28 cm

7 : 3

1300 cm²

(i) We have, radius of cylindrical tank
r = 1 m
and height of cylindrical tank h = 3.5 m
Volume of cylindrical tank = \(\pi\)r²h
\(=\frac { 22 }{ 7 } \times 1\times 1\times 3.5\)
= 11 m³
Let the rainfall be h m, then
Volume of water on the roof = Volume of cylindrical tank
\(\Rightarrow\) 22 \(\times\) 20 \(\times\) h = 11
\(\Rightarrow \ h=\frac { 11 }{ 22\times 20 } \)
\(=\frac { 1 }{ 40 } m\)
\(=\frac { 100 }{ 40 } \)
= 2.5 cm
(ii) Save water to save earth.

Vol. of hemispherical bowl = \(\frac { 2 }{ 3 } \pi r^{ 3 }\)
\(=\frac { 2 }{ 3 } \pi { \left( \frac { 36 }{ 2 } \right) }^{ 3 }{ cm }^{ 3 }\)
Vol. of cylindrical bottle = \(\pi\)r²h
= \(\pi\)(3)² \(\times\)6 cm³
Suppose required bottles are x.
\(\therefore \ x=\frac { \frac { 2 }{ 3 } \pi { (18) }^{ 3 } }{ \pi ({ 3) }^{ 3 }\times 6 } =24\)

Height of cylindrical reservoir = 12 m
Total cost of plastering it from inside = Rs 5652
Cost of 1 m² = Rs 15
\(\therefore\) Surface area plastered = \(\frac { Total\ cost }{ cost/{ m }^{ 2 } } \)
\(2\pi rh=\frac { 5652 }{ 15 } \)
\(\therefore \ 2\times \frac { 22 }{ 7 } \times r\times 12=\frac { 5652 }{ 15 } \)
\(\therefore \ r=\frac { 5652\times 7 }{ 15\times 2\times 22\times 12 } \)
= 4.99 m \(\approx \)5 m
\(\therefore\) Volume = \(\pi\)r²h
\(=\frac { 22 }{ 7 } \times 5\times 5\times 12\)
= 942.857 m³
= 942857 litre

l:b = 4:3, h = 550 cm,
l = 4y, b = 3y (taking y as constant)
h = 550 cm = 5.5 m
Total cost = L.S.A \(\times\)Rate per square metres
5082 = 2h(l + b)\(\times\)6.60
=2 \(\times\)5.5(4y + 3y) \(\times\)6.60
5082 = 2\(\times\)5.5\(\times\)7y\(\times\)6.60
10 = y
l = 4y = 4\(\times\)10 = 40 m
b = 3y = 3\(\times\)10 = 30 m

For resulting cuboid
Length (l)    = 6 + 6 = 12 cm
Breadth (b) = 6 cm
Height (h)   = 6 cm
\(\therefore \) Surface area 
= 2(lb + bh + hl) 
= 2(12 \(\times\) 6 + 6 \(\times\) 6 + 6 \(\times\) 12)
= 360 cm²    

(i)    2r = 4.2 m
\(\therefore \)    \(r=\frac { 4.2 }{ 2 } m=2.1\quad m\)
h = 4.5 m
\(\therefore \)    Lateral or curved surface area = \(2\pi rh\)
 \(=2\times \frac { 22 }{ 7 } \times 2.1\times 4.5=59.4{ m }^{ 2 }\)
(ii) Total surface area = \(2\pi r\left( h+r \right) \)
 \(=2\times \frac { 22 }{ 7 } \times 2.1\times \left( 4.5+2.1 \right) \)
\(\\ =2\times \frac { 22 }{ 7 } \times 2.1\times 6.6=87.12{ m }^{ 2 }\)
Let the actual area of steel used be x m².
Since \(\frac { 1 }{ 12 } \) of the actual steel used was wasted, the area of the steel which has gone into the tank \(=\frac { 11 }{ 12 } \)  of x.
\(\therefore \)   \(\frac { 11 }{ 12 } x=87.12\)
\(\therefore \)   \(x=\frac { 87.12\times 12 }{ 11 } =95.04{ m }^{ 2 }\)
\(\therefore \)  Steel actually used = 95.04 m².

h = 28 m
2r = 5 cm
\(\therefore \)  \(r=\frac { 5 }{ 2 } cm=\frac { 5 }{ 2\times 100 } m\)
\(=\frac { 5 }{ 200 } m=\frac { 1 }{ 40 } m\)
\(\therefore \) Total radiating surface in the system 
\(=2\pi rh\)
\(\\ =2\times \frac { 22 }{ 7 } \times \frac { 1 }{ 40 } \times 28=4.4{ m }^{ 2 }.\)

h = 1 m = 100 cm
2r = 140 cm
\(\Rightarrow\)  \(r=\frac { 140 }{ 2 } cm=70cm\)
 \(\therefore\)  Total surface area of the closed cylindrical tank   
 \(=2\pi r\left( h+r \right) \)
\(\\ =2\times \frac { 22 }{ 7 } \times 70\left( 100+70 \right) \)
\(\\ =74800{ cm }^{ 2 }=\frac { 74800 }{ 100\times 100 } { m }^{ 2 }\)
\(\\ =7.48{ m }^{ 2 }\)
Hence, 7.48 square metres of the sheet are required.

Let the radius of the base of the cylinder be r cm.
h = 14 cm
 Curved surface area = 88 cm²   Given
\(\Rightarrow\)   \(2\pi rh=88\)
\(\Rightarrow\)   \(2\times \frac { 22 }{ 7 } \times r\times 14=88\)
\(\Rightarrow\)   \(r=\frac { 88\times 7 }{ 2\times 22\times 14 } \)
\(\Rightarrow\)   r = 1
\(\Rightarrow\)  2r = 2
Hence, the diameter of the base of the cylinder is 2 cm.

Number of cubes = \(\frac { { \left( 20 \right) }^{ 3 } }{ { \left( 5 \right) }^{ 3 } } =64\)

Volume = 15 \(\times\) 10 \(\times\) 8 = 1200 cm³