Curved surface area = \(\pi\)rl
\(=\frac{22}{7} \times 7 \times 10 \mathrm{~cm}^{2}\) 
= 220 cm²

Here, h = 16 cm and r = 12 cm.
So, from l² = h² + r², we have
\(l=\sqrt{16^{2}+12^{2}} \mathrm{~cm}=20 \mathrm{~cm}\)
So, curved surface area = \(\pi\)rl
= 3.14 x 12 x 20 cm²
= 753.6 cm²
Further, total surface area = \(\pi\)rl + \(\pi\)r²
= (753.6 + 3.14 x 12 x 12) cm²
= (753.6 + 452.16) cm²
= 1205.76 cm²

Radius of the base of the cylindrical kaleidoscope (r) = 3.5 cm.
Height (length) of kaleidoscope (h) = 25 cm
Area of chart paper required = curved surface area of the kaleidoscope
= 2\(\pi\)rh
\(=2 \times \frac{22}{7} \times 3.5 \times 25 \mathrm{~cm}^{2}\)
= 550 cm²

Since Hameed is getting the five outer faces of the tank covered with tiles, he would need to know the surface area of the tank, to decide on the number of tiles required.
Edge of the cubical tank = 1.5 m = 150 cm (= a)
So, surface area of the tank = 5 x 150 x 150 cm²
Area of each square tile = side x side = 25 x 25 cm²
So, the number of tiles required \(=\frac{\text { surface area of the tank }}{\text { area of each tile }}\)
\(=\frac{5 \times 150 \times 150}{25 \times 25}=180\)
Cost of 1 dozen tiles, i.e., cost of 12 tiles = RS. 360
Therefore, cost of one tile \(=\text { RS. } \frac{360}{12}=\text { RS. } 30\)
So, the cost of 180 tiles = 180 x RS.30 = RS. 5400

Since Mary wants to paste the paper on the outer surface of the box; the quantity of paper required would be equal to the surface area of the box which is of the shape of a cuboid. The dimensions of the box are:
Length =80 cm, Breadth = 40 cm, Height = 20 cm.
The surface area of the box = 2(lb + bh + hl)
= 2[(80 x 40) + (40 x 20) + (20 x 80)] cm²
= 2[3200 + 800 + 1600] cm²
= 2 x 5600 cm² = 11200 cm²
The area of each sheet of the paper = 40 x 40 cm²
= 1600 cm²
Therefore, number of sheets required \(=\frac{\text { surface area of box }}{\text { area of one sheet of paper }}\)
\(=\frac{11200}{1600}=7\)
So, she would require 7 sheets

(i) Each edge of the cubical box (a) = 10 cm
\(\therefore \)  Lateral surface area of the cubical box
= 4a² = 4(10)² = 400 cm².
For cuboidal box
l = 12.5 cm, b = 10 cm,
h = 8 cm
\(\therefore \)  Lateral surface area of the cuboidal box 
= 2(l + b)h
= 2(12.5 + 10)(8) = 360 cm².
Cubical box has the greater lateral surface area than the cuboidal box by (400 - 360)cm²,
i.e.,  40 cm².
(ii) Total surface area of the cubical box = 6a²
= 6(10)² = 600 cm²
Total surface area of the cuboidal box
= 2(lb + bh + hl)
= 2[(12.5)(10) + (10)(8) + (8)(12.5)] 
= 2[125 + 80 + 100] = 610 cm².
Cubical box has the smaller total surface area than the cuboidal box by (610 - 600) cm², i.e., 10 cm².

For a brick
l = 22.5 cm, b = 10 cm,
h = 7.5 cm
\(\therefore \)  Total surface area of a brick 
= 2 (lb + bh + hl)
= 2 (22.5 \(\times\) 10 + 10 \(\times\) 7.5 + 7.5 \(\times\) 22.5)
= 2 (225 + 75 + 168.75)
= 2(468.75) = 937.5 cm² = .09375 m²
\(\therefore \)  Number of bricks that can be painted out 
\(=\frac { 9.375 }{ .09375 } =100.\)

Let the length, breadth and height of the rectangular hall be l m, b m and h m respectively.
Perimeter = 250 m
\(\Rightarrow\) 2(l + b) = 250
\(\Rightarrow\)  l + b = 125                ...(1)
Area of the four walls
\(=\frac { 15000 }{ 10 } =1500\)m²      
\(\Rightarrow\) 2(l + b)h = 1500
\(\Rightarrow\)  (l + b)h  = 750                  
\(\Rightarrow\)  125 h  = 750                Using (1)
\(\Rightarrow\)   \(h=\frac { 750 }{ 125 } \)
\(\Rightarrow\)   h = 6 m
Hence, the height of the hall is 6 m.            

l = 5 m, b = 4 m,
h = 3 m
Area of the walls of the room = 2(l + b)h
= 2(5 + 4)3 = 54 m²
Area of the ceiling = lb
= (5) (4) = 20 m2
\(\therefore \) Total area of the walls of the room and the ceiling = 54 m² + 20 m² = 74 m²
\(\therefore \) Cost of white washing the walls of the room and the ceiling = 74 \(\times\) 7.50 = Rs 555.

(i) l = 1.5 m, b = 1.25 m,
h = 65 cm = 0.65 m.
\(\therefore \)  The area of the sheet required for making the box = lb + 2(bh + hl)
= (1.5)(1.25) + 2{(1.25)(0.65) + (0.65)(1.5)}
= 1.875 + 2{0.8125+0.975}
= 1.875 + 2(1.7875) = 1.875 + 3.575
= 5.45 m².
(ii) The cost of sheet for it = Rs 5.45 \(\times\) 20
= Rs 109.

