Mix Examples - Surface Areas and Volumes Questions in English

(B) For a single cylindrical coin:
Radius $(r) = 1.4 \, cm = \frac{14}{10} \, cm$
Height $(h) = 0.1 \, cm = \frac{1}{10} \, cm$
Volume of one coin $= \pi r^2 h = \frac{22}{7} \times (\frac{14}{10})^2 \times \frac{1}{10} \, cm^3$
$= \frac{22}{7} \times \frac{196}{100} \times \frac{1}{10} = \frac{22 \times 28}{1000} = \frac{616}{1000} \, cm^3 = 0.616 \, cm^3$
Since $60$ such coins are stacked to form a larger cylinder,the total volume is:
Total Volume $= 60 \times 0.616 \, cm^3 = 36.96 \, cm^3$.

(C) The formula for the volume of a cylinder is $V = \pi r^2 h$.
Given:
Radius $(r) = 10 \, cm$
Height $(h) = 20 \, cm$
$\pi = 3.14$
Substituting the values into the formula:
$V = 3.14 \times (10)^2 \times 20$
$V = 3.14 \times 100 \times 20$
$V = 314 \times 20$
$V = 6280 \, cm^3$
Therefore,the volume of the cylinder is $6280 \, cm^3$.

(A) The volume $V$ of a cylinder is given by the formula $V = \pi r^2 h$,where $r$ is the radius and $h$ is the height.
Given: $r = 3.5\,cm = \frac{7}{2}\,cm$ and $V = 2310\,cm^3$.
Substituting the values into the formula:
$2310 = \frac{22}{7} \times (\frac{7}{2})^2 \times h$
$2310 = \frac{22}{7} \times \frac{49}{4} \times h$
$2310 = \frac{11 \times 7}{2} \times h$
$2310 = 38.5 \times h$
$h = \frac{2310}{38.5} = 60\,cm$.
Therefore,the height of the cylinder is $60\,cm$.

(B) Given: Radius $r = 70 \,cm = 0.7 \,m$ and height $h = 1.5 \,m$.
The volume $V$ of a cylinder is given by the formula $V = \pi r^2 h$.
Substituting the values: $V = \frac{22}{7} \times (0.7)^2 \times 1.5$.
$V = \frac{22}{7} \times 0.49 \times 1.5 = 22 \times 0.07 \times 1.5 = 1.54 \times 1.5 = 2.31 \,m^3$.
Since $1 \,m^3 = 1000 \,litres$,the capacity is $2.31 \times 1000 = 2310 \,litres$.

(C) Given:
Curved Surface Area $(CSA)$ of the cylinder = $8800 \, cm^2$
Radius $(r)$ = $28 \, cm$
Formula for $CSA$ of a cylinder = $2 \pi r h$
$2 \times \frac{22}{7} \times 28 \times h = 8800$
$2 \times 22 \times 4 \times h = 8800$
$176 \times h = 8800$
$h = \frac{8800}{176} = 50 \, cm$
Now,Volume of the cylinder = $\pi r^2 h$
Volume = $\frac{22}{7} \times 28 \times 28 \times 50$
Volume = $22 \times 4 \times 28 \times 50$
Volume = $88 \times 1400 = 1,23,200 \, cm^3$.

(D) The volume of the submerged cuboid is $V = 22\, cm \times 16\, cm \times 7\, cm = 2464\, cm^3$.
When the cuboid is submerged,the volume of water displaced is equal to the volume of the cuboid.
Let the rise in the water level be $h\, cm$.
The volume of the displaced water in the cylindrical container is given by $\pi r^2 h$,where $r = 14\, cm$.
So,$\pi r^2 h = 2464$.
Using $\pi = \frac{22}{7}$,we have $\frac{22}{7} \times 14 \times 14 \times h = 2464$.
$22 \times 2 \times 14 \times h = 2464$.
$616 \times h = 2464$.
$h = \frac{2464}{616} = 4\, cm$.
Therefore,the rise in the water level is $4\, cm$.

