Volume of a Combination of Solids Questions in English

(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.

(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 $\frac{1}{3}\pi h(r_1^2 + r_2^2 + r_1r_2)$.
$V = \frac{1}{3} \times 3.14 \times 8 \times (9^2 + 3^2 + 9 \times 3) = \frac{1}{3} \times 3.14 \times 8 \times (81 + 9 + 27) = \frac{1}{3} \times 3.14 \times 8 \times 117$.
$V = 3.14 \times 8 \times 39 = 979.68\, cm^3$.

(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 = $\frac{1}{3}\pi r^2h + \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^2(h + 2r) = \frac{1}{3} \times 3.14 \times 6^2 \times (8 + 2 \times 6) = \frac{1}{3} \times 3.14 \times 36 \times (8 + 12) = 3.14 \times 12 \times 20 = 37.68 \times 20 = 753.6 \, cm^3$.

(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 = $66528 / 1000 = 66.528\, liters$.

(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 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:
$V = \frac{1}{3} \times \frac{22}{7} \times 12 \times (14^2 + 7^2 + 14 \times 7)$
$V = \frac{1}{3} \times \frac{22}{7} \times 12 \times (196 + 49 + 98)$
$V = \frac{1}{3} \times \frac{22}{7} \times 12 \times (343)$
$V = 22 \times 4 \times 49 = 4312 \, cm^3$.

(A) Given: $R = 28 \, cm$,$r = 7 \, cm$,$h = 72 \, cm$.
First,calculate the slant height $l = \sqrt{h^2 + (R - r)^2} = \sqrt{72^2 + (28 - 7)^2} = \sqrt{5184 + 441} = \sqrt{5625} = 75 \, cm$.
Curved Surface Area $(CSA)$ $= \pi(R + r)l = \frac{22}{7} \times (28 + 7) \times 75 = 22 \times 5 \times 75 = 8250 \, cm^2$.
Total Surface Area $(TSA)$ $= \text{CSA} + \pi R^2 + \pi r^2 = 8250 + \frac{22}{7} \times (28^2 + 7^2) = 8250 + \frac{22}{7} \times (784 + 49) = 8250 + 22 \times 119 = 8250 + 2618 = 10868 \, cm^2$.
Volume $= \frac{1}{3} \pi h (R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 72 \times (28^2 + 7^2 + 28 \times 7) = \frac{22 \times 24}{7} \times (784 + 49 + 196) = \frac{528}{7} \times 1029 = 528 \times 147 = 77616 \, cm^3$.

(B) The volume of a cone with base radius $r$ and height $h$ is given by the formula $V = \frac{1}{3} \pi r^{2} h$.

(A) sphere is a three-dimensional object where every point on the surface is at an equal distance $r$ from the center. The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^{3}$.
$A$ hemisphere is exactly half of a sphere. Therefore,the volume of a hemisphere is half the volume of a sphere.
Volume of a hemisphere $= \frac{1}{2} \times (\text{Volume of a sphere})$
Volume of a hemisphere $= \frac{1}{2} \times \frac{4}{3} \pi r^{3} = \frac{2}{3} \pi r^{3}$.

(C) The volume $V$ of a sphere with radius $r$ is given by the formula:
$V = \frac{4}{3} \pi r^{3}$
Therefore,the correct option is $C$.

(A) The volume of a cylinder with base radius $r$ and height $h$ is calculated by multiplying the area of the circular base by the height of the cylinder.
Area of the base $= \pi r^{2}$.
Therefore,Volume of the cylinder $= \text{Area of base} \times \text{height} = \pi r^{2} h$.

(C) The diameter of the sphere is $d = 10 \, cm$.
Therefore,the radius $r$ is $r = \frac{d}{2} = \frac{10}{2} = 5 \, cm$.
The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Substituting the value of $r$ into the formula:
$V = \frac{4}{3} \pi (5)^3$
$V = \frac{4}{3} \pi (125)$
$V = \frac{500}{3} \pi \, cm^3$.

(A) The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Given the radius $r = 1.5 \text{ cm} = \frac{3}{2} \text{ cm}$.
Substituting the value of $r$ into the formula:
$V = \frac{4}{3} \pi \left( \frac{3}{2} \right)^3$
$V = \frac{4}{3} \pi \left( \frac{27}{8} \right)$
$V = \left( \frac{4}{3} \times \frac{27}{8} \right) \pi$
$V = \left( \frac{1}{1} \times \frac{9}{2} \right) \pi$
$V = 4.5 \pi \text{ cm}^3$.

