Volume of a Combination of Solids Questions in English

(C) Let the edges of the three cubes be $3x, 4x,$ and $5x \, \text{cm}$ respectively.
The volume of the three cubes is $V = (3x)^3 + (4x)^3 + (5x)^3 = 27x^3 + 64x^3 + 125x^3 = 216x^3 \, \text{cm}^3$.
Let the side of the new single cube be $a$. The volume of this new cube is $a^3 = 216x^3$,which implies $a = 6x$.
The diagonal of a cube with side $a$ is given by $d = a\sqrt{3}$.
Given that the diagonal is $12\sqrt{3} \, \text{cm}$,we have $a\sqrt{3} = 12\sqrt{3}$,so $a = 12 \, \text{cm}$.
Since $a = 6x$,we have $6x = 12$,which gives $x = 2$.
Therefore,the edges of the three cubes are $3(2) = 6 \, \text{cm}$,$4(2) = 8 \, \text{cm}$,and $5(2) = 10 \, \text{cm}$.

(D) Given,the edges of three solid cubes are $3 \, cm$,$4 \, cm$,and $5 \, cm$ respectively.
The volume of a cube is given by the formula $V = (\text{edge})^3$.
Volume of the first cube $= (3)^3 = 27 \, cm^3$.
Volume of the second cube $= (4)^3 = 64 \, cm^3$.
Volume of the third cube $= (5)^3 = 125 \, cm^3$.
Total volume of the three cubes $= 27 + 64 + 125 = 216 \, cm^3$.
Let the edge of the new single cube be $R \, cm$.
Then,the volume of the new cube $= R^3$.
Since the total volume remains constant,$R^3 = 216$.
Taking the cube root on both sides,$R = \sqrt[3]{216} = 6 \, cm$.
Thus,the edge of the resulting cube is $6 \, cm$.

(A) Given,dimensions of the cuboid $= 9 \, cm \times 11 \, cm \times 12 \, cm$.
Volume of the cuboid $= 9 \times 11 \times 12 = 1188 \, cm^3$.
Diameter of the spherical shot $= 3 \, cm$.
Radius of the shot,$r = \frac{3}{2} = 1.5 \, cm$.
Volume of one spherical shot $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.5)^3$.
Volume of one shot $= \frac{4}{3} \times \frac{22}{7} \times 3.375 = \frac{99}{7} \approx 14.143 \, cm^3$.
Number of shots $= \frac{\text{Volume of cuboid}}{\text{Volume of one shot}} = \frac{1188}{99/7} = \frac{1188 \times 7}{99} = 12 \times 7 = 84$.
Thus,$84$ shots can be made.

(B) Given, volume of the frustum $= 28.49 \, \text{L} = 28.49 \times 1000 \, \text{cm}^3 = 28490 \, \text{cm}^3$ (since $1 \, \text{L} = 1000 \, \text{cm}^3$).
Radius of the top $(r_1) = 28 \, \text{cm}$.
Radius of the bottom $(r_2) = 21 \, \text{cm}$.
Let the height of the bucket be $h \, \text{cm}$.
The volume of a frustum of a cone is given by $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.
Substituting the values: $28490 = \frac{1}{3} \times \frac{22}{7} \times h \times (28^2 + 21^2 + 28 \times 21)$.
$28490 = \frac{22}{21} \times h \times (784 + 441 + 588)$.
$28490 = \frac{22}{21} \times h \times 1813$.
$h = \frac{28490 \times 21}{22 \times 1813}$.
$h = \frac{598290}{39886} = 15 \, \text{cm}$.

(C) Let the cone be $ORN$ with base radius $r_1 = 8 \, cm$ and height $OM = 12 \, cm$.
Let $P$ be the mid-point of $OM$,then $OP = PM = \frac{12}{2} = 6 \, cm$.
Since the plane is parallel to the base,$\triangle OPD \sim \triangle OMN$.
By properties of similar triangles,$\frac{OP}{OM} = \frac{PD}{MN}$.
$\frac{6}{12} = \frac{PD}{8} \Rightarrow \frac{1}{2} = \frac{PD}{8} \Rightarrow PD = 4 \, cm$.
The plane divides the cone into a smaller cone (top) and a frustum (bottom).
Volume of smaller cone $V_1 = \frac{1}{3} \pi (PD)^2 (OP) = \frac{1}{3} \pi (4)^2 (6) = 32 \pi \, cm^3$.
Volume of the original cone $V = \frac{1}{3} \pi (MN)^2 (OM) = \frac{1}{3} \pi (8)^2 (12) = 256 \pi \, cm^3$.
Volume of the frustum $V_2 = V - V_1 = 256 \pi - 32 \pi = 224 \pi \, cm^3$.
The ratio of the volume of the smaller cone to the frustum is $V_1 : V_2 = 32 \pi : 224 \pi = 1 : 7$.