(A) Given that the sheet is a square with side length $s = 44 \, cm$.
When it is rolled along its length to form a cylinder,the length of the sheet becomes the circumference of the base of the cylinder.
Therefore,the circumference $C = 2 \pi r = 44 \, cm$.
Using $\pi = 22/7$,we have $2 \times (22/7) \times r = 44$.
This simplifies to $(44/7) \times r = 44$,which gives $r = 7 \, cm$.
The height $h$ of the cylinder is equal to the side of the square,so $h = 44 \, cm$.
The volume $V$ of the cylinder is given by $V = \pi r^2 h$.
$V = (22/7) \times (7)^2 \times 44$.
$V = (22/7) \times 49 \times 44 = 22 \times 7 \times 44$.
$V = 154 \times 44 = 6776 \, cm^3$.

(B) Given: Circumference of the base $(C)$ = $440\, cm$,Height $(h)$ = $80\, cm$.
Step $1$: Find the radius $(r)$ using $C = 2\pi r$.
$440 = 2 \times (22/7) \times r$
$r = (440 \times 7) / 44 = 70\, cm$.
Step $2$: Convert dimensions to meters $(m)$: $r = 0.7\, m$,$h = 0.8\, m$.
Step $3$: Calculate volume $(V)$ using $V = \pi r^2 h$.
$V = (22/7) \times (0.7)^2 \times 0.8$
$V = (22/7) \times 0.49 \times 0.8$
$V = 22 \times 0.07 \times 0.8 = 1.232\, m^3$.

(C) Given: Thickness of the tube = $5 \, mm = 0.5 \, cm$. Inner radius $(r)$ = $10 \, cm / 2 = 5 \, cm$. Length $(h)$ = $70 \, cm$. Outer radius $(R)$ = $r + \text{thickness} = 5 \, cm + 0.5 \, cm = 5.5 \, cm$. Volume of the metal = $\pi(R^2 - r^2)h = \frac{22}{7} \times ((5.5)^2 - (5)^2) \times 70 = 22 \times (30.25 - 25) \times 10 = 22 \times 5.25 \times 10 = 1155 \, cm^3$. Mass = $\text{Volume} \times \text{Density} = 1155 \, cm^3 \times 8 \, g/cm^3 = 9240 \, g$.

(D) $1$. The volume of the earth taken out from the well is equal to the volume of the cylinder: $V = \pi r^2 h$.
$2$. Given diameter $d = 14\, m$,so radius $r = 7\, m$. Depth $h = 50\, m$.
$3$. $V = \frac{22}{7} \times 7 \times 7 \times 50 = 22 \times 7 \times 50 = 7700\, m^3$.
$4$. This earth is spread over a rectangular field of area $A = 110\, m \times 100\, m = 11000\, m^2$.
$5$. Let the increase in height be $H$. Then,$V = A \times H$.
$6$. $7700 = 11000 \times H$.
$7$. $H = \frac{7700}{11000} = 0.7\, m$.
$8$. Converting to $cm$: $0.7 \times 100 = 70\, cm$.

(B) Given for a cone: radius $(r) = 7 \, cm$ and curved surface area $= 550 \, cm^2$.
The formula for the curved surface area of a cone is $\pi r l$.
Therefore,$550 = \frac{22}{7} \times 7 \times l$.
$550 = 22 \times l \implies l = \frac{550}{22} = 25 \, cm$.
Using the relation $l^2 = r^2 + h^2$,we find the height $(h)$:
$h^2 = l^2 - r^2 = (25)^2 - (7)^2 = 625 - 49 = 576$.
$h = \sqrt{576} = 24 \, cm$.
The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24$.
$V = 22 \times 7 \times 8 = 1232 \, cm^3$.

