(i) 57.750 L
(ii) Social cohension 

577.5 m³

The centre O of sphere will be the centroid of the  ABCD 
∴   OA= \(\frac { 1 }{ 3 } \) AB
∴  AB=3(OA)=3a
or sin 30^o = [in rt ∠dΔOKB]
⇒  \(\frac { 1 }{ 2 } =\frac { a }{ OB } \) \(\frac { 1 }{ 2 } =\frac { a }{ OB } \)OB=2a
⇒ AB=OA+OB=a+2a=3a
Now, In rt. ∠dΔABC  ,∠ABC=30^o ,∠BAC=90^o
 ⇒ \(\frac { AC }{ AB } =tan\quad { 30 }^{ o }\) 
⇒ Volume of cone BCD= \(\frac { 1 }{ 3 } \pi (AC)^{ 2 }\times AB\)
= \(\frac { 1 }{ 3 } \pi (a\sqrt { 3 } )^{ 2 }\times 3a\)
=3 \(\pi \) a²
and volume of sphere =\(\frac { 4 }{ 3 } \)  \(\pi \) a³
water remaining in the cone
=\(\frac { 4 }{ 3 } \)  \(\pi \) a³ =\(\frac { 5\pi }{ 3 } \) a³

For upper portion :
 r= \(\frac { 6 }{ 2 } \) m=3m
Height (h)= (28-24)m=4 m
∴  Slant height (l) = \(\sqrt { { h }^{ 2 }+r^{ 2 } } =\sqrt { { 4 }^{ 2 }+3^{ 2 } } \)
=5 m
C.S.A =\(\pi \) rl =\(\frac { 22 }{ 7 } \times 3\times 5{ m }^{ 2 }\) 
 =\(\frac { 330 }{ 7 } \) m²
For lower conical part (frustum)
R= \(\frac { 20 }{ 2 } \) m=10 m
r=\(\frac { 6 }{ 2 } \) m=3 m
and  h=24 m
∴ Slant height l= \(\sqrt { { (R-r) }^{ 2 }+{ h }^{ 2 } } \)
= \(\sqrt { { (10-3) }^{ 2 }+{ 24 }^{ 2 } } \)
=25 m
C.S.A  = \(\pi \)(R+r)l
= \(\frac { 22 }{ 7 } \times (10+3)\times 5\) m²
=  \(\frac { 1430 }{ 7 } \) m²
∴ The quantity of canvas required
=C.S.A of the cone + C.S.A of the frustum
=  \(\frac { 330 }{ 7 } { m }^{ 2 }+\frac { 1430 }{ 7 } { m }^{ 2 }=251\frac { 3 }{ 7 } \) m²

R₁ = 7m ; R² =13cm
Slant height of cone L = 12m
Slant height of frustum, I = \(\sqrt { (8)^{ 2 }+6^{ 2 } } =10m\)
= \(\sqrt { 8^{ 2 }+6^{ 2 } } =10m\)
Area of canvas required = curved surface area of frustum + curved surface area of cone 
= \(\pi (R_{ 1 }-R_{ 2 })l+\pi R_{ 1 }L\)
\(=\pi (13+7)\times 10+\pi \times 7\times 12\)
\(=200\pi +84\pi =284\pi m^{ 2 }\)

Volume of liquid = volume of box - volume of 16 spheres
= 16 x 8 x 8 - 16 x \(\frac{4}{3}\) x 3.14 x 2 x 2 x2
 = 1024 cm³ - 535.89 cm³
 = 488.10 cm³ 

Internal radius of pipe = 1cm
Rate = 80 cm/s
Quality of liquid passed in 1 sec = \(\pi r^{ 2 }h\)
\(=\frac { 22 }{ 7 } \times 1\times 1\times 80cm^{ 3 }=\frac { 1760 }{ 7 } cm^{ 3 }\)
Quantity of liquid passed in \(\frac { 1 }{ 2 } \) hr i.e. 30 min = \(\left( \frac { 1760 }{ 7 } \times 1800 \right) cm^{ 3 }\)
Volume of tank filled in \(\frac { 1 }{ 2 } \) hour = \(\pi \times (40)^{ 2 }\times h\)
Volume of tank filled in \(\frac { 1 }{ 2 } \) hour = quantity of water passed through pipe in  \(\frac { 1 }{ 2 } \) hour
\(\Rightarrow \frac { 22 }{ 7 } \times (40)^{ 2 }\times h=\frac { 1760 }{ 7 } \times 1800\Rightarrow h=90cm\)

Volume of hollow sphere = volume of cone
 \(\frac{4}{3}\pi (R^3 - r^3)=\frac{1}{3}\pi r^2h\)
\(\Rightarrow \ \frac{4}{3}(5^3 - r^3)=\frac{1}{3}\times(7)^2 \times8\)
\(\Rightarrow \ \frac{4}{3}(125 - r^3)=\frac{1}{3}\times49 \times8\)
\(​​\Rightarrow \ 500 -4r^3 = 49\times 8\)
\(\Rightarrow \ -4r^3 = 392 - 500\)
\(\Rightarrow \ -4r^3 = -108\)
\(\Rightarrow \ r^3 =27\)
\(\Rightarrow \ r =3\) cm
Diameter, d = 6 cm

Apparent capacity Of glass =  \(\pi r^{ 2 }h\)
= 3.14 \(\times \left( \frac { 5 }{ 2 } \right) ^{ 2 }\times 10cm^{ 3 }=196.25cm^{ 3 }\)
Actual capacity of glass = apparent capacity — volume of hemispherical part 
= 196.25 cm³ - \(\frac { 2 }{ 3 } \times 3.14\times \left( \frac { 5 }{ 2 } \right) ^{ 3 }=196.25cm^{ 3 }=32.70\quad cm^{ 3 }=163.55cm^{ 2 }\)

Diameter of cylinder = 4m \(\Rightarrow \) Radius of cylinder = 2m
Height of cylinder = 2.1 m
Slant height of cone = 2.8 m
Area of canvas used = C.S.A of cylinder + C.S.A of cone
\(=2\pi rh+\pi rl=\pi r\left[ 2h+l \right] \)
\(=\frac { 22 }{ 7 } \times 2\left[ 2\times 2.1+2.8 \right] =\frac { 22 }{ 7 } \times 2\left[ 4.2+2.8 \right] \)
\(=\frac { 22 }{ 7 } \times 2\times 7=44m^{ 2 }\)
Cost of 1 m² of canvas = Rs 500
\(\therefore \) Total cost = Rs 500 X 44 = Rs 22000
Volume of air = volume of cone + volume of cylinder 
= \(\frac { 1 }{ 3 } \pi r^{ 2 }H+\pi r^{ 2 }h\left[ \frac { 1 }{ 3 } H+h \right] \)
Height of cone = \(\sqrt { l^{ 2 }-r^{ 2 } } =\sqrt { (2.8)^{ 2 }-2^{ 2 } } =\sqrt { 7.84-4 } =\sqrt { 3.84 } =1.95\)
\(=\frac { 22 }{ 7 } \times 4\times 2.75=34.57m^{ 2 }\)​​​​​​​