(A) Given that,the side of the solid cube $(a) = 7 \, cm$.
The height of the conical cavity (cone),$h = 7 \, cm$.
The radius of the conical cavity (cone),$r = 3 \, cm$.
Volume of the cube $= a^3 = (7)^3 = 343 \, cm^3$.
Volume of the conical cavity (cone) $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3)^2 \times 7$.
$= \frac{1}{3} \times 22 \times 9 = 66 \, cm^3$.
Volume of the remaining solid $=$ Volume of the cube $-$ Volume of the conical cavity.
$= 343 - 66 = 277 \, cm^3$.
Thus,the volume of the remaining solid is $277 \, cm^3$.

(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: $A = \pi l (r_1 + r_2) + \pi r_2^2$.
$A = 3.14 \times 31.62 \times (20 + 10) + 3.14 \times 10^2 = 3.14 \times 31.62 \times 30 + 314 = 2978.56 + 314 = 3292.56 \, cm^2 \approx 3292.6 \, cm^2$.

(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 $= \frac{\text{Volume of vessel}}{\text{Volume per minute}} = \frac{3200 \pi}{62.5 \pi} = \frac{3200}{62.5} = 51.2 \, minutes$.
$51.2 \, minutes = 51 \, minutes + 0.2 \times 60 \, seconds = 51 \, minutes \, 12 \, seconds$.

(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 $43.27 \,cm$.

(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 $2.5 \, cm$.

(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 $= 22 \times 75 \times 10 = 16500 \, cm^3$.

(A) Radius of the cylinder $=$ radius of each hemisphere $= r = 2.1 \,m$.
Height of the cylinder $h =$ Total height of the tank $- 2 \times$ Radius of the hemisphere $= 9.2 - 2 \times 2.1 = 5 \,m$.
Volume of the combined tank $=$ Volume of the cylinder $+ 2 \times$ Volume of the hemisphere
$= \pi r^{2} h + 2 \times (\frac{2}{3} \pi r^{3}) = \pi r^{2} (h + \frac{4}{3} r)$
$= \frac{22}{7} \times 2.1 \times 2.1 \times (5 + \frac{4}{3} \times 2.1)$
$= \frac{22}{7} \times 2.1 \times 2.1 \times (5 + 2.8) = \frac{22}{7} \times 2.1 \times 2.1 \times 7.8 = 108.108 \,m^{3}$.
Since $1 \,m^{3} = 1000$ litres,
Volume in litres $= 108.108 \times 1000 = 1,08,108$ litres.
Thus,the tank can contain $1,08,108$ litres of petrol.

(B) Radius of the cylinder $=$ radius of each cone $= r = 10 \,cm$.
Height of the cylinder $H = 40 \,cm$.
Slant height of the cone $l = 26 \,cm$.
For the cone,$l^2 = r^2 + h^2$,where $h$ is the height of the cone.
$h = \sqrt{l^2 - r^2} = \sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24 \,cm$.
Volume of the combined container $=$ Volume of the cylinder $+ 2 \times$ Volume of the cone.
Volume $= \pi r^2 H + 2 \times (\frac{1}{3} \pi r^2 h) = \pi r^2 (H + \frac{2}{3} h)$.
Substituting the values: Volume $= \frac{22}{7} \times 10^2 \times (40 + \frac{2}{3} \times 24) = \frac{22}{7} \times 100 \times (40 + 16) = \frac{22}{7} \times 100 \times 56$.
Volume $= 22 \times 100 \times 8 = 17600 \,cm^3$.
Since $1000 \,cm^3 = 1 \,litre$,the volume in litres is $\frac{17600}{1000} = 17.6 \,litres$.
Thus,the model can contain $17.6 \,litres$ of water.