(C) For a given cone,the curved surface area $= 423.9 \, cm^2$.
Radius : slant height $= 3:5$.
Let the radius be $r = 3x \, cm$ and the slant height be $l = 5x \, cm$.
The curved surface area of a cone is given by $\pi rl$.
Therefore,$423.9 = 3.14 \times 3x \times 5x$.
$423.9 = 3.14 \times 15x^2$.
$x^2 = \frac{423.9}{47.1} = 9$.
So,$x = 3$.
Thus,radius $r = 3(3) = 9 \, cm$ and slant height $l = 5(3) = 15 \, cm$.
Using the relation $l^2 = h^2 + r^2$,we find the height $h$:
$h^2 = l^2 - r^2 = 15^2 - 9^2 = 225 - 81 = 144$.
$h = \sqrt{144} = 12 \, cm$.
The volume of a cone is $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times 3.14 \times 9^2 \times 12$.
$V = 3.14 \times 81 \times 4 = 1017.36 \, cm^3$.

(D) Let the radius $r = 5x$ and slant height $l = 13x$.
The curved surface area $(CSA)$ of a cone is given by $\pi rl = 1836.9 \, cm^2$.
Substituting the values: $3.14 \times 5x \times 13x = 1836.9$.
$3.14 \times 65x^2 = 1836.9$.
$204.1x^2 = 1836.9$.
$x^2 = \frac{1836.9}{204.1} = 9$.
$x = 3$.
So,$r = 5 \times 3 = 15 \, cm$ and $l = 13 \times 3 = 39 \, cm$.
The height $h$ of the cone is $\sqrt{l^2 - r^2} = \sqrt{39^2 - 15^2} = \sqrt{1521 - 225} = \sqrt{1296} = 36 \, cm$.
The volume $V$ of the cone is $\frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times 15^2 \times 36$.
$V = 3.14 \times 225 \times 12 = 3.14 \times 2700 = 8478 \, cm^3$.

(A) For the given sphere,radius $(r) = 6\, cm$.
For the given cylindrical wire,the diameter is $1\, cm$,so the radius $(R) = \frac{1}{2}\, cm$.
Let the length of the wire be $H\, cm$.
Since the volume remains constant when the sphere is melted and recast into a wire:
Volume of cylinder = Volume of sphere
$\pi R^2 H = \frac{4}{3} \pi r^3$
Substituting the values:
$R^2 H = \frac{4}{3} r^3$
$(\frac{1}{2})^2 \times H = \frac{4}{3} \times (6)^3$
$\frac{1}{4} \times H = \frac{4}{3} \times 216$
$\frac{1}{4} \times H = 4 \times 72$
$\frac{1}{4} \times H = 288$
$H = 288 \times 4 = 1152\, cm$.
To convert the length into meters:
$H = \frac{1152}{100} = 11.52\, m$.
Thus,the length of the wire is $11.52\, m$.

(D) $1$. Radius of a wooden sphere $(R) = \frac{21}{2} = 10.5\, cm$.
$2$. Surface area of one sphere $= 4\pi R^2 = 4 \times \frac{22}{7} \times 10.5 \times 10.5 = 1386\, cm^2$.
$3$. Area covered by the cylindrical support on the sphere $= \pi r^2 = \frac{22}{7} \times 1.5 \times 1.5 \approx 7.07\, cm^2$.
$4$. Surface area to be painted silver for one sphere $= 1386 - 7.07 = 1378.93\, cm^2$.
$5$. Total area for $8$ spheres $= 8 \times 1378.93 = 11031.44\, cm^2$.
$6$. Cost of silver paint $= 11031.44 \times 0.25 = ₹ 2757.86$.
$7$. Curved surface area of one cylindrical support $= 2\pi rh = 2 \times \frac{22}{7} \times 1.5 \times 7 = 66\, cm^2$.
$8$. Total area to be painted black for $8$ supports $= 8 \times 66 = 528\, cm^2$.
$9$. Cost of black paint $= 528 \times 0.05 = ₹ 26.40$.
$10$. Total cost $= 2757.86 + 26.40 = ₹ 2784.26$.

