Mix Examples - Surface Areas and Volumes Questions in English

(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 $n = \frac{V_{cylinder}}{V_{ball}} = \frac{144\pi}{\frac{1}{6}\pi} = 144 \times 6 = 864$.

(C) The volume of the metallic cylinder is given by $V_{cylinder} = \pi r^2 h$. Substituting $r = 2\, cm$ and $h = 8\, cm$,we get $V_{cylinder} = \pi \times (2)^2 \times 8 = 32\pi\, cm^3$.
The volume of one spherical ball with diameter $0.5\, cm$ (radius $r_b = 0.25\, cm = 1/4\, cm$) is $V_{ball} = \frac{4}{3} \pi r_b^3 = \frac{4}{3} \pi (1/4)^3 = \frac{4}{3} \pi \times \frac{1}{64} = \frac{\pi}{48}\, cm^3$.
The number of balls produced is $n = \frac{V_{cylinder}}{V_{ball}} = \frac{32\pi}{\pi/48} = 32 \times 48 = 1536$.

(D) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
When three spheres are melted and recast into a new sphere,the total volume remains conserved.
Let the radii of the three spheres be $r_1 = 3 \, cm$,$r_2 = 4 \, cm$,and $r_3 = 5 \, cm$.
Let the radius of the new sphere be $R$.
The total volume of the three spheres is $V_{total} = \frac{4}{3} \pi (r_1^3 + r_2^3 + r_3^3)$.
The volume of the new sphere is $V_{new} = \frac{4}{3} \pi R^3$.
Equating the volumes: $\frac{4}{3} \pi (3^3 + 4^3 + 5^3) = \frac{4}{3} \pi R^3$.
$R^3 = 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216$.
$R = \sqrt[3]{216} = 6 \, cm$.

(A) The volume of the metallic sphere is given by $V = \frac{4}{3} \pi r^3$,where $r = 6 \, cm$.
$V = \frac{4}{3} \times \pi \times (6)^3 = \frac{4}{3} \times \pi \times 216 = 288 \pi \, cm^3$.
The wire is in the shape of a cylinder with diameter $1 \, cm$,so the radius $R = 0.5 \, cm = \frac{1}{2} \, cm$.
Let the length of the wire be $h$. The volume of the wire is $V = \pi R^2 h$.
Equating the volumes: $288 \pi = \pi \times (\frac{1}{2})^2 \times h$.
$288 = \frac{1}{4} \times h$.
$h = 288 \times 4 = 1152 \, cm$.
Since $100 \, cm = 1 \, m$,the length in metres is $1152 / 100 = 11.52 \, m$.

(B) The volume of the soil taken out from the well is equal to the volume of the cylinder formed by the well.
Volume of soil = $\pi r^2 h = \pi \times (2)^2 \times 50 = 200\pi \, m^3$.
Let the increase in the level of the circular plot be $H$ meters.
The soil is spread over a circular plot of radius $R = 40 \, m$.
The volume of the soil spread is equal to the volume of the cylinder of radius $40 \, m$ and height $H$.
Volume = $\pi R^2 H = \pi \times (40)^2 \times H = 1600\pi H \, m^3$.
Equating the two volumes: $1600\pi H = 200\pi$.
$H = \frac{200}{1600} = \frac{1}{8} \, m$.
To convert the height into $cm$,multiply by $100$: $H = \frac{1}{8} \times 100 = 12.5 \, cm$.

(C) The volume of the metallic cone is given by the formula $V_{cone} = \frac{1}{3} \pi r^2 h$. Substituting the given values $r = 5 \, cm$ and $h = 8 \, cm$,we get $V_{cone} = \frac{1}{3} \pi (5)^2 (8) = \frac{200}{3} \pi \, cm^3$.
The volume of each spherical ball with diameter $0.5 \, cm$ (radius $r_b = 0.25 \, cm = \frac{1}{4} \, cm$) is $V_{ball} = \frac{4}{3} \pi r_b^3 = \frac{4}{3} \pi (\frac{1}{4})^3 = \frac{4}{3} \pi (\frac{1}{64}) = \frac{1}{48} \pi \, cm^3$.
The number of balls produced is $n = \frac{V_{cone}}{V_{ball}} = \frac{\frac{200}{3} \pi}{\frac{1}{48} \pi} = \frac{200}{3} \times 48 = 200 \times 16 = 3200$.

