Mix Examples - Surface Areas and Volumes Questions in English

(C) Let the volume of cone $A$ be $2V$ and the volume of cone $B$ be $V$. Let the height of cone $A$ be $h_1 \, cm$,then the height of cone $B$ is $(21 - h_1) \, cm$.
Given,the diameter of the cones $= 6 \, cm$.
Therefore,the radius of the cones $r = \frac{6}{2} = 3 \, cm$.
Volume of cone $A = 2V = \frac{1}{3} \pi r^2 h_1 = \frac{1}{3} \pi (3)^2 h_1 = 3 \pi h_1$.
Thus,$V = \frac{3}{2} \pi h_1$ ...... $(i)$
Volume of cone $B = V = \frac{1}{3} \pi r^2 (21 - h_1) = \frac{1}{3} \pi (3)^2 (21 - h_1) = 3 \pi (21 - h_1)$ ...... $(ii)$
Equating $(i)$ and $(ii)$:
$\frac{3}{2} \pi h_1 = 3 \pi (21 - h_1)$
$\frac{1}{2} h_1 = 21 - h_1$
$\frac{3}{2} h_1 = 21$
$h_1 = 14 \, cm$.
Height of cone $A = 14 \, cm$,Height of cone $B = 21 - 14 = 7 \, cm$.
Volume of cone $A = 3 \pi (14) = 3 \times \frac{22}{7} \times 14 = 132 \, cm^3$.
Volume of cone $B = 3 \pi (7) = 3 \times \frac{22}{7} \times 7 = 66 \, cm^3$.
Volume of the cylinder $= \pi r^2 H = \frac{22}{7} \times (3)^2 \times 21 = 22 \times 9 \times 3 = 594 \, cm^3$.
Volume of the remaining portion $=$ Volume of cylinder $-$ (Volume of cone $A$ $+$ Volume of cone $B$)
$= 594 - (132 + 66) = 594 - 198 = 396 \, cm^3$.

(D) The ice cream cone is a combination of a hemisphere and a cone.
Given,radius of the hemisphere $(r)$ $= 5 \, cm$.
Volume of the hemisphere $= \frac{2}{3} \pi r^{3} = \frac{2}{3} \times \frac{22}{7} \times (5)^{3} = \frac{5500}{21} \approx 261.90 \, cm^{3}$.
Now,the radius of the cone $(r)$ $= 5 \, cm$.
The total height of the ice cream cone is $10 \, cm$. Since the hemisphere has a radius of $5 \, cm$,the height of the conical part $(h)$ $= 10 - 5 = 5 \, cm$.
Volume of the cone $= \frac{1}{3} \pi r^{2} h = \frac{1}{3} \times \frac{22}{7} \times (5)^{2} \times 5 = \frac{2750}{21} \approx 130.95 \, cm^{3}$.
Total volume of the ice cream cone $= 261.90 + 130.95 = 392.85 \, cm^{3}$.
Since $\frac{1}{6}$ part is left unfilled,the volume of ice cream filled is $\frac{5}{6}$ of the total volume.
Required volume of ice cream $= \frac{5}{6} \times 392.85 = 5 \times 65.475 = 327.375 \approx 327.4 \, cm^{3}$.

(A) Given,diameter of a marble $= 1.4 \,cm$.
$\therefore$ Radius of a marble $(r) = \frac{1.4}{2} = 0.7 \,cm$.
Volume of one spherical marble $= \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.7)^3 = \frac{4}{3} \pi (0.343) = \frac{1.372}{3} \pi \,cm^3$.
Given,diameter of the cylindrical beaker $= 7 \,cm$.
$\therefore$ Radius of the beaker $(R) = \frac{7}{2} = 3.5 \,cm$.
Height of the water level raised $(h) = 5.6 \,cm$.
Volume of the water displaced (raised) in the beaker $= \pi R^2 h = \pi (3.5)^2 (5.6) = \pi (12.25) (5.6) = 68.6 \pi \,cm^3$.
Let the number of marbles be $n$.
Then,$n \times (\text{Volume of one marble}) = \text{Volume of raised water}$.
$n \times \frac{1.372}{3} \pi = 68.6 \pi$.
$n = \frac{68.6 \times 3}{1.372} = \frac{205.8}{1.372} = 150$.
Thus,$150$ marbles are required.

(B) Given that,spherical lead shots are made from a solid rectangular lead piece.
$\therefore$ Number of spherical lead shots $= \frac{\text{Volume of solid rectangular lead piece}}{\text{Volume of a spherical lead shot}}$ ......$(i)$
Given,diameter of a spherical lead shot $= 4.2 \, cm$.
$\therefore$ Radius of a spherical lead shot,$r = \frac{4.2}{2} = 2.1 \, cm$.
Volume of a spherical lead shot $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3 = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1 = 4 \times 22 \times 0.1 \times 2.1 \times 2.1 = 38.808 \, cm^3$.
Volume of the rectangular lead piece (cuboid) $= l \times b \times h = 66 \times 42 \times 21 = 58212 \, cm^3$.
From Eq. $(i)$,
Number of spherical lead shots $= \frac{58212}{38.808} = 1500$.
Hence,the required number of spherical lead shots is $1500$.

(C) Given that,spherical lead shots are made from a solid cube of lead.
$\therefore$ Number of spherical lead shots $= \frac{\text{Volume of the solid cube}}{\text{Volume of one spherical lead shot}}$ ...........$(i)$
Given that,the diameter of a spherical lead shot $= 4 \,cm$.
$\Rightarrow$ Radius of a spherical lead shot $(r) = \frac{4}{2} = 2 \,cm$.
Volume of one spherical lead shot $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (2)^3 = \frac{4 \times 22 \times 8}{3 \times 7} = \frac{704}{21} \,cm^3$.
Given that,the edge of the solid cube $(a) = 44 \,cm$.
Volume of the solid cube $= a^3 = (44)^3 = 44 \times 44 \times 44 = 85184 \,cm^3$.
From Eq. $(i)$,
Number of spherical lead shots $= \frac{44 \times 44 \times 44}{\frac{4}{3} \times \frac{22}{7} \times 8} = \frac{44 \times 44 \times 44 \times 3 \times 7}{4 \times 22 \times 8} = 11 \times 21 \times 11 = 2541$.
Hence,the required number of spherical lead shots is $2541$.

