Mix Examples - Surface Areas and Volumes Questions in English

(C) In the Indus Valley Civilisation,the baked bricks used for construction were standardized. Archaeological findings confirm that these bricks typically maintained a length-to-width-to-thickness ratio of $4: 2: 1$. This ratio provided structural stability for the buildings constructed during that period.

(D) pyramid is a solid figure in three-dimensional geometry. It is formed by connecting a polygonal base to a point,called the apex. The base can be any polygon (such as a triangle,square,pentagon,etc.),and the side faces are triangles that meet at the apex.

(A) pyramid is a polyhedron whose base is a polygon and whose side faces are triangles that meet at a common vertex called the apex.

(FALSE) The statement is False.
Justification: $A$ pyramid is a solid object with a polygonal base and triangular side faces that meet at a common vertex (apex). While the base can be any polygon,the side faces are not necessarily equilateral triangles; they can be any type of triangle depending on the dimensions of the base and the height of the pyramid.

(A) Since Mary wants to paste the paper on the outer surface of the box,the quantity of paper required is equal to the total surface area of the box,which is in the shape of a cuboid.
The dimensions of the box are: Length $(l)$ = $80 \, cm$,Breadth $(b)$ = $40 \, cm$,Height $(h)$ = $20 \, cm$.
The total surface area of the cuboid = $2(lb + bh + hl)$.
Surface area = $2[(80 \times 40) + (40 \times 20) + (20 \times 80)] \, cm^2$.
Surface area = $2[3200 + 800 + 1600] \, cm^2$.
Surface area = $2 \times 5600 \, cm^2 = 11200 \, cm^2$.
The area of each square sheet of paper = side $\times$ side = $40 \, cm \times 40 \, cm = 1600 \, cm^2$.
The number of sheets required = (Total surface area of the box) / (Area of one sheet of paper).
Number of sheets = $11200 / 1600 = 7$.
Therefore,she would require $7$ sheets.

(B) Since Hameed is getting the five outer faces of the tank covered with tiles,he needs to calculate the surface area of these five faces to determine the number of tiles required.
Edge of the cubical tank $(a)$ $= 1.5\, m = 150\, cm$.
Surface area of the five faces $= 5 \times a^2 = 5 \times 150 \times 150\, cm^2 = 112500\, cm^2$.
Area of each square tile $= 25\, cm \times 25\, cm = 625\, cm^2$.
Number of tiles required $= \frac{\text{Total surface area}}{\text{Area of one tile}} = \frac{112500}{625} = 180$.
Cost of $1$ dozen $(12)$ tiles $= Rs. 360$.
Cost of $1$ tile $= \frac{360}{12} = Rs. 30$.
Total cost for $180$ tiles $= 180 \times 30 = Rs. 5400$.

(109) Given:
Length $(l) = 1.5\, m$
Breadth $(b) = 1.25\, m$
Depth $(h) = 65\, cm = 0.65\, m$
$(i)$ Since the box is open at the top,the area of the sheet required is the sum of the area of the four walls and the base:
Area $= 2lh + 2bh + lb$
$= [2 \times 1.5 \times 0.65 + 2 \times 1.25 \times 0.65 + 1.5 \times 1.25]\, m^2$
$= (1.95 + 1.625 + 1.875)\, m^2 = 5.45\, m^2$
$(ii)$ Cost of sheet per $m^2 = Rs. 20$
Total cost for $5.45\, m^2 = 5.45 \times 20 = Rs. 109$

(D) Given: Length $(l) = 5\, m$,Breadth $(b) = 4\, m$,Height $(h) = 3\, m$.
We need to whitewash the four walls and the ceiling.
Area to be whitewashed = Area of four walls + Area of ceiling
$= 2(l + b)h + (l \times b)$
$= 2(5 + 4) \times 3 + (5 \times 4)$
$= 2(9) \times 3 + 20$
$= 54 + 20 = 74\, m^2$.
Cost of whitewashing per $m^2 = Rs. 7.50$.
Total cost = $74 \times 7.50 = Rs. 555$.

