Here, Volume of a hemisphere = 2425 \(\frac{1}{2}\) cm³
\(\Rightarrow\)   \(\frac { 2 }{ 3 } \times \frac { 22 }{ 7 } \times { r }^{ 3 }=\frac { 4851 }{ 2 } \)
\(\Rightarrow\)    \({ r }^{ 3 }=\frac { 4851 }{ 2 } \times \frac { 3\times 7 }{ 2\times 22 } \)
\(=\frac { 441\times 21 }{ 2\times 2\times 2 } \)
\(\Rightarrow\)      r = \(\frac{21}{2} cm\)
Now, curved surface area = 2\(\pi\)r²
= 2 x \(\frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } \)
= 693 cm²

Radius of hemisphere portion = Radius of cylindrical portion = \(\frac{7}{2}m\)
Total length of the tanker = 20 m
\(\therefore\) Length of cylindrical portion = 20 - 7 = 13 m

Now, Capacity (volume) of the oil tanker
= Volume of cylinder + 2 x Volume of hemisphere
= \(\pi r^2h+2\times\frac{2}{3}\pi r^3\)
= \(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times13+2\times\frac{2}{3}\times\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\times\frac{7}{2}\)
= 500.5 + 179.67 = 680.17 m³

Radius of the base of the cone and hemisphere
= \(\frac{18}{2}\) = 9 cm
Height of cone = 12 cm
Slant height l = \(\sqrt{9^2+12^2}\) = \(\sqrt{81+144}\)
= \(\sqrt{225}\) = 15 cm
 
Total surface area of toy
= C.S.A. of hemisphere + C.S.A. of cone
= \(2\pi r^2+\pi rl=\pi r(2r+l)\)
= 3.14 x 9(2 x 9 + 15)
= 3.14 x 9 x 33 = 932.58 cm²

Surface area of remaining block =
surface area of cuboid + curved surface area of cylinder - surface area of two circular end of cylinder hole
= 2[lb + bh + lh] + \(2\pi rh-2\pi r^2\)
= 2[15 x 10 + 10 x 5 + 15 x 5] + 2 x \(\frac{22}{7}\times\frac{7}{2}\) x 5 - 2 x \(\frac{22}{7}\times\frac{7}{2}\times\frac{7}{2}\)
= 2(150 + 50 + 75) + 110 - 77 cm²
= 583 cm² 

Given that, tent is a combination of a cylinder and a cone.

For conical portion,
Slant height, l = 2.8 m
Radius, r = Radius of cylinder  \(=\frac{\text { Diameter }}{2}=\frac{4}{2}=2 \mathrm{~m}\)
For cylindrical portion,
Radius, r = \(\frac{4}{2}=2 \mathrm{~cm}\)
Height, h = 2.1 m
\(\therefore\) Required surface area of the tent = CSA of cone + CSA of cylinder
\(\begin{aligned} & =\pi r l+2 \pi r h=\pi r(l+2 h) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 2 \times(2.8+2 \times 2.1) \\ \end{aligned}\)
\(\begin{aligned} & =\frac{44}{7}(2.8+4.2)=\frac{44}{7} \times 7=44 \mathrm{~m}^2 \end{aligned}\)
Now, cost of the canvas of the tent at the rate of Rs 500 per m² = Surface area \(\times\) Cost per m² = 44 \(\times\) 500 = Rs 22000

Total surface area of a solid hemisphere
=2πr²+πr²=3πr²

Now, 3πr²=462
⇒ r²=\(\frac{462\times{7}}{3\times{22}}\)=49⇒ r=7
Volume of hemisphere=\(\frac{2}{3}\pi{r}^{3}=\frac{2}{3}\times{\frac{22}{7}}\times\left(7\right)^{3}\)
=\(\frac{2}{3}\)x22x49 cm³=718.67 cm³

Coins are cylindrical in shape.
Height (h₁) of cylindrical coins = 2 mm = 0.2 cm
Radius (r) of circular end of coins = 1.75/2 =0.875 cm
Let n coins be melted to form the required cuboids.
Volume of n coins = Volume of cuboids
nxπxr²xh₁ = lxbxh
n x π x (0.875)² x 0.2 = 5.5 x 10 x 3.5
\(n=\frac{5.5 \times 10 \times 3.5 \times 7}{(0.875)^{2} \times 0.2 \times 22}=400\)
Therefore, the number of coins melted to form such a cuboid is 400.

Radius of cone=\(\frac{1}{2}\)x edge of cube
=\(\frac{9}{2}\)cm

Height of cone, h=9 cm
Volume of cone=\(\frac{1}{3}\pi{r}^{2}h\)
=\(\frac{1}{3}\times{\frac{22}{7}}\times{\frac{9}{2}}\times{\frac{9}{2}}\times{9}\)
=190.928 cm³

Given, solid is a combination of a cone and a hemisphere.
Also, radius of the cone, r = radius of the hemisphere
= 1 cm

Height of the cone, h = 1 cm
\(\therefore\) Required volume of the solid = Volume of the cone + Volume of the hemisphere
\(\begin{aligned} & =\frac{1}{3} \pi r^2 h+\frac{2}{3} \pi r^3=\frac{1}{3} \pi(1)^2(1)+\frac{2}{3} \pi(1)^3 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{\pi}{3}+\frac{2}{3} \pi=\frac{(\pi+2 \pi)}{3}=\frac{3 \pi}{3}=\pi \mathrm{cm}^3 \end{aligned}\)

Let the length of each edge of the cube be x cm
Volume = x³ cm³⇒27=x³ 
Length of the cuboid formed = 3 cm + 3 cm = 6 cm
breadth, b = 3 cm, height, h = 3 cm
Surface area of the cuboid fomed = 2(lb + bh + hl) 
=2(6x3+3x3+3x6)
=2(18+9+18)=2x45 = 90 cm² 