(C) The solid consists of a cylinder and two cones at its ends.
Given:
Radius of cylinder and cones $(r)$ = $7\, cm$.
Height of each cone $(h_1)$ = $18\, cm$.
Total height of the solid $(H)$ = $64\, cm$.
Height of the cylinder $(h_2)$ = $H - 2 \times h_1 = 64 - 2(18) = 64 - 36 = 28\, cm$.
Volume of the solid = Volume of cylinder + $2 \times$ Volume of cone.
Volume = $\pi r^2 h_2 + 2 \times (\frac{1}{3} \pi r^2 h_1)$.
Volume = $\pi r^2 (h_2 + \frac{2}{3} h_1)$.
Volume = $\frac{22}{7} \times 7^2 \times (28 + \frac{2}{3} \times 18)$.
Volume = $22 \times 7 \times (28 + 12) = 154 \times 40 = 6160\, cm^3$.

(D) The volume of the solid is the sum of the volume of the cone and the volume of the hemisphere.
Given: Radius $r = 14 \, cm$,Height of cone $h = 32 \, cm$.
Volume of cone $V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times 32 = \frac{1}{3} \times 22 \times 2 \times 14 \times 32 = \frac{19712}{3} \approx 6570.67 \, cm^3$.
Volume of hemisphere $V_2 = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 = \frac{2}{3} \times 22 \times 2 \times 14 \times 14 = \frac{17248}{3} \approx 5749.33 \, cm^3$.
Total Volume $V = V_1 + V_2 = \frac{19712 + 17248}{3} = \frac{36960}{3} = 12320 \, cm^3$.

(A) The solid consists of a cylinder and two cones at its ends.
Radius of cylinder $(r)$ = $21 \, cm$.
Height of each cone $(h_{cone})$ = $30 \, cm$.
Total height of the solid = $140 \, cm$.
Height of the cylinder $(h_{cyl})$ = Total height - $2 \times h_{cone} = 140 - 2(30) = 140 - 60 = 80 \, cm$.
Volume of the solid = Volume of cylinder + $2 \times$ Volume of cone.
Volume of cylinder = $\pi r^2 h_{cyl} = \frac{22}{7} \times 21 \times 21 \times 80 = 22 \times 3 \times 21 \times 80 = 1,10,880 \, cm^3$.
Volume of two cones = $2 \times (\frac{1}{3} \pi r^2 h_{cone}) = 2 \times \frac{1}{3} \times \frac{22}{7} \times 21 \times 21 \times 30 = 2 \times 22 \times 21 \times 30 = 27,720 \, cm^3$.
Total Volume = $1,10,880 + 27,720 = 1,38,600 \, cm^3$.

(B) The solid consists of a cylinder and two hemispheres at its ends.
Radius of the cylinder $(r)$ = $4.2 \, cm$.
Total height of the solid = $15 \, cm$.
Height of the cylindrical part $(h)$ = Total height - $2 \times$ radius of hemisphere = $15 - 2(4.2) = 15 - 8.4 = 6.6 \, cm$.
Volume of the solid = Volume of cylinder + $2 \times$ Volume of hemisphere.
Volume of solid = $\pi r^2 h + 2 \times (\frac{2}{3} \pi r^3) = \pi r^2 (h + \frac{4}{3} r)$.
Volume = $\frac{22}{7} \times (4.2)^2 \times (6.6 + \frac{4}{3} \times 4.2)$.
Volume = $\frac{22}{7} \times 17.64 \times (6.6 + 5.6) = 22 \times 2.52 \times 12.2$.
Volume = $55.44 \times 12.2 = 676.368 \, cm^3$.

(C) The top consists of a cone and a hemisphere joined at their bases. The radius of both the cone and the hemisphere is $r = 2.8 \, cm$. The height of the cone is $h = 6.4 \, cm$.
Volume of the cone $V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (2.8)^2 \times 6.4 = \frac{1}{3} \times \frac{22}{7} \times 7.84 \times 6.4 \approx 52.53 \, cm^3$.
Volume of the hemisphere $V_{hemi} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (2.8)^3 = \frac{2}{3} \times \frac{22}{7} \times 21.952 \approx 46.03 \, cm^3$.
Total volume $V = V_{cone} + V_{hemi} = 52.53 + 46.03 = 98.56 \, cm^3$.