(D) Given that,a wall is constructed with the help of bricks and mortar.
Number of bricks $= \frac{\text{Volume of wall} - (\frac{1}{10} \text{th volume of wall})}{\text{Volume of a brick}}$ .........$(i)$
Length of wall $(l) = 24 \, m$,Thickness $(b) = 0.4 \, m$,Height $(h) = 6 \, m$.
Volume of wall $= l \times b \times h = 24 \times 0.4 \times 6 = 57.6 \, m^3$.
Volume occupied by mortar $= \frac{1}{10} \times 57.6 = 5.76 \, m^3$.
Volume occupied by bricks $= 57.6 - 5.76 = 51.84 \, m^3$.
Dimensions of a brick $= 25 \, cm \times 16 \, cm \times 10 \, cm = 0.25 \, m \times 0.16 \, m \times 0.10 \, m$.
Volume of one brick $= 0.25 \times 0.16 \times 0.10 = 0.004 \, m^3$.
Number of bricks $= \frac{51.84}{0.004} = \frac{51840}{4} = 12960$.
Hence,the required number of bricks used in constructing the wall is $12960$.

(A) metallic circular disc is essentially a thin right circular cylinder.
For the metallic circular disc:
Diameter $= 1.5\, cm$,so radius $(r) = \frac{1.5}{2} = 0.75\, cm$.
Height $(h) = 0.2\, cm$.
Volume of one disc $= \pi r^2 h = \pi \times (0.75)^2 \times 0.2 = \pi \times 0.5625 \times 0.2 = 0.1125\pi\, cm^3$.
For the large right circular cylinder:
Diameter $= 4.5\, cm$,so radius $(R) = \frac{4.5}{2} = 2.25\, cm$.
Height $(H) = 10\, cm$.
Volume of the cylinder $= \pi R^2 H = \pi \times (2.25)^2 \times 10 = \pi \times 5.0625 \times 10 = 50.625\pi\, cm^3$.
Number of discs required $= \frac{\text{Volume of cylinder}}{\text{Volume of one disc}} = \frac{50.625\pi}{0.1125\pi} = \frac{506250}{1125} = 450$.
Thus,the required number of metallic circular discs is $450$.

(N/A) The capacity (volume) of the bucket is given by the formula $V = \frac{\pi h}{3} [r_1^2 + r_2^2 + r_1 r_2]$.
Given $h = 30 \, cm$,$r_1 = 20 \, cm$,and $r_2 = 10 \, cm$.
Capacity $= \frac{3.14 \times 30}{3} [20^2 + 10^2 + 20 \times 10] = 31.4 [400 + 100 + 200] = 31.4 \times 700 = 21980 \, cm^3$.
Since $1000 \, cm^3 = 1 \, litre$,the capacity is $21.98 \, litres$.
The cost of milk at $Rs. \, 25$ per $litre$ is $21.98 \times 25 = Rs. \, 549.50$.
The slant height $l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{30^2 + (20 - 10)^2} = \sqrt{900 + 100} = \sqrt{1000} \approx 31.62 \, cm$.
The surface area of the bucket (open at the top) is the curved surface area plus the area of the base: $A = \pi l (r_1 + r_2) + \pi r_2^2$.
$A = 3.14 \times 31.62 \times (20 + 10) + 3.14 \times 10^2 = 3.14 \times 31.62 \times 30 + 314 = 2978.56 + 314 = 3292.56 \, cm^2 \approx 3292.6 \, cm^2$.

(N/A) Let $r$ be the radius of the hemisphere and the cone,and $h$ be the height of the cone. Given $h = 4\, cm$ and diameter $d = 8\, cm$,so $r = 4\, cm$.
Volume of the toy $=$ Volume of the hemisphere $+$ Volume of the cone
$= \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r + h)$
$= \frac{1}{3} \times \frac{22}{7} \times 4^2 \times (2 \times 4 + 4) = \frac{1}{3} \times \frac{22}{7} \times 16 \times 12 = \frac{1408}{7} \approx 201.14\, cm^3$.
$A$ cube circumscribes the toy. The height of the toy is $h + r = 4 + 4 = 8\, cm$ and the width is $8\, cm$. Thus,the edge of the cube is $a = 8\, cm$.
Volume of the cube $= a^3 = 8^3 = 512\, cm^3$.
Difference in volumes $= 512 - 201.14 = 310.86\, cm^3$.
Total surface area of the toy $=$ Curved surface area of cone $+$ Curved surface area of hemisphere
$= \pi r l + 2 \pi r^2$,where slant height $l = \sqrt{h^2 + r^2} = \sqrt{4^2 + 4^2} = 4\sqrt{2}\, cm$.
$= \pi r (l + 2r) = \frac{22}{7} \times 4 \times (4\sqrt{2} + 8) = \frac{88}{7} \times 4(\sqrt{2} + 2) = \frac{352}{7} (1.414 + 2) \approx 171.68\, cm^2$.

(D) Let the radius of the hemispherical dome be $r$ meters and the total height of the building be $h$ meters.
Since the base diameter of the dome is equal to $\frac{2}{3}$ of the total height,therefore $2r = \frac{2}{3}h$,which implies $r = \frac{h}{3}$.
Let $H$ be the height of the cylindrical portion. Then $H = h - r = h - \frac{h}{3} = \frac{2}{3}h$ meters.
The volume of air inside the building is the sum of the volume of the hemispherical dome and the volume of the cylindrical portion.
Volume $= \frac{2}{3}\pi r^3 + \pi r^2 H = \frac{2}{3}\pi (\frac{h}{3})^3 + \pi (\frac{h}{3})^2 (\frac{2}{3}h) = \frac{2}{3}\pi (\frac{h^3}{27}) + \pi (\frac{h^2}{9}) (\frac{2}{3}h) = \frac{2\pi h^3}{81} + \frac{2\pi h^3}{27} = \frac{2\pi h^3 + 6\pi h^3}{81} = \frac{8\pi h^3}{81}$.
Given the volume is $67 \frac{1}{21} = \frac{1408}{21} \, m^3$.
So,$\frac{8}{81} \times \frac{22}{7} \times h^3 = \frac{1408}{21}$.
$h^3 = \frac{1408}{21} \times \frac{81 \times 7}{8 \times 22} = \frac{1408}{21} \times \frac{567}{176} = 8 \times 27 = 216$.
$h = \sqrt[3]{216} = 6 \, m$.

