Let r₁ and h₁ be the radius and height of the cone - I and let r₂ and h₂ be the radius and height of the cone-II.
Given h₂ = 2h₁ = 2 and \(\frac { Volume\ of\ the\ cone\ I }{ Volume\ of\ the\ cone\ II } =\frac { 2 }{ 3 } \)
\(\frac { \frac { 1 }{ 3 } { \pi r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \frac { 1 }{ 3 } { \pi r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } \times \frac { { h }_{ 1 } }{ 2{ h }_{ 2 } } =\frac { 2 }{ 3 } \)
\(\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } =\frac { 4 }{ 3 } \text {gives} \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 2 }{ \sqrt { 3 } } \)
Therefore, ratio of their radii = 2 : \(\sqrt3\)

Let the radius of the sphere be 'r' cm
Surface area = \(4 \pi r^{2}\)
when radius is increased by 25% ,then new diameter = r + 25% + r
\(=r+\frac{25 r}{100}=\frac{5 r}{4}\)
Surface area of new sphere
\(=4 \pi\left(\frac{5 r}{4}\right)^{2} \)
\(=4 \pi\left(\frac{25 r^{2}}{16}\right) \)
\(=\frac{25 \pi r^{2}}{4} \)
Increase in surface area = \(\frac{25 \pi r^{2}}{4}-4 \pi r^{2}\)
\(=\frac{25 \pi r^{2}-16 \pi r^{2}}{4} \)
\(=\frac{9 \pi r^{2}}{4} \)
Percentage increase in surface area
\(=\frac{9 \pi r^{2} / 4}{4 \pi r^{2}} \times 100 \% \)
\(=\frac{900}{16} \%=56.25 \% \)

Let l, R and r be the slant height, top radius and bottom radius of the frustum.
Given that, l = 5 cm, R = 4 cm, r = 1 cm
Now, C.S.A. of the frustum \(\pi\)(R + r)l sq.units
\(\frac { 22 }{ 7 } \times (4+1)\times 5\)
\(=\frac { 550 }{ 7 } \)
Therefore, C.S.A. = 78.57 cm²

Let r be the radius of the sphere. Given that, surface area of sphere = 154 m²
4\(\pi\)r² = 154
\(4\times \frac { 22 }{ 7 } \times { r }^{ 2 }=154\)
gives \({ r }^{ 2 }=154\times \frac { 1 }{ 4 } \times \frac { 7 }{ 22 } \)
hence, \({ r }^{ 2 }=\frac { 49 }{ 4 } \)We get r = \(\frac{7}{2}\)
Therefore, diameter is 7 m

Given that, height h = 25 Given that, height h = 25 m; thickness = 2 m.
internal radius r = 6 m
Now, external radius R = 6 + 2 = 8m
C.S.A. of the cylindrical tunnel = C.S.A. of the hollow cylinder
C.S.A. of the hollow cylinder = 2\(\pi\)(R + r)h sq.units 
\(2\times \frac { 22 }{ 7 } (8+6)\times 25\)
Hence, C.S.A. of the cylindrical tunnel = 2200 m²
Area covered by one litre of paint = 10 m²
Number of litres required to paint the tunnel \(=\frac { 2200 }{ 10 } =220\)
Therefore, 220 litres of paint is needed to paint the tunnel.

Given that, diameter d = 2.8 m and height = 3 m
radius r = 1.4 m
Area covered in one revolution = curved surface area of the cylinder
= 2\(\pi\)rh sq. units
\(2\times \frac { 22 }{ 7 } \times 1.4\times 3=26.4\)
Area covered in 1 revolution = 26.4 m²
Area covered in 8 revolutions = 8 x 26.4 = 211.2
Therefore, area covered is 211.2 m²

Given that, height of the cylinder h = 20 cm ; radius r =14 cm
Now, C.S.A. of the cylinder = 2p\(\pi\)h sq. units
C.S.A. of the cylinder = \(2\times \frac { 22 }{ 7 } \times 14\times 20=2\times 22\times 2\times 20\)
T.S.A. of the cylinder \(=2\pi r(h+r)\)sq.units
\(=2\times \frac { 22 }{ 7 } \times 14\times (20+14)=2\times \frac { 22 }{ 7 } \times 14\times 34\)
= 2992 cm²
Therefore, C.S.A. = 1760 cm² and T.S.A. = 2992 cm²

Let the number of small spheres obtained be n.
Let r be the radius of each small sphere and R be the radius of metallic sphere.
Here, R = 16 cm, r = 2 cm
Now, n\(\times\)(Volume of a small sphere) = Volume of big metallic sphere
\(n\left( \frac { 4 }{ 3 } { \pi r }^{ 2 } \right) =\frac { 4 }{ 3 } \pi { R }^{ 3 }\)
\(n\left( \frac { 4 }{ 3 } \pi \times { 2 }^{ 3 } \right) =\frac { 4 }{ 3 } \pi \times { 16 }^{ 3 }\)
8n = 4096 gives n = 512
Therefore, there will be 512 small spheres.

