Textbook - Surface Areas and Volumes Questions in English

(A) The total surface area to be coloured is the sum of the curved surface area of the hemisphere and the curved surface area of the cone.
Given:
Diameter of the top $= 3.5 \, cm$
Radius $(r) = \frac{3.5}{2} = 1.75 \, cm$
Total height of the top $= 5 \, cm$
Height of the cone $(h) = \text{Total height} - \text{Radius of hemisphere} = 5 - 1.75 = 3.25 \, cm$
Slant height of the cone $(l) = \sqrt{r^2 + h^2} = \sqrt{(1.75)^2 + (3.25)^2} = \sqrt{3.0625 + 10.5625} = \sqrt{13.625} \approx 3.69 \, cm \approx 3.7 \, cm$
Curved surface area of hemisphere $= 2 \pi r^2 = 2 \times \frac{22}{7} \times 1.75 \times 1.75 = 19.25 \, cm^2$
Curved surface area of cone $= \pi r l = \frac{22}{7} \times 1.75 \times 3.7 = 20.35 \, cm^2$
Total area to be coloured $= 19.25 + 20.35 = 39.6 \, cm^2$.

(B) The total surface area of the cube $= 6 \times (\text{edge})^2 = 6 \times 5 \times 5 \, cm^2 = 150 \, cm^2$.
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 $= \text{TSA of cube} - \text{base area of hemisphere} + \text{CSA of hemisphere}$.
$= 150 - \pi r^2 + 2\pi r^2 = (150 + \pi r^2) \, cm^2$.
Given diameter $= 4.2 \, cm$,so radius $r = 2.1 \, cm$.
$= 150 + \left(\frac{22}{7} \times 2.1 \times 2.1\right) \, cm^2$.
$= 150 + (22 \times 0.3 \times 2.1) \, cm^2 = 150 + 13.86 \, cm^2 = 163.86 \, cm^2$.

(C) Let the radius of the cone be $r$, slant height of the cone be $l$, height of the cone be $h$, radius of the cylinder be $r^{\prime}$, and height of the cylinder be $h^{\prime}$.
Given: $r = 2.5\, cm$, $h = 6\, cm$, $r^{\prime} = 1.5\, cm$, and $h^{\prime} = 26 - 6 = 20\, cm$.
The slant height $l = \sqrt{r^2 + h^2} = \sqrt{2.5^2 + 6^2} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5\, cm$.
The conical portion has its circular base resting on the base of the cylinder. Since the base of the cone is larger than the base of the cylinder, a ring-shaped area at the base of the cone is also to be painted orange.
Area to be painted orange = ($CSA$ of the cone) + (Base area of the cone - Base area of the cylinder)
$= \pi r l + (\pi r^2 - \pi (r^{\prime})^2)$
$= 3.14 \times 2.5 \times 6.5 + 3.14 \times (2.5^2 - 1.5^2)$
$= 51.025 + 3.14 \times (6.25 - 2.25)$
$= 51.025 + 3.14 \times 4 = 51.025 + 12.56 = 63.585\, cm^2$.
Area to be painted yellow = ($CSA$ of the cylinder) + (Area of the bottom base of the cylinder)
$= 2 \pi r^{\prime} h^{\prime} + \pi (r^{\prime})^2$
$= \pi r^{\prime} (2h^{\prime} + r^{\prime})$
$= 3.14 \times 1.5 \times (2 \times 20 + 1.5)$
$= 4.71 \times 41.5 = 195.465\, cm^2$.
Thus, the area painted orange is $63.585\, cm^2$ and the area painted yellow is $195.465\, cm^2$.

(D) Let $h$ be the height of the cylinder and $r$ be the common radius of the cylinder and the hemisphere.
Given: $h = 1.45\, m = 145\, cm$ and $r = 30\, cm$.
The total surface area of the bird-bath is the sum of the curved surface area of the cylinder and the curved surface area of the hemisphere.
Total Surface Area $= \text{CSA of cylinder} + \text{CSA of hemisphere}$
$= 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$
$= 2 \times \frac{22}{7} \times 30 \times (145 + 30)\, cm^2$
$= 2 \times \frac{22}{7} \times 30 \times 175\, cm^2$
$= 2 \times 22 \times 30 \times 25\, cm^2$
$= 33000\, cm^2$
Since $1\, m^2 = 10000\, cm^2$,we have:
Total Surface Area $= \frac{33000}{10000}\, m^2 = 3.3\, m^2$.

(A) Given that,the volume of each cube is $64 \, cm^3$.
Let the edge of the cube be $a$.
Then,$a^3 = 64 \, cm^3$.
$a = \sqrt[3]{64} = 4 \, cm$.
When two such cubes are joined end to end,they form a cuboid.
The dimensions of the resulting cuboid are:
Length $(l)$ = $4 \, cm + 4 \, cm = 8 \, cm$
Breadth $(b)$ = $4 \, cm$
Height $(h)$ = $4 \, cm$
The surface area of a cuboid is given by the formula: $2(lb + bh + lh)$.
Surface area = $2(8 \times 4 + 4 \times 4 + 8 \times 4)$
$= 2(32 + 16 + 32)$
$= 2(80)$
$= 160 \, cm^2$.
Thus,the surface area of the resulting cuboid is $160 \, cm^2$.

(B) It can be observed that the radius $(r)$ of the cylindrical part and the hemispherical part is the same (i.e.,$7 \, cm$).
Height of the hemispherical part = Radius $= 7 \, cm$.
Height of the cylindrical part $(h) = 13 - 7 = 6 \, cm$.
Inner surface area of the vessel = $CSA$ of the cylindrical part + $CSA$ of the hemispherical part.
Inner surface area $= 2 \pi r h + 2 \pi r^2 = 2 \pi r (h + r)$.
Inner surface area $= 2 \times \frac{22}{7} \times 7 \times (6 + 7)$.
Inner surface area $= 44 \times 13 = 572 \, cm^2$.

(C) It can be observed that the radius of the conical part and the hemispherical part is the same (i.e.,$r = 3.5 \, cm = \frac{7}{2} \, cm$).
Height of the hemispherical part = Radius $(r) = 3.5 \, cm = \frac{7}{2} \, cm$.
Height of the conical part $(h) = 15.5 \, cm - 3.5 \, cm = 12 \, cm$.
Slant height $(l)$ of the conical part is given by $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{\left( \frac{7}{2} \right)^2 + (12)^2} = \sqrt{\frac{49}{4} + 144} = \sqrt{\frac{49 + 576}{4}} = \sqrt{\frac{625}{4}} = \frac{25}{2} \, cm = 12.5 \, cm$.
Total surface area of the toy = $CSA$ of the conical part + $CSA$ of the hemispherical part.
Total surface area = $\pi r l + 2 \pi r^2 = \pi r (l + 2r)$.
Total surface area = $\frac{22}{7} \times \frac{7}{2} \times \left( 12.5 + 2 \times 3.5 \right) = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5 \, cm^2$.