Diameter of tent = 24m radius = 12m
Slant height=13m
Curved surface area =\(\pi r l=\frac{22}{7} \times 12 \times=\frac{3432}{7} m^2\)

Given, diameter of base of the greenhouse is circular = 12 m
radius = 6 m
Height of greenhouse cylindrical wall = 2 m
\(\therefore\) CSA of the hemispherical roof = 2 \(\times\) \(\pi\) \(\times\) 6²m²
= 72 \(\pi\)m²
and CSA of the cylindrical wall = 2\(\times\)\(\pi\)\(\times\)6\(\times\)2 m²
= 24\(\pi\)m²
TSA of the greenhouse to be covered = 72\(\pi\) + 24  \(\pi\)
= 96\(\pi\)m²
The cost of the plastic sheet required to cover the entire greenhouse = 96\(\pi\) \(\times\) 77= Rs 23232

Length of the capsule = 14 mm
Diameter of the capsule = 4 mm
Radius of the capsule \(=\frac{4}{2}=2 \mathrm{~mm}\)

Length of the cylinder = Total length of capsule - Radius of left hemisphere - Radius of right hemisphere
  = 14 - 2 - 2 = 10 mm
Surface area of capsule = Curved surface area of cylinder + Surface area of left hemisphere + Surface area of right hemisphere
\(\begin{aligned} =2 \pi r h+2 \pi r^2+2 \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} =2 \pi r h+4 \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} =2 \times \frac{22}{7} \times 2 \times 10+4 \times \frac{22}{7} \times 2 \times 2 \end{aligned}\)
= 125.71 + 50.29
= 176
Thus, the required surface area of the capsule is 176 mm².
Volume of capsule = Volume of cylinder + Volume of left hemisphere + Volume of right hemisphere
\(\begin{aligned} & =\pi r^2 h+\frac{2}{3} \pi r^3+\frac{2}{3} \pi r^3 \\ \end{aligned}\)
\(\begin{aligned} & =\pi r^2 h+\frac{4}{3} \pi r^3 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 2 \times 2 \times 10+\frac{4}{3} \times \frac{22}{7} \times 2 \times 2 \times 2 \end{aligned}\)
= 125.71 + 33.52
= 159.23 mm³

This top is exactly like the object we have discussed. So, we can conveniently use the result we have arrived at there. That is :
TSA of the toy = CSA of hemisphere + CSA of cone
Now, the curved surface area of the hemisphere = \(\frac{1}{2}\left(4 \pi r^{2}\right)=2 \pi r^{2} \)
\(=\left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \mathrm{cm}^{2}\)
Also, the height of the cone = height of the top – height (radius) of the hemispherical part
\(=\left(5-\frac{3.5}{2}\right) \mathrm{cm}=3.25 \mathrm{~cm}\)
So, the slant height of the cone \((l)=\sqrt{r^{2}+h^{2}}=\sqrt{\left(\frac{3.5}{2}\right)^{2}+(3.25)^{2}} \mathrm{~cm}=3.7 \mathrm{~cm}(\text { approx. })\)
Therefore, CSA of cone \(=\pi r l=\left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \mathrm{cm}^{2}\)
This gives the surface area of the top as
\(=\left(2 \times \frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2}\right) \mathrm{cm}^{2}+\left(\frac{22}{7} \times \frac{3.5}{2} \times 3.7\right) \mathrm{cm}^{2}\)
\(=\frac{22}{7} \times \frac{3.5}{2}(3.5+3.7) \mathrm{cm}^{2}=\frac{11}{2} \times(3.5+3.7) \mathrm{cm}^{2}=39.6 \mathrm{~cm}^{2}(\text { approx. })\)
You may note that ‘total surface area of the top’ is not the sum of the total surface areas of the cone and hemisphere

Total surface area (TSA) of toy = CSA of hemisphere + CSA of cone
This is because the toy is obtained by joining the plane surfaces of hemisphere and cone.

(c) Given, side of cube = a cm
\(\Rightarrow\)   Diameter of sphere =  a cm
\(\Rightarrow\)   Radius of sphere = \(\frac{a}{2}\)cm
We know that surface area of sphere = 4\(\pi\)r²
\(=4 \pi \times \frac{a}{2} \times \frac{a}{2}\)
= a²\(\pi\)cm²

Given dimension of conical part of a silo is radius,

of base = 1.5 m

Height =2 m

And dimension of the cylindrical part of a silo is radius of base = 1.5 m

Height =7m

(i) We know that slant height

\(l=\sqrt{r^2+h^2}=\sqrt{(1.5)^2+(2)^2}=2.5 \mathrm{~m}\)

(ii) Curved surface area of cone = πrl

\(=\frac{22}{7} \times 1.5 \times 2.5=11.78 \mathrm{~m}^2\)

(iii) (a) Curved surface area of cylinder = 2πrh

\(=2 \times \frac{22}{7} \times 1.5 \times 7=66 \mathrm{~m}^2\)

Cost of 1 m² of metal sheet =₹2000

∴ Cost of metal sheet used = 66 × 2000=₹132000

Or

(b) Volume of cylinder = πr²h

\(=\frac{22}{7} \times(1.5)^2 \times 7=49.5 \mathrm{~m}^3\)

Volume of cone \(=\frac{1}{3} \pi r^2 h\)

\(=\frac{1}{3} \times \frac{22}{7} \times(1.5)^2 \times 2=4.71 \mathrm{~m}^3\)

Thus, total capacity of 1 silo= 49.5 + 4.71= 54.21 m³.