(A) Let the height of the cone be $h \, cm$.
Given,the radius of the base of the cone $r = 6 \, cm$.
The volume of the cone is given by $V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 h = \frac{36 \pi h}{3} = 12 \pi h \, cm^3$.
Given,the radius of the hemisphere $R = 8 \, cm$.
The volume of the hemisphere is given by $V_{hemisphere} = \frac{2}{3} \pi R^3 = \frac{2}{3} \pi (8)^3 = \frac{2}{3} \pi (512) = \frac{1024 \pi}{3} \, cm^3$.
Since the hemisphere is melted and recast into a cone,their volumes must be equal:
$12 \pi h = \frac{1024 \pi}{3}$
$h = \frac{1024 \pi}{3 \times 12 \pi} = \frac{1024}{36} = \frac{256}{9} \approx 28.44 \, cm$.
Thus,the height of the cone is $28.44 \, cm$.

(B) Given,dimensions of the base of the rectangular tank $= 11 \,m \times 6 \,m$ and the height of water $= 5 \,m$.
Volume of the water in the rectangular tank $= 11 \times 6 \times 5 = 330 \,m^3$.
Also,the radius of the cylindrical tank is given as $r = 3.5 \,m$.
Let the height of the water level in the cylindrical tank be $h$. The volume of water in the cylindrical tank is given by the formula $V = \pi r^2 h$.
$V = \frac{22}{7} \times (3.5)^2 \times h = \frac{22}{7} \times 3.5 \times 3.5 \times h = 22 \times 0.5 \times 3.5 \times h = 38.5 h \,m^3$.
Since the volume of water remains the same when transferred:
$38.5 h = 330$.
$h = \frac{330}{38.5} = \frac{3300}{385} \approx 8.571 \,m$.
Rounding to one decimal place,the height of the water level is $8.6 \,m$.

(N/A) Let the external length $(l) = 36 \, cm$,external breadth $(b) = 25 \, cm$,and external height $(h) = 16.5 \, cm$. The thickness of the iron is $x = 1.5 \, cm$.
External volume of the open box $= l \times b \times h = 36 \times 25 \times 16.5 = 14850 \, cm^3$.
For an open box,the internal dimensions are:
Internal length $(l_i) = l - 2x = 36 - 2(1.5) = 36 - 3 = 33 \, cm$.
Internal breadth $(b_i) = b - 2x = 25 - 2(1.5) = 25 - 3 = 22 \, cm$.
Internal height $(h_i) = h - x = 16.5 - 1.5 = 15 \, cm$ (since it is an open box,only the base thickness is subtracted).
Internal volume $= l_i \times b_i \times h_i = 33 \times 22 \times 15 = 10890 \, cm^3$.
Volume of iron required = External volume - Internal volume
$= 14850 - 10890 = 3960 \, cm^3$.
Weight of the box = Volume of iron $\times$ density
$= 3960 \, cm^3 \times 7.5 \, g/cm^3 = 29700 \, g = 29.7 \, kg$.

(D) Given,length of the barrel of a fountain pen $h = 7 \, cm$.
Diameter $d = 5 \, mm = 0.5 \, cm$,so radius $r = 0.25 \, cm$.
Volume of the cylindrical barrel $V = \pi r^2 h = \frac{22}{7} \times (0.25)^2 \times 7 = 22 \times 0.0625 = 1.375 \, cm^3$.
It is given that $1.375 \, cm^3$ of ink writes $3300$ words.
Volume of ink in the bottle $= \frac{1}{5} \, L = \frac{1}{5} \times 1000 \, cm^3 = 200 \, cm^3$.
Number of words written with $200 \, cm^3$ of ink $= \frac{3300}{1.375} \times 200$.
$= \frac{3300 \times 200}{1.375} = \frac{660000}{1.375} = 480000$ words.

(A) Given,speed of water flow $= 10 \, m/min = 1000 \, cm/min$.
Radius of the pipe $r = \frac{5 \, mm}{2} = 2.5 \, mm = 0.25 \, cm$.
Area of the cross-section of the pipe $= \pi r^2 = \pi \times (0.25)^2 = 0.0625 \pi \, cm^2$.
Volume of water flowing per minute $= \text{Area} \times \text{Speed} = 0.0625 \pi \times 1000 = 62.5 \pi \, cm^3$.
Volume of the conical vessel $= \frac{1}{3} \pi R^2 H$,where $R = 20 \, cm$ and $H = 24 \, cm$.
Volume $= \frac{1}{3} \pi \times (20)^2 \times 24 = 8 \pi \times 400 = 3200 \pi \, cm^3$.
Time required $= \frac{\text{Volume of vessel}}{\text{Volume per minute}} = \frac{3200 \pi}{62.5 \pi} = \frac{3200}{62.5} = 51.2 \, minutes$.
$51.2 \, minutes = 51 \, minutes + 0.2 \times 60 \, seconds = 51 \, minutes \, 12 \, seconds$.

(N/A) Given that,a heap of rice is in the form of a cone.
Height of the cone $(h) = 3.5 \, m$.
Diameter of the cone $= 9 \, m$,so the radius $(r) = \frac{9}{2} = 4.5 \, m$.
Volume of the rice $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (4.5)^2 \times 3.5 = \frac{1}{3} \times \frac{22}{7} \times 20.25 \times 3.5 = 74.25 \, m^3$.
To cover the heap,we need the curved surface area of the cone,which is $\pi r l$,where $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{(4.5)^2 + (3.5)^2} = \sqrt{20.25 + 12.25} = \sqrt{32.5} \approx 5.70 \, m$.
Curved surface area $= \frac{22}{7} \times 4.5 \times 5.70 \approx 80.61 \, m^2$.
Thus,the volume is $74.25 \, m^3$ and the required canvas cloth is $80.61 \, m^2$.

(C) The curved surface area of a cylinder is given by $CSA = 2 \pi r h$,where $2 \pi r$ is the circumference of the base and $h$ is the length (height).
Given,circumference of the base $= 1.5 \,cm$ and length $h = 25 \,cm$.
Curved surface area of one pencil $= 1.5 \,cm \times 25 \,cm = 37.5 \,cm^2$.
Since $1 \,dm = 10 \,cm$,then $1 \,dm^2 = 100 \,cm^2$. Therefore,$1 \,cm^2 = 0.01 \,dm^2$.
Curved surface area of one pencil in $dm^2 = 37.5 \times 0.01 = 0.375 \,dm^2$.
Total curved surface area for $120000$ pencils $= 120000 \times 0.375 \,dm^2 = 45000 \,dm^2$.
Cost of colouring $= 45000 \,dm^2 \times \operatorname{Rs.} 0.05/dm^2 = \operatorname{Rs.} 2250$.