(D) The volume of the metallic sphere is given by the formula $V = \frac{4}{3} \pi r^3$,where $r = 6 \, cm$.
$V = \frac{4}{3} \pi (6)^3 = \frac{4}{3} \pi (216) = 288 \pi \, cm^3$.
The wire is in the shape of a cylinder with radius $R = 0.4 \, cm$ and length $h$. The volume of the wire is $V = \pi R^2 h$.
Since the volume remains constant during the process of melting and recasting,we have:
$288 \pi = \pi (0.4)^2 h$
$288 = 0.16 \times h$
$h = \frac{288}{0.16} = \frac{28800}{16} = 1800 \, cm$.
To convert the length into metres,we divide by $100$:
$h = \frac{1800}{100} = 18 \, m$.

(A) The volume of the metallic sphere with diameter $21\,cm$ (radius $R = 10.5\,cm$ or $21/2\,cm$) is given by $V_s = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (\frac{21}{2})^3 = \frac{4}{3} \pi \times \frac{9261}{8} = \frac{3087}{2} \pi \, cm^3$.
The volume of one cone with diameter $7\,cm$ (radius $r = 3.5\,cm$ or $7/2\,cm$) and height $h = 3\,cm$ is given by $V_c = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (\frac{7}{2})^2 \times 3 = \pi \times \frac{49}{4} = 12.25 \pi \, cm^3$.
The number of cones $n$ is calculated as $n = \frac{V_s}{V_c} = \frac{3087 \pi / 2}{49 \pi / 4} = \frac{3087}{2} \times \frac{4}{49} = \frac{3087 \times 2}{49} = 63 \times 2 = 126$.
Thus,$126$ cones can be produced.

(B) For the bucket in the shape of a frustum of a cone,the larger radius $r_{1} = 28\,cm$,the smaller radius $r_{2} = 7\,cm$,and the height $h = 45\,cm$.
The volume of the frustum of a cone is given by the formula:
$V = \frac{1}{3} \pi h (r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})$
Substituting the values:
$V = \frac{1}{3} \times \frac{22}{7} \times 45 \times (28^{2} + 7^{2} + 28 \times 7)$
$V = \frac{22 \times 15}{7} \times (784 + 49 + 196)$
$V = \frac{22 \times 15}{7} \times 1029$
$V = 22 \times 15 \times 147$
$V = 48510\,cm^{3}$
Since $1000\,cm^{3} = 1\,litre$:
$V = \frac{48510}{1000}\,litres = 48.51\,litres$
Thus,the bucket can hold $48.51\,litres$ of water.

(N/A) For the given frustum of a cone,
bigger radius $r_{1} = 14 \,cm$,smaller radius $r_{2} = 7 \,cm$,and height $h = 24 \,cm$.
Slant height of the frustum of a cone $l = \sqrt{h^{2} + (r_{1} - r_{2})^{2}}$
$= \sqrt{24^{2} + (14 - 7)^{2}}$
$= \sqrt{576 + 49}$
$= \sqrt{625} = 25 \,cm$.
Curved Surface Area $(CSA)$ $= \pi l(r_{1} + r_{2})$
$= \frac{22}{7} \times 25 \times (14 + 7)$
$= \frac{22}{7} \times 25 \times 21 = 22 \times 25 \times 3 = 1650 \,cm^{2}$.
Total Surface Area $(TSA)$ $= \pi l(r_{1} + r_{2}) + \pi r_{1}^{2} + \pi r_{2}^{2}$
$= 1650 + \frac{22}{7} \times (14^{2} + 7^{2})$
$= 1650 + \frac{22}{7} \times (196 + 49)$
$= 1650 + \frac{22}{7} \times 245$
$= 1650 + 22 \times 35 = 1650 + 770 = 2420 \,cm^{2}$.
Volume $= \frac{1}{3} \pi h(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})$
$= \frac{1}{3} \times \frac{22}{7} \times 24 \times (196 + 49 + 98)$
$= \frac{22 \times 8}{7} \times 343$
$= 22 \times 8 \times 49 = 8624 \,cm^{3}$.
Thus,the $CSA$ is $1650 \,cm^{2}$,$TSA$ is $2420 \,cm^{2}$,and volume is $8624 \,cm^{3}$.