(B) Given diameter $d = 28 \, cm$,so radius $r = \frac{d}{2} = 14 \, cm$.
Surface area of a sphere $= 4 \pi r^2 = 4 \times \frac{22}{7} \times 14 \times 14 = 88 \times 28 = 2464 \, cm^2$.
Volume of a sphere $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 = \frac{4}{3} \times 22 \times 2 \times 14 \times 14 = \frac{34496}{3} = 11498 \frac{2}{3} \, cm^3$.

(N/A) Given: Diameter $d = 42\, cm$,so radius $r = \frac{42}{2} = 21\, cm$.
$1$. Curved Surface Area $(CSA)$ of a hemisphere $= 2\pi r^2$
$= 2 \times \frac{22}{7} \times 21 \times 21 = 2 \times 22 \times 3 \times 21 = 2772\, cm^2$.
$2$. Total Surface Area $(TSA)$ of a hemisphere $= 3\pi r^2$
$= 3 \times \frac{22}{7} \times 21 \times 21 = 3 \times 22 \times 3 \times 21 = 4158\, cm^2$.
$3$. Volume of a hemisphere $= \frac{2}{3}\pi r^3$
$= \frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 2 \times 22 \times 21 \times 21 = 19404\, cm^3$.

(A) The surface area of a sphere is given by the formula $A = 4 \pi r^2$.
Given $A = 5544 \, cm^2$,we have $4 \times \frac{22}{7} \times r^2 = 5544$.
$r^2 = \frac{5544 \times 7}{4 \times 22} = \frac{5544 \times 7}{88} = 63 \times 7 = 441$.
Thus,$r = \sqrt{441} = 21 \, cm$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times (21)^3 = \frac{4}{3} \times \frac{22}{7} \times 21 \times 21 \times 21$.
$V = 4 \times 22 \times 21 \times 21 = 88 \times 441 = 38808 \, cm^3$.

(C) The volume of the large sphere of radius $R = 20\, cm$ is given by $V_s = \frac{4}{3} \pi R^3 = \frac{4}{3} \pi (20)^3 = \frac{4}{3} \pi (8000)\, cm^3$.
The volume of one small hemisphere of radius $r = 5\, cm$ is given by $V_h = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (5)^3 = \frac{2}{3} \pi (125)\, cm^3$.
The number of hemispheres $n$ is given by the ratio of the volume of the sphere to the volume of one hemisphere:
$n = \frac{V_s}{V_h} = \frac{\frac{4}{3} \pi (8000)}{\frac{2}{3} \pi (125)} = \frac{4 \times 8000}{2 \times 125} = \frac{32000}{250} = 128$.
Thus,the total number of hemispheres obtained is $128$.

(A) The volume of the hemispherical bowl is given by $V_h = \frac{2}{3} \pi r^3$,where $r = 18\, cm$.
$V_h = \frac{2}{3} \times \pi \times (18)^3 = \frac{2}{3} \times \pi \times 5832 = 3888\pi\, cm^3$.
The volume of one cylindrical bottle is given by $V_c = \pi R^2 h$,where $R = 2\, cm$ and $h = 4\, cm$.
$V_c = \pi \times (2)^2 \times 4 = \pi \times 4 \times 4 = 16\pi\, cm^3$.
The number of bottles required is $n = \frac{V_h}{V_c} = \frac{3888\pi}{16\pi} = 243$.
Therefore,$243$ bottles are required.

(C) The volume of the metal sphere is given by the formula $V_s = \frac{4}{3} \pi r^3$,where $r = 12\, cm$.
$V_s = \frac{4}{3} \pi (12)^3 = \frac{4}{3} \pi (1728) = 2304 \pi\, cm^3$.
The volume of one cone is given by $V_c = \frac{1}{3} \pi r^2 h$,where $r = 3\, cm$ and $h = 4\, cm$.
$V_c = \frac{1}{3} \pi (3)^2 (4) = \frac{1}{3} \pi (9)(4) = 12 \pi\, cm^3$.
The number of cones obtained is the ratio of the volume of the sphere to the volume of one cone:
$N = \frac{V_s}{V_c} = \frac{2304 \pi}{12 \pi} = 192$.
Thus,$192$ cones are obtained.