(D) Given: Length of the pond $= 50\, m$,Width of the pond $= 44\, m$.
Required rise in water level $= 21\, cm = 0.21\, m$.
Volume of water required in the pond $= 50 \times 44 \times 0.21 = 462\, m^3$.
Radius of the pipe $(r) = \frac{14}{2} = 7\, cm = 0.07\, m$.
Speed of water $= 15\, km/h = 15000\, m/h$.
Volume of water flowing through the pipe in $1\, h = \pi r^2 \times \text{speed} = \frac{22}{7} \times (0.07)^2 \times 15000$.
$= \frac{22}{7} \times 0.0049 \times 15000 = 22 \times 0.0007 \times 15000 = 231\, m^3$.
Time required to fill $462\, m^3$ of water $= \frac{\text{Total volume required}}{\text{Volume flow per hour}} = \frac{462}{231} = 2\, h$.
Thus,the required time is $2\, h$.

(A) Given that,a solid iron cuboidal block is recast into a hollow cylindrical pipe.
Volume of the cuboidal block $= l \times b \times h = 4.4 \, m \times 2.6 \, m \times 1 \, m = 11.44 \, m^3$.
Internal radius of the hollow cylindrical pipe $(r_i) = 30 \, cm = 0.3 \, m$.
Thickness of the pipe $= 5 \, cm = 0.05 \, m$.
External radius $(r_e) = r_i + \text{thickness} = 0.3 \, m + 0.05 \, m = 0.35 \, m$.
Volume of the hollow cylindrical pipe $= \pi (r_e^2 - r_i^2) h_1$,where $h_1$ is the length of the pipe.
Volume $= \frac{22}{7} \times (0.35^2 - 0.3^2) \times h_1 = \frac{22}{7} \times (0.1225 - 0.09) \times h_1 = \frac{22}{7} \times 0.0325 \times h_1$.
Since the volume remains constant: $11.44 = \frac{22}{7} \times 0.0325 \times h_1$.
$h_1 = \frac{11.44 \times 7}{22 \times 0.0325} = \frac{80.08}{0.715} = 112 \, m$.
Thus,the length of the pipe is $112 \, m$.

(B) Let the rise of water level in the pond be $h \, m$,when $500$ persons are taking a dip into a cuboidal pond.
Given that,
Length of the cuboidal pond $= 80 \, m$
Breadth of the cuboidal pond $= 50 \, m$
Now,the volume of the water rise in the pond is given by:
Volume $= \text{Length} \times \text{Breadth} \times \text{Height}$
Volume $= 80 \times 50 \times h = 4000h \, m^3$
The average displacement of water by one person $= 0.04 \, m^3$
So,the total displacement of water by $500$ persons $= 500 \times 0.04 \, m^3 = 20 \, m^3$
According to the problem,the volume of the rise in water level equals the total displacement of water by $500$ persons:
$4000h = 20$
$h = \frac{20}{4000} \, m$
$h = \frac{1}{200} \, m = 0.005 \, m$
To convert the height into centimeters $(cm)$:
$h = 0.005 \times 100 \, cm = 0.5 \, cm$
Hence,the required rise of water level in the pond is $0.5 \, cm$.

(C) Given,dimensions of the cuboidal box $= 16 \,cm \times 8 \,cm \times 8 \,cm$.
Volume of the cuboidal box $= 16 \times 8 \times 8 = 1024 \,cm^3$.
Radius of one glass sphere $(r) = 2 \,cm$.
Volume of one glass sphere $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (2)^3 = \frac{4}{3} \times \frac{22}{7} \times 8 = \frac{704}{21} \approx 33.524 \,cm^3$.
Volume of $16$ glass spheres $= 16 \times 33.524 = 536.384 \,cm^3$.
Volume of water filled $= \text{Volume of cuboidal box} - \text{Volume of } 16 \text{ glass spheres}$.
Volume of water $= 1024 - 536.384 = 487.616 \,cm^3$.
Rounding to one decimal place,the volume is $487.6 \,cm^3$.

(D) Given: Height of the milk container $(h) = 16 \, cm$,radius of the lower end $(r) = 8 \, cm$,and radius of the upper end $(R) = 20 \, cm$.
The volume of a frustum of a cone is given by the formula: $V = \frac{\pi h}{3} (R^2 + r^2 + Rr)$.
Substituting the values:
$V = \frac{1}{3} \times \frac{22}{7} \times 16 \times (20^2 + 8^2 + 20 \times 8)$
$V = \frac{22 \times 16}{21} \times (400 + 64 + 160)$
$V = \frac{352}{21} \times 624$
$V = \frac{219648}{21} \approx 10459.43 \, cm^3$.
Since $1000 \, cm^3 = 1 \, litre$,the volume in litres is $V = \frac{10459.43}{1000} = 10.45943 \, litres$.
The cost of milk at the rate of $Rs. \, 22$ per $litre$ is:
Cost $= 10.45943 \times 22 = Rs. \, 230.107 \approx Rs. \, 230.12$ (rounding to the nearest option).

(N/A) Given,radius of the base of the bucket $(R) = 18 \,cm$.
Height of the bucket $(H) = 32 \,cm$.
Volume of the sand in the cylindrical bucket $= \pi R^2 H = \pi \times (18)^2 \times 32 = 10368 \pi \,cm^3$.
Let the radius of the conical heap be $r \,cm$ and its height be $h = 24 \,cm$.
Volume of the conical heap $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 \times 24 = 8 \pi r^2 \,cm^3$.
Since the volume of sand remains the same,$8 \pi r^2 = 10368 \pi$.
$r^2 = \frac{10368}{8} = 1296$.
$r = \sqrt{1296} = 36 \,cm$.
The slant height $l$ of the cone is given by $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{36^2 + 24^2} = \sqrt{1296 + 576} = \sqrt{1872}$.
$l = 12\sqrt{13} \approx 43.27 \,cm$.
Thus,the radius of the heap is $36 \,cm$ and the slant height is approximately $43.27 \,cm$.