(D) For the frustum of a cone,the circumference of the base is $C_1 = 2 \pi r_1 = 48 \, cm$,so $r_1 = \frac{24}{\pi} \, cm$.
The circumference of the top is $C_2 = 2 \pi r_2 = 36 \, cm$,so $r_2 = \frac{18}{\pi} \, cm$.
The height of the frustum is $h = 11 \, cm$.
The volume $V$ of a frustum of a cone is given by $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.
Substituting the values: $V = \frac{1}{3} \pi (11) \left[ (\frac{24}{\pi})^2 + (\frac{18}{\pi})^2 + (\frac{24}{\pi})(\frac{18}{\pi}) \right]$.
$V = \frac{11 \pi}{3} \left[ \frac{576}{\pi^2} + \frac{324}{\pi^2} + \frac{432}{\pi^2} \right]$.
$V = \frac{11 \pi}{3} \left[ \frac{1332}{\pi^2} \right] = \frac{11 \times 1332}{3 \pi} = \frac{11 \times 444}{\pi} = \frac{4884}{\pi}$.
Using $\pi \approx \frac{22}{7}$,$V = \frac{4884 \times 7}{22} = 222 \times 7 = 1554 \, cm^3$.

(A) Given: Diameter $D_1 = 32 \,cm$,so radius $r_1 = 16 \,cm$. Diameter $D_2 = 20 \,cm$,so radius $r_2 = 10 \,cm$. Height $h = 8 \,cm$.
$1$. Slant height $l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{8^2 + (16 - 10)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \,cm$.
$2$. Curved Surface Area $(CSA)$ = $\pi(r_1 + r_2)l = 3.14 \times (16 + 10) \times 10 = 3.14 \times 26 \times 10 = 816.4 \,cm^2$.
$3$. Volume $V = \frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1r_2) = \frac{1}{3} \times 3.14 \times 8 \times (16^2 + 10^2 + 16 \times 10) = \frac{1}{3} \times 3.14 \times 8 \times (256 + 100 + 160) = \frac{1}{3} \times 3.14 \times 8 \times 516 = 3.14 \times 8 \times 172 = 4320.64 \,cm^3$.

(B) The volume $V$ of a frustum of a cone is given by the formula $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr)$.
Given: $R = 28 \, cm$, $r = 21 \, cm$, and $V = 28.490 \, \text{litres} = 28490 \, cm^3$ (since $1 \, \text{litre} = 1000 \, cm^3$).
Substituting the values into the formula:
$28490 = \frac{1}{3} \times \frac{22}{7} \times h \times (28^2 + 21^2 + 28 \times 21)$.
$28490 = \frac{22}{21} \times h \times (784 + 441 + 588)$.
$28490 = \frac{22}{21} \times h \times (1813)$.
$28490 = \frac{22 \times 1813}{21} \times h$.
$28490 = \frac{39886}{21} \times h$.
$h = \frac{28490 \times 21}{39886} = \frac{598290}{39886} = 15 \, cm$.

(A) Given: Radii $r_1 = 9\, cm$,$r_2 = 3\, cm$,and height $h = 8\, cm$.
First,calculate the slant height $l$ using the formula $l = \sqrt{h^2 + (r_1 - r_2)^2}$.
$l = \sqrt{8^2 + (9 - 3)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\, cm$.
Curved Surface Area $(CSA)$ of the frustum is given by $\pi(r_1 + r_2)l$.
$CSA = 3.14 \times (9 + 3) \times 10 = 3.14 \times 12 \times 10 = 376.8\, cm^2$.
Volume of the frustum is given by $\frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1r_2)$.
$V = \frac{1}{3} \times 3.14 \times 8 \times (9^2 + 3^2 + 9 \times 3) = \frac{1}{3} \times 3.14 \times 8 \times (81 + 9 + 27) = \frac{1}{3} \times 3.14 \times 8 \times 117$.
$V = 3.14 \times 8 \times 39 = 979.68\, cm^3$.