(A) The container consists of a cylinder and a hemisphere.
Radius of the cylinder $(r)$ = $8.4 \,cm$.
Radius of the hemisphere $(r)$ = $8.4 \,cm$.
Total height of the container $(H)$ = $52.8 \,cm$.
Height of the cylinder $(h)$ = $H - r = 52.8 - 8.4 = 44.4 \,cm$.
Volume of the container = Volume of cylinder + Volume of hemisphere.
Volume = $\pi r^2 h + \frac{2}{3} \pi r^3 = \pi r^2 (h + \frac{2}{3} r)$.
Volume = $\frac{22}{7} \times (8.4)^2 \times (44.4 + \frac{2}{3} \times 8.4)$.
Volume = $\frac{22}{7} \times 70.56 \times (44.4 + 5.6) = \frac{22}{7} \times 70.56 \times 50$.
Volume = $22 \times 10.08 \times 50 = 11088 \,cm^3$.
Since $1000 \,cm^3 = 1 \,litre$,the volume in litres = $11088 / 1000 = 11.088 \,litres$.

(A) Given: Radius of the cone $(r)$ = $15 \, cm$,Slant height $(l)$ = $25 \, cm$.
First,find the height $(h)$ of the cone using the relation $l^2 = r^2 + h^2$.
$h^2 = 25^2 - 15^2 = 625 - 225 = 400$.
$h = \sqrt{400} = 20 \, cm$.
Volume of the cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times 15^2 \times 20 = 3.14 \times 225 \times \frac{20}{3} = 3.14 \times 75 \times 20 = 3.14 \times 1500 = 4710 \, cm^3$.
Volume of the hemisphere = $\frac{2}{3} \pi r^3 = \frac{2}{3} \times 3.14 \times 15^3 = \frac{2}{3} \times 3.14 \times 3375 = 2 \times 3.14 \times 1125 = 7065 \, cm^3$.
Total volume of the solid = Volume of cone + Volume of hemisphere = $4710 + 7065 = 11775 \, cm^3$.

(B) The solid consists of a cone mounted on a hemisphere.
Given: Radius of the hemisphere $(r)$ = $15 \, cm$.
Total height of the solid $(H)$ = $55 \, cm$.
Height of the cone $(h)$ = $H - r = 55 - 15 = 40 \, cm$.
Volume of the solid = Volume of the cone + Volume of the hemisphere.
Volume of the cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times (15)^2 \times 40 = 3000 \pi \, cm^3$.
Volume of the hemisphere = $\frac{2}{3} \pi r^3 = \frac{2}{3} \times \pi \times (15)^3 = 2250 \pi \, cm^3$.
Total Volume = $3000 \pi + 2250 \pi = 5250 \pi \, cm^3$.
Using $\pi \approx 3.14$,Total Volume = $5250 \times 3.14 = 16485 \, cm^3$. Rounding to the nearest option,the answer is $16500 \, cm^3$.

(C) For the cylindrical well,the radius $r = \frac{5.4}{2} = 2.7 \, m$ and the depth $h = 20 \, m$.
The volume of the earth dug out is $V = \pi r^2 h = \pi \times (2.7)^2 \times 20 \, m^3$.
This earth is spread on a circular ground of radius $R = 15 \, m$ to a height $H$. The volume of this spread earth is $V = \pi R^2 H = \pi \times (15)^2 \times H \, m^3$.
Equating the volumes: $\pi \times (2.7)^2 \times 20 = \pi \times (15)^2 \times H$.
$H = \frac{2.7 \times 2.7 \times 20}{15 \times 15} = \frac{7.29 \times 20}{225} = \frac{145.8}{225} = 0.648 \, m$.
Converting the height to centimeters: $H = 0.648 \times 100 = 64.8 \, cm$.
Thus,the increase in the level of the ground is $64.8 \, cm$.

(D) Radius of the sphere $R = 21 \,cm$.
Volume of the sphere $= \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (21)^3 \,cm^3$.
For the cylindrical wire,diameter $= 0.5 \,cm$,so radius $r = \frac{0.5}{2} = 0.25 = \frac{1}{4} \,cm$.
Let the length of the wire be $h \,cm$.
Volume of the wire $= \pi r^2 h = \pi (\frac{1}{4})^2 h \,cm^3$.
Since the sphere is recast into a wire,their volumes are equal:
$\pi (\frac{1}{4})^2 h = \frac{4}{3} \pi (21)^3$
$\frac{1}{16} h = \frac{4}{3} \times 21 \times 21 \times 21$
$h = 16 \times 4 \times 7 \times 21 \times 21$
$h = 64 \times 7 \times 441 = 197568 \,cm$.
To convert into meters,divide by $100$:
$h = \frac{197568}{100} = 1975.68 \,m$.
Thus,the length of the wire is $1975.68 \,m$.

