times 20 \times h =8800 \)
\(h =\frac{8800 \times 3 \times 7}{22 \times 20 \times 20} \)
Height = 21 cm

Given R = 2 cm, r = 1 cm
Capacity of glass = Volume of frustum
\(=\frac{\pi h}{3}\left(R^{2}+R r+r^{2}\right) c u \text {. units. } \)
\(=\frac{22}{7} \times \frac{14}{3}\left((2)^{2} \times(2 \times 1)+(1)^{2}\right) \)
\(=\frac{44}{3}(4+2+1) \)
\(=\frac{308}{3} \)
\(=102.67 \mathrm{~cm}^{3} \)

Diameter = 0.21 cm;
\(\text { Radius }=\frac{0.21}{2}\)
= 0.105 cm
Amount of water displaced = Volume of the ball
\(=\frac{4}{3} \pi r^{3}\)
\(=\frac{4}{3} \times \frac{22}{7} \times(0.105)^{3} \)
\(=0.004851 \mathrm{~cm}^{3} \)

h = 15 cm, radius = r
\(\text { Volume } \frac{1}{3} \pi r^{2} h=1570\)
\(\frac{1}{3} \times \frac{22}{7} \times r^{2} \times 15=1570\)
\(r=\sqrt{100}\)
\(\text { radius of the base } \mathrm{r}=\sqrt{100}\)
\(= 10 cm\)

Radius = r, height = h.
\(\text { Circumference } 2 \pi r=132\)
\(\Rightarrow r=21 \mathrm{~cm}\)
\(\text { Volume }=\pi r^{2} h \text { cu. units }\)
\(=\frac{22}{7} \times 21 \times 21 \times 25 \)
\(=34650 \mathrm{~cm}^{3} \)
\(\text { Quantity of water }=\frac{34650}{1000} \text { litres }\)
\(=34.65 \text { litres. }\)
\(\left[\because 1000 \mathrm{~cm}^{3}=1 \mathrm{ltr} .\right]\)

Given R = 15cm, r = 5cm, h = 24cm
\(l =\sqrt{h^{2}+(R-r)^{2}} \)
\(=\sqrt{(24)^{2}+(15-5)^{2}} \)
\(=\sqrt{676}=26 \mathrm{~cm} \)
\(\text { C.S. } A =\pi l(R+r) \)
\(=\frac{22}{7} \times 26 \times(15+5) \)
\(=\frac{11440}{7}=1634.28 \mathrm{~cm}^{2} \)

Inner radius r = 5cm
Thickness = 0.25 cm
outer radius R = 5 + 0.25 = 5.25cm
\(\therefore \text { Outer curved surface area }=2 \pi R^{2} \text { Sq. units }\)
\(=2 \times \frac{22}{7} \times(5.25)^{2} \)
\(=173.25 \mathrm{~cm}^{2} \)

\(\text { Total surface area }=4 \pi r^{2}\)
\(4 \pi r^{2} =154 \)
\(4 \times \frac{22}{7} \times r^{2} =154 \)
\(r^{2} =\frac{154 \times 7}{4 \times 22}=12.25 \)
\(r =\sqrt{12.25}=3.5 \)
\(\text { Radius } =3.5 \mathrm{~cm} \)

Diameter = 24 m,
\(\text { radius } r=\frac{24}{2}=12 \mathrm{~m}\)
\(\text { Slant height } l=21 \mathrm{~m}\)
\(\text { Total surface area }=\pi r(l+r) \text { sq. units }\)
\(=\frac{22}{7} \times 12 \times(21+12) \)
\(=1244.57 \mathrm{~m}^{2}(\mathrm{app}) \)

Total radiating surface - curved surface area of cylindrical pipe.
\(=2 \pi r h\)
\(\text { Length } h =28 \mathrm{~m} \)
\(\text { Radius } r =\frac{5}{2} \mathrm{~cm} \)
\(=2.5 \mathrm{~cm}=0.025 \mathrm{~m} \)
\(\therefore \text { Total radiating surface }=2 \pi r h\)
\(=2 \times \frac{22}{7} \times 0.025 \times 28 \)
\(=4.4 \mathrm{~m}^{2} \)

Radius of the cylinder = r cm
Height 'h' = 14 cm
\(\text { C.S.A } =2 \pi r h \text { sq. units } \)
\(\therefore 2 \pi r h =88 \)
\(2 \times \frac{22}{7} \times r \times 14 =88 \)
r = cm
Diameter of the base = 2 x 1 = 2cm

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)
\(\Rightarrow 330\times 147{ cm }^{ 2 }\Rightarrow 48510{ cm }^{ 2 }\)