Let the diameter of earth be d.
\(\therefore\) The radius of the earth will be r₁=\(\frac { d }{ 2 } \)
Diameter of moon will be \(\frac { d }{ 4 } \) and radius of moon (r₂)=\(\frac { d }{ 4 } \)
Volume of moon=\(\frac { 4 }{ 3 } \pi { r }_{ 2 }^{ 3 }\)
\(=\frac { 4 }{ 3 } \pi { \left( \frac { d }{ 8 } \right) }^{ 3 }\)
\(=\frac { 1 }{ 512 } \times { \pi d }^{ 3 }\times \frac { 4 }{ 3 } \)
Volume of earth = \(\frac { 4 }{ 3 } \pi { r }_{ 1 }^{ 3 }\)
\(=\frac { 4 }{ 3 } p{ \left( \frac { d }{ 2 } \right) }^{ 3 }\)
\(=\frac { 1 }{ 8 } \times \frac { 4 }{ 3 } \pi { d }^{ 3 }\)
\(\frac { Volume\ of\ moon }{ Volume\ of\ earth } =\frac { \frac { 1 }{ 512 } \times { \pi d }^{ 2 }\times \frac { 4 }{ 3 } }{ \frac { 1 }{ 8 } \times \frac { 4 }{ 3 } \pi { d }^{ 3 } } \)
\(=\frac { 1 }{ 64 } \)
\(\Rightarrow \ Volume\ of\ moon=\frac { 1 }{ 64 } (Volume\ of\ earth)\)
\(\Rightarrow \ \frac { Volume\ of\ moon }{ volume\ of\ earth } =\frac { 64 }{ 1 } \)

Let radius of the base be r cm.
Given, volume of a cone = 1570 cm³
\(\Rightarrow \ \frac { 1 }{ 3 } \pi { r }^{ 2 }h=1570[\because volume\ of\ a\ cone=\frac { 1 }{ 3 } \pi { r }^{ 2 }h]\)
\(\Rightarrow \ \frac { 1 }{ 3 } \times 3.14\times { r }^{ 2 }\times 15=1570\)
\(\Rightarrow \ { r }^{ 2 }=\frac { 1570 }{ 3.14\times 5 } =\frac { 15700 }{ 157 } \)
\(\Rightarrow\)  r = 10 cm
Hence, radius of the base is 10 cm.

(i) Given, radius (r) = 6 cm and height (h) = 7 cm
\(\because\) Volume of the right circular cone =\(\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times { (6) }^{ 2 }\times 7=\frac { 22 }{ 3 } \times 6\times 6\)
\(=22\times 12=264{ cm }^{ 3 }\)
(ii) 154 cm³

Given, diameter = 50 cm
\(\therefore\) Radius (r) = \(\frac { 50 }{ 2\times 100 } =0.25\quad m\quad \left[ \because 1cm=\frac { 1 }{ 100 } m \right] \)
and height (h) = 3.5 m
Curved surface area of the pillar = 2\(\pi\)rh
 = 2 x \(\frac { 22 }{ 7 } \) x 0.25 x 3.5 = 2 x 22 x 0.25 x 0.5 = 55 m²
Now, cost of painting per m² = Rs 12.50
\(\therefore\) Cost of painting 5.5 m² = 12.50 x 5.5 = Rs 68. 75

Thickness of bowl  = 0.25 cm
Inner radius of bowl  = 5 cm
\(\therefore\)  Outer radius of bowl
= 5 cm + 0.25 cm = 5.25 cm
\(\therefore\)  Outer curved surface area of the bowl
 \(=2\pi { r }^{ 2 }\)
\(\\ =2\times \frac { 22 }{ 7 } \times { \left( 5.25 \right) }^{ 2 }{ cm }^{ 2 }\)
\(\\ =173.25{ cm }^{ 2 }\)

(i) Here, r = 7 and I = 25 cm
⇒ h = \(\sqrt{l^{2}-r^{2}}=\sqrt{25^{2}-7^{2}}=\sqrt{625-49}\) = 24cm
∴ Volume of the conical vessel = \(\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \times \frac{22}{7} \times\) (7)² ｘ 24 cm³
\(=\frac{1}{3} \times \frac{22}{7} \times\) 7 ｘ 7 ｘ 24 cm³
= 22 х 7 х 8 cm³ = 1232 cm³
\(=\frac{1232}{1000} l\) = 1.232 1=
[∵ 1000 cm³ = 1 l]
Thus, the required capacity of the conical vessel is 1.232 I.
(ii) Here, height (h) = 12 cm and I = 13 cm
r = \(\sqrt{l^{2}-\mathrm{h}^{2}}=\sqrt{13^{2}-12^{2}}=\sqrt{169-144}\)
= \(\sqrt{25}\) = 5cm
Now, volume of the conical vessel = \(\frac{1}{3} \pi r^{2} h\)
\(=\frac{1}{3} \times \frac{22}{7}\) ✕ (5)² ✕ 12 cm³
=\(\frac{22 \times 5 \times 5 \times 4}{7} \mathrm{~cm}^{3}=\frac{2200}{7} \mathrm{~cm}^{3}\)
∴ Capacity of the conical vessel = \(\frac{2200}{7} \times \frac{1}{1000} l=\frac{22}{70} l=\frac{11}{35} l\)
Thus, the required capacity of the conical vessel is \(\frac{11}{35} l\).