(A) Let the length,breadth,and height of the rectangular hall be $l\, m, b\, m,$ and $h\, m$ respectively.
The area of the four walls is given by the formula: $\text{Area} = 2lh + 2bh = 2(l + b)h$.
The perimeter of the floor of the hall is $2(l + b) = 250\, m$.
Substituting the perimeter into the area formula,we get: $\text{Area of four walls} = 250 \times h = 250h\, m^2$.
The cost of painting is given as $Rs. 10$ per $m^2$.
Total cost of painting = $\text{Area} \times \text{Rate} = 250h \times 10 = 2500h$.
Given that the total cost is $Rs. 15000$,we set up the equation: $2500h = 15000$.
Solving for $h$: $h = \frac{15000}{2500} = 6\, m$.
Therefore,the height of the hall is $6\, m$.

(B) The total surface area of one brick is given by the formula $2(lb + bh + lh)$.
Substituting the given dimensions: $l = 22.5 \,cm$,$b = 10 \,cm$,$h = 7.5 \,cm$.
Surface area of one brick $= 2(22.5 \times 10 + 10 \times 7.5 + 22.5 \times 7.5) \,cm^2$.
$= 2(225 + 75 + 168.75) \,cm^2 = 2(468.75) \,cm^2 = 937.5 \,cm^2$.
Convert the total paintable area from $m^2$ to $cm^2$: $9.375 \,m^2 = 9.375 \times 10000 \,cm^2 = 93750 \,cm^2$.
Let $n$ be the number of bricks that can be painted.
$n \times 937.5 = 93750$.
$n = \frac{93750}{937.5} = 100$.
Thus,$100$ bricks can be painted.

(C) Lateral surface area of a cube $= 4 \times (\text{edge})^2$.
Given edge of the cube $= 10\, cm$.
Lateral surface area of the cubical box $= 4 \times (10\, cm)^2 = 4 \times 100\, cm^2 = 400\, cm^2$.
Lateral surface area of a cuboid $= 2 \times h \times (l + b)$.
Given length $(l) = 12.5\, cm$,breadth $(b) = 10\, cm$,and height $(h) = 8\, cm$.
Lateral surface area of the cuboidal box $= 2 \times 8\, cm \times (12.5\, cm + 10\, cm) = 16\, cm \times 22.5\, cm = 360\, cm^2$.
Comparing the two areas,$400\, cm^2 > 360\, cm^2$.
The cubical box has a greater lateral surface area.
The difference is $400\, cm^2 - 360\, cm^2 = 40\, cm^2$.

(D) Total surface area of the cubical box $= 6 \times (\text{edge})^2 = 6 \times (10\, cm)^2 = 600\, cm^2$.
Total surface area of the cuboidal box $= 2(lh + bh + lb) = 2(12.5 \times 8 + 10 \times 8 + 12.5 \times 10) = 2(100 + 80 + 125) = 2(305) = 610\, cm^2$.
Comparing the two, the total surface area of the cubical box $(600\, cm^2)$ is smaller than that of the cuboidal box $(610\, cm^2)$.
The difference in surface area $= 610\, cm^2 - 600\, cm^2 = 10\, cm^2$.
Thus, the cubical box has a smaller total surface area by $10\, cm^2$.

(A) The greenhouse is in the shape of a cuboid.
Length $(l) = 30 \, cm$
Breadth $(b) = 25 \, cm$
Height $(h) = 25 \, cm$
The total surface area of the glass is equal to the total surface area of the cuboid.
Total Surface Area $= 2(lb + lh + bh)$
$= 2(30 \times 25 + 30 \times 25 + 25 \times 25) \, cm^2$
$= 2(750 + 750 + 625) \, cm^2$
$= 2(2125) \, cm^2$
$= 4250 \, cm^2$
Thus,the total area of the glass is $4250 \, cm^2$.

(B) cuboid has $12$ edges,consisting of $4$ lengths $(l)$,$4$ widths $(b)$,and $4$ heights $(h)$.
Given dimensions are length $(l)$ = $30 \, cm$,width $(b)$ = $25 \, cm$,and height $(h)$ = $25 \, cm$.
The total length of tape required for all $12$ edges is given by the formula:
Total length $= 4(l + b + h)$
$= 4(30 + 25 + 25) \, cm$
$= 4(80) \, cm$
$= 320 \, cm$
Therefore,$320 \, cm$ of tape is required.