(N/A) The rocket is a combination of a right circular cylinder and a cone.
Given: Diameter of the cylinder $= 6 \, cm$.
Radius of the cylinder $(r) = \frac{6}{2} = 3 \, cm$.
Height of the cylinder $(H) = 12 \, cm$.
Volume of the cylinder $= \pi r^2 H = 3.14 \times (3)^2 \times 12 = 3.14 \times 9 \times 12 = 339.12 \, cm^3$.
Curved surface area of the cylinder $= 2 \pi r H = 2 \times 3.14 \times 3 \times 12 = 226.08 \, cm^2$.
For the cone,slant height $(l) = 5 \, cm$ and radius $(r) = 3 \, cm$.
Height of the cone $(h) = \sqrt{l^2 - r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \, cm$.
Volume of the cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times (3)^2 \times 4 = 3.14 \times 3 \times 4 = 37.68 \, cm^3$.
Curved surface area of the cone $= \pi r l = 3.14 \times 3 \times 5 = 47.1 \, cm^2$.
Total volume of the rocket $= \text{Volume of cylinder} + \text{Volume of cone} = 339.12 + 37.68 = 376.8 \, cm^3$.
Total surface area of the rocket $= \text{CSA of cone} + \text{CSA of cylinder} + \text{Area of base of cylinder} = 47.1 + 226.08 + (\pi r^2) = 47.1 + 226.08 + (3.14 \times 3^2) = 47.1 + 226.08 + 28.26 = 301.44 \, cm^2$.

(C) Let the total height of the building be equal to the internal diameter of the dome,which is $2r \, m$.
Therefore,the radius of the building (or dome) is $\frac{2r}{2} = r \, m$.
The height of the cylindrical part is $2r - r = r \, m$.
Volume of the cylinder $= \pi r^2 h = \pi r^2 (r) = \pi r^3 \, m^3$.
Volume of the hemispherical dome $= \frac{2}{3} \pi r^3 \, m^3$.
Total volume of the building $=$ Volume of the cylinder $+$ Volume of the hemispherical dome.
Total volume $= \pi r^3 + \frac{2}{3} \pi r^3 = \frac{5}{3} \pi r^3 \, m^3$.
Given that the volume of air inside is $41 \frac{19}{21} \, m^3 = \frac{880}{21} \, m^3$.
Equating the volumes: $\frac{5}{3} \pi r^3 = \frac{880}{21}$.
Using $\pi = \frac{22}{7}$,we get $\frac{5}{3} \times \frac{22}{7} \times r^3 = \frac{880}{21}$.
$\frac{110}{21} r^3 = \frac{880}{21}$.
$r^3 = \frac{880}{110} = 8$.
$r = \sqrt[3]{8} = 2 \, m$.
Total height of the building $= 2r = 2 \times 2 = 4 \, m$.

(D) Given,radius of the hemispherical bowl,$r = 9 \,cm$.
The volume of the hemispherical bowl is $V_{bowl} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \pi \times 9^3 = \frac{2}{3} \times \pi \times 729 = 486 \pi \,cm^3$.
Radius of each cylindrical bottle,$R = 1.5 \,cm$ and height,$h = 4 \,cm$.
The volume of one cylindrical bottle is $V_{bottle} = \pi R^2 h = \pi \times (1.5)^2 \times 4 = \pi \times 2.25 \times 4 = 9 \pi \,cm^3$.
Number of bottles required $= \frac{V_{bowl}}{V_{bottle}} = \frac{486 \pi}{9 \pi} = 54$.
Therefore,$54$ bottles are needed to empty the bowl.

(N/A) $(i)$ When a solid object is submerged in a container full of water,the volume of water displaced is equal to the volume of the submerged part of the object.
$(ii)$ The total volume of water initially in the cylinder is equal to the volume of the cylinder.
$(iii)$ The volume of water left in the cylinder $=$ (Total volume of the cylinder) $-$ (Volume of the submerged cone).
Given:
Height of the cone $(h_1)$ $= 120 \,cm$
Radius of the cone $(r)$ $= 60 \,cm$
Height of the cylinder $(h_2)$ $= 180 \,cm$
Radius of the cylinder $(r)$ $= 60 \,cm$
Volume of the cone $= \frac{1}{3} \pi r^2 h_1 = \frac{1}{3} \times \pi \times (60)^2 \times 120 = 144000 \pi \,cm^3$.
Volume of the cylinder $= \pi r^2 h_2 = \pi \times (60)^2 \times 180 = 648000 \pi \,cm^3$.
Since the cone is fully submerged,the volume of water displaced is equal to the volume of the cone,which is $144000 \pi \,cm^3$.
Volume of water left in the cylinder $= 648000 \pi - 144000 \pi = 504000 \pi \,cm^3$.
Using $\pi \approx \frac{22}{7}$:
Volume $= 504000 \times \frac{22}{7} = 72000 \times 22 = 1584000 \,cm^3$.
Converting to cubic meters $(1 \,m^3 = 10^6 \,cm^3)$:
Volume $= \frac{1584000}{1000000} \,m^3 = 1.584 \,m^3$.
Thus,the volume of water left in the cylinder is $1.584 \,m^3$.

(B) Given,radius of the tank,$r_{1} = 40 \,cm$.
Let the rise in the height of the water level in the tank in half an hour be $h_{1} \,cm$.
Internal radius of the cylindrical pipe,$r_{2} = 1 \,cm$.
Speed of water flow $= 80 \,cm/s$.
Time $= 30 \,minutes = 30 \times 60 = 1800 \,seconds$.
Length of water column flowing out of the pipe in $1800 \,seconds$,$h_{2} = 80 \times 1800 = 144000 \,cm$.
According to the principle of conservation of volume:
Volume of water in the cylindrical tank $=$ Volume of water flowed from the pipe in half an hour.
$\pi r_{1}^{2} h_{1} = \pi r_{2}^{2} h_{2}$
$40^2 \times h_{1} = 1^2 \times 144000$
$1600 \times h_{1} = 144000$
$h_{1} = \frac{144000}{1600} = 90 \,cm$.
Thus,the water level in the tank rises by $90 \,cm$ in half an hour.