(D) From the figure,it can be observed that the greatest diameter possible for such a hemisphere is equal to the cube's edge,i.e.,$7 \, cm$.
Radius $(r)$ of the hemispherical part $= \frac{7}{2} = 3.5 \, cm$.
Total surface area of the solid $=$ Surface area of the cubical part $+$ $CSA$ of the hemispherical part $-$ Area of the base of the hemispherical part.
Total surface area $= 6(\text{Edge})^2 + 2\pi r^2 - \pi r^2 = 6(\text{Edge})^2 + \pi r^2$.
Total surface area of the solid $= 6(7)^2 + \frac{22}{7} \times 3.5 \times 3.5$.
$= 6(49) + \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} = 294 + 38.5 = 332.5 \, cm^2$.

(N/A) Diameter of hemisphere $=$ Edge of cube $= l$
Radius of hemisphere $r = \frac{l}{2}$
Total surface area of the remaining solid $=$ Total surface area of the cube $+$ Curved surface area of the hemisphere $-$ Area of the circular base of the hemisphere
$= 6(\text{Edge})^2 + 2\pi r^2 - \pi r^2$
$= 6l^2 + \pi r^2$
Substituting $r = \frac{l}{2}$:
$= 6l^2 + \pi \left(\frac{l}{2}\right)^2$
$= 6l^2 + \frac{\pi l^2}{4}$
$= \frac{24l^2 + \pi l^2}{4}$
$= \frac{l^2}{4}(24 + \pi) \text{ unit}^2$

(B) It can be observed that:
Radius $(r)$ of the cylindrical part $=$ Radius $(r)$ of the hemispherical part $= \frac{\text{Diameter of the capsule}}{2} = \frac{5}{2} \, mm = 2.5 \, mm$.
Length of the cylindrical part $(h) = \text{Length of the entire capsule} - 2 \times r$
$= 14 \, mm - 2 \times (2.5 \, mm) = 14 \, mm - 5 \, mm = 9 \, mm$.
Surface area of the capsule $= 2 \times \text{CSA of hemispherical part} + \text{CSA of cylindrical part}$
$= 2 \times (2 \pi r^2) + 2 \pi r h$
$= 4 \pi r^2 + 2 \pi r h$
$= 2 \pi r (2r + h)$
$= 2 \times \frac{22}{7} \times 2.5 \times (2 \times 2.5 + 9)$
$= 2 \times \frac{22}{7} \times 2.5 \times (5 + 9)$
$= 2 \times \frac{22}{7} \times 2.5 \times 14$
$= 2 \times 22 \times 2.5 \times 2$
$= 220 \, mm^2$.

(C) Given that,
Height $(h)$ of the cylindrical part $= 2.1 \, m$
Diameter of the cylindrical part $= 4 \, m$
Radius $(r)$ of the cylindrical part $= 2 \, m$
Slant height $(l)$ of the conical part $= 2.8 \, m$
Area of canvas used = $CSA$ of conical part + $CSA$ of cylindrical part
$= \pi rl + 2\pi rh$
$= \pi r(l + 2h)$
$= \frac{22}{7} \times 2 \times (2.8 + 2 \times 2.1)$
$= \frac{44}{7} \times (2.8 + 4.2)$
$= \frac{44}{7} \times 7 = 44 \, m^2$
Cost of $1 \, m^2$ canvas $= Rs. \, 500$
Cost of $44 \, m^2$ canvas $= 44 \times 500 = Rs. \, 22000$
Therefore,the cost of the canvas for making the tent is $Rs. \, 22000$.

(D) Given that,
Height $(h)$ of the conical part $=$ Height $(h)$ of the cylindrical part $= 2.4 \, cm$
Diameter of the cylindrical part $= 1.4 \, cm$
Therefore,radius $(r)$ of the cylindrical part $= 0.7 \, cm$
Slant height $(l)$ of the conical part $= \sqrt{r^2 + h^2}$
$= \sqrt{(0.7)^2 + (2.4)^2} = \sqrt{0.49 + 5.76}$
$= \sqrt{6.25} = 2.5 \, cm$
Total surface area of the remaining solid $= \text{CSA of cylindrical part} + \text{CSA of conical part} + \text{Area of cylindrical base}$
$= 2 \pi rh + \pi rl + \pi r^2$
$= 2 \times \frac{22}{7} \times 0.7 \times 2.4 + \frac{22}{7} \times 0.7 \times 2.5 + \frac{22}{7} \times 0.7 \times 0.7$
$= 4.4 \times 2.4 + 2.2 \times 2.5 + 2.2 \times 0.7$
$= 10.56 + 5.50 + 1.54 = 17.60 \, cm^2$
The total surface area of the remaining solid to the nearest $cm^2$ is $18 \, cm^2$.

(A) Given that,
Radius $(r)$ of the cylindrical part $=$ Radius $(r)$ of the hemispherical part $= 3.5 \, cm$.
Height of the cylindrical part $(h) = 10 \, cm$.
The total surface area of the article is the sum of the curved surface area of the cylinder and the curved surface areas of the two hemispheres.
Total Surface Area $= \text{CSA of cylinder} + 2 \times \text{CSA of hemisphere}$
$= 2 \pi r h + 2 \times (2 \pi r^2)$
$= 2 \pi r h + 4 \pi r^2$
$= 2 \pi r (h + 2r)$
$= 2 \times \frac{22}{7} \times 3.5 \times (10 + 2 \times 3.5)$
$= 2 \times \frac{22}{7} \times 3.5 \times (10 + 7)$
$= 2 \times \frac{22}{7} \times 3.5 \times 17$
$= 44 \times 0.5 \times 17$
$= 22 \times 17 = 374 \, cm^2$.