(N/A) Given: $R = 35 \, cm$,$r = 21 \, cm$,$h = 48 \, cm$.
First,calculate the slant height $l = \sqrt{h^2 + (R - r)^2} = \sqrt{48^2 + (35 - 21)^2} = \sqrt{2304 + 196} = \sqrt{2500} = 50 \, cm$.
Curved Surface Area $(CSA)$ $= \pi(R + r)l = \frac{22}{7} \times (35 + 21) \times 50 = \frac{22}{7} \times 56 \times 50 = 22 \times 8 \times 50 = 8800 \, cm^2$.
Total Surface Area $(TSA)$ $= \pi l(R + r) + \pi R^2 + \pi r^2 = 8800 + \frac{22}{7} \times (35^2 + 21^2) = 8800 + \frac{22}{7} \times (1225 + 441) = 8800 + \frac{22}{7} \times 1666 = 8800 + 22 \times 238 = 8800 + 5236 = 14036 \, cm^2$.
Volume $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 48 \times (35^2 + 21^2 + 35 \times 21) = \frac{22 \times 16}{7} \times (1225 + 441 + 735) = \frac{352}{7} \times 2401 = 352 \times 343 = 120736 \, cm^3$.

(A) Given: Radius of cylinder $(r)$ = $8 \, cm$,Height of cylinder $(h_1)$ = $9 \, cm$,Height of cone $(h_2)$ = $15 \, cm$.
Slant height of cone $(l)$ = $\sqrt{r^2 + h_2^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \, cm$.
Total Surface Area = Curved surface area of cylinder + Curved surface area of cone + Base area of cylinder.
$= 2\pi rh_1 + \pi rl + \pi r^2 = \pi r(2h_1 + l + r) = \frac{22}{7} \times 8 \times (2 \times 9 + 17 + 8) = \frac{176}{7} \times (18 + 17 + 8) = \frac{176}{7} \times 43 \approx 1081.14 \, cm^2$.
(Note: Using $\pi = 3.14$,Area $\approx 3.14 \times 8 \times 43 = 1079.68 \, cm^2$).
Volume = Volume of cylinder + Volume of cone = $\pi r^2 h_1 + \frac{1}{3} \pi r^2 h_2 = \pi r^2 (h_1 + \frac{h_2}{3}) = \frac{22}{7} \times 8^2 \times (9 + \frac{15}{3}) = \frac{22}{7} \times 64 \times 14 = 22 \times 64 \times 2 = 2816 \, cm^3$.

(A) Given: Radius of cone and hemisphere $r = 6 \, cm$,Height of cone $h = 8 \, cm$.
First,find the slant height $l$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \, cm$.
Curved Surface Area $(CSA)$ of the solid = $CSA$ of cone + $CSA$ of hemisphere.
$CSA$ = $\pi rl + 2\pi r^2 = \pi r(l + 2r) = 3.14 \times 6 \times (10 + 2 \times 6) = 18.84 \times (10 + 12) = 18.84 \times 22 = 414.48 \, cm^2$.
Volume of the solid = Volume of cone + Volume of hemisphere.
Volume = $\frac{1}{3}\pi r^2h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2(h + 2r) = \frac{1}{3} \times 3.14 \times 6^2 \times (8 + 2 \times 6) = \frac{1}{3} \times 3.14 \times 36 \times (8 + 12) = 3.14 \times 12 \times 20 = 37.68 \times 20 = 753.6 \, cm^3$.

(C) The total surface area of the tent is the sum of the curved surface area of the cylinder and the curved surface area of the cone.
Radius $(r)$ = $1.2\, m$.
Height of cylinder $(h_1)$ = $4\, m$.
Height of cone $(h_2)$ = $3.5\, m$.
Slant height of the cone $(l)$ = $\sqrt{r^2 + h_2^2} = \sqrt{(1.2)^2 + (3.5)^2} = \sqrt{1.44 + 12.25} = \sqrt{13.69} = 3.7\, m$.
Curved surface area of cylinder = $2\pi rh_1 = 2 \times \frac{22}{7} \times 1.2 \times 4 \approx 30.17\, m^2$.
Curved surface area of cone = $\pi rl = \frac{22}{7} \times 1.2 \times 3.7 \approx 13.95\, m^2$.
Total area = $30.17 + 13.95 = 44.12\, m^2$.
Rounding to the nearest whole number,the cloth used is $44\, m^2$.