(C) Given,length of roof $= 22 \, m$ and breadth of roof $= 20 \, m$.
Let the rainfall be $a \, cm = \frac{a}{100} \, m$.
Volume of water collected on the roof $= \text{length} \times \text{breadth} \times \text{height of rainfall} = 22 \times 20 \times \frac{a}{100} = \frac{22a}{5} \, m^3$.
Radius of the cylindrical vessel $r = \frac{\text{diameter}}{2} = \frac{2}{2} = 1 \, m$.
Height of the cylindrical vessel $h = 3.5 \, m = \frac{7}{2} \, m$.
Volume of the cylindrical vessel $= \pi r^2 h = \frac{22}{7} \times (1)^2 \times \frac{7}{2} = 11 \, m^3$.
Since the water collected from the roof fills the vessel,the volumes are equal:
$\frac{22a}{5} = 11$.
$a = \frac{11 \times 5}{22} = 2.5 \, cm$.
Thus,the rainfall is $2.5 \, cm$.

(D) Given:
Length of the cuboid $(l) = 10 \, cm$,breadth $(b) = 5 \, cm$,and height $(h) = 4 \, cm$.
Volume of the cuboid $= l \times b \times h = 10 \times 5 \times 4 = 200 \, cm^3$.
Radius of each conical depression $(r) = 0.5 \, cm$ and depth $(h_1) = 2.1 \, cm$.
Volume of one conical depression $= \frac{1}{3} \pi r^2 h_1 = \frac{1}{3} \times \frac{22}{7} \times (0.5)^2 \times 2.1 = \frac{1}{3} \times \frac{22}{7} \times 0.25 \times 2.1 = 0.55 \, cm^3$.
Volume of $4$ conical depressions $= 4 \times 0.55 = 2.2 \, cm^3$.
Edge of the cubical depression $(a) = 3 \, cm$.
Volume of the cubical depression $= a^3 = 3^3 = 27 \, cm^3$.
Volume of wood in the stand $=$ Volume of cuboid $-$ (Volume of $4$ conical depressions $+$ Volume of cubical depression)
$= 200 - (2.2 + 27) = 200 - 29.2 = 170.8 \, cm^3$.

(A) For both the cones:
Radius $r = 5 \, cm$ and height $h = 12 \, cm$.
For the cylinder:
Radius $r = 5 \, cm$.
Height of the cylinder $H = \text{Total height of the article} - 2 \times \text{Height of cone}$.
$H = 41 - 2 \times 12 = 41 - 24 = 17 \, cm$.
Slant height of the cone $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.
Total Surface Area $(TSA)$ of the combined article:
$TSA = \text{CSA of the cylinder} + 2 \times \text{CSA of the cone}$.
$TSA = 2 \pi r H + 2 \times (\pi r l) = 2 \pi r (H + l)$.
$TSA = 2 \times 3.14 \times 5 \times (17 + 13)$.
$TSA = 3.14 \times 10 \times 30 = 3.14 \times 300 = 942 \, cm^2$.
Thus,the total surface area of the given article is $942 \, cm^2$.

(B) Radius of the cone $=$ radius of the hemisphere $= r = 5 \, cm$.
Height of the cone $= h = 12 \, cm$.
The slant height of the cone is given by $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.
The total surface area of the article is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.
Total Surface Area $= \pi rl + 2\pi r^2 = \pi r(l + 2r)$.
Substituting the values: $3.14 \times 5 \times (13 + 2 \times 5) = 3.14 \times 5 \times (13 + 10) = 3.14 \times 5 \times 23$.
$= 15.7 \times 23 = 361.1 \, cm^2$.

(C) The radius of the cylinder is equal to the radius of both the hemispheres,$r = 7 \, cm$.
The height of the cylinder is calculated by subtracting the radii of the two hemispheres from the total height of the article:
$h = 34 \, cm - (2 \times 7 \, cm) = 34 - 14 = 20 \, cm$.
The total surface area $(TSA)$ of the combined article is the sum of the curved surface area $(CSA)$ of the cylinder and the curved surface areas of the two hemispheres:
$TSA = CSA_{\text{cylinder}} + 2 \times CSA_{\text{hemisphere}}$
$TSA = 2 \pi r h + 2 \times (2 \pi r^2)$
$TSA = 2 \pi r (h + 2r)$
Substituting the values:
$TSA = 2 \times \frac{22}{7} \times 7 \times (20 + 2 \times 7)$
$TSA = 44 \times (20 + 14)$
$TSA = 44 \times 34 = 1496 \, cm^2$.
Thus,the total surface area of the given article is $1496 \, cm^2$.

(D) Radius of the cylinder = radius of the cone = radius of the hemisphere = $r = 14 \, cm$.
Height of the cylinder $H = 22 \, cm$.
Height of the cone $h = 48 \, cm$.
Slant height of the cone $l = \sqrt{r^2 + h^2} = \sqrt{14^2 + 48^2} = \sqrt{196 + 2304} = \sqrt{2500} = 50 \, cm$.
The total surface area $(TSA)$ of the combined article is the sum of the curved surface areas $(CSA)$ of the cylinder,the cone,and the hemisphere.
$TSA = CSA_{\text{cylinder}} + CSA_{\text{cone}} + CSA_{\text{hemisphere}}$
$TSA = 2\pi rH + \pi rl + 2\pi r^2 = \pi r(2H + l + 2r)$.
Substituting the values: $TSA = \frac{22}{7} \times 14 \times (2 \times 22 + 50 + 2 \times 14)$.
$TSA = 44 \times (44 + 50 + 28) = 44 \times 122 = 5368 \, cm^2$.
Thus,the total surface area of the article is $5368 \, cm^2$.

(A) cuboid is formed by arranging six cubes touching each other face to face in a row.
For the cuboid so formed,the length $l = 6 \times 3 \, cm = 18 \, cm$.
The breadth $b = 3 \, cm$ and the height $h = 3 \, cm$.
The total surface area $(TSA)$ of the cuboid is given by the formula $2(lb + bh + hl)$.
Substituting the values: $TSA = 2(18 \times 3 + 3 \times 3 + 3 \times 18)$.
$TSA = 2(54 + 9 + 54) = 2(117) = 234 \, cm^2$.
Thus,the total surface area of the solid formed is $234 \, cm^2$.