(C) Length $(l_1)$ of bigger box $= 25 \,cm$,Breadth $(b_1) = 20 \,cm$,Height $(h_1) = 5 \,cm$.
Total surface area of bigger box $= 2(l_1b_1 + l_1h_1 + b_1h_1) = 2(25 \times 20 + 25 \times 5 + 20 \times 5) \,cm^2 = 2(500 + 125 + 100) \,cm^2 = 1450 \,cm^2$.
Extra area for overlaps $= 5\% \text{ of } 1450 \,cm^2 = (1450 \times 0.05) \,cm^2 = 72.5 \,cm^2$.
Total area for $1$ bigger box $= 1450 + 72.5 = 1522.5 \,cm^2$.
Area for $250$ bigger boxes $= 250 \times 1522.5 = 380625 \,cm^2$.
Length $(l_2) = 15 \,cm$,Breadth $(b_2) = 12 \,cm$,Height $(h_2) = 5 \,cm$.
Total surface area of smaller box $= 2(15 \times 12 + 15 \times 5 + 12 \times 5) \,cm^2 = 2(180 + 75 + 60) \,cm^2 = 630 \,cm^2$.
Extra area for overlaps $= 5\% \text{ of } 630 \,cm^2 = (630 \times 0.05) \,cm^2 = 31.5 \,cm^2$.
Total area for $1$ smaller box $= 630 + 31.5 = 661.5 \,cm^2$.
Area for $250$ smaller boxes $= 250 \times 661.5 = 165375 \,cm^2$.
Total cardboard area $= 380625 + 165375 = 546000 \,cm^2$.
Cost $= (546000 / 1000) \times 4 = 546 \times 4 = Rs. 2184$.

(D) The shelter is a cuboidal structure without a base.
Length $(l) = 4 \, m$,Breadth $(b) = 3 \, m$,and Height $(h) = 2.5 \, m$.
The tarpaulin is required for the four walls and the top.
Area of tarpaulin required $= \text{Lateral Surface Area} + \text{Area of the top}$
$= 2(lh + bh) + (l \times b)$
$= [2(4 \times 2.5 + 3 \times 2.5) + (4 \times 3)] \, m^2$
$= [2(10 + 7.5) + 12] \, m^2$
$= [2(17.5) + 12] \, m^2$
$= (35 + 12) \, m^2 = 47 \, m^2$.
Thus,$47 \, m^2$ of tarpaulin is required.

(A) The kaleidoscope is cylindrical in shape.
Radius of the base of the cylinder $(r) = 3.5\, cm$.
Height (length) of the cylinder $(h) = 25\, cm$.
The area of the chart paper required is equal to the curved surface area of the cylinder.
Curved Surface Area $= 2\pi rh$
$= 2 \times \frac{22}{7} \times 3.5 \times 25\, cm^{2}$
$= 2 \times 22 \times 0.5 \times 25\, cm^{2}$
$= 44 \times 12.5\, cm^{2}$
$= 550\, cm^{2}$

(B) Given:
Height $(h) = 14\, cm$
Curved surface area $(CSA) = 88\, cm^2$
Let the radius of the base be $r$ and the diameter be $d = 2r$.
The formula for the curved surface area of a cylinder is $CSA = 2\pi rh$.
Since $d = 2r$,we can write $CSA = \pi dh$.
Substituting the given values:
$88 = \frac{22}{7} \times d \times 14$
$88 = 22 \times d \times 2$
$88 = 44 \times d$
$d = \frac{88}{44} = 2\, cm$.
Thus,the diameter of the base of the cylinder is $2\, cm$.

(C) The height $(h)$ of the cylindrical tank is $1 \,m$.
The base diameter is $140 \,cm$,so the base radius $(r)$ is $\frac{140}{2} \,cm = 70 \,cm = 0.7 \,m$.
The area of the sheet required is equal to the total surface area of the closed cylindrical tank,which is given by the formula $2\pi r(r + h)$.
Substituting the values: $\text{Area} = 2 \times \frac{22}{7} \times 0.7 \times (0.7 + 1) \,m^2$.
$\text{Area} = 2 \times 22 \times 0.1 \times 1.7 \,m^2$.
$\text{Area} = 4.4 \times 1.7 \,m^2 = 7.48 \,m^2$.
Thus,$7.48 \,m^2$ of the metal sheet is required.

(D) Given:
Length (height) of the cylindrical pipe,$h = 77\, cm$.
Inner diameter,$d_{1} = 4\, cm$.
Inner radius,$r_{1} = \frac{d_{1}}{2} = \frac{4}{2} = 2\, cm$.
The formula for the curved surface area $(CSA)$ of a cylinder is $2\pi rh$.
Inner curved surface area $= 2 \times \pi \times r_{1} \times h$
$= 2 \times \frac{22}{7} \times 2 \times 77$
$= 2 \times 22 \times 2 \times 11$
$= 968\, cm^{2}$.
Thus,the inner curved surface area of the metal pipe is $968\, cm^{2}$.