(A) $1$. Volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
$2$. Total volume of the five spheres = $\frac{4}{3}\pi (1^3 + 3^3 + 4^3 + 5^3 + 8^3) = \frac{4}{3}\pi (1 + 27 + 64 + 125 + 512) = \frac{4}{3}\pi (729)$.
$3$. Let the radius of the new large sphere be $R$. Then $\frac{4}{3}\pi R^3 = \frac{4}{3}\pi (729)$,which implies $R^3 = 729$,so $R = 9\, cm$.
$4$. Volume of one cylinder = $\pi r^2 h = \pi (3^2)(4) = 36\pi\, cm^3$.
$5$. Number of cylinders = $\frac{\text{Total Volume}}{\text{Volume of one cylinder}} = \frac{\frac{4}{3}\pi (729)}{36\pi} = \frac{972\pi}{36\pi} = 27$ cylinders.

(A) The surface area of a sphere is given by the formula $S = 4\pi r^{2}$.
Given $S = 1256 \, cm^{2}$ and $\pi = 3.14$.
$1256 = 4 \times 3.14 \times r^{2}$
$1256 = 12.56 \times r^{2}$
$r^{2} = \frac{1256}{12.56} = 100$
$r = 10 \, cm$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^{3}$.
$V = \frac{4}{3} \times 3.14 \times (10)^{3}$
$V = \frac{4}{3} \times 3.14 \times 1000$
$V = \frac{12560}{3} = 4186 \frac{2}{3} \, cm^{3}$.

(B) Let the radius of the sphere be $r_1$ and the radius of the hemisphere be $r_2$.
The surface area of a sphere is $4\pi r_1^2$ and the curved surface area of a hemisphere is $2\pi r_2^2$.
Given the ratio: $\frac{4\pi r_1^2}{2\pi r_2^2} = \frac{9}{2}$.
Simplifying this,we get $\frac{2r_1^2}{r_2^2} = \frac{9}{2}$,which implies $\frac{r_1^2}{r_2^2} = \frac{9}{4}$.
Taking the square root,we find $\frac{r_1}{r_2} = \frac{3}{2}$.
The volume of a sphere is $V_1 = \frac{4}{3}\pi r_1^3$ and the volume of a hemisphere is $V_2 = \frac{2}{3}\pi r_2^3$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{2}{3}\pi r_2^3} = 2 \times \left(\frac{r_1}{r_2}\right)^3$.
Substituting the ratio $\frac{r_1}{r_2} = \frac{3}{2}$,we get $\frac{V_1}{V_2} = 2 \times \left(\frac{3}{2}\right)^3 = 2 \times \frac{27}{8} = \frac{27}{4}$.
Thus,the ratio of their volumes is $27:4$.

(C) The volume of the cubic tank is $V_{total} = 15.625 \, m^3$.
Since the tank is a cube, the side length $s$ is given by $s = \sqrt[3]{15.625} = 2.5 \, m$.
The current depth of water is $h = 1.3 \, m$.
The volume of water currently in the tank is $V_{current} = \text{side} \times \text{side} \times \text{depth} = 2.5 \times 2.5 \times 1.3 = 6.25 \times 1.3 = 8.125 \, m^3$.
The volume of water already used is $V_{used} = V_{total} - V_{current} = 15.625 - 8.125 = 7.5 \, m^3$.