(B) $1$. The article consists of a cylinder and two cones at its ends.
$2$. Radius of cylinder $(r)$ = $30 \, cm$.
$3$. Height of each cone $(h_{cone})$ = $40 \, cm$.
$4$. Total height of the article = $180 \, cm$.
$5$. Height of the cylinder $(h_{cyl})$ = $180 - (40 + 40) = 100 \, cm$.
$6$. Slant height of the cone $(l)$ = $\sqrt{r^2 + h_{cone}^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \, cm$.
$7$. Total Surface Area = (Curved Surface Area of Cylinder) + $2$ $\times$ (Curved Surface Area of Cone).
$8$. Total Surface Area = $(2 \pi r h_{cyl}) + 2 \times (\pi r l) = 2 \pi r (h_{cyl} + l)$.
$9$. Total Surface Area = $2 \times 3.14 \times 30 \times (100 + 50) = 188.4 \times 150 = 28260 \, cm^2$.

(C) The article consists of a cylinder and two hemispheres at its ends.
Given: Radius $r = 21 \, cm$,Total height $H = 92 \, cm$.
The height of the cylindrical part $h = H - 2r = 92 - 2(21) = 92 - 42 = 50 \, cm$.
The total surface area of the article = (Curved surface area of cylinder) + $2$ $\times$ (Curved surface area of hemisphere).
Total surface area = $2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$.
Substituting the values: $2 \times \frac{22}{7} \times 21 \times (50 + 2(21)) = 2 \times 22 \times 3 \times (50 + 42) = 132 \times 92 = 12,144 \, cm^2$.

(D) The radius $(r)$ of the hemispherical dome is $21\, m$.
The curved surface area $(CSA)$ of a hemisphere is given by the formula $2\pi r^2$.
$CSA = 2 \times \frac{22}{7} \times 21 \times 21$
$CSA = 2 \times 22 \times 3 \times 21$
$CSA = 44 \times 63 = 2772\, m^2$.
The cost of painting is ₹ $80$ per $m^2$.
Total cost = $CSA \times \text{Rate} = 2772 \times 80 = 2,21,760$.
Thus,the total cost of painting the outer surface is ₹ $2,21,760$.

(A) Given: Radius $(r)$ = $7\, cm$,Height $(h)$ = $24\, cm$.
First,find the slant height $(l)$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25\, cm$.
The surface area to be painted consists of the curved surface area of the cone and the curved surface area of the hemisphere.
Curved Surface Area of Cone = $\pi rl = \frac{22}{7} \times 7 \times 25 = 550\, cm^2$.
Curved Surface Area of Hemisphere = $2\pi r^2 = 2 \times \frac{22}{7} \times 7 \times 7 = 308\, cm^2$.
Total Surface Area = $550 + 308 = 858\, cm^2$.
Cost of painting = Total Surface Area $\times$ Rate = $858 \times 2 = ₹ 1716$.

(B) The article consists of a cylinder with two cones attached at its ends.
Radius of cylinder $(r)$ = $21 \, cm$.
Height of cylinder $(h)$ = $50 \, cm$.
Height of cone $(H)$ = $28 \, cm$.
Slant height of cone $(l)$ = $\sqrt{r^2 + H^2} = \sqrt{21^2 + 28^2} = \sqrt{441 + 784} = \sqrt{1225} = 35 \, cm$.
The total surface area of the article = (Curved surface area of cylinder) + $2 \times$ (Curved surface area of cone).
Curved surface area of cylinder = $2 \pi r h = 2 \times \frac{22}{7} \times 21 \times 50 = 44 \times 3 \times 50 = 6600 \, cm^2$.
Curved surface area of one cone = $\pi r l = \frac{22}{7} \times 21 \times 35 = 22 \times 3 \times 35 = 2310 \, cm^2$.
Total surface area = $6600 + 2 \times 2310 = 6600 + 4620 = 11220 \, cm^2$.

(C) Given: Radius $(r)$ = $3.5\, m$,Height $(h)$ = $12\, m$.
First,find the slant height $(l)$ of the cone using the formula $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{(3.5)^2 + (12)^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\, m$.
The curved surface area $(CSA)$ of the cone is given by $\pi rl$.
$CSA = \frac{22}{7} \times 3.5 \times 12.5 = 22 \times 0.5 \times 12.5 = 11 \times 12.5 = 137.5\, m^2$.
The cost of painting is at the rate of ₹ $30/m^2$.
Total cost = $137.5 \times 30 = ₹ 4125$.

(D) The article consists of a cylinder and two hemispheres at its ends.
Given: Radius of cylinder $(r)$ = $20 \, cm$,Total height $(H)$ = $80 \, cm$.
The height of the cylindrical part $(h)$ = $H - 2r = 80 - 2(20) = 80 - 40 = 40 \, cm$.
The total surface area of the article = (Curved surface area of cylinder) + $2$ $\times$ (Curved surface area of hemisphere).
$= 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r)$.
$= 2 \times 3.14 \times 20 \times (40 + 2 \times 20)$.
$= 125.6 \times (40 + 40) = 125.6 \times 80 = 10048 \, cm^2$.

(A) The article consists of a cylinder,a hemisphere,and a cone.
Radius of cylinder $(r)$ = $28 \, cm$,Height of cylinder $(h_1)$ = $54.5 \, cm$.
Radius of hemisphere $(r)$ = $28 \, cm$.
Radius of cone $(r)$ = $28 \, cm$,Height of cone $(h_2)$ = $21 \, cm$.
Slant height of cone $(l)$ = $\sqrt{r^2 + h_2^2} = \sqrt{28^2 + 21^2} = \sqrt{784 + 441} = \sqrt{1225} = 35 \, cm$.
The total surface area is the sum of the curved surface area of the cylinder,the curved surface area of the hemisphere,and the curved surface area of the cone.
Total Surface Area = $2\pi rh_1 + 2\pi r^2 + \pi rl$.
Total Surface Area = $\pi r (2h_1 + 2r + l) = \frac{22}{7} \times 28 \times (2 \times 54.5 + 2 \times 28 + 35)$.
Total Surface Area = $22 \times 4 \times (109 + 56 + 35) = 88 \times 200 = 17600 \, cm^2$.

(B) $1$. Identify the dimensions: Radius $(r)$ = $5 \, cm$,Height of each cone $(h_{cone})$ = $12 \, cm$,Total height $(H)$ = $41 \, cm$.
$2$. Calculate the height of the cylinder $(h_{cyl})$: $h_{cyl} = H - 2 \times h_{cone} = 41 - 2(12) = 41 - 24 = 17 \, cm$.
$3$. Calculate the slant height $(l)$ of the cone: $l = \sqrt{r^2 + h_{cone}^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.
$4$. The total surface area of the article consists of the curved surface area of the cylinder and the curved surface areas of the two cones.
$5$. Total Surface Area = $2 \pi r h_{cyl} + 2 \times (\pi r l) = 2 \pi r (h_{cyl} + l)$.
$6$. Substitute the values: $2 \times 3.14 \times 5 \times (17 + 13) = 31.4 \times 30 = 942 \, cm^2$.