(B) Given:
Length of the pipe $(h)$ = $77 \, cm$
Inner diameter $(d_1)$ = $4 \, cm$,so inner radius $(r_1)$ = $2 \, cm$
Outer diameter $(d_2)$ = $4.4 \, cm$,so outer radius $(r_2)$ = $2.2 \, cm$
Total surface area of the pipe = (Inner curved surface area) + (Outer curved surface area) + (Area of two circular rings at the ends)
Inner curved surface area = $2 \pi r_1 h = 2 \times \frac{22}{7} \times 2 \times 77 = 2 \times 22 \times 2 \times 11 = 968 \, cm^2$
Outer curved surface area = $2 \pi r_2 h = 2 \times \frac{22}{7} \times 2.2 \times 77 = 2 \times 22 \times 2.2 \times 11 = 1064.8 \, cm^2$
Area of two circular rings = $2 \times \pi (r_2^2 - r_1^2) = 2 \times \frac{22}{7} \times (2.2^2 - 2^2) = 2 \times \frac{22}{7} \times (4.84 - 4) = 2 \times \frac{22}{7} \times 0.84 = 2 \times 22 \times 0.12 = 5.28 \, cm^2$
Total surface area = $968 + 1064.8 + 5.28 = 2038.08 \, cm^2$.

(C) The roller is cylindrical in shape.
Height $(h)$ of the cylindrical roller = Length of the roller = $120\, cm$.
Radius $(r)$ of the circular end of the roller = $\frac{84}{2}\, cm = 42\, cm$.
Curved Surface Area $(CSA)$ of the roller = $2\pi rh$.
$CSA = 2 \times \frac{22}{7} \times 42 \times 120\, cm^2$.
$CSA = 2 \times 22 \times 6 \times 120\, cm^2 = 31680\, cm^2$.
The area of the playground is equal to the total area covered in $500$ revolutions.
Area of the playground = $500 \times CSA$ of the roller.
Area = $500 \times 31680\, cm^2 = 15840000\, cm^2$.
Since $1\, m^2 = 10000\, cm^2$,we convert the area to $m^2$:
Area = $\frac{15840000}{10000}\, m^2 = 1584\, m^2$.

(D) Given: Diameter of the cylindrical pillar $= 50 \, cm = 0.5 \, m$.
Radius $(r) = \frac{0.5}{2} = 0.25 \, m$.
Height $(h) = 3.5 \, m$.
Curved Surface Area $(CSA)$ of the cylinder $= 2 \pi rh$.
$CSA = 2 \times \frac{22}{7} \times 0.25 \times 3.5 \, m^2$.
$CSA = 2 \times \frac{22}{7} \times 0.25 \times \frac{35}{10} = 2 \times 22 \times 0.25 \times 0.5 = 5.5 \, m^2$.
Cost of painting per $m^2 = Rs. \, 12.50$.
Total cost of painting $= 5.5 \times 12.50 = Rs. \, 68.75$.

(A) Let the height of the right circular cylinder be $h$.
The radius $(r)$ of the base of the cylinder is given as $0.7 \, m$.
The curved surface area $(CSA)$ of the cylinder is given as $4.4 \, m^2$.
The formula for the curved surface area of a cylinder is $CSA = 2 \pi r h$.
Substituting the given values into the formula:
$2 \times \frac{22}{7} \times 0.7 \times h = 4.4$
Simplify the expression:
$2 \times 22 \times 0.1 \times h = 4.4$
$4.4 \times h = 4.4$
Solving for $h$:
$h = \frac{4.4}{4.4} = 1 \, m$.
Therefore,the height of the cylinder is $1 \, m$.