(N/A) $1$. Volume of the container: The container consists of a cylinder and a cone. The radius $r = 35 \,cm$,height of cylinder $h_1 = 52 \,cm$,and height of cone $h_2 = 12 \,cm$.
Volume of cylinder $V_1 = \pi r^2 h_1 = \frac{22}{7} \times 35 \times 35 \times 52 = 200,200 \,cm^3$.
Volume of cone $V_2 = \frac{1}{3} \pi r^2 h_2 = \frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 12 = 15,400 \,cm^3$.
Total volume $V = V_1 + V_2 = 200,200 + 15,400 = 215,600 \,cm^3$.
Since $1,000 \,cm^3 = 1 \,litre$,the volume is $215.6 \,litres$.
$2$. Total Surface Area: The surface area includes the base of the cylinder,the curved surface area of the cylinder,and the curved surface area of the cone.
Slant height of cone $l = \sqrt{r^2 + h_2^2} = \sqrt{35^2 + 12^2} = \sqrt{1225 + 144} = \sqrt{1369} = 37 \,cm$.
Base area of cylinder $= \pi r^2 = \frac{22}{7} \times 35 \times 35 = 3,850 \,cm^2$.
Curved surface area of cylinder $= 2 \pi r h_1 = 2 \times \frac{22}{7} \times 35 \times 52 = 11,440 \,cm^2$.
Curved surface area of cone $= \pi r l = \frac{22}{7} \times 35 \times 37 = 4,070 \,cm^2$.
Total surface area $= 3,850 + 11,440 + 4,070 = 19,360 \,cm^2$.

(A) The tank consists of a cylinder and two hemispheres at the ends.
Radius of the cylinder $(r)$ = $21\, cm$.
Radius of the hemispheres $(r)$ = $21\, cm$.
Total height of the tank = $62\, cm$.
Height of the cylindrical part $(h)$ = Total height - $2 \times$ radius of hemisphere = $62 - 2(21) = 62 - 42 = 20\, cm$.
Volume of the tank = Volume of cylinder + $2 \times$ Volume of hemisphere.
Volume = $\pi r^2 h + 2 \times (2/3) \pi r^3 = \pi r^2 (h + 4/3 r)$.
Volume = $(22/7) \times (21)^2 \times (20 + (4/3) \times 21) = (22/7) \times 441 \times (20 + 28) = 22 \times 63 \times 48 = 66528\, cm^3$.
Since $1000\, cm^3 = 1\, liter$, the volume in liters = $66528 / 1000 = 66.528\, liters$.

(N/A) $1$. Slant height $(l)$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{21^2 + 20^2} = \sqrt{441 + 400} = \sqrt{841} = 29 \, cm$.
$2$. Curved Surface Area $(CSA)$ of the cone: $CSA_{cone} = \pi rl = \frac{22}{7} \times 21 \times 29 = 22 \times 3 \times 29 = 1914 \, cm^2$.
$3$. Curved Surface Area $(CSA)$ of the hemisphere: $CSA_{hemi} = 2\pi r^2 = 2 \times \frac{22}{7} \times 21 \times 21 = 2 \times 22 \times 3 \times 21 = 2772 \, cm^2$.
$4$. Total Curved Surface Area: $1914 + 2772 = 4686 \, cm^2$.
$5$. Volume of the cone $(V_{cone})$: $V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 21 \times 21 \times 20 = 22 \times 21 \times 20 = 9240 \, cm^3$.
$6$. Volume of the hemisphere $(V_{hemi})$: $V_{hemi} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 2 \times 22 \times 21 \times 21 = 19404 \, cm^3$.
$7$. Total Volume: $9240 + 19404 = 28644 \, cm^3$.

(C) The tank consists of a cylinder and two hemispheres at the ends.
Radius of the cylinder and hemispheres, $r = 24 / 2 = 12 \, cm$.
Height of the cylindrical part, $h = 78 - (12 + 12) = 78 - 24 = 54 \, cm$.
Volume of the tank = Volume of cylinder + $2 \times$ Volume of hemisphere.
Volume = $\pi r^2 h + 2 \times (2/3 \pi r^3) = \pi r^2 (h + 4/3 r)$.
Volume = $3.14159 \times (12)^2 \times (54 + 4/3 \times 12) = 3.14159 \times 144 \times 70$.
Volume = $31667.2$
approx $31667 \, cm^3$.
Since $1000 \, cm^3 = 1 \, litre$, the volume in litres is $31667 / 1000 = 31.667 \, litres \approx 31.68 \, litres$ (using $\pi \approx 22/7$ gives $31.68 \, litres$).

