Given, T.S.A = 1540 cm² 
\(\therefore\) 2\(\pi\)r(h + r) = 1540 cm²
Also,  h = 4r
\(\therefore\) 2\(\pi\)r(4r + r) = 1540
\(\Rightarrow\) 2\(\pi\)\(\times\)5r ²= 1540
\(\Rightarrow \ { r }^{ 2 }=\frac { 1540\times 7 }{ 2\times 5\times 22 } \)
\(\Rightarrow\) r²= 49
\(\Rightarrow\) r = 7 cm
Now h = 4r
\(\Rightarrow\) h = 28 cm.

24640 cm³

Since the concrete mixture that is to be used to build up the pillars is going to occupy the entire space of the pillar, what we need to find here is the volume of the cylinders.
Radius of base of a cylinder = 20 cm
Height of the cylindrical pillar = 10 m = 1000 cm
So, volume of each cylinder = \(\pi\)r²h
\(=\frac{22}{7} \times 20 \times 20 \times 1000 \mathrm{~cm}^{3}\)
\(=\frac{8800000}{7} \mathrm{~cm}^{3}\)
\(\left.=\frac{8.8}{7} \mathrm{~m}^{3} \text { (Since } 1000000 \mathrm{~cm}^{3}=1 \mathrm{~m}^{3}\right)\)
Therefore, volume of 14 pillars = volume of each cylinder x 14
\(=\frac{8.8}{7} \times 14 \mathrm{~m}^{3}\)
= 17.6 m³
So, 14 pillars would need 17.6 m³ of concrete mixture.

99

5400

6 days

130 m

\(\frac { 8712 }{ 7 } { cm }^{ 2 }\)

21.6 m

28 cm

126 cm²

r_cone = 3 cm
h_cone = 4 cm
l_cone= 5 cm

Above given cone is formed with radius 3 cm, height 4 cm and slant height 5 cm when revolved about the fixed side of 4 cm.
\(V=\frac { 1 }{ 3 } \pi { r }^{ 2 }h\)
\(=\frac { 1 }{ 3 } .\frac { 22 }{ 7 } .(3)(3)(4)\)
= 37.71 cm³
Total surface area=\(\pi\)rl+\(\pi\)r²
\(=\frac { 22 }{ 7 } \times 3(5+3)\)
= 75.43 cm²

Radius of the wall (r) =  \(\frac{10}{2}\) m = 5 m
Depth of the wall (h )= 10 m
ஃ  Volume of the earth dug out
= Volume of the well
=\(\pi r^2h=\pi (5)^2(10)=250\pi m^3\)
Radius of the well with embankment (R)
= 5 + 5 = 10 m
ஃ  Area of the embankment
= Area of the well with embankment-Area of the well without embankment
\(=\pi R^2-\pi r^2=\pi (R+r)(R-r)\)
\(= \pi(10+5)(10-5)=75 \pi m^2\)
ஃ Height of the embankment  =\(\frac{Volume\ of \ the\ earth\ dug\ out }{Area \ of \ the \ embankment}\)
\(=\frac{250 \pi}{75\pi}=\frac{10}{3}m\)

Length of the pipe (h) = 21 dm =210 cm
∵  Outer diameter = 10 cm
ஃ  Outer radius (R) =\(\frac{10}{2}\)cm = 5 cm
ஃ  Inner diameter =6 cm
ஃ  Inner radius (r) = \(\frac{6}{2}\) cm = 3 cm
Volume of copper used in making the pipe
= Volume of the outer cylinder- Volume of the inner cylinder
\(=\pi R^2h-\pi r^2h\)
\(=\frac{22}{7}\times(5)^2\times210-\frac{22}{7}\times(3)^2\times210\)
\(=\frac{22}{7}\times210\times(5)^2-(3)^2\)
\(=\frac{22}{7}\times210\times16=10560\ cm^3\)