(A) Let the radius of the large metallic sphere be $r = 42\, cm$ and the radius of each small ball be $R\, cm$.
The volume of the large sphere is equal to the total volume of the $8000$ small balls.
Volume of sphere $= \frac{4}{3} \pi r^3$
Volume of $8000$ balls $= 8000 \times \frac{4}{3} \pi R^3$
Equating the volumes: $8000 \times \frac{4}{3} \pi R^3 = \frac{4}{3} \pi r^3$
$8000 R^3 = r^3$
$8000 R^3 = (42)^3$
$R^3 = \frac{42 \times 42 \times 42}{8000}$
$R^3 = \left(\frac{42}{20}\right)^3$
$R = \frac{42}{20} = 2.1\, cm$
Thus,the radius of each small ball is $2.1\, cm$.

(B) Radius of the cylinder $r = 15\, cm$ and height $h = 20\, cm$.
Radius of each ball $R = 3\, cm$.
Suppose $n$ balls are produced.
Since the volume remains constant during melting and recasting:
Volume of $n$ balls $=$ Volume of the cylinder
$n \times \frac{4}{3} \pi R^3 = \pi r^2 h$
$n \times \frac{4}{3} \times (3)^3 = (15)^2 \times 20$
$n \times \frac{4}{3} \times 27 = 225 \times 20$
$n \times 36 = 4500$
$n = \frac{4500}{36} = 125$
Thus,$125$ balls are produced.

(C) Radius of the sphere $R = 18 \,cm$.
For each cylinder,radius $r = 3 \,cm$ and height $h = 12 \,cm$.
Suppose $n$ cylinders are produced.
Volume of $n$ cylinders = Volume of the sphere.
$n \times \pi r^{2} h = \frac{4}{3} \pi R^{3}$.
$n \times r^{2} h = \frac{4}{3} R^{3}$.
$n \times 3^{2} \times 12 = \frac{4}{3} \times 18^{3}$.
$n \times 9 \times 12 = \frac{4}{3} \times 5832$.
$n \times 108 = 4 \times 1944$.
$n \times 108 = 7776$.
$n = \frac{7776}{108} = 72$.
Thus,$72$ cylinders are produced.

(D) The volume of the metallic sphere is given by the formula $V = \frac{4}{3} \pi r^3$,where $r = 21 \,cm$.
$V = \frac{4}{3} \times \pi \times (21)^3 = 4 \times \pi \times 441 \times 7 = 12348 \pi \,cm^3$.
The wire is in the shape of a cylinder with radius $R = 0.5 \,cm$ and length $h$. The volume of the wire is $V = \pi R^2 h$.
Equating the volumes: $12348 \pi = \pi \times (0.5)^2 \times h$.
$12348 = 0.25 \times h$.
$h = \frac{12348}{0.25} = 12348 \times 4 = 49392 \,cm$.
To convert the length into metres,divide by $100$: $h = \frac{49392}{100} = 493.92 \,m$.

(A) Given: Diameter of the metallic sphere $D = 12 \, cm$,so radius $R = 6 \, cm$.
Radius of the smaller ball $r = 0.3 \, cm$.
Let the number of balls produced be $n$.
Since the volume remains constant during melting and recasting,the volume of the large sphere equals the sum of the volumes of $n$ small balls.
$V_{\text{large}} = n \times V_{\text{small}}$
$\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$
$R^3 = n \times r^3$
$n = \frac{R^3}{r^3} = \left( \frac{R}{r} \right)^3$
$n = \left( \frac{6}{0.3} \right)^3 = (20)^3$
$n = 8000$.
Thus,the number of balls produced is $8000$.

(B) The volume of the metallic cylinder is given by $V_{cylinder} = \pi r^2 h = \pi \times (6)^2 \times 4 = 144\pi \, cm^3$.
The diameter of each ball is $1 \, cm$,so the radius $r_{ball} = 0.5 \, cm$.
The volume of one spherical ball is $V_{ball} = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (0.5)^3 = \frac{4}{3} \pi (0.125) = \frac{0.5}{3} \pi = \frac{1}{6} \pi \, cm^3$.
The number of balls produced is $n = \frac{V_{cylinder}}{V_{ball}} = \frac{144\pi}{\frac{1}{6}\pi} = 144 \times 6 = 864$.