(B) The volume of air inside the shed (when there are no people or machinery) is given by the sum of the volume of the cuboid and the volume of the half cylinder.
The dimensions of the cuboid are $length = 15\,m,$ $breadth = 7\,m,$ and $height = 8\,m.$
Volume of cuboid $= length \times breadth \times height = 15 \times 7 \times 8 = 840\,m^3.$
The half cylinder has a diameter of $7\,m,$ so its radius $r = 3.5\,m = \frac{7}{2}\,m.$ The length of the cylinder is $15\,m.$
Volume of half cylinder $= \frac{1}{2} \times (\pi r^2 h) = \frac{1}{2} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 15 = 288.75\,m^3.$
Total volume of the shed $= 840 + 288.75 = 1128.75\,m^3.$
Space occupied by machinery $= 300\,m^3.$
Space occupied by $20$ workers $= 20 \times 0.08 = 1.6\,m^3.$
Total space occupied $= 300 + 1.6 = 301.6\,m^3.$
Volume of air in the shed $= 1128.75 - 301.6 = 827.15\,m^3.$

(C) Given,inner diameter of the cylindrical glass $= 5 \, cm$,so radius $r = 2.5 \, cm$. Height $h = 10 \, cm$.
The apparent capacity of the glass is the volume of the cylinder:
Apparent capacity $= \pi r^2 h = 3.14 \times (2.5)^2 \times 10 = 3.14 \times 6.25 \times 10 = 196.25 \, cm^3$.
The actual capacity is reduced by the volume of the hemispherical raised portion at the base:
Volume of hemisphere $= \frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times (2.5)^3 = \frac{2}{3} \times 3.14 \times 15.625 = 32.70833... \approx 32.71 \, cm^3$.
Actual capacity $= \text{Apparent capacity} - \text{Volume of hemisphere} = 196.25 - 32.71 = 163.54 \, cm^3$.

(D) Let $BPC$ be the hemisphere and $ABC$ be the cone standing on the base of the hemisphere.
The radius $BO$ of the hemisphere (as well as of the cone) $= \frac{1}{2} \times 4 \, cm = 2 \, cm$.
So,the volume of the toy $= \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h$.
$= \left[ \frac{2}{3} \times 3.14 \times (2)^3 + \frac{1}{3} \times 3.14 \times (2)^2 \times 2 \right] \, cm^3 = 25.12 \, cm^3$.
Now,let the right circular cylinder $EFGH$ circumscribe the given solid. The radius of the base of the right circular cylinder $= HP = BO = 2 \, cm$,and its height is $EH = AO + OP = (2 + 2) \, cm = 4 \, cm$.
So,the required volume difference $=$ volume of the right circular cylinder $-$ volume of the toy.
$= (3.14 \times 2^2 \times 4 - 25.12) \, cm^3$.
$= (50.24 - 25.12) \, cm^3 = 25.12 \, cm^3$.
Hence,the required difference of the two volumes $= 25.12 \, cm^3$.

(N/A) Given that,
Height $(h)$ of the conical part $=$ Radius $(r)$ of the conical part $= 1\, cm$.
Radius $(r)$ of the hemispherical part $=$ Radius of the conical part $(r)$ $= 1\, cm$.
Volume of the solid $=$ Volume of the conical part $+$ Volume of the hemispherical part.
Volume of the solid $= \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3$.
Substituting the values:
Volume of the solid $= \frac{1}{3} \pi (1)^2 (1) + \frac{2}{3} \pi (1)^3$.
Volume of the solid $= \frac{1}{3} \pi + \frac{2}{3} \pi = \frac{3}{3} \pi = \pi\, cm^3$.

(B) From the figure,it can be observed that:
Height $(h_1)$ of each conical part $= 2\, cm$.
Height $(h_2)$ of the cylindrical part $= 12 - (2 \times \text{height of conical part}) = 12 - (2 \times 2) = 12 - 4 = 8\, cm$.
Radius $(r)$ of the cylindrical part $=$ Radius of the conical part $= \frac{3}{2} = 1.5\, cm$.
Volume of air present in the model $=$ Volume of cylinder $+ 2 \times$ Volume of cones.
Volume $= \pi r^2 h_2 + 2 \times (\frac{1}{3} \pi r^2 h_1)$.
Volume $= \pi r^2 (h_2 + \frac{2}{3} h_1) = \pi \times (\frac{3}{2})^2 \times (8 + \frac{2}{3} \times 2)$.
Volume $= \pi \times \frac{9}{4} \times (8 + \frac{4}{3}) = \pi \times \frac{9}{4} \times (\frac{24+4}{3}) = \pi \times \frac{9}{4} \times \frac{28}{3}$.
Volume $= \pi \times 3 \times 7 = 21 \pi$.
Using $\pi = \frac{22}{7}$,Volume $= 21 \times \frac{22}{7} = 3 \times 22 = 66\, cm^3$.

(C) It can be observed that:
Radius $(r)$ of cylindrical part $=$ Radius $(r)$ of hemispherical part $= \frac{2.8}{2} = 1.4\,cm$.
Length of each hemispherical part $=$ Radius of hemispherical part $= 1.4\,cm$.
Length $(h)$ of cylindrical part $= 5 - 2 \times$ (Length of hemispherical part) $= 5 - 2 \times 1.4 = 2.2\,cm$.
Volume of one gulab jamun $=$ Volume of cylindrical part $+ 2 \times$ Volume of hemispherical part
$= \pi r^2 h + 2 \times (\frac{2}{3} \pi r^3) = \pi r^2 h + \frac{4}{3} \pi r^3$
$= \frac{22}{7} \times (1.4)^2 \times 2.2 + \frac{4}{3} \times \frac{22}{7} \times (1.4)^3$
$= 13.552 + 11.49866... \approx 25.05\,cm^3$.
Volume of $45$ gulab jamuns $= 45 \times 25.05 = 1127.25\,cm^3$.
Volume of sugar syrup $= 30\%$ of total volume
$= \frac{30}{100} \times 1127.25 = 338.175\,cm^3$.
Rounding to the nearest whole number,the volume is approximately $338\,cm^3$.

(D) The volume of the wooden stand is equal to the volume of the cuboid minus the volume of the four conical depressions.
Dimensions of the cuboid: Length $(l) = 15 \, cm$,Breadth $(b) = 10 \, cm$,Height $(H) = 3.5 \, cm$.
Volume of the cuboid $= l \times b \times H = 15 \times 10 \times 3.5 = 525 \, cm^3$.
Dimensions of each conical depression: Radius $(r) = 0.5 \, cm$,Depth $(h) = 1.4 \, cm$.
Volume of one cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 1.4$.
Volume of four cones $= 4 \times \left( \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 1.4 \right) = 4 \times \left( \frac{1}{3} \times 22 \times 0.25 \times 0.2 \right) = 4 \times \left( \frac{1.1}{3} \right) = \frac{4.4}{3} \approx 1.4667 \, cm^3$.
Volume of wood $= 525 - 1.4667 = 523.5333 \, cm^3$.
Rounding to two decimal places,the volume is $523.53 \, cm^3$.