(B) The inner diameter of the circular well is $d = 3.5 \, m$.
The inner radius $(r)$ is given by $r = \frac{d}{2} = \frac{3.5}{2} \, m = 1.75 \, m$.
The depth $(h)$ of the circular well is $10 \, m$.
The inner curved surface area of a cylinder is given by the formula $2 \pi r h$.
Substituting the values,we get:
Inner curved surface area $= 2 \times \frac{22}{7} \times 1.75 \times 10 \, m^2$.
$= 2 \times \frac{22}{7} \times \frac{175}{100} \times 10 \, m^2$.
$= 2 \times 22 \times \frac{1}{7} \times \frac{7}{4} \times 10 \, m^2$.
$= 44 \times 0.25 \times 10 \, m^2 = 110 \, m^2$.
Thus,the inner curved surface area of the circular well is $110 \, m^2$.

$(C)$ Given: Diameter of the well $(d)$ $= 3.5\, m$, Depth $(h)$ $= 10\, m$.
Radius $(r)$ $= d/2 = 3.5/2 = 1.75\, m$.
The curved surface area $(CSA)$ of the cylindrical well $= 2\pi rh$.
$CSA = 2 \times (22/7) \times 1.75 \times 10$.
$CSA = 2 \times (22/7) \times 17.5 = 44 \times 2.5 = 110\, m^2$.
Cost of plastering $= \text{Area} \times \text{Rate} = 110\, m^2 \times Rs. 40/m^2 = Rs. 4400$.

(D) The height $(h)$ of the cylindrical pipe is equal to its length,so $h = 28 \, m$.
The radius $(r)$ of the circular end of the pipe is half of the diameter,so $r = \frac{5}{2} \, cm = 2.5 \, cm$.
To maintain consistent units,convert the radius into meters: $r = \frac{2.5}{100} \, m = 0.025 \, m$.
The total radiating surface of the pipe is its Curved Surface Area $(CSA)$,which is given by the formula $CSA = 2 \pi r h$.
Substituting the values: $CSA = 2 \times \frac{22}{7} \times 0.025 \times 28$.
$CSA = 2 \times 22 \times 0.025 \times 4 = 44 \times 0.1 = 4.4 \, m^2$.
Thus,the total radiating surface in the system is $4.4 \, m^2$.

(A) Height $(h)$ of the cylindrical tank $= 4.5\, m$.
Radius $(r)$ of the circular end of the cylindrical tank $= \frac{4.2}{2}\, m = 2.1\, m$.
Lateral or curved surface area $(CSA)$ of the cylinder is given by the formula $2\pi rh$.
Substituting the values: $CSA = 2 \times \frac{22}{7} \times 2.1 \times 4.5\, m^2$.
$CSA = 2 \times 22 \times 0.3 \times 4.5\, m^2$.
$CSA = 44 \times 1.35\, m^2$.
$CSA = 59.4\, m^2$.
Therefore,the lateral or curved surface area of the tank is $59.4\, m^2$.

(B) First, calculate the Total Surface Area $(TSA)$ of the closed cylindrical tank:
$TSA = 2\pi r(r + h)$
Given $r = 2.1 \text{ m}$ and $h = 4.5 \text{ m}$,
$TSA = 2 \times \frac{22}{7} \times 2.1 \times (2.1 + 4.5) = 2 \times \frac{22}{7} \times 2.1 \times 6.6 = 87.12 \text{ m}^2$.
Let $A$ be the actual steel used. Since $1/12$ of the steel was wasted, the steel used for the tank is $A - \frac{1}{12}A = \frac{11}{12}A$.
Equating this to the surface area of the tank:
$\frac{11}{12}A = 87.12$
$A = \frac{87.12 \times 12}{11} = 7.92 \times 12 = 95.04 \text{ m}^2$.
Thus, the actual steel used is $95.04 \text{ m}^2$.

(C) The frame of the lampshade is cylindrical in shape.
Given:
Diameter of the base $= 20 \, cm$,so the radius $(r) = \frac{20}{2} = 10 \, cm$.
Height of the frame $= 30 \, cm$.
Margin for folding at the top and bottom $= 2.5 \, cm$ each.
Total height $(h)$ of the cloth required $= 30 \, cm + 2.5 \, cm + 2.5 \, cm = 35 \, cm$.
The cloth required to cover the lampshade is equal to the curved surface area of the cylinder.
Curved Surface Area $= 2 \pi rh$
$= 2 \times \frac{22}{7} \times 10 \times 35 \, cm^2$
$= 2 \times 22 \times 10 \times 5 \, cm^2$
$= 2200 \, cm^2$.
Thus,$2200 \, cm^2$ of cloth is required to cover the lampshade.