Radius of hemisphere = 6 cm
Volume of hemisphere \(=\frac{2}{3} \pi r^{3} \text { cu. units }\)
\(=\frac{2}{3} \pi(6)^{3} \)
\(=\frac{2}{3} \pi(216) \)
\(=144 \pi \mathrm{cm}^{3} \)
base of cone = 6 cm
Height of the cone = 12 cm
Volume of the cone \(=\frac{1}{3} \pi r^{2} h \text { cu. units } \)
\(=\frac{1}{3} \pi(6)^{2}(12) \)
= 144 cm³
volume of the solid = Volume of cone + Volume of hemisphere
\(=144 \pi+144 \pi=288 \pi\)
Volume of water displaced
= Volume of the solid placed in the cylinder
\(=288 \pi=288 \times \frac{22}{7}\)
= 905.14 cm³

Let R, r be the top and bottom radii of the frustum.
Let h₁, h₂ be the heights of the frustum and cylinder respectively.
Given that, R = 12 cm, r = 6 cm, h₂ = 12 cm
Now, h₁ = 20 – 12 = 8 cm
Here, Slant height of the frustum l = \(\sqrt { \left( R-r \right) ^{ 2 }+{ h }_{ 1 }^{ 2 } } units\)
\(=\sqrt { 36+64 } \)
l = 10 cm
Outer surface area = \(2\pi r{ h }_{ 2 }+\pi (R+r)\quad l\quad sq.units\)
\(=\pi [2r{ h }_{ 2 }+(R+r)l]\)
\(=\pi [(2\times 6\times 12)+(18\times 10)]\)
\(=\pi [144+180]\)
\(=\frac { 22 }{ 7 } \times 324=1018.28\)
Therefore, outer surface area of the funnel is 1018.28 cm².

Let r and h be the radius and height of the cylinder respectively.
Given that, volume of the tank = 1.078 x 10⁶ = 1078000 litre
1078 m³ (since 1l = \(\frac{1}{1000}m^3\))
diameter = 7m gives radius = \(\frac{7}{2}\)m
volume of the tank = \(\pi\)r h 2 cu. units
1078 = \(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times h\)
Therefore, height of the tank is 28 m

Given total surface area of cylinder
= 1848 sq.m
(h + r) = 1848
It is given that C.S.A \(=\frac{5}{6}(\text { T.S.A })\)
\(\text { C.S.A }=\frac{5}{6}(1848)=1540 \)
\(\text { C.S.A }=\frac{5}{6}(\text { T.S.A }) \)
\(2 \pi r h=\frac{5}{6}(2 \pi r(h+r)) \)
h = 5r
We have C.S.A = 1540
2rh = 1540
\(2 \times \frac{22}{7} \times r \times 5 r =1540 \)
\(r^{2} =\frac{1540 \times 7}{44 \times 5}=49 \)
r = 7
h = 5r = 5(7) = 35
radius = 7 m , height = 35 m

Let the internal and external radii of the hemispherical shell be r and R
respectively.
Given that, R = 5 m, r = 3 m
C.S.A. of the shell = 2\(\pi\)(R² + r²) sq. units
\(=2\times \frac { 22 }{ 7 } \times \left( 25+9 \right) =213.71\)
T.S.A. of the shell = \(\pi\)(3R² + r²) sq. units
\(=\frac { 22 }{ 7 } (75+9)=264\)
Therefore, C.S.A. = 213.71 m² and T.S.A. = 264 m².

Let h and r be the height and radius of the cone and cylinder.
Let l be the slant height of the cone.
Given that, h = 2.4 cm and d = 1.4 cm ; r = 0.7 cm
Here, total surface area of the remaining solid} C.S.A. of the cylinder + C.S.A. of the cone + area of the bottom
= 2\(\pi\)rh + \(\pi\)rl + \(\pi\)r² sq.units
Now, \(\\ \\ l=\sqrt { { r }^{ 2 }+{ h }^{ 2 } } =\sqrt { 0.49+5.76 } =\sqrt { 6.25 } =2.5cm\)
Area of the remaining solid = 2\(\pi\)rh + \(\pi\)rl + \(\pi\)r² sq.units
= \(\pi\)r(2h + l + r)
\(\frac { 22 }{ 7 } \times 0.7\times [(2\times 2.4)+2.5+0.7]\)
Therefore, total surface area of the remaining solid is 17.6 m²