(D) The tank consists of one cylinder and two identical cones.
Radius of the cylinder and cones $(r)$ = $14 / 2 = 7 \, cm$.
Height of the cylinder $(h_1)$ = $20 \, cm$.
Height of each cone $(h_2)$ = $12 \, cm$.
Volume of the tank = (Volume of cylinder) + $2$ $\times$ (Volume of cone).
Volume of cylinder = $\pi r^2 h_1 = (22/7) \times 7^2 \times 20 = 22 \times 7 \times 20 = 3080 \, cm^3$.
Volume of one cone = $(1/3) \pi r^2 h_2 = (1/3) \times (22/7) \times 7^2 \times 12 = 22 \times 7 \times 4 = 616 \, cm^3$.
Total volume = $3080 + 2 \times 616 = 3080 + 1232 = 4312 \, cm^3$.

(A) The solid consists of a cone and a hemisphere joined at their bases.
Radius of the cone $(r)$ = $0.6 \ m$.
Height of the cone $(h)$ = $1.6 \ m$.
Radius of the hemisphere $(r)$ = $0.6 \ m$.
Volume of the cone = $\frac{1}{3} \pi r^{2} h = \frac{1}{3} \times \pi \times (0.6)^{2} \times 1.6 = 0.192 \pi \ m^{3}$.
Volume of the hemisphere = $\frac{2}{3} \pi r^{3} = \frac{2}{3} \times \pi \times (0.6)^{3} = 0.144 \pi \ m^{3}$.
Total volume = $0.192 \pi + 0.144 \pi = 0.336 \pi \ m^{3}$.
Using $\pi \approx 3.14159$,Total volume $\approx 0.336 \times 3.14159 \approx 1.05557 \ m^{3}$,which rounds to $1.056 \ m^{3}$.

(B) The volume of the cylinder is given by $V_{cylinder} = \pi r^2 h$. Given diameter $= 1 \, cm$,so radius $r = 0.5 \, cm = \frac{1}{2} \, cm$ and height $h = 4 \, cm$.
$V_{cylinder} = \pi \times (\frac{1}{2})^2 \times 4 = \pi \times \frac{1}{4} \times 4 = \pi \, cm^3$.
The volume of one spherical ball is $V_{sphere} = \frac{4}{3} \pi r^3$. Given radius $r = \frac{1}{8} \, cm$.
$V_{sphere} = \frac{4}{3} \times \pi \times (\frac{1}{8})^3 = \frac{4}{3} \times \pi \times \frac{1}{512} = \frac{\pi}{3 \times 128} = \frac{\pi}{384} \, cm^3$.
The number of balls produced is $n = \frac{V_{cylinder}}{V_{sphere}} = \frac{\pi}{\frac{\pi}{384}} = 384$.

(C) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Let $R$ be the radius of the large metallic sphere $(R = 42 \, cm)$ and $r$ be the radius of the small ball $(r = 1.4 \, cm)$.
The number of balls $n$ that can be produced is given by the ratio of the volume of the large sphere to the volume of one small ball:
$n = \frac{\frac{4}{3} \pi R^3}{\frac{4}{3} \pi r^3} = \left( \frac{R}{r} \right)^3$.
Substituting the values:
$n = \left( \frac{42}{1.4} \right)^3 = (30)^3$.
$n = 27,000$.
Therefore,$27,000$ balls can be produced.

(D) The volume of the metallic cylinder is given by $V_{cylinder} = \pi r^2 h = \pi \times (4)^2 \times 3 = 48\pi\, cm^3$.
The volume of one spherical ball is given by $V_{ball} = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \pi \times (0.5)^3 = \frac{4}{3} \times \pi \times 0.125 = \frac{0.5}{3}\pi = \frac{1}{6}\pi\, cm^3$.
The number of balls produced is $n = \frac{V_{cylinder}}{V_{ball}} = \frac{48\pi}{\frac{1}{6}\pi} = 48 \times 6 = 288$.

(A) The volume of the hemispherical bowl is given by $V_h = \frac{2}{3} \pi r^3$,where $r = 9 \, cm$.
$V_h = \frac{2}{3} \times \pi \times (9)^3 = \frac{2}{3} \times \pi \times 729 = 486 \pi \, cm^3$.
The volume of one cylindrical bottle is given by $V_c = \pi r_c^2 h$,where diameter $d = 3 \, cm$,so radius $r_c = 1.5 \, cm$ and height $h = 4 \, cm$.
$V_c = \pi \times (1.5)^2 \times 4 = \pi \times 2.25 \times 4 = 9 \pi \, cm^3$.
The number of bottles that can be filled is $n = \frac{V_h}{V_c} = \frac{486 \pi}{9 \pi} = 54$.
Therefore,$54$ bottles can be filled.