(D) The penholder is a cylinder with a base,so its surface area is the sum of the curved surface area and the area of the circular base.
Radius $(r) = 3 \,cm$
Height $(h) = 10.5 \,cm$
Surface area of $1$ penholder $= 2\pi rh + \pi r^2 = \pi r(2h + r)$
$= \frac{22}{7} \times 3 \times (2 \times 10.5 + 3) \,cm^2$
$= \frac{22}{7} \times 3 \times (21 + 3) \,cm^2$
$= \frac{22}{7} \times 3 \times 24 \,cm^2 = \frac{1584}{7} \,cm^2$
Total cardboard required for $35$ competitors $= 35 \times \left(\frac{1584}{7}\right) \,cm^2$
$= 5 \times 1584 \,cm^2 = 7920 \,cm^2$
Thus,$7920 \,cm^2$ of cardboard is required.

(N/A) Given: Curved surface area $= 308 \, cm^2$,Slant height $(l) = 14 \, cm$.
$(i)$ Let the radius of the base be $r \, cm$.
We know that the curved surface area of a cone is given by $\pi r l$.
$\therefore \pi r l = 308$
$\Rightarrow \frac{22}{7} \times r \times 14 = 308$
$\Rightarrow 44 \times r = 308$
$\Rightarrow r = \frac{308}{44} = 7 \, cm$.
Thus,the radius of the base is $7 \, cm$.
$(ii)$ Total surface area of a cone $= \text{Curved surface area} + \text{Base area}$.
Base area $= \pi r^2 = \frac{22}{7} \times 7^2 = 154 \, cm^2$.
Total surface area $= 308 \, cm^2 + 154 \, cm^2 = 462 \, cm^2$.

(A) The volume of a cylinder is given by the formula $V = \pi r^2 h$,where $r$ is the radius and $h$ is the height.
Let the initial radius be $r$ and the initial height be $h$. The initial volume is $V_1 = \pi r^2 h$.
According to the problem,the new radius $r' = \frac{r}{2}$ and the new height $h' = 2h$.
The new volume $V_2$ is calculated as:
$V_2 = \pi (r')^2 (h')$
$V_2 = \pi (\frac{r}{2})^2 (2h)$
$V_2 = \pi (\frac{r^2}{4}) (2h)$
$V_2 = \frac{2}{4} \pi r^2 h$
$V_2 = \frac{1}{2} \pi r^2 h$
Since $V_1 = \pi r^2 h$,we have $V_2 = \frac{1}{2} V_1$.
Therefore,the volume is halved.

(C) The total surface area of a cube is given by the formula $6 \times (\text{edge})^2$.
Given that the total surface area is $96 \, cm^2$, we have $6 \times (\text{edge})^2 = 96$.
Dividing both sides by $6$, we get $(\text{edge})^2 = 16$.
Taking the square root of both sides, the edge length is $\text{edge} = \sqrt{16} = 4 \, cm$.
The volume of a cube is given by the formula $(\text{edge})^3$.
Therefore, the volume $= (4)^3 = 64 \, cm^3$.

(D) The volume of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
Given,radius $r = 2.1 \, cm$ and height $h = 8.4 \, cm$.
Volume of cone $= \frac{1}{3} \pi (2.1)^2 \times 8.4$.
When the cone is melted and recast into a sphere,the volume remains constant.
Let the radius of the sphere be $R$. The volume of a sphere is $V = \frac{4}{3} \pi R^3$.
Equating the volumes: $\frac{4}{3} \pi R^3 = \frac{1}{3} \pi (2.1)^2 \times 8.4$.
Dividing both sides by $\frac{1}{3} \pi$: $4 R^3 = (2.1)^2 \times 8.4$.
$R^3 = \frac{(2.1)^2 \times 8.4}{4} = (2.1)^2 \times 2.1 = (2.1)^3$.
Taking the cube root on both sides,$R = 2.1 \, cm$.
Thus,the radius of the sphere is $2.1 \, cm$.

(A) The formula for the curved surface area $(CSA)$ of a cylinder is $CSA = 2 \pi r h$,where $r$ is the radius and $h$ is the height.
When the radius is doubled $(r' = 2r)$ and the height is halved $(h' = h/2)$,the new curved surface area $(CSA')$ becomes:
$CSA' = 2 \pi (2r) \times (h/2)$
$CSA' = 2 \pi r h$
Since the new curved surface area is equal to the original curved surface area,the curved surface area remains the same.

