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 r be the radius of the hemisphere.
Given that, base area = \(\pi\)r² = 1386 sq. m
T.S.A. = 3 \(\pi\)r² sq.m
= 3 x 1386 = 4158
Therefore, T.S.A. of the hemispherical solid is 4158 m².

Let r and h be the radius and height of the cone respectively.
Given that, radius r = 7 m and height h = 24 m
Hence, l = \(\sqrt { { r }^{ 2 }+{ h }^{ 2 } } \)
\(=\sqrt { 49+576 } \)
\(l=\sqrt { 625 } =25m\)
C.S.A. of the conical tent = \(\pi\)rl sq. units
Area of the canvas \(=\frac { 22 }{ 7 } \times 7\times 25={ 550 }m^{ 2 }\)
Now, length of the canvas \(\frac{Area\ of\ the\ canvas}{width}=\frac{550}{4}=137.5m\)
Therefore, the length of the canvas is 137.5 m

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

Let h, r and R be the height, top and bottom radii of the frustum.
Given that, h = 45 cm, R = 28 cm, r = 7 cm
Now, Volume \(=\frac { 1 }{ 3 } \pi \left[ { R }^{ 2 }+Rr+{ r }^{ 2 } \right] h\quad cu.units\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times \left[ { 28 }^{ 2 }+(28\times 7)+{ 7 }^{ 2 } \right] \times 45\)
\(=\frac { 1 }{ 3 } \times \frac { 22 }{ 7 } \times 1029\times 45=48510\)
Therefore, volume of the frustum is 48510 cm³