(A) Height $(h)$ of the conical vessel $= 8 \, cm$.
Radius $(r_1)$ of the conical vessel $= 5 \, cm$.
Radius $(r_2)$ of each lead shot $= 0.5 \, cm$.
Let $n$ be the number of lead shots dropped into the vessel.
According to the problem,the volume of water that flows out is equal to the total volume of the lead shots dropped.
Volume of water spilled $= \frac{1}{4} \times$ Volume of the cone.
$\frac{1}{4} \times (\frac{1}{3} \pi r_1^2 h) = n \times (\frac{4}{3} \pi r_2^3)$.
Canceling $\frac{1}{3} \pi$ from both sides:
$\frac{1}{4} r_1^2 h = 4 n r_2^3$.
$r_1^2 h = 16 n r_2^3$.
Substituting the values:
$5^2 \times 8 = 16 \times n \times (0.5)^3$.
$25 \times 8 = 16 \times n \times (\frac{1}{2})^3$.
$200 = 16 \times n \times \frac{1}{8}$.
$200 = 2n$.
$n = 100$.
Thus,the number of lead shots dropped in the vessel is $100$.

(B) From the figure,it can be observed that:
Height $(h_1)$ of the larger cylinder $= 220 \,cm$
Radius $(r_1)$ of the larger cylinder $= \frac{24}{2} = 12 \,cm$
Height $(h_2)$ of the smaller cylinder $= 60 \,cm$
Radius $(r_2)$ of the smaller cylinder $= 8 \,cm$
Total volume of the pole $=$ Volume of the larger cylinder $+$ Volume of the smaller cylinder
$= \pi r_1^2 h_1 + \pi r_2^2 h_2$
$= \pi(12)^2 \times 220 + \pi(8)^2 \times 60$
$= \pi(144 \times 220 + 64 \times 60)$
$= 3.14 \times (31680 + 3840)$
$= 3.14 \times 35520 = 111532.8 \,cm^3$
Mass of $1 \,cm^3$ of iron $= 8 \,g$
Total mass of the pole $= 111532.8 \times 8 = 892262.4 \,g$
Since $1 \,kg = 1000 \,g$,the mass in $kg$ is $\frac{892262.4}{1000} = 892.2624 \,kg \approx 892.262 \,kg$.

(C) Radius $(r)$ of the hemispherical part = Radius $(r)$ of the conical part = $60\, cm$.
Height $(h_2)$ of the conical part of the solid = $120\, cm$.
Height $(h_1)$ of the cylinder = $180\, cm$.
Radius $(r)$ of the cylinder = $60\, cm$.
The volume of water left in the cylinder = Volume of the cylinder - Volume of the solid.
Volume of the solid = Volume of the cone + Volume of the hemisphere = $\frac{1}{3}\pi r^2 h_2 + \frac{2}{3}\pi r^3$.
Volume of water left = $\pi r^2 h_1 - (\frac{1}{3}\pi r^2 h_2 + \frac{2}{3}\pi r^3)$.
Substituting the values: $\pi(60)^2(180) - (\frac{1}{3}\pi(60)^2 \times 120 + \frac{2}{3}\pi(60)^3)$.
$= \pi(60)^2 [180 - (40 + 40)] = \pi(3600)(100) = 360,000\pi\, cm^3$.
Using $\pi = \frac{22}{7}$,Volume $\approx 360,000 \times 3.14159 = 1,130,973.36\, cm^3$.
Since $1\, m^3 = 1,000,000\, cm^3$,the volume in $m^3$ is $\approx 1.131\, m^3$.

(D) Height $(h)$ of the cylindrical part $= 8 \, cm$.
Radius $(r_{2})$ of the cylindrical part $= \frac{2}{2} = 1 \, cm$.
Radius $(r_{1})$ of the spherical part $= \frac{8.5}{2} = 4.25 \, cm$.
Volume of the vessel $=$ Volume of the sphere $+$ Volume of the cylinder.
Volume $= \frac{4}{3} \pi r_{1}^{3} + \pi r_{2}^{2} h$.
Volume $= \frac{4}{3} \times 3.14 \times (4.25)^{3} + 3.14 \times (1)^{2} \times 8$.
Volume $= \frac{4}{3} \times 3.14 \times 76.765625 + 3.14 \times 8$.
Volume $= 321.392 + 25.12 = 346.512 \, cm^{3}$.
Since the calculated volume is $346.512 \, cm^{3}$,the child's measurement of $345 \, cm^{3}$ is incorrect.

(A) The volume of the cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
Substituting the given values,$r = 6\, cm$ and $h = 24\, cm$:
$V = \frac{1}{3} \times \pi \times 6^2 \times 24 = \frac{1}{3} \times \pi \times 36 \times 24 = 288\pi\, cm^3$.
Let the radius of the sphere be $R$. The volume of the sphere is given by $V = \frac{4}{3} \pi R^3$.
Since the volume of the clay remains the same after reshaping,we equate the two volumes:
$\frac{4}{3} \pi R^3 = 288\pi$.
Dividing both sides by $\pi$ and multiplying by $\frac{3}{4}$:
$R^3 = 288 \times \frac{3}{4} = 72 \times 3 = 216$.
Taking the cube root of both sides:
$R = \sqrt[3]{216} = 6\, cm$.
Thus,the radius of the sphere is $6\, cm$.

(D) The volume of water in the overhead tank equals the volume of water removed from the sump.
Volume of the overhead tank (cylinder) $= \pi r^2 h = 3.14 \times 0.6 \, m \times 0.6 \, m \times 0.95 \, m = 1.07388 \, m^3$.
Volume of the sump (cuboid) $= l \times b \times h = 1.57 \, m \times 1.44 \, m \times 0.95 \, m = 2.14776 \, m^3$.
Volume of water left in the sump $= 2.14776 \, m^3 - 1.07388 \, m^3 = 1.07388 \, m^3$.
Let the height of the water left in the sump be $H$. Since the base area of the sump remains constant,$l \times b \times H = 1.07388 \, m^3$.
$1.57 \, m \times 1.44 \, m \times H = 1.07388 \, m^3$.
$2.2608 \, m^2 \times H = 1.07388 \, m^3$.
$H = \frac{1.07388}{2.2608} \, m = 0.475 \, m = 47.5 \, cm$.
Ratio of capacities $= \frac{\text{Capacity of tank}}{\text{Capacity of sump}} = \frac{1.07388}{2.14776} = \frac{1}{2}$.
Thus,the capacity of the tank is half the capacity of the sump.

(A) The volume of the copper rod is given by the formula for the volume of a cylinder: $V = \pi r^2 h$.
Given diameter $= 1 \,cm$,so radius $r = 0.5 \,cm$. Length $h = 8 \,cm$.
Volume $= \pi \times (0.5)^2 \times 8 = \pi \times 0.25 \times 8 = 2\pi \,cm^3$.
When the rod is drawn into a wire,the volume remains constant.
The length of the new wire is $18 \,m = 1800 \,cm$.
Let the radius of the wire be $r_w$. The volume of the wire is $\pi \times r_w^2 \times 1800$.
Equating the volumes: $\pi \times r_w^2 \times 1800 = 2\pi$.
$r_w^2 = \frac{2\pi}{1800\pi} = \frac{1}{900}$.
$r_w = \sqrt{\frac{1}{900}} = \frac{1}{30} \,cm$.
The thickness (diameter) of the wire is $2 \times r_w = 2 \times \frac{1}{30} = \frac{1}{15} \,cm$.
$\frac{1}{15} \,cm \approx 0.0667 \,cm = 0.67 \,mm$.