(C) Let the radii of the two cylinders be $r_1 = 2r$ and $r_2 = 3r$,and their heights be $h_1 = 5h$ and $h_2 = 3h$.
The volume of a cylinder is given by the formula $V = \pi r^2 h$.
Therefore,the ratio of their volumes is:
$\frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} = \frac{\pi (2r)^2 (5h)}{\pi (3r)^2 (3h)}$
$\frac{V_1}{V_2} = \frac{4r^2 \times 5h}{9r^2 \times 3h} = \frac{20r^2h}{27r^2h} = \frac{20}{27}$
Thus,the ratio of their volumes is $20:27$.

(D) The lateral surface area of a cube is given by the formula: $4 \times (\text{edge})^2$.
Given that the lateral surface area is $256 \, m^2$,we have:
$4 \times (\text{edge})^2 = 256$
$(\text{edge})^2 = 256 \div 4 = 64$
$\text{edge} = \sqrt{64} = 8 \, m$.
The volume of a cube is given by the formula: $(\text{edge})^3$.
Therefore,the volume $= (8)^3 = 512 \, m^3$.

(A) First,convert all dimensions to the same unit (meters).
Dimensions of the plank: $4 \,m$,$50 \,cm = 0.5 \,m$,and $20 \,cm = 0.2 \,m$.
Volume of the pit $= 16 \,m \times 12 \,m \times 4 \,m = 768 \,m^3$.
Volume of one plank $= 4 \,m \times 0.5 \,m \times 0.2 \,m = 0.4 \,m^3$.
Number of planks $= \frac{\text{Volume of pit}}{\text{Volume of one plank}} = \frac{768}{0.4} = 1920$.

(B) The longest pole that can be placed in a room is equal to the length of the space diagonal of the cuboid.
The formula for the space diagonal of a cuboid with dimensions $l$,$b$,and $h$ is given by $\sqrt{l^2 + b^2 + h^2}$.
Given dimensions are $l = 10 \text{ m}$,$b = 10 \text{ m}$,and $h = 5 \text{ m}$.
Substituting these values into the formula:
$\text{Length} = \sqrt{10^2 + 10^2 + 5^2}$
$= \sqrt{100 + 100 + 25}$
$= \sqrt{225}$
$= 15 \text{ m}$.

(TRUE) True.
Here,the radius of the sphere $=$ radius of the cylinder $= r$.
The diameter of the sphere $=$ height of the cylinder $= 2r$.
The surface area of the sphere $= 4 \pi r^2$.
The curved surface area of the cylinder $= 2 \pi r h = 2 \pi r (2r) = 4 \pi r^2$.
Since both areas are equal to $4 \pi r^2$,the statement is true.

(B) In a right circular cone,the height $(h)$,radius $(r)$,and slant height $(l)$ always form the sides of a right-angled triangle where the slant height is the hypotenuse.
According to the Pythagorean theorem,the relationship is $l^{2} = h^{2} + r^{2}$.
Since this relationship holds true for all right circular cones,the given statement is False.

(A) Let $r$ be the radius and $h$ be the height of the cylinder.
The formula for the curved surface area $(CSA)$ of a cylinder is $CSA = 2 \pi r h$.
According to the problem,the new radius $r' = 2r$ and the new height $h' = h/2$.
Calculating the new curved surface area $(CSA')$:
$CSA' = 2 \pi r' h' = 2 \pi (2r) (h/2) = 2 \pi r h$.
Since $CSA' = CSA$,the curved surface area remains unchanged when the radius is doubled and the height is halved.
Therefore,the given statement is True.

(A) The edge of the cube is $2r$. The largest right circular cone that can be fitted inside this cube will have a height $h = 2r$ and a base diameter equal to the edge of the cube,so the radius of the cone is $R = r$.
The volume of this cone is given by $V_{cone} = \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi (r)^2 (2r) = \frac{2}{3} \pi r^3$.
The volume of a hemisphere of radius $r$ is given by $V_{hemisphere} = \frac{2}{3} \pi r^3$.
Since the volume of the cone is $\frac{2}{3} \pi r^3$ and the volume of the hemisphere is $\frac{2}{3} \pi r^3$,the volumes are equal.
Hence,the given statement is True.

