Given, model is a combination of a cylinder and two cones. clearly, volume of the air will be equal to the sum of the volumes of two cones and one cylinder.
We have, diameter of the model, BC = ED = 3 cm
\(\therefore\) Radius of cone = radius of cylinder, \(r=\frac{3}{2}=1.5 \mathrm{~cm}\)
Length of cone, h₁ = 2 cm
Total length of the model, AF = 12 cm

\(\therefore\) Length of the cylinder, OO' = AF - (AO + O' F)
= 12 - (2 + 2) = 8 cm = h₂
Now, volume of the air inside the model = Volume of air inside (cone + cylinder + cone)
\(\begin{aligned} & =\left(\frac{1}{3} \pi r^2 h_1+\pi r^2 h_2+\frac{1}{3} \pi r^2 h_1\right) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{3} \pi r^2\left(h_1+3 h_2+h_1\right) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times 1.5 \times 1.5(2+3 \times 8+2) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times \frac{2.25}{3} \times(2+24+2) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 0.75 \times 28=22 \times 3=66 \mathrm{~cm}^3 \end{aligned}\)

Area of the field= area of the rectangle + Area of the two semicircular ends on the shorter sides
Length l of rectangle = 80m and breath of rectangle=42cm
d of the semicircle=42m ⇒ radius (r)=\({42\over 2}m=21m\)
Area of the field = \([80\times42+2\times{\pi\over 2}r^2]m^2\)
\(=\left(80\times42-{22\over 7}\times21\times21\right)m^2\)
=(3360+1386)m²=4746m²

l=length of the arc=\({\theta\pi r\over 180^0}\)
\(={45^0\over 18060}\times{22\over 7}\times7={11\over 2}cm\)

Permiter of the sector
\(=(r+r+l)=\left(7+7+{11\over2}\right)cm\)
\(\left(14+{11\over2}\right)cm=\left(28+11\over 2\right)cm\)
\(={39\over 2}cm=19.5cm\)

Given, sector angle, \(\theta\)= 80°
and distance or radius, r = 16.5 km
\(\begin{aligned} \therefore \text { Area of sector } & =\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{80^{\circ}}{360^{\circ}} \times 3.14 \times(16.5)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{2 \times 3.14 \times 272.25}{9} \end{aligned}\)
\(\begin{aligned} =\frac{1709.73}{9} \end{aligned}\)
= 189.97 km²
which is the required area of the sea over which the ships are warned.

We know that in 1 hour (i.e., 60 minutes), the minute hand rotates 360°.
In 5 minutes, minute hand will rotate = 360^@/60xx5 = 30^@
Therefore, the area swept by the minute hand in 5 minutes will be the area of a sector of 30° in a circle of 14 cm radius.
Area of sector of angle θ = \(\frac{\theta}{360^{\circ}} \times \pi r^{2}\)
Area of sector of 30° \(=\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 14 \times 14\)
\(\begin{array}{l} =\frac{22}{12} \times 2 \times 14 \\ =\frac{11 \times 14}{3} \end{array}\)
=154/3 cm²
Therefore, the area swept by the minute hand in 5 minutes is 154/3 cm²

Let BPC be the hemisphere and ABC be the cone standing on the base of the hemisphere (see Figure). The radius BO of the hemisphere (as well as of the cone) \(=\frac{1}{2} \times 4 \mathrm{~cm}=2 \mathrm{~cm} .\)
So, volume of the toy = \(\begin{aligned} & =\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h \end{aligned}\)
\(\begin{aligned} & =\left[\frac{2}{3} \times 3.14 \times(2)^3+\frac{1}{3} \times 3.14 \times(2)^2 \times 2\right] \mathrm{cm}^3=25.12 \mathrm{~cm}^3 \end{aligned}\)
Now, let the right circular cylinder EFGH circumscribe the given solid. The radius of the base of the right circular cylinder = HP = BO = 2 cm, and its height is EH = AO + OP = (2 + 2) cm = 4 cm
So, the colume required = volume of the right circular cylinder - volume of the toy
= 3.14 \(\times\)2² \(\times\)4 - 25.12) cm³
= 25.12 cm³
hence, the required difference of the two volumes = 25.12 cm³.

Capacity of the vessel = Volume of hemispherical bowl + Volume of the cylinder
= 2480.7619 cm³

Given sector is OAPB 
Area of the sector \(=\frac{\theta}{360} \times \pi r^{2}\)
\(=\frac{30}{360} \times 3.14 \times 4 \times 4 \mathrm{~cm}^{2}\)
\(=\frac{12.56}{3} \mathrm{~cm}^{2}=4.19 \mathrm{~cm}^{2}(\text { approx. })\)
Area of the corresponding major sector
\(\begin{aligned} &=\pi r^{2}-\text { area of sector } \mathrm{OAPB}\\ &=(3.14 \times 16-4.19) \mathrm{cm}^{2}\\ &=46.05 \mathrm{~cm}^{2}=46.1 \mathrm{~cm}^{2}(\text { approx. }) \end{aligned}\)
Alternatively, area of the major sector  \(=\frac{(360-\theta)}{360} \times \pi r^{2}\)
\(\begin{array}{l} =\left(\frac{360-30}{360}\right) \times 3.14 \times 16 \mathrm{~cm}^{2} \\ =\frac{330}{360} \times 3.14 \times 16 \mathrm{~cm}^{2}=46.05 \mathrm{~cm}^{2} \\ =46.1 \mathrm{~cm}^{2}(\text { approx. }) \end{array}\)

42 cm

Diameter of cylinder = diameter of the hemisphere = 7 cm 
\(\therefore \) Radius of cylinder = \(\frac { 7 }{ 2 } \) cm
Total height of the solid = 19 cm
Height of the cylinder = 19 - \(\left( \frac { 7 }{ 2 } +\frac { 7 }{ 2 } \right) =12cm\)
Volume of the solid = volume of the cylinder + 2 X volume of one hemisphere
= \(\pi r^{ 2 }h+2\times \frac { 2 }{ 3 } \pi r^{ 3 }=\pi r^{ 2 }\left( h+\frac { 4 }{ 3 } r \right) \)
\(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \left( 12\times \frac { 4 }{ 3 } \times \frac { 7 }{ 2 } \right) cm^{ 3 }\)
\(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \left( 12+\frac { 14 }{ 3 } \right) =641.666cm^{ 2 }=641.67cm^{ 3 }\)
Surface area of the solid = curved surface area of the cylinder + 2 x curved surface area of a hemisphere
= \(2\pi rh+2\times 2\pi r^{ 2 }=2\pi r(h+2r)\)
\(=2\times \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \left( 12+2\times \frac { 7 }{ 2 } \right) =418cm^{ 2 }\)  

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
In ΔOVS,
OV/OS = cos 60º
OV/12 = 1/2
OV = 6 cm
\(S \frac{V}{S} O=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\)
\(\frac{S V}{12}=\frac{\sqrt{3}}{2} \)
\(S V=6 \sqrt{3} \mathrm{~cm} \)
\(S T=2 S V=2 \times 6 \sqrt{3}=12 \sqrt{3} \mathrm{~cm}\)
Area of ΔOST = 1/2 x ST x OV
\(\frac{1}{2} \times 12 \sqrt{3} \times 6 \)
\(=36 \sqrt{3}=36 \times 1.73=62.28 \mathrm{~cm}^{2}\)
Area of sector OSUT \(=\frac{120^{\circ}}{360^{\circ}} \times \pi(12)^{2}\)
Area of segment SUT = Area of sector OSUT − Area of ΔOST
= 150.72 − 62.28
= 88.44 cm²