(A) The diameter of the cylindrical glass is $d = 7\, cm$, so the radius is $r = 3.5\, cm$.
The height of the milk in the glass is $h = 12\, cm$.
The volume of milk in one glass is given by the formula $V = \pi r^2 h$.
$V = \frac{22}{7} \times 3.5 \times 3.5 \times 12 = 22 \times 0.5 \times 3.5 \times 12 = 462\, cm^3$.
Since $1000\, cm^3 = 1\, litre$, the volume of milk in one glass is $0.462\, litres$.
For $1600$ students, the total milk required is $1600 \times 0.462 = 739.2\, litres$.

(A) When a right triangle with sides $6 \, cm, 8 \, cm$ and $10 \, cm$ is revolved about the side $8 \, cm$,the side $8 \, cm$ becomes the height $(h)$ of the cone,and the side $6 \, cm$ becomes the radius $(r)$ of the base. The hypotenuse $10 \, cm$ becomes the slant height $(l)$ of the cone.
Given: $r = 6 \, cm, h = 8 \, cm, l = 10 \, cm$.
Volume of the cone $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times (6)^2 \times 8 = \frac{1}{3} \times 3.14 \times 36 \times 8 = 3.14 \times 12 \times 8 = 301.44 \, cm^3$.
Curved surface area of the cone $CSA = \pi r l = 3.14 \times 6 \times 10 = 188.4 \, cm^2$.

(C) $1$. The internal diameter of the hemispherical tank is $14\, m$,so the radius $r = 7\, m$.
$2$. The volume of a hemispherical tank is given by $V = \frac{2}{3} \pi r^3$.
$3$. $V = \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 = \frac{2}{3} \times 22 \times 49 = \frac{2156}{3} = 718 \frac{2}{3}\, m^3$.
$4$. We know that $1\, m^3 = 1\, kilolitre$. Therefore,the tank contains $50\, m^3$ of water.
$5$. The volume of water to be pumped = (Total capacity) - (Volume already present).
$6$. Volume to be pumped = $718 \frac{2}{3} - 50 = 668 \frac{2}{3}\, m^3$.

(D) Let the radius of the sphere and the cylinder be $r$. Let the height of the cylinder be $h$.
Given that the volumes are equal:
Volume of sphere = $\frac{4}{3} \pi r^3$
Volume of cylinder = $\pi r^2 h$
Equating the two: $\frac{4}{3} \pi r^3 = \pi r^2 h$
$h = \frac{4}{3} r$
The diameter of the cylinder is $d = 2r$.
We need to find the percentage by which the diameter exceeds the height:
Difference = $d - h = 2r - \frac{4}{3} r = \frac{2}{3} r$
Percentage = $\frac{d - h}{h} \times 100 = \frac{(2/3)r}{(4/3)r} \times 100 = \frac{2}{4} \times 100 = 50\%$.

(B) The statement is False.
In a right circular cone,the radius $(r)$,height $(h)$,and slant height $(l)$ form a right-angled triangle where the slant height is the hypotenuse.
According to the Pythagorean theorem,the relationship is $l^2 = r^2 + h^2$.
Given: $r = 3 \, cm$,$h = 5 \, cm$,and $l = 4 \, cm$.
Calculating $r^2 + h^2 = 3^2 + 5^2 = 9 + 25 = 34$.
Calculating $l^2 = 4^2 = 16$.
Since $16 \neq 34$,the given dimensions cannot form a right circular cone because the slant height must be the longest side of the right-angled triangle formed by the radius and the height.

(B) The lateral surface area $(LSA)$ of a cube is given by the formula $LSA = 4a^2$,where $a$ is the length of the edge.
Given,edge $a = 5 \, cm$.
Substituting the value in the formula:
$LSA = 4 \times (5)^2$
$LSA = 4 \times 25$
$LSA = 100 \, cm^2$.
The statement claims the lateral surface area is $150 \, cm^2$,which is incorrect.
Therefore,the statement is False.

(TRUE) The volume of a cylinder is given by the formula: $V = \text{Base Area} \times \text{Height}$.
Given:
Base Area = $154 \, cm^{2}$
Height = $20 \, cm$
Calculating the volume:
$V = 154 \, cm^{2} \times 20 \, cm = 3080 \, cm^{3}$.
Since the calculated volume matches the given value, the statement is True.