(TRUE) Let $r$ be the radius and $h$ be the height of both the cylinder and the right circular cone.
The volume of a cylinder is given by the formula $V_{cylinder} = \pi r^{2} h$.
The volume of a right circular cone is given by the formula $V_{cone} = \frac{1}{3} \pi r^{2} h$.
Comparing the two volumes,we can see that $V_{cylinder} = 3 \times (\frac{1}{3} \pi r^{2} h) = 3 \times V_{cone}$.
Therefore,the volume of the cylinder is indeed three times the volume of the cone.
Hence,the given statement is True.

(TRUE) Let the radius of the base of the cone,hemisphere,and cylinder be $r$. Since they stand on equal bases,their radii are equal.
Given that they have the same height,and the height of a hemisphere is equal to its radius $(r)$,the height of the cone and the cylinder must also be $r$.
$1$. Volume of the cone $(V_1)$: $V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (r) = \frac{1}{3} \pi r^3$.
$2$. Volume of the hemisphere $(V_2)$: $V_2 = \frac{2}{3} \pi r^3$.
$3$. Volume of the cylinder $(V_3)$: $V_3 = \pi r^2 h = \pi r^2 (r) = \pi r^3 = \frac{3}{3} \pi r^3$.
Comparing the volumes: $V_1 : V_2 : V_3 = \frac{1}{3} \pi r^3 : \frac{2}{3} \pi r^3 : \frac{3}{3} \pi r^3 = 1 : 2 : 3$.
Thus,the ratio of their volumes is $1 : 2 : 3$. The statement is True.

(B) Let the length of the edge of the cube be $a$.
The formula for the length of the diagonal of a cube is $\sqrt{3} a$.
Given that the length of the diagonal of the cube is $6 \sqrt{3} \, \text{cm}$.
Equating the two,we get $\sqrt{3} a = 6 \sqrt{3}$.
Dividing both sides by $\sqrt{3}$,we get $a = 6 \, \text{cm}$.
Therefore,the length of the edge of the cube is $6 \, \text{cm}$,not $3 \, \text{cm}$.
Hence,the given statement is False.

(A) Let $a$ be the edge length of the cube.
Since the sphere is inscribed in the cube,the diameter of the sphere is equal to the edge length of the cube,$a$. Therefore,the radius of the sphere is $r = \frac{a}{2}$.
Volume of the cube $(V_1)$ = $(\text{edge})^3 = a^3$.
Volume of the sphere $(V_2)$ = $\frac{4}{3} \pi r^3 = \frac{4}{3} \pi (\frac{a}{2})^3 = \frac{4}{3} \pi (\frac{a^3}{8}) = \frac{\pi a^3}{6}$.
Ratio of the volume of the cube to the volume of the sphere = $\frac{V_1}{V_2} = \frac{a^3}{\frac{\pi a^3}{6}} = \frac{6}{\pi}$.
Thus,the ratio is $6: \pi$.
Hence,the given statement is true.

(TRUE) Let $r$ be the radius and $h$ be the height of the cylinder.
The original volume of the cylinder is given by $V_{1} = \pi r^{2} h$.
When the radius is doubled,the new radius becomes $2r$. When the height is halved,the new height becomes $\frac{h}{2}$.
The new volume $V_{2}$ is calculated as:
$V_{2} = \pi (2r)^{2} \times \left(\frac{h}{2}\right)$
$V_{2} = \pi (4r^{2}) \times \left(\frac{h}{2}\right)$
$V_{2} = 2 \pi r^{2} h$
$V_{2} = 2 V_{1}$
Since the new volume is twice the original volume,the statement is True.

(N/A) Surface area of the sphere $= 4\pi r^2 = 4 \times \pi \times 5^2 = 100\pi \, cm^2$.
Curved surface area of the cone $= \pi rl = \pi \times 4 \times l = 4\pi l \, cm^2$,where $l$ is the slant height of the cone.
According to the problem,the surface area of the sphere $= 5 \times$ curved surface area of the cone.
$100\pi = 5 \times 4\pi l$
$100\pi = 20\pi l$
$l = \frac{100}{20} = 5 \, cm$.
To find the height $h$ of the cone,use the formula $l^2 = h^2 + r^2$:
$5^2 = h^2 + 4^2$
$25 = h^2 + 16$
$h^2 = 25 - 16 = 9$
$h = 3 \, cm$.
Volume of the cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 4^2 \times 3$
$= \frac{22}{7} \times 16 = \frac{352}{7} \approx 50.29 \, cm^3$.