For cone
Base radius (r) = 8 m
Height (h)        = 6 m
\(\therefore \)  Slant height (l) = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { { \left( 8 \right) }^{ 2 }+{ \left( 6 \right) }^{ 2 } } =10m\)
\(\therefore \)  Curved surface area = \(\pi rl\)
\(=\pi \left( 8 \right) \left( 10 \right) \)
\(\\ =80\pi { m }^{ 2 }\)
For cylinder
Base radius (R) = 8 m
Height (H) = 14 m
\(\therefore \)  Curved surface area = \(2\pi RH\)
 \(=2\pi \left( 8 \right) \left( 14 \right) \)
\(\\ =224\pi { m }^{ 2 }\)
\(\therefore \)  Total curved surface area = Curved surface area of the cone + Curved surface area of the cylinder
\(=80\pi +224\pi =304\pi { m }^{ 2 }\)
\(\\ =304\times 3.14{ m }^{ 2 }=954.56{ m }^{ 2 }\ \)
\(\therefore \)  Cost of canvas = 954.56 \(\times\) 50
= Rs 47728

Let the outer radius be R and inner radius be r.
Outer Surface Area-Inner Surface Area = 44 cm²
2\(\pi\)Rh - 2\(\pi\)rh = 44
2\(\pi\)\(\times\)14(R-r) = 44
\(\therefore \ R-r=\frac { 44\times 7 }{ 2\times 22\times 14 } \)
\(=\frac { 1 }{ 2 } cm\quad \quad .....(i)\)
Volume of metal = 99 cm³
\(\pi\)R²h -\(\pi\)r²h = 99
14\(\pi\)(R² - r²) = 99
\({ R }^{ 2 }-{ r }^{ 2 }=\frac { 99\times 7 }{ 22\times 14 } \)
\(=\frac { 9 }{ 4 } { cm }^{ 2 }\)
\((R+r)(R-r)=\frac { 9 }{ 4 } \)
\(R+r=\frac { 9 }{ 4 } \times 2\)
\(=\frac { 9 }{ 2 } \quad \quad \quad .......(ii)\)
Adding (i) and (ii), we get
\(2R=\frac { 10 }{ 2 } \)
\(\therefore \ R=\frac { 5 }{ 2 } \)
\(r=\frac { 9 }{ 2 } -\frac { 5 }{ 2 } \)
\(=\frac { 4 }{ 2 } =2\)
\(\therefore\) Outer radius = 2.5 cm and inner radius = 2 cm.

Let r and h be the radius and height of the cylinder given r : h = 5 : 7
\(\therefore\) The radius of the cylinder(r) = 5x
and The height of the cylinder (h) = 7x
Volume of the cylinder = \(\pi\)r²h
\(\therefore\) 4400=\(\frac { 22 }{ 7 } \times { (5x) }^{ 2 }\times 7x\)
\(\left( \because \pi =\frac { 22 }{ 7 } \right) \)
\(\Rightarrow \ 4400=\frac { 22 }{ 7 } \times 5x\times 5x\times 7x\)
\(\Rightarrow \ { x }^{ 3 }=\frac { 4400\times 7 }{ 22\times 5\times 5\times 7 } \)
\(\Rightarrow\) x³ = 8 = 2³
\(\therefore\) x = 2
Hence, the radius of the cylinder = 5x
= 5(2)
= 10 cm.

\(\frac { 2673 }{ 14 } c{ m }^{ 3 }\)

Required number \(=\frac { 60\times 30\times 30 }{ 15\times 6\times 4 } =150\)

Let r and R be the radii of the smaller and larger spheres respectively, we have
\(r=\frac { 5 }{ 2 } cm\)
Volume of the smaller sphere \(=\frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi { \left( \frac { 5 }{ 2 } \right) }^{ 3 }\)
\(=\frac { 4 }{ 3 } \times \pi \times \frac { 125 }{ 8 } { cm }^{ 3 }\)
Density of metal\(=\frac { mass }{ Valume } \)
\(=\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } g\quad { cm }^{ 3 }\)      ...........(i)
Volume of larger sphere = \(\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
Density of metal=\(\frac { mass }{ Volume } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)    ......(ii)
From (i) and (ii), we have
\(\frac { 740 }{ \frac { 4 }{ 3 } \pi \times \frac { 125 }{ 8 } } =\frac { 5920 }{ \frac { 4 }{ 3 } \pi { R }^{ 3 } } \)
\(\Rightarrow \quad { R }^{ 3 }=\frac { 5920\times 125 }{ 740\times 8 } \)
= 125
\(\Rightarrow\) R = 5 cm.