(16.5) The radius of the hemispherical tank is $r = \frac{3}{2} \, m = 1.5 \, m$.
The total volume of the hemispherical tank is $V = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (1.5)^3 = \frac{2}{3} \times \frac{22}{7} \times \frac{27}{8} = \frac{99}{14} \, m^3$.
Since $1 \, m^3 = 1000 \, litres$,the total volume in litres is $\frac{99}{14} \times 1000 = \frac{99000}{14} \, litres$.
We need to empty half of the tank,so the volume to be emptied is $\frac{1}{2} \times \frac{99000}{14} = \frac{99000}{28} \, litres$.
The rate of emptying is $3 \frac{4}{7} = \frac{25}{7} \, litres/second$.
Time taken in seconds $= \frac{\text{Volume}}{\text{Rate}} = \frac{99000}{28} \div \frac{25}{7} = \frac{99000}{28} \times \frac{7}{25} = \frac{99000}{100} = 990 \, seconds$.
To convert into minutes,divide by $60$: $990 \div 60 = 16.5 \, minutes$.

(A) Radius of the sphere $(r_1) = 4.2 \, cm$.
Radius of the cylinder $(r_2) = 6 \, cm$.
Let the height of the cylinder be $h$.
Since the sphere is melted and recast into a cylinder,their volumes remain equal.
Volume of sphere = Volume of cylinder
$\frac{4}{3} \pi r_1^3 = \pi r_2^2 h$
$\frac{4}{3} \times (4.2)^3 = (6)^2 \times h$
$\frac{4}{3} \times 4.2 \times 4.2 \times 4.2 = 36 \times h$
$h = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 36}$
$h = \frac{4 \times 74.088}{108}$
$h = \frac{296.352}{108} = 2.744 \, cm$.
Thus,the height of the cylinder is $2.744 \, cm$.

(B) Let the radii of the three metallic spheres be $r_1 = 6 \, cm$,$r_2 = 8 \, cm$,and $r_3 = 10 \, cm$.
Let the radius of the new solid sphere be $r$.
Since the spheres are melted to form a single sphere,the volume of the new sphere is equal to the sum of the volumes of the three individual spheres.
Volume of a sphere $= \frac{4}{3} \pi r^3$.
Therefore,$\frac{4}{3} \pi r^3 = \frac{4}{3} \pi r_1^3 + \frac{4}{3} \pi r_2^3 + \frac{4}{3} \pi r_3^3$.
Dividing both sides by $\frac{4}{3} \pi$,we get $r^3 = r_1^3 + r_2^3 + r_3^3$.
Substituting the values: $r^3 = 6^3 + 8^3 + 10^3$.
$r^3 = 216 + 512 + 1000 = 1728$.
Taking the cube root on both sides: $r = \sqrt[3]{1728} = 12 \, cm$.
Thus,the radius of the resulting sphere is $12 \, cm$.

(C) The shape of the well is cylindrical.
Depth $(h)$ of the well $= 20\, m$.
Radius $(r)$ of the circular end of the well $= \frac{7}{2}\, m = 3.5\, m$.
Area of the platform $= \text{Length} \times \text{Breadth} = 22\, m \times 14\, m = 308\, m^2$.
Let the height of the platform be $H$.
Since the volume of the soil dug from the well is equal to the volume of the soil spread on the platform:
Volume of soil from well $= \pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 20 = \frac{22}{7} \times \frac{49}{4} \times 20 = 22 \times 7 \times 5 = 770\, m^3$.
Volume of soil on platform $= \text{Area of platform} \times H = 308 \times H$.
Equating the volumes: $308 \times H = 770$.
$H = \frac{770}{308} = 2.5\, m$.
Therefore,the height of the platform is $2.5\, m$.

(D) The shape of the well is cylindrical.
Depth $(h_1)$ of the well $= 14 \, m$.
Radius $(r_1)$ of the circular end of the well $= \frac{3}{2} \, m = 1.5 \, m$.
Width of the embankment $= 4 \, m$.
The embankment is in the shape of a hollow cylinder (circular ring) with inner radius $r_1 = 1.5 \, m$ and outer radius $r_2 = r_1 + \text{width} = 1.5 + 4 = 5.5 \, m = \frac{11}{2} \, m$.
Let the height of the embankment be $h_2$.
Volume of soil dug from the well = Volume of earth used to form the embankment.
$\pi \times r_1^2 \times h_1 = \pi \times (r_2^2 - r_1^2) \times h_2$
$\pi \times (1.5)^2 \times 14 = \pi \times ((5.5)^2 - (1.5)^2) \times h_2$
$2.25 \times 14 = (30.25 - 2.25) \times h_2$
$31.5 = 28 \times h_2$
$h_2 = \frac{31.5}{28} = 1.125 \, m$.
Therefore,the height of the embankment is $1.125 \, m$.

(A) Height $(h_1)$ of the cylindrical container $= 15 \, cm$.
Radius $(r_1)$ of the cylindrical container $= \frac{12}{2} = 6 \, cm$.
Volume of the cylindrical container $= \pi r_1^2 h_1 = \pi \times (6)^2 \times 15 = 540\pi \, cm^3$.
For the ice-cream cone,radius $(r_2) = \frac{6}{2} = 3 \, cm$ and height $(h_2) = 12 \, cm$.
The cone has a hemispherical top with the same radius $r_2 = 3 \, cm$.
Volume of one ice-cream cone $= \text{Volume of cone} + \text{Volume of hemisphere} = \frac{1}{3} \pi r_2^2 h_2 + \frac{2}{3} \pi r_2^3$.
Volume of one ice-cream cone $= \frac{1}{3} \pi (3)^2 (12) + \frac{2}{3} \pi (3)^3 = 36\pi + 18\pi = 54\pi \, cm^3$.
Let $n$ be the number of cones that can be filled.
$n = \frac{\text{Volume of cylinder}}{\text{Volume of one ice-cream cone}} = \frac{540\pi}{54\pi} = 10$.
Therefore,$10$ ice-cream cones can be filled.