(A) The article consists of a cylinder and two hemispheres attached at both ends.
Radius of the cylinder $(r)$ = $21 \, cm$.
Height of the cylinder $(h)$ = $18 \, cm$.
Radius of the hemispheres $(r)$ = $21 \, cm$.
The total surface area of the article = (Curved surface area of the cylinder) + $2 \times$ (Curved surface area of the hemisphere).
Curved surface area of the cylinder = $2 \pi r h = 2 \times \frac{22}{7} \times 21 \times 18 = 2 \times 22 \times 3 \times 18 = 2376 \, cm^2$.
Curved surface area of two hemispheres = $2 \times (2 \pi r^2) = 4 \pi r^2 = 4 \times \frac{22}{7} \times 21 \times 21 = 4 \times 22 \times 3 \times 21 = 5544 \, cm^2$.
Total surface area = $2376 + 5544 = 7920 \, cm^2$.

(C) Given: Radius of hemisphere $(r)$ = $3.5\, cm = \frac{7}{2}\, cm$.
Total height of the toy = $15.5\, cm$.
Height of the cone $(h)$ = Total height - Radius of hemisphere = $15.5 - 3.5 = 12\, cm$.
Slant height of the cone $(l)$ = $\sqrt{r^2 + h^2} = \sqrt{(3.5)^2 + 12^2} = \sqrt{12.25 + 144} = \sqrt{156.25} = 12.5\, cm$.
Total surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere.
$= \pi rl + 2\pi r^2 = \pi r(l + 2r)$.
$= \frac{22}{7} \times 3.5 \times (12.5 + 2 \times 3.5) = 11 \times (12.5 + 7) = 11 \times 19.5 = 214.5\, cm^2$.

(D) The article consists of a cone placed on a hemisphere. The total surface area is the sum of the curved surface area of the cone and the curved surface area of the hemisphere.
$1$. Slant height of the cone $(l)$: $l = \sqrt{r^2 + h^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, cm$.
$2$. Curved surface area of the cone $(CSA_{cone})$: $\pi rl = 3.14 \times 5 \times 13 = 204.1 \, cm^2$.
$3$. Curved surface area of the hemisphere $(CSA_{hemi})$: $2\pi r^2 = 2 \times 3.14 \times 5^2 = 2 \times 3.14 \times 25 = 157 \, cm^2$.
$4$. Total surface area: $CSA_{cone} + CSA_{hemi} = 204.1 + 157 = 361.1 \, cm^2$.

(A) The volume of a cylinder is given by the formula $V = \pi r^2 h$.
For the first cylinder: $r_1 = 3.5 \, cm$,$h_1 = 10 \, cm$. Volume $V_1 = \pi (3.5)^2 (10) = \pi (12.25)(10) = 122.5 \pi \, cm^3$.
For the second cylinder: $r_2 = 7 \, cm$,$h_2 = 10 \, cm$. Volume $V_2 = \pi (7)^2 (10) = \pi (49)(10) = 490 \pi \, cm^3$.
Total volume of the two cylinders = $V_1 + V_2 = 122.5 \pi + 490 \pi = 612.5 \pi \, cm^3$.
Let the radius of the new cylinder be $R$ and its height be $H = 50 \, cm$.
The volume of the new cylinder is $V_{new} = \pi R^2 H = \pi R^2 (50)$.
Since the volume remains constant,$50 \pi R^2 = 612.5 \pi$.
$R^2 = \frac{612.5}{50} = 12.25$.
$R = \sqrt{12.25} = 3.5 \, cm$.

(B) The volume of a frustum of a cone is given by the formula: $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.
Given: $r_1 = 14 \, cm$,$r_2 = 7 \, cm$,and $h = 12 \, cm$.
Substituting the values into the formula:
$V = \frac{1}{3} \times \frac{22}{7} \times 12 \times (14^2 + 7^2 + 14 \times 7)$
$V = \frac{1}{3} \times \frac{22}{7} \times 12 \times (196 + 49 + 98)$
$V = \frac{1}{3} \times \frac{22}{7} \times 12 \times (343)$
$V = 22 \times 4 \times 49 = 4312 \, cm^3$.