(D) The formula for the volume of a frustum of a cone is $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.
Given: $r_1 = 6 \, cm$,$r_2 = 4 \, cm$,and $h = 9 \, cm$.
Substituting the values into the formula:
$V = \frac{1}{3} \pi (9) (6^2 + 4^2 + 6 \times 4)$
$V = 3 \pi (36 + 16 + 24)$
$V = 3 \pi (76)$
$V = 228 \pi \, cm^3$.
Thus,the volume is $228 \pi \, cm^3$.

(A) The volume of a cylinder is given by the formula $V = \pi r^2 h$,where $V$ is the volume,$r$ is the radius,and $h$ is the height.
Given: $V = 550 \, cm^3$,$r = 5 \, cm$,and $\pi \approx \frac{22}{7}$.
Substituting the values into the formula: $550 = \frac{22}{7} \times (5)^2 \times h$.
$550 = \frac{22}{7} \times 25 \times h$.
$550 = \frac{550}{7} \times h$.
$h = 550 \times \frac{7}{550}$.
$h = 7 \, cm$.

(B) The formula for the volume of a cone is $V = \frac{1}{3} \pi r^{2} h$.
Given that the radius $r = 5 \, cm$ and the height $h = 12 \, cm$.
Substituting these values into the formula:
$V = \frac{1}{3} \times \pi \times (5)^{2} \times 12$
$V = \frac{1}{3} \times \pi \times 25 \times 12$
$V = \pi \times 25 \times 4$
$V = 100 \pi \, cm^{3}$.
Therefore,the volume is $100 \, \pi \, cm^{3}$.

(C) Let the diameters of the two cones be $d_1$ and $d_2$,and their heights be $h_1$ and $h_2$.
The ratio of diameters is $d_1:d_2 = 2:3$,so the ratio of their radii is $r_1:r_2 = 2:3$.
The ratio of their heights is $h_1:h_2 = 9:4$.
The volume of a cone is given by $V = \frac{1}{3}\pi r^2 h$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right)$.
Substituting the given ratios: $\frac{V_1}{V_2} = \left(\frac{2}{3}\right)^2 \times \left(\frac{9}{4}\right) = \frac{4}{9} \times \frac{9}{4} = 1$.
Thus,the ratio of their volumes is $1:1$.

(D) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$,where $r$ is the radius.
Let the original radius be $r_1 = r$.
Then the original volume is $V_1 = \frac{4}{3} \pi r^3$.
If the radius is doubled,the new radius becomes $r_2 = 2r$.
The new volume is $V_2 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = 8 \times (\frac{4}{3} \pi r^3)$.
Therefore,$V_2 = 8 \times V_1$.
Thus,the volume becomes $8$ times the volume of the original sphere.

(B) The formula for the volume of a cylinder is $V = \pi r^{2} h$.
Given that the radius $r = 5 \, cm$ and the height $h = 5 \, cm$.
Substituting these values into the formula:
$V = \pi \times (5)^{2} \times 5$
$V = \pi \times 25 \times 5$
$V = 125\pi \, cm^{3}$.
Therefore,the volume is $125\pi \, cm^{3}$.

(C) The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^{3}$.
Given the radius $r = 3 \, cm$.
Substituting the value of $r$ into the formula:
$V = \frac{2}{3} \times \pi \times (3)^{3}$
$V = \frac{2}{3} \times \pi \times 27$
$V = 2 \times \pi \times 9$
$V = 18 \pi \, cm^{3}$.
Therefore,the volume is $18 \pi \, cm^{3}$.

(D) Let the diameters of the two cylinders be $d_1$ and $d_2$,and their heights be $h_1$ and $h_2$.
Given,$d_1:d_2 = 3:4$,so the radii $r_1:r_2 = 3:4$.
Given,$h_1:h_2 = 4:5$.
The volume of a cylinder is given by $V = \pi r^2 h$.
The ratio of the volumes is $\frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right)$.
Substituting the given ratios: $\frac{V_1}{V_2} = \left(\frac{3}{4}\right)^2 \times \left(\frac{4}{5}\right) = \frac{9}{16} \times \frac{4}{5} = \frac{9 \times 1}{4 \times 5} = \frac{9}{20}$.
Thus,the ratio of their volumes is $9:20$.

(B) The volume $V$ of a sphere with radius $r$ is given by the formula $V = \frac{4}{3} \pi r^3$.
If the radius is doubled,the new radius $r' = 2r$.
The new volume $V'$ is given by $V' = \frac{4}{3} \pi (r')^3$.
Substituting $r' = 2r$ into the formula,we get $V' = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = 8 \times (\frac{4}{3} \pi r^3)$.
Therefore,$V' = 8V$.
Thus,the volume becomes $8$ times the original volume.