(B) Coins are cylindrical in shape.
Height $(h_1)$ of cylindrical coins $= 2 \,mm = 0.2 \,cm$.
Radius $(r)$ of circular end of coins $= \frac{1.75}{2} = 0.875 \,cm$.
Let $n$ coins be melted to form the required cuboid.
Volume of $n$ coins $=$ Volume of cuboid
$n \times \pi \times r^2 \times h_1 = l \times b \times h$
$n \times \frac{22}{7} \times (0.875)^2 \times 0.2 = 5.5 \times 10 \times 3.5$
$n = \frac{5.5 \times 10 \times 3.5 \times 7}{22 \times (0.875)^2 \times 0.2}$
$n = \frac{192.5 \times 7}{22 \times 0.765625 \times 0.2} = \frac{1347.5}{3.36875} = 400$.
Therefore,the number of coins melted to form such a cuboid is $400$.

(N/A) Height $(h_1)$ of the cylindrical bucket $= 32 \, cm$.
Radius $(r_1)$ of the base of the bucket $= 18 \, cm$.
Height $(h_2)$ of the conical heap $= 24 \, cm$.
Let the radius of the base of the conical heap be $r_2$.
Since the volume of sand remains the same:
Volume of sand in the cylindrical bucket $=$ Volume of sand in the conical heap.
$\pi \times r_1^2 \times h_1 = \frac{1}{3} \pi \times r_2^2 \times h_2$
$\pi \times 18^2 \times 32 = \frac{1}{3} \pi \times r_2^2 \times 24$
$18^2 \times 32 = r_2^2 \times 8$
$r_2^2 = \frac{324 \times 32}{8} = 324 \times 4 = 1296$
$r_2 = \sqrt{1296} = 36 \, cm$.
Now,the slant height $(l)$ of the cone is given by $l = \sqrt{r_2^2 + h_2^2}$.
$l = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872}$.
$l = \sqrt{144 \times 13} = 12\sqrt{13} \, cm$.
Thus,the radius is $36 \, cm$ and the slant height is $12\sqrt{13} \, cm$.

(N/A) The cross-sectional area of the canal is $6 \, m \times 1.5 \, m = 9 \, m^2$.
The speed of water is $10 \, km/h = \frac{10000 \, m}{60 \, min} = \frac{500}{3} \, m/min$.
The volume of water that flows from the canal in $30 \, minutes$ is:
$V = \text{Area} \times \text{Speed} \times \text{Time} = 9 \, m^2 \times \frac{500}{3} \, m/min \times 30 \, min = 45000 \, m^3$.
Let the irrigated area be $A$. The volume of water required for irrigation is $A \times \text{depth}$.
Given depth $= 8 \, cm = 0.08 \, m$.
Equating the volumes:
$A \times 0.08 \, m = 45000 \, m^3$
$A = \frac{45000}{0.08} \, m^2 = 562500 \, m^2$.
Thus,the area irrigated in $30 \, minutes$ is $562500 \, m^2$.

(A) Radius of the pipe $(r_1) = \frac{20 \,cm}{2} = 10 \,cm = 0.1 \,m$.
Area of cross-section of the pipe $= \pi r_1^2 = \pi \times (0.1)^2 = 0.01 \pi \,m^2$.
Speed of water $= 3 \,km/h = \frac{3000 \,m}{60 \,min} = 50 \,m/min$.
Volume of water flowing through the pipe in $1 \,minute = \text{Area} \times \text{Speed} = 0.01 \pi \times 50 = 0.5 \pi \,m^3$.
Radius of the cylindrical tank $(r_2) = \frac{10 \,m}{2} = 5 \,m$.
Depth of the cylindrical tank $(h_2) = 2 \,m$.
Volume of the cylindrical tank $= \pi r_2^2 h_2 = \pi \times (5)^2 \times 2 = 50 \pi \,m^3$.
Let the time taken to fill the tank be $t \,minutes$.
Volume of water in $t \,minutes = t \times 0.5 \pi \,m^3$.
Equating the volumes: $t \times 0.5 \pi = 50 \pi$.
$t = \frac{50}{0.5} = 100 \,minutes$.
Thus,the tank will be filled in $100 \,minutes$.

(N/A) $(i)$ Volume of the frustum $= \frac{1}{3} \pi h (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2})$
$= \frac{1}{3} \cdot \frac{22}{7} \cdot 45 \cdot [(28)^{2} + (7)^{2} + (28)(7)] \, cm^{3}$
$= \frac{1}{3} \cdot \frac{22}{7} \cdot 45 \cdot [784 + 49 + 196] \, cm^{3}$
$= \frac{1}{3} \cdot \frac{22}{7} \cdot 45 \cdot 1029 \, cm^{3} = 48510 \, cm^{3}$
$(ii)$ Slant height $l = \sqrt{h^{2} + (r_{1} - r_{2})^{2}} = \sqrt{(45)^{2} + (28 - 7)^{2}} \, cm$
$= \sqrt{2025 + (21)^{2}} = \sqrt{2025 + 441} = \sqrt{2466} \approx 49.66 \, cm$
Curved surface area $= \pi (r_{1} + r_{2}) l = \frac{22}{7} (28 + 7) (49.66) = \frac{22}{7} \cdot 35 \cdot 49.66 = 110 \cdot 49.66 = 5462.6 \, cm^{2}$
$(iii)$ Total surface area $= \pi (r_{1} + r_{2}) l + \pi r_{1}^{2} + \pi r_{2}^{2}$
$= 5462.6 + \frac{22}{7} (28)^{2} + \frac{22}{7} (7)^{2}$
$= 5462.6 + 2464 + 154 = 8080.6 \, cm^{2}$

(C) The mould is in the shape of a frustum of a cone.
Given: Diameters of the circular faces are $d_1 = 35 \,cm$ and $d_2 = 30 \,cm$.
Radii are $r_1 = \frac{35}{2} = 17.5 \,cm$ and $r_2 = \frac{30}{2} = 15 \,cm$.
Vertical height $h = 14 \,cm$.
The volume $V$ of a frustum of a cone is given by $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.
$V = \frac{1}{3} \times \frac{22}{7} \times 14 \times (17.5^2 + 15^2 + 17.5 \times 15) \,cm^3$.
$V = \frac{1}{3} \times 22 \times 2 \times (306.25 + 225 + 262.5) \,cm^3$.
$V = \frac{44}{3} \times 793.75 \,cm^3 = 11641.66... \,cm^3 \approx 11641.67 \,cm^3$.
Given that the mass of $1 \,cm^3$ of molasses is $1.2 \,g$.
Total mass $= 11641.67 \times 1.2 \,g = 13970.004 \,g$.
Converting to $kg$: $13970.004 \div 1000 = 13.97 \,kg$.
Rounding to the nearest whole number,the mass is approximately $14 \,kg$.

