Volume of one cube=125 cm³
(edge)³=125
edge=\(\sqrt [ 3 ]{ 125 } \)=5 cm
Length of the cuboid = 20 cm and
height = 5 cm
Surface area of cuboid =2(lb+bh+lh)
=2[20x5+5x5+5x20]
= 2 x 225 = 450 cm²
Volume of the cuboid =lxbxh
= 20 x 5 x 5
= 500 cm³ 

Let r₁ be 3x and r₂ be 5x, h₁=y(say) and h₂=y(say)
Volume of cone I=\(\frac{1}{3}\)π x 9x² x y
Volume of cone II=\(\frac{1}{3}\)π x 25x² x y
Volume of cone I: volume of cone II=\(\frac{1}{3}\) π.9x².y:\(\frac{1}{3}\)π.25x².y=9:25

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 

Surface area=154 cm²
\(\Rightarrow\ \ 4\pi r^2=154\ \ \Rightarrow r^2={154\times7\over4\times22}\)
\(\Rightarrow\ \ r=\sqrt{7\times7\over2\times2}={7\over2}=3.5\)

Here, vessel is a combination of a hollow hemisphere and a hollow cylinder.

For cylindrical portion,
Diameter = AB = DC = 14 cm
\(\therefore\) Radius = OB = O'C = O' P
\(=\frac{A B}{2}=\frac{14}{2}=7 \mathrm{~cm}\)
Total length of vessel, PO = 13 cm
\(\therefore\) Length of cylinder, OO' = PO - O'P = 13 - 7 = 6 cm
For hemispherical portion,
Radius of hemisphere = Height of hemisphere = 7 cm
Now, the inner surface area of the vessel = Curved surface area of cylinder + Curved surface area of hemisphere
\(\begin{aligned} & =2 \pi r h+2 \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =2 \times \frac{22}{7} \times 7 \times 6+2 \times \frac{22}{7} \times(7)^2 \\ \end{aligned}\)
\(\begin{aligned} & =2 \times 22 \times 6+2 \times 22 \times 7 \\ \end{aligned}\)
\(\begin{aligned} & =44(6+7)=44 \times 13=572 \mathrm{~cm}^2 \end{aligned}\)

The total surface area of the cube = 6 × (edge)² = 6 × 5 × 5 cm² = 150 cm².
Note that the part of the cube where the hemisphere is attached is not included in the surface area.
So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere
\(=150-\pi r^{2}+2 \pi r^{2}=\left(150+\pi r^{2}\right) \mathrm{cm}^{2} \)
\(=150 \mathrm{~cm}^{2}+\left(\frac{22}{7} \times \frac{4.2}{2} \times \frac{4.2}{2}\right) \mathrm{cm}^{2} \)
\(=(150+13.86) \mathrm{cm}^{2}=163.86 \mathrm{~cm}^{2}\)

Height of cylinder = 20 cm
radius of cylinder = 3.5 cm = radius of each hemisphere
Total surface area of the article = 2X C.S.A of a hemisphere
= \(2\times 2\pi r^{ 2 }+2\pi rh\rightleftharpoons 2\pi r(2r+h)\)
= \(2\times \frac { 22 }{ 7 } \times 3.5\left[ 2\times 3.5+20 \right] \)
= 44 x 0.5[7+20] = 44 x 0.5 x 27 cm² = 594.0 cm²

Here, toy is a combination of a hemisphere and a cone.

Given, total height of toy,
AD = 15.5 cm
For hemispherical portion,
Radius, OC = OD = OB = 3.5 cm
For conical portion,
Height, OA = AD - OD
= 15.5 - 3.5 = 12 cm
and radius = 3.5 cm
Now, total surface area of the toy = Curved surface area of cone + curved surface area of hemisphere
\(\begin{aligned} & =\pi r l+2 \pi r^2=\pi r \sqrt{h^2+r^2}+2 \pi r^2 \quad\left[\because l=\sqrt{h^2+r^2}\right] \\ \end{aligned}\)
\(\begin{aligned} & =\frac{22}{7} \times 3.5 \times \sqrt{(12)^2+(3.5)^2}+2 \times \frac{22}{7} \times(3.5)^2 \\ \end{aligned}\)
\(\begin{aligned} & =11 \sqrt{144+12.25}+22 \times 3.5 \\ \end{aligned}\)
\(\begin{aligned} & =11 \sqrt{156.25}+11 \times 7 \end{aligned}\)
= 11(12.5) + 77
=137.5 + 77
= 214.5 cm²

Total surface area of the article = CSA of cylinder + CSA of cone + CSA of hemisphere
Ans. 316.94 cm²

583 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

Let radius of the tank be x dm then
height of the tank be 6x dm
Total cost of painting = Rs 237.60
Area to be painted = \(\frac { 237.60\times 100 }{ 60 } \) =396 sq dm
\(\Rightarrow 2\pi r(r+h)=396\Rightarrow 2\times \frac { 22 }{ 7 } \times x(x+6x)=396\Rightarrow x\times 7x=\frac { 396\times 7 }{ 2\times 22 } \)
\(\Rightarrow x^{ 2 }=9\Rightarrow x=3\) dm
\(\Rightarrow \) Radius = 3dm , height = 6X3 dm = 18dm
Volume = \(\frac { 22 }{ 7 } \times 3^{ 2 }\times 18dm^{ 2 }=509.14\quad dm^{ 2 }\)
We have R= external radius = 12.5 cm
r = internal radius =(external radius - thickness ) = (12.5-1) cm = 11.5 cm
h= height of the pipe = 20 cm
\(\therefore \) Total surface area of the pipe = external curved surface + internal curved surface + 2 (area of the base of the ring )
\(2\pi Rh+2\pi rh+2(\pi R^{ 2 }-\pi r^{ 2 })=2\pi (R+r)h+2\pi (R^{ 2 }-r^{ 2 })\\ \)
= \(2\pi (R+r)h+2\pi (R+r)h+2\pi (R^{ 2 }-r^{ 2 })\)
\(=2\times \frac { 22 }{ 7 } \times (12.5+11.5)\times (20+12.5-11.5)cm^{ 2 }\)
\(=2\times \frac { 22 }{ 7 } \times 24\times 21cm^{ 2 }=3168\quad cm^{ 2 }\)

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