(B) The volume $V$ of a sphere with radius $r$ is given by the formula $V = \frac{4}{3} \pi r^{3}$.
Therefore,the correct option is $B$.

(A) sphere has a volume of $V = \frac{4}{3} \pi r^{3}$.
Since a hemisphere is exactly half of a sphere,its volume is half of the volume of the sphere.
Therefore,the volume of a hemisphere $= \frac{1}{2} \times (\frac{4}{3} \pi r^{3}) = \frac{2}{3} \pi r^{3}$.

(B) $1-$ rupee coin is cylindrical in shape.
The volume of a cylinder is given by the formula $V = \pi r^{2} h$,where $r$ is the radius of the base and $h$ is the thickness (height) of the coin.
Therefore,the volume of a $1-$ rupee coin is $V = \pi r^{2} h$.

(B) frustum of a cone is formed when a cone is cut by a plane parallel to its base.
Let $h$ be the height of the frustum,and $r_{1}$ and $r_{2}$ be the radii of the two circular bases.
The formula for the volume $V$ of a frustum of a cone is given by:
$V = \frac{1}{3} \pi h(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})$
Therefore,option $B$ is the correct formula.

(C) We know that $1 \text{ litre}$ is defined as the volume of a cube with side length $10 \text{ cm}$.
Since the volume of a cube is given by $(\text{side})^3$,we have:
$1 \text{ litre} = (10 \text{ cm})^3 = 10 \times 10 \times 10 \text{ cm}^3 = 1000 \text{ cm}^3$.
Therefore,$1 \text{ litre} = 1000 \text{ cm}^3$.

(C) We know that $1 \, m = 100 \, cm$.
Therefore,$1 \, m^3 = (100 \, cm) \times (100 \, cm) \times (100 \, cm) = 1,000,000 \, cm^3$.
We also know that $1 \, \text{liter} = 1000 \, cm^3$.
To convert $cm^3$ to liters,we divide by $1000$.
Thus,$1,000,000 \, cm^3 / 1000 = 1000 \, \text{liters}$.
So,$1 \, m^3 = 1000 \, \text{liters}$.

(C) We know that $1 \ m^{3} = 1000 \ \text{litres}$.
Also,$1 \ \text{kilolitre} = 1000 \ \text{litres}$.
Therefore,$1 \ m^{3} = 1 \ \text{kilolitre}$.

(D) The $CSA$ (Curved Surface Area) of a cylinder is given by the formula $2 \pi r h$.
The volume of a cylinder is given by the formula $\pi r^{2} h$.
According to the problem,the $CSA$ and the volume are numerically equal:
$2 \pi r h = \pi r^{2} h$
Dividing both sides by $\pi r h$ (assuming $r \neq 0$ and $h \neq 0$):
$2 = r$
Therefore,the radius of the cylinder is $2$ units.

(C) The radius of the cylinder $r$ is given by $r = \frac{\text{diameter}}{2} = \frac{14\, cm}{2} = 7\, cm$.
The formula for the volume of a cylinder is $V = \pi r^2 h$.
Substituting the values $r = 7\, cm$ and $h = 10\, cm$ into the formula:
$V = \frac{22}{7} \times (7)^2 \times 10$
$V = \frac{22}{7} \times 49 \times 10$
$V = 22 \times 7 \times 10$
$V = 1540\, cm^3$.
Therefore,the volume of the cylinder is $1540\, cm^3$.

(D) The area of the base of a cylinder is given by the formula $A = \pi r^{2}$.
Given that the area of the base is $20 \, cm^{2}$,we have $\pi r^{2} = 20 \, cm^{2}$.
The volume $V$ of a cylinder is calculated using the formula $V = \pi r^{2} h$.
Substituting the given values,$V = (20 \, cm^{2}) \times (5 \, cm) = 100 \, cm^{3}$.
Therefore,the volume of the cylinder is $100 \, cm^{3}$.

(B) The formula for the volume of a cylinder is $V = \pi r^2 h$.
Given that the volume $V = 20 \pi \, cm^3$ and the height $h = 5 \, cm$.
Substituting these values into the formula: $20 \pi = \pi r^2 \times 5$.
Dividing both sides by $5 \pi$,we get $r^2 = 4$.
Taking the square root,$r = 2 \, cm$.
The diameter $d$ is twice the radius: $d = 2r = 2 \times 2 = 4 \, cm$.