Volume of cylinder = \(\pi\)r²h cu. units
Volume of cylinder \(A=\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 } \times 211\)
= 808.5cm³
Volume of cylinder B \(=\frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times \frac { 21 }{ 2 } \times 7\)
= 2425.5 cm³
Therefore, volume of cylinder B is greater than volume of cylinder A.
(ii) T.S.A. of cylinder = 2\(\pi\)r(h + r) sq. units
T.S.A. of cylinder \(A=2\times\frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times (21+3.5)\) = 539 cm²
T.S.A. of cylinder \(B=2\times \frac { 22 }{ 7 } \times \frac { 21 }{ 2 } \times (7+10.5)\) = 1155 cm²
Hence verified that cylinder B with greater volume has a greater surface area.
(iii) \(\frac { Volume\ of\ cylinderA }{ Volume\ of\ cylinder\ B } =\frac { 808.5 }{ 2425.5 } =\frac { 1 }{ 3 } \)
Therefore, ratio of the volumes of cylinders A and B is 1:3.

Let r, R and h be the internal radius, external radius and height of the hollow cylinder respectively.
Given that, r = 21cm, R = 28 cm, h = 9 cm
Now, volume of hollow cylinder = \(\pi\)(R² − r²)h cu. units
\(=\frac { 22 }{ 7 } \left( { 28 }^{ 2 }-21^{ 2 } \right) \times 9\)
\(=\frac { 22 }{ 7 } (784-441)\times 9=9702\)
Therefore, volume of iron used = 9702 cm³

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

From the figure
r = 6 cm
R = 12 cm
h = 8 cm
\(l =\sqrt{h^{2}+(\mathrm{R}-\mathrm{r})^{2}} \)
\(=\sqrt{8^{2}+(12-6)^{2}} \)
\(=\sqrt{64+36} \)
\(=\sqrt{100}=10 \mathrm{~cm} \)
Area to be painted = C.S.A + area of top circular region
\(=\pi(R+r) l+\pi r^{2} \)
\(=\frac{22}{7}(12+6)(10)+\frac{22}{7}(6)^{2} \)
\(=\frac{22}{7}(180)+\frac{22}{7}(36) \)
\(=\frac{22}{7}(180+36) \)
\(=\frac{22}{7}(216)=\frac{4752}{7}=678.86 \)
Cost of painting per sq. cm = Rs. 2
Total cost = 678.86 x 2 = Rs. 1357.72

Internal diameter = 20 cm
External diameter = 28 cm
Internal radius = 10 cm
External radius = 14 cm
Total surface area \(=\pi\left(3 \mathrm{R}^{2}+\mathrm{r}^{2}\right) \text { sq. units } \)
\(=\frac{22}{7}\left(3(14)^{2}+(10)^{2}\right) \)
\(=\frac{22}{7}[588+100] \)
\(=\frac{22}{7} \times 688 \)
\(=\frac{15136}{7} \mathrm{~cm}^{2} \)
Cost of painting per sq.cm = Rs 0.14
Total cost \(=\frac{15136}{7} \times 0.14\)
= Rs. 302.72

Area of the paper = 5720 cm²
Given radius of birthday cap r = 5 cm
height of birthday cap 'h' = 12 cm
slant height \(l=\sqrt{h^{2}+r^{2}} \)
\(=\sqrt{12^{2}+5^{2}}=\sqrt{144+25} \)
\(=\sqrt{169} \ =13 \mathrm{~cm} \)
CSA of conical cap = \(\pi r l\) sq. units
\(=\frac{22}{7} \times 5 \times 13=\frac{1430}{7}\)
Number of birthday caps
\(=\frac{\text { Area of paper sheet }}{\text { CSA of conical cap }} \)
\(=\frac{5720}{1430} \times 7=28 \text { caps } \)

Right triangle PQR, right angled at Q and
PR = 20 cm, QR = 16 cm
PQ² = PR² - QR²
= (20)² - (16)²
= 400 - 256 = 144
PQ = 12 cm

When right triangle PQR, rotates about QR, a right circular cone is formed with PQ = 12 cm as base radius and PR = 20 cm as
slant height.
C.S.A of the Cone = \(\pi r l\) sq. units
\(=\frac{22}{7} \times 12 \times 20=754.29 \mathrm{~cm}^{2}\)
When right triangle PQR, rotates about PQR, a right circular cone is formed with
QR = 16 cm as base radius and PR = 20 cm as slant height

C.S.A of the Cone \(=\pi r l \text { sq.units } \)
\(=\frac{22}{7} \times 16 \times 20 \)
= 1005.71 cm²
Hence, C.S.A of the cone when rotates about PQ is larger.

Let h, l, R and r be the height, slant height, outer radius and inner radius of the frustum.
Given that, diameter of the top = 10 m; radius of the top R = 5 m.
diameter of the bottom = 4 m; radius of the bottom r = 2 m, height h = 4 m
Now, \(l=\sqrt { { h }^{ 2 }+\left( R-{ r } \right) ^{ 2 } } \)
\(=\sqrt { { 4 }^{ 2 }+(5-2)^{ 2 } } \)
\(l=\sqrt { 16+9 } =\sqrt { 25 } =5m\)
Here, C.S.A. = \(\pi\)(R + r)l sq. units
\(\frac { 22 }{ 7 } (5+2)\times 5={ 110m }^{ 2 }\)
T.S.A. = \(\pi\)(R + r)l + \(\pi\)R² + \(\pi\)r² sq. units
\(\frac { 22 }{ 7 } \left[ (5+2)5+25+4 \right] =\frac { 1408 }{ 7 } =201.14\)
Therefore, C.S.A. = 110 m² and T.S.A. = 201.14 m²

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