(N/A) The total height of the bucket is $40 \, cm$,which includes the height of the base. So,the height of the frustum of the cone is $h = 40 - 6 = 34 \, cm$.
The radii are $r_1 = \frac{45}{2} = 22.5 \, cm$ and $r_2 = \frac{25}{2} = 12.5 \, cm$.
The slant height of the frustum is $l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{34^2 + (22.5 - 12.5)^2} = \sqrt{1156 + 100} = \sqrt{1256} \approx 35.44 \, cm$.
The area of the metallic sheet used is the sum of the curved surface area of the frustum,the area of the circular base (bottom),and the curved surface area of the cylindrical base.
Area $= \pi l(r_1 + r_2) + \pi r_2^2 + 2 \pi r_2 h_{base}$
$= \frac{22}{7} \times 35.44 \times (22.5 + 12.5) + \frac{22}{7} \times (12.5)^2 + 2 \times \frac{22}{7} \times 12.5 \times 6$
$= \frac{22}{7} \times (1240.4 + 156.25 + 150) = \frac{22}{7} \times 1546.65 \approx 4860.9 \, cm^2$.
The volume of water the bucket can hold is the volume of the frustum:
$V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$
$= \frac{1}{3} \times \frac{22}{7} \times 34 \times (22.5^2 + 12.5^2 + 22.5 \times 12.5)$
$= \frac{1}{3} \times \frac{22}{7} \times 34 \times (506.25 + 156.25 + 281.25) = \frac{1}{3} \times \frac{22}{7} \times 34 \times 943.75 \approx 33615.48 \, cm^3$.

(N/A) Radius $(r_1)$ of the upper base of the glass $= \frac{4}{2} = 2 \, cm$.
Radius $(r_2)$ of the lower base of the glass $= \frac{2}{2} = 1 \, cm$.
Height $(h)$ of the glass $= 14 \, cm$.
Capacity of the glass $=$ Volume of the frustum of a cone
$= \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$
$= \frac{1}{3} \times \frac{22}{7} \times 14 \times (2^2 + 1^2 + 2 \times 1)$
$= \frac{1}{3} \times 22 \times 2 \times (4 + 1 + 2)$
$= \frac{44}{3} \times 7 = \frac{308}{3} \, cm^3 = 102 \frac{2}{3} \, cm^3$.
Therefore,the capacity of the glass is $102 \frac{2}{3} \, cm^3$.

(B) Let the radii of the circular ends of the frustum be $r_1$ and $r_2$ respectively.
Given that the perimeters of the circular ends are $18 \,cm$ and $6 \,cm$.
So,$2 \pi r_1 = 18 \implies r_1 = \frac{9}{\pi} \,cm$.
And $2 \pi r_2 = 6 \implies r_2 = \frac{3}{\pi} \,cm$.
The slant height $(l)$ of the frustum is given as $4 \,cm$.
The formula for the curved surface area $(CSA)$ of a frustum of a cone is $CSA = \pi (r_1 + r_2) l$.
Substituting the values:
$CSA = \pi \left( \frac{9}{\pi} + \frac{3}{\pi} \right) \times 4$
$CSA = \pi \left( \frac{12}{\pi} \right) \times 4$
$CSA = 12 \times 4 = 48 \,cm^2$.
Thus,the curved surface area of the frustum is $48 \,cm^2$.

(N/A) Radius $(r_2)$ at the upper circular end $= 4 \, cm$.
Radius $(r_1)$ at the lower circular end $= 10 \, cm$.
Slant height $(l)$ of the frustum $= 15 \, cm$.
Area of material used for making the fez $=$ $CSA$ of the frustum $+$ Area of the upper circular end.
Area $= \pi(r_1 + r_2)l + \pi r_2^2$.
Area $= \pi(10 + 4) \times 15 + \pi(4)^2$.
Area $= \pi(14) \times 15 + 16\pi$.
Area $= 210\pi + 16\pi = 226\pi$.
Using $\pi = \frac{22}{7}$,Area $= 226 \times \frac{22}{7} = \frac{4972}{7} \, cm^2$.
Area $= 710 \frac{2}{7} \, cm^2$.
Therefore,the area of material used for making it is $710 \frac{2}{7} \, cm^2$.

(N/A) Radius $(r_1)$ of upper end of container $= 20 \, cm$
Radius $(r_2)$ of lower end of container $= 8 \, cm$
Height $(h)$ of container $= 16 \, cm$
Slant height $(l)$ of frustum $= \sqrt{(r_1 - r_2)^2 + h^2} = \sqrt{(20 - 8)^2 + 16^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20 \, cm$
Capacity of container $=$ Volume of frustum $= \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$
$= \frac{1}{3} \times 3.14 \times 16 \times (20^2 + 8^2 + 20 \times 8) = \frac{1}{3} \times 3.14 \times 16 \times (400 + 64 + 160) = \frac{1}{3} \times 3.14 \times 16 \times 624 = 10449.92 \, cm^3 = 10.44992 \, litres \approx 10.45 \, litres$
Cost of $1 \, litre$ milk $= Rs. \, 20$
Cost of $10.45 \, litres$ milk $= 10.45 \times 20 = Rs. \, 209$
Area of metal sheet used $= \pi (r_1 + r_2) l + \pi r_2^2 = 3.14 \times (20 + 8) \times 20 + 3.14 \times 8^2 = 3.14 \times 28 \times 20 + 3.14 \times 64 = 1758.4 + 200.96 = 1959.36 \, cm^2$
Cost of $100 \, cm^2$ metal sheet $= Rs. \, 8$
Cost of $1959.36 \, cm^2$ metal sheet $= \frac{1959.36 \times 8}{100} = Rs. \, 156.7488 \approx Rs. \, 156.75$
Therefore,the cost of the milk is $Rs. \, 209$ and the cost of the metal sheet is $Rs. \, 156.75$.

(A) In $\triangle AEG$,$\frac{EG}{AG} = \tan 30^{\circ}$. Since $AG = 10 \, cm$,$EG = 10 \tan 30^{\circ} = \frac{10}{\sqrt{3}} \, cm$. This is the radius $r_1$ of the top of the frustum.
In $\triangle ABD$,$\frac{BD}{AD} = \tan 30^{\circ}$. Since $AD = 20 \, cm$,$BD = 20 \tan 30^{\circ} = \frac{20}{\sqrt{3}} \, cm$. This is the radius $r_2$ of the base of the frustum.
The height $h$ of the frustum is $10 \, cm$.
The volume of the frustum is $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.
$V = \frac{1}{3} \times \frac{22}{7} \times 10 \left[ \left( \frac{10}{\sqrt{3}} \right)^2 + \left( \frac{20}{\sqrt{3}} \right)^2 + \left( \frac{10}{\sqrt{3}} \times \frac{20}{\sqrt{3}} \right) \right]$
$V = \frac{220}{21} \left[ \frac{100}{3} + \frac{400}{3} + \frac{200}{3} \right] = \frac{220}{21} \times \frac{700}{3} = \frac{22000}{9} \, cm^3$.
The wire has a radius $r = \frac{1}{32} \, cm$. Let its length be $l$.
Volume of wire = $\pi r^2 l = \frac{22}{7} \times \left( \frac{1}{32} \right)^2 \times l$.
Equating volumes: $\frac{22000}{9} = \frac{22}{7} \times \frac{1}{1024} \times l$.
$l = \frac{22000}{9} \times \frac{7 \times 1024}{22} = \frac{1000 \times 7168}{9} = \frac{7168000}{9} \approx 796444.44 \, cm$.
Converting to meters: $l = 7964.44 \, m$.