(C) The article consists of a cylinder and a cone joined together.
Given: Radius of cylinder and cone $(r) = 6 \, cm$,Height of cone $(h_{cone}) = 8 \, cm$,Total height of article $= 20 \, cm$.
Height of cylinder $(h_{cyl}) = 20 \, cm - 8 \, cm = 12 \, cm$.
Slant height of cone $(l) = \sqrt{r^2 + h_{cone}^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm$.
Total Surface Area = (Curved Surface Area of Cylinder) + (Curved Surface Area of Cone) + (Area of base of cylinder).
Total Surface Area = $(2 \pi r h_{cyl}) + (\pi r l) + (\pi r^2)$.
Total Surface Area = $\pi r (2 h_{cyl} + l + r) = 3.14 \times 6 \times (2 \times 12 + 10 + 6)$.
Total Surface Area = $18.84 \times (24 + 10 + 6) = 18.84 \times 40 = 753.60 \, cm^2$.
Wait,re-evaluating: The article is open at the other end,so the base area is not included.
Surface Area = (Curved Surface Area of Cylinder) + (Curved Surface Area of Cone) = $2 \pi r h_{cyl} + \pi r l$.
Surface Area = $\pi r (2 h_{cyl} + l) = 3.14 \times 6 \times (2 \times 12 + 10) = 18.84 \times (24 + 10) = 18.84 \times 34 = 640.56 \, cm^2$.

(D) Let the radius of the hemispherical base be $r$ and the slant height of the cone be $l = 5 \, cm$.
The total surface area of the article is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.
Total Surface Area $= \pi rl + 2\pi r^2 = 103.62 \, cm^2$.
Given $\pi = 3.14$,we have $3.14 \times r \times 5 + 2 \times 3.14 \times r^2 = 103.62$.
$15.7r + 6.28r^2 = 103.62$.
Dividing by $3.14$: $5r + 2r^2 = 33$.
$2r^2 + 5r - 33 = 0$.
Solving the quadratic equation: $2r^2 + 11r - 6r - 33 = 0 \implies r(2r + 11) - 3(2r + 11) = 0$.
$(r - 3)(2r + 11) = 0$. Since $r > 0$,$r = 3 \, cm$.
The height of the cone $h$ is given by $h = \sqrt{l^2 - r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \, cm$.
The total height of the article is the sum of the height of the cone and the radius of the hemisphere: $H = h + r = 4 + 3 = 7 \, cm$.

(A) Given: Radius of the cone $(r)$ = $15 \, cm$,Height of the cone $(h)$ = $20 \, cm$.
First,calculate the slant height $(l)$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \, cm$.
The total surface area of the article consists of the curved surface area of the cone and the curved surface area of the hemisphere.
Curved surface area of the cone = $\pi rl = 3.14 \times 15 \times 25 = 1177.5 \, cm^2$.
Curved surface area of the hemisphere = $2\pi r^2 = 2 \times 3.14 \times 15^2 = 2 \times 3.14 \times 225 = 1413 \, cm^2$.
Total surface area = $1177.5 + 1413 = 2590.5 \, cm^2$.

(B) Radius of the cone $=$ radius of the hemisphere $= r = 15\, cm$ and slant height of the cone $l = 25\, cm$.
For a cone,$l^{2} = r^{2} + h^{2}$.
$\therefore h^{2} = l^{2} - r^{2} = 25^{2} - 15^{2} = 625 - 225 = 400 = (20)^{2}$.
So,$h = 20\, cm$.
Volume of the combined solid $=$ Volume of the cone $+$ Volume of the hemisphere.
Volume $= \frac{1}{3} \pi r^{2} h + \frac{2}{3} \pi r^{3} = \frac{1}{3} \pi r^{2} (h + 2r)$.
Substituting the values: Volume $= \frac{1}{3} \times 3.14 \times 15 \times 15 \times (20 + 2 \times 15)$.
Volume $= 3.14 \times 5 \times 15 \times (20 + 30) = 3.14 \times 75 \times 50 = 3.14 \times 3750 = 11775\, cm^{3}$.
Thus,the volume of the given solid is $11775\, cm^{3}$.

(C) Radius of the cylinder $=$ radius of the hemisphere $= r = 8.4 \,cm$.
Height of the cylinder $h =$ Total height $-$ Radius of the hemisphere.
$h = 52.8 - 8.4 = 44.4 \,cm$.
Volume of the combined tank $=$ Volume of the cylinder $+$ Volume of the hemisphere.
$V = \pi r^2 h + \frac{2}{3} \pi r^3 = \pi r^2 (h + \frac{2}{3} r)$.
$V = \frac{22}{7} \times 8.4 \times 8.4 \times (44.4 + \frac{2}{3} \times 8.4)$.
$V = \frac{22}{7} \times 8.4 \times 8.4 \times (44.4 + 5.6)$.
$V = \frac{22}{7} \times 8.4 \times 8.4 \times 50$.
$V = 22 \times 1.2 \times 8.4 \times 50 = 11088 \,cm^3$.
Since $1000 \,cm^3 = 1 \,litre$,the volume in litres is $\frac{11088}{1000} = 11.088 \,litres$.

(D) Radius of the cone $=$ radius of the hemisphere $= r = 15 \, cm$.
Height of the cone $h = \text{Total height} - \text{Radius of the hemisphere} = 55 - 15 = 40 \, cm$.
Volume of the combined solid $=$ Volume of the cone $+$ Volume of the hemisphere.
Volume $= \frac{1}{3} \pi r^2 h + \frac{2}{3} \pi r^3 = \frac{1}{3} \pi r^2 (h + 2r)$.
Substituting the values: Volume $= \frac{1}{3} \times \frac{22}{7} \times 15 \times 15 \times (40 + 2 \times 15)$.
Volume $= \frac{1}{3} \times \frac{22}{7} \times 225 \times (40 + 30) = \frac{1}{3} \times \frac{22}{7} \times 225 \times 70$.
Volume $= 22 \times 75 \times 10 = 16500 \, cm^3$.