(B) The length of the longest rod that can be placed in a rectangular box of dimensions $l \times b \times h$ is given by the space diagonal formula: $d = \sqrt{l^2 + b^2 + h^2}$.
Given dimensions are $l = 12\, cm$,$b = 4\, cm$,and $h = 3\, cm$.
Calculating the diagonal: $d = \sqrt{12^2 + 4^2 + 3^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13\, cm$.
The maximum length of a bar that can fit inside the box is $13\, cm$.
Since the bar is $15\, cm$ long and $15\, cm > 13\, cm$,the bar cannot fit in the box.
Therefore,the statement is False.

(B) The formula for the curved surface area of a cylinder is $A = 2\pi rh$,where $r$ is the radius and $h$ is the height.
Given: $r = 10\, cm$,$A = 440\, cm^{2}$,and taking $\pi = 22/7$.
Substituting the values: $440 = 2 \times (22/7) \times 10 \times h$.
$440 = (440/7) \times h$.
$h = (440 \times 7) / 440$.
$h = 7\, cm$.
Therefore,the height of the cylinder is $7\, cm$.

(C) The formula for the total surface area $(TSA)$ of a cube with edge length '$a$' is given by:
$TSA = 6a^2$
Given that $TSA = 864 \, cm^2$,we can set up the equation:
$6a^2 = 864$
Divide both sides by $6$:
$a^2 = \frac{864}{6}$
$a^2 = 144$
Taking the square root of both sides:
$a = \sqrt{144}$
$a = 12 \, cm$
Therefore,the edge of the cube is $12 \, cm$.

(D) The volume of a cuboid is given by the formula: $V = \text{length} \times \text{breadth} \times \text{height}$.
Given:
Length $(l)$ = $30 \, cm$
Breadth $(b)$ = $20 \, cm$
Height $(h)$ = $15 \, cm$
Substituting these values into the formula:
$V = 30 \, cm \times 20 \, cm \times 15 \, cm$
$V = 600 \, cm^2 \times 15 \, cm$
$V = 9000 \, cm^3$.
Therefore, the correct option is $D$.

(D) Let the radii of the two cylinders be $r_1$ and $r_2$,and their heights be $h_1$ and $h_2$. Given that $h_1 = h_2 = h$ and $r_1:r_2 = 4:5$.
The volume of a cylinder is given by the formula $V = \pi r^2 h$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{\pi r_1^2 h}{\pi r_2^2 h} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^2$.
Substituting the given ratio: $\frac{V_1}{V_2} = \left(\frac{4}{5}\right)^2 = \frac{16}{25}$.
Therefore,the ratio of their volumes is $16:25$.

(A) We know that $1 \, m = 100 \, cm$.
Therefore,$1 \, m^3 = (100 \, cm) \times (100 \, cm) \times (100 \, cm) = 1,000,000 \, cm^3$.
Since $1000 \, cm^3 = 1 \, \text{litre}$,we can write $1 \, cm^3 = \frac{1}{1000} \, \text{litre}$.
Thus,$1,000,000 \, cm^3 = \frac{1,000,000}{1000} \, \text{litres} = 1000 \, \text{litres}$.
Hence,$1 \, m^3 = 1000 \, \text{litres}$.

(B) 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:
$1 \text{ litre} = (10 \text{ cm})^3 = 10 \text{ cm} \times 10 \text{ cm} \times 10 \text{ cm} = 1000 \text{ cm}^3$.
Therefore,$1 \text{ litre} = 1000 \text{ cm}^3$.

