The inner diameter of hemispherical bowel=8 cm
Then its inner radius(r) = 4 cm
Thickness of steel = 0.2 cm
Hence, outer radius of bowl(R) = 4 + 0.2 = 4.2 cm
Hence, the outer curved surface area of the bowl = 2\(\pi\)R²
\(=2\times \frac { 22 }{ 7 } \times { (4.2) }^{ 2 }\)
\(=\frac { 44 }{ 7 } \times \frac { 42 }{ 10 } \times \frac { 42 }{ 10 } \)
= 110.88 cm²
Hence, the cost of polishing its outer surface area
= 2\(\times\)110.88
= Rs 221.76.

Let the length, breadth and height of the cuboid be l, b and h units respectively. then,
p = lb
q = bh
r = hl
ஃ pqr = (lb)(bh)(hl)
 =l²b²h²     .........(1)
Again   v = lbh
ஃ  v² = (lbh)²
= l²b²h²    .......(2) 
(1) and (2) give
v² = pqr

L.H.S = \(\frac{1}{v}=\frac{1}{abc} \ \ \ \ \ \ \ \ \ .....(1)\)
R.H.S = \(\frac{2}{s}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\)
=\(\frac{2}{2(ab+bc+ca)}(\frac{bc+ca+ab}{abc})\)
\(=\frac{1}{abc}\ \ \ \ \ \ \ ....(2)\)
(1) and (2) give
\(\frac{1}{v}=\frac{2}{s}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\)

For room
l = 12 m,
b = 9 m,
h = 8 m
\(\therefore\)   Length of the longest rod that can be placed in the room
= Length of the diagonal 
\(=\sqrt { { l }^{ 2 }+{ b }^{ 2 }+{ h }^{ 2 } } \) 
\(\\ =\sqrt { { \left( 12 \right) }^{ 2 }+{ \left( 9 \right) }^{ 2 }+{ \left( 8 \right) }^{ 2 } } \)
\(\\ =\sqrt { 144+81+64 } \)
\(\\ =\sqrt { 289 } \)
\(\\ =17m.\)

Inner Radius (r)   = 6 cm
Outer Radius (R) = 6 cm + 1 cm = 7 cm
Total surface area of the bowl 
\(=2\pi { r }^{ 2 }+2\pi { R }^{ 2 }+\pi \left( { R }^{ 2 }-{ r }^{ 2 } \right) \)
\(\\ =2\pi { \left( 6 \right) }^{ 2 }+2\pi { \left( 7 \right) }^{ 2 }+\pi \left( { 7 }^{ 2 }-{ 6 }^{ 2 } \right) \)
\(\\ =72\pi +98\pi +13\pi \)
\(\\ =183\pi { cm }^{ 2 }\)

\(\frac { { \left( CSA \right) }_{ 1 } }{ { \left( CSA \right) }_{ 2 } } =\frac { \pi r\left( 4k \right) }{ \pi r\left( 3k \right) } =\frac { 4 }{ 3 } =4\ :\ 3\)

For roller
\(r=\frac { 1.5 }{ 2 } m=0.75m\)
\(\\ h=84cm=0.84m\)
\(\therefore \)  Curved surface area = \(2\pi rh\)  
\(=2\times \frac { 22 }{ 7 } \times 0.75\times 0.84\)
\(\\ =3.96{ m }^{ 2 }\)
\(\therefore \)  Area of the ground levelled in 1 revolution 
= 3.96 m²
\(\therefore \)  Area of the ground levelled in 100 revolutions 
= 3.96  100 m² = 396 m²
\(\therefore \)  Cost of levelling
= Rs \(396\times \frac { 50 }{ 100 } =\) Rs 198

For tank
\(r=\frac { 3.8 }{ 2 } m=1.9m\)
\(\\ h=4.9m\)
 Curved surface area
\(=2\pi r\left( h+r \right) \)
\(\\ =2.\frac { 22 }{ 7 } .\frac { 19 }{ 10 } \left( 4.9+1.9 \right) { m }^{ 2 }\)
\(\\ =2.\frac { 22 }{ 7 } .\frac { 19 }{ 10 } .\frac { 68 }{ 10 } { m }^{ 2 }\)
\(\\ =81.21{ m }^{ 2 }\)

External diameter = 75 mm
\(\therefore \)  External radius (r) \(=\frac { 75 }{ 2 } mm=37.5mm\)
\(=\frac { 37.5 }{ 10 } cm=3.75cm\)
Length of the pipe (h)
= 4.2 m = 4.2 \(\times\) 100 cm = 420 cm
\(\therefore \)  Area of the outer surface = \(2\pi rh\)
\(=2\times \frac { 22 }{ 7 } \times 3.75\times 420=9900{ cm }^{ 2 }.\)