(B) It can be observed that $1$ round of wire will cover $3 \,mm$ $(0.3 \,cm)$ height of the cylinder.
Number of rounds $= \frac{\text{Height of cylinder}}{\text{Diameter of wire}} = \frac{12 \,cm}{0.3 \,cm} = 40$ rounds.
Length of wire required in $1$ round $=$ Circumference of the base of the cylinder $= 2 \pi r = 2 \pi \times 5 = 10 \pi \,cm$.
Total length of wire in $40$ rounds $= 40 \times 10 \pi = 400 \pi \,cm = 400 \times 3.14159 \approx 1256.64 \,cm$.
Radius of the wire $= \frac{0.3 \,cm}{2} = 0.15 \,cm$.
Volume of the wire $=$ Area of cross-section of wire $\times$ Total length of wire $= \pi (0.15)^2 \times 1256.64 \approx 88.826 \,cm^3$.
Mass of the wire $=$ Volume $\times$ Density $= 88.826 \,cm^3 \times 8.88 \,g/cm^3 \approx 788.77 \,g$.
Using $\pi \approx \frac{22}{7}$ for calculation: Length $= 400 \times \frac{22}{7} = \frac{8800}{7} \approx 1257.14 \,cm$. Volume $= \pi \times (0.15)^2 \times \frac{8800}{7} = \frac{22}{7} \times 0.0225 \times \frac{8800}{7} \approx 88.93 \,cm^3$. Mass $= 88.93 \times 8.88 \approx 789.7 \,g$. The closest option is $789.41 \,g$.

(C) Let the right triangle be $ABC$ with sides $AB = 3\, cm$,$BC = 4\, cm$,and $\angle B = 90^\circ$.
Hypotenuse $AC = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\, cm$.
When the triangle is revolved about the hypotenuse $AC$,a double cone is formed where the radius $r$ of the common base is the altitude $OB$ of the triangle $ABC$ to the hypotenuse $AC$.
Area of $\triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 3 \times 4 = 6\, cm^2$.
Also,Area of $\triangle ABC = \frac{1}{2} \times AC \times OB = \frac{1}{2} \times 5 \times OB$.
So,$\frac{1}{2} \times 5 \times OB = 6 \implies OB = \frac{12}{5} = 2.4\, cm$.
The surface area of the double cone is the sum of the curved surface areas of the two cones: $S = \pi r l_1 + \pi r l_2$,where $l_1 = AB = 3\, cm$ and $l_2 = BC = 4\, cm$.
$S = \pi r (l_1 + l_2) = 3.14 \times 2.4 \times (3 + 4) = 3.14 \times 2.4 \times 7 = 52.752\, cm^2 \approx 52.75\, cm^2$.

(D) The total volume of the cistern is $V_{cistern} = 150 \times 120 \times 110 = 1980000 \,cm^3$.
The volume of water already present is $129600 \,cm^3$.
The volume to be filled is $V_{fill} = 1980000 - 129600 = 1850400 \,cm^3$.
Let $n$ be the number of bricks. The volume of one brick is $V_{brick} = 22.5 \times 7.5 \times 6.5 = 1096.875 \,cm^3$.
Each brick absorbs $\frac{1}{17}$ of its volume,so the effective volume occupied by one brick in the water is $V_{eff} = V_{brick} - \frac{1}{17} V_{brick} = \frac{16}{17} V_{brick}$.
To fill the cistern,the total volume of $n$ bricks must equal the volume to be filled: $n \times V_{eff} = V_{fill}$.
$n \times \frac{16}{17} \times 1096.875 = 1850400$.
$n \times \frac{16}{17} \times \frac{8775}{8} = 1850400$.
$n \times \frac{2 \times 8775}{17} = 1850400$.
$n = \frac{1850400 \times 17}{17550} = 1792$.

(N/A) Area of the valley $= 7280 \, km^2 = 7280 \times (1000 \, m)^2 = 7.28 \times 10^9 \, m^2$.
Rainfall depth $= 10 \, cm = 0.1 \, m$.
Total volume of rainfall $= \text{Area} \times \text{Depth} = 7.28 \times 10^9 \, m^2 \times 0.1 \, m = 7.28 \times 10^8 \, m^3$.
Volume of one river $= \text{length} \times \text{width} \times \text{depth} = 1072 \, km \times 75 \, m \times 3 \, m = 1072000 \, m \times 75 \, m \times 3 \, m = 2.412 \times 10^8 \, m^3$.
Volume of three such rivers $= 3 \times 2.412 \times 10^8 \, m^3 = 7.236 \times 10^8 \, m^3$.
Since $7.28 \times 10^8 \, m^3 \approx 7.236 \times 10^8 \, m^3$,the total rainfall is approximately equivalent to the volume of three such rivers.

(N/A) Radius $(r_1)$ of the upper circular end of the frustum part $= \frac{18}{2} = 9 \, cm$.
Radius $(r_2)$ of the lower circular end of the frustum part $=$ Radius of the circular end of the cylindrical part $= \frac{8}{2} = 4 \, cm$.
Height $(h_1)$ of the frustum part $= 22 - 10 = 12 \, cm$.
Height $(h_2)$ of the cylindrical part $= 10 \, cm$.
Slant height $(l)$ of the frustum part $= \sqrt{(r_1 - r_2)^2 + h_1^2} = \sqrt{(9 - 4)^2 + (12)^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.
Area of the tin sheet required $=$ Curved Surface Area $(CSA)$ of the frustum part $+$ $CSA$ of the cylindrical part.
Area $= \pi(r_1 + r_2)l + 2\pi r_2 h_2$.
Area $= \frac{22}{7} \times (9 + 4) \times 13 + 2 \times \frac{22}{7} \times 4 \times 10$.
Area $= \frac{22}{7} \times 13 \times 13 + \frac{22}{7} \times 80$.
Area $= \frac{22}{7} \times (169 + 80) = \frac{22}{7} \times 249 = \frac{5478}{7} \, cm^2$.
Area $= 782 \frac{4}{7} \, cm^2$.