(C) Let the radii of the two cylinders be $r_1$ and $r_2$,and their heights be $h_1$ and $h_2$.
Given that $r_1:r_2 = 3:2$ and $h_1:h_2 = 8:11$.
The volume of a cylinder is given by the formula $V = \pi r^2 h$.
The ratio of the volumes of the two cylinders 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 values: $\frac{V_1}{V_2} = \left(\frac{3}{2}\right)^2 \times \left(\frac{8}{11}\right) = \frac{9}{4} \times \frac{8}{11}$.
Calculating the result: $\frac{9 \times 8}{4 \times 11} = \frac{72}{44} = \frac{18}{11}$.
Thus,the ratio of their volumes is $18:11$.

(A) Given,the diameter of the solid hemisphere $d = 21 \, cm$.
Therefore,the radius $r = \frac{d}{2} = \frac{21}{2} \, cm = 10.5 \, cm$.
The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^3$.
Substituting the values,we get $V = \frac{2}{3} \times \frac{22}{7} \times (10.5)^3$.
$V = \frac{2}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5$.
$V = \frac{44}{21} \times 1157.625$.
$V = 2425.5 \, cm^3$.

(B) Let the edge length of the cube be $x \ cm$.
The surface area of a cube is given by the formula $S = 6x^2 \ cm^2$.
The volume of a cube is given by the formula $V = x^3 \ cm^3$.
According to the problem,the numerical values of the surface area and volume are equal:
$6x^2 = x^3$
Dividing both sides by $x^2$ (since $x \neq 0$):
$6 = x$
Therefore,the edge length of the cube is $6 \ cm$.

(A) cube has $6$ identical square faces.
If the length of each edge of the cube is $a$,then the area of one face is $a \times a = a^{2}$.
Since there are $6$ such faces,the total surface area of the cube is $6 \times a^{2} = 6 a^{2}$.

(B) cuboid has $6$ rectangular faces.
Let the length be $l$,breadth be $b$,and height be $h$.
The area of the opposite faces are $(l \times b)$,$(b \times h)$,and $(h \times l)$.
Since there are two of each face,the total surface area is $2(lb + bh + hl)$.

(D) cube has $6$ faces in total. The lateral surface area refers to the area of the four side faces,excluding the top and bottom faces. If the side length of the cube is $a$,the area of one face is $a^{2}$. Therefore,the lateral surface area is $4 \times a^{2} = 4 a^{2}$.

(A) The total surface area of a cylinder consists of the area of the two circular bases and the curved surface area.
Area of two circular bases $= 2 \times (\pi r^2) = 2 \pi r^2$.
Curved surface area of the cylinder $= 2 \pi rh$.
Total surface area $= 2 \pi r^2 + 2 \pi rh$.
Factoring out $2 \pi r$,we get $2 \pi r(r + h)$ or $2 \pi r(h + r)$.

(D) The lateral surface area of a cuboid is defined as the sum of the areas of its four vertical faces (excluding the top and bottom faces).
For a cuboid with length $l$,breadth $b$,and height $h$,the area of the four vertical faces is given by the perimeter of the base multiplied by the height.
Lateral Surface Area $= (2l + 2b) \times h = 2h(l + b)$.

(B) The curved surface area of a cylinder is calculated by multiplying the circumference of its base $(2 \pi r)$ by its height $(h)$.
Therefore,the formula for the curved surface area of a cylinder is $2 \pi r h$.

(D) We know that $1 \, m = 100 \, cm$.
Therefore,$1 \, m^{2} = 1 \, m \times 1 \, m$.
Substituting the value of $1 \, m$ in centimeters,we get:
$1 \, m^{2} = (100 \, cm) \times (100 \, cm) = 10000 \, cm^{2}$.
Thus,$1 \, m^{2} = 10000 \, cm^{2}$.

(B) We know that $1\, m = 100\, cm$.
Therefore,$1\, m^{3} = (1\, m) \times (1\, m) \times (1\, m)$.
Substituting the value of $1\, m$ in centimeters: $1\, m^{3} = (100\, cm) \times (100\, cm) \times (100\, cm)$.
$1\, m^{3} = 100 \times 100 \times 100\, cm^{3}$.
$1\, m^{3} = 1000000\, cm^{3}$.