Mix Examples - Surface Areas and Volumes Questions in English

(B) Given that the edge length of the cube is $a = 16\, cm$.
The formula for the total surface area of a cube is $6a^2$.
Substituting the value of $a$ into the formula:
Total Surface Area $= 6 \times (16)^2\, cm^2$
$= 6 \times 256\, cm^2$
$= 1536\, cm^2$.
Therefore,the total surface area of the cube is $1536\, cm^2$.

(C) For the cuboidal box: length $(l) = 30 \, cm$,breadth $(b) = 25 \, cm$,and height $(h) = 20 \, cm$.
Total surface area of the cuboidal box $= 2(lb + bh + hl) = 2(30 \times 25 + 25 \times 20 + 20 \times 30) \, cm^2 = 2(750 + 500 + 600) \, cm^2 = 2(1850) \, cm^2 = 3700 \, cm^2$.
For the cubical box: edge $(a) = 25 \, cm$.
Total surface area of the cubical box $= 6a^2 = 6(25)^2 \, cm^2 = 6(625) \, cm^2 = 3750 \, cm^2$.
Comparing the two,the cubical box requires more metal sheet.
The difference in the area is $3750 \, cm^2 - 3700 \, cm^2 = 50 \, cm^2$.
Thus,the cubical box uses $50 \, cm^2$ more metal sheet.

(D) Given: Length $(l) = 2\, m = 200\, cm$,breadth $(b) = 1\, m = 100\, cm$,and height $(h) = 80\, cm$.
The tank has five surfaces to be covered: two sides of $(l \times h)$,two sides of $(b \times h)$,and one top surface of $(l \times b)$.
Area of one square tile $= 20\, cm \times 20\, cm = 400\, cm^2$.
$1$. Tiles for two surfaces of $(l \times h) = 2 \times \frac{200 \times 80}{400} = 2 \times \frac{16000}{400} = 2 \times 40 = 80$.
$2$. Tiles for two surfaces of $(b \times h) = 2 \times \frac{100 \times 80}{400} = 2 \times \frac{8000}{400} = 2 \times 20 = 40$.
$3$. Tiles for the top surface of $(l \times b) = \frac{200 \times 100}{400} = \frac{20000}{400} = 50$.
Total number of tiles $= 80 + 40 + 50 = 170$.

(A) The formula for the total surface area of a cuboid is $2(lb + bh + lh)$,where $l$ is the length,$b$ is the breadth,and $h$ is the height.
Given: $l = 20\, cm$,$b = 15\, cm$,$h = 12\, cm$.
Substituting the values into the formula:
Total Surface Area $= 2(20 \times 15 + 15 \times 12 + 20 \times 12)$
$= 2(300 + 180 + 240)$
$= 2(720)$
$= 1440\, cm^2$.

(B) The formula for the total surface area of a cube is $6a^2$,where $a$ is the length of the edge of the cube.
Given,$a = 35\, cm$.
Total surface area $= 6 \times (35)^2$
$= 6 \times 1225$
$= 7350\, cm^2$.
Therefore,the correct option is $B$.

(C) The dimensions of the cuboid are length $(l)$ = $50 \, cm$,breadth $(b)$ = $40 \, cm$,and height $(h)$ = $30 \, cm$.
The formula for the lateral surface area of a cuboid is $2h(l + b)$.
Substituting the given values into the formula:
Lateral Surface Area = $2 \times 30 \times (50 + 40)$
= $60 \times 90$
= $5400 \, cm^2$.

(D) When two cubes of edge $a = 8 \, cm$ are joined end to end,they form a cuboid.
The dimensions of the resulting cuboid are:
Length $(l)$ = $8 \, cm + 8 \, cm = 16 \, cm$
Breadth $(b)$ = $8 \, cm$
Height $(h)$ = $8 \, cm$
The total surface area of a cuboid is given by the formula: $TSA = 2(lb + bh + lh)$.
Substituting the values:
$TSA = 2(16 \times 8 + 8 \times 8 + 16 \times 8)$
$TSA = 2(128 + 64 + 128)$
$TSA = 2(320)$
$TSA = 640 \, cm^{2}$.

(A) The lateral surface area of a cube is given by the formula $4a^{2}$,where $a$ is the length of the edge of the cube.
Given,$a = 15 \, cm$.
Lateral Surface Area $= 4 \times (15)^{2} \, cm^{2}$.
Lateral Surface Area $= 4 \times 225 \, cm^{2}$.
Lateral Surface Area $= 900 \, cm^{2}$.

(B) The dimensions of the cuboidal tank are: length $(l)$ = $10 \, m$,breadth $(b)$ = $8 \, m$,and height $(h)$ = $3 \, m$.
The total surface area of a closed cuboidal tank is given by the formula: $2(lb + bh + lh)$.
Substituting the values: $2(10 \times 8 + 8 \times 3 + 10 \times 3) = 2(80 + 24 + 30) = 2(134) = 268 \, m^2$.
The cost of the iron sheet is calculated by multiplying the total surface area by the rate per $m^2$.
Cost = $268 \, m^2 \times ₹ 50/m^2 = ₹ 13,400$.

(C) $1$. Calculate the area of the four walls: $Area = 2(l + b) \times h = 2(7 + 5) \times 4 = 2(12) \times 4 = 96\, m^2$.
$2$. Calculate the area of the ceiling: $Area = l \times b = 7 \times 5 = 35\, m^2$.
$3$. Total area to be whitewashed (excluding doors and windows): $Total\, Area = (Area\, of\, walls + Area\, of\, ceiling) - (Area\, of\, 2\, doors + Area\, of\, 4\, windows)$.
$4$. $Area\, of\, 2\, doors = 2 \times (3 \times 2) = 12\, m^2$.
$5$. $Area\, of\, 4\, windows = 4 \times (2 \times 1) = 8\, m^2$.
$6$. $Total\, Area = (96 + 35) - (12 + 8) = 131 - 20 = 111\, m^2$.
$7$. $Cost = Total\, Area \times Rate = 111 \times 200 = ₹22,200$.

(N/A) Let the dimensions of the cuboidal box be $6x$,$4x$,and $3x$ meters.
The total surface area of the cuboid is $SA = 2(lb + bh + lh) = 2((6x)(4x) + (4x)(3x) + (6x)(3x)) = 2(24x^2 + 12x^2 + 18x^2) = 2(54x^2) = 108x^2 \, m^2$.
The cost of covering with white paper at $₹ 2.00/m^2$ is $2 \times 108x^2 = 216x^2$.
The cost of covering with coloured paper at $₹ 2.50/m^2$ is $2.50 \times 108x^2 = 270x^2$.
The difference in cost is $270x^2 - 216x^2 = 54x^2$.
Given that the difference is $₹ 1350$,we have $54x^2 = 1350$.
$x^2 = 1350 / 54 = 25$.
$x = 5$.
Thus,the dimensions are $6(5) = 30 \, m$,$4(5) = 20 \, m$,and $3(5) = 15 \, m$.

Given for the cylinder:
Radius $(r) = 28 \, cm$
Height $(h) = 90 \, cm$
Curved surface area of a cylinder $= 2 \pi r h$
$= 2 \times \frac{22}{7} \times 28 \times 90 \, cm^2$
$= 2 \times 22 \times 4 \times 90 \, cm^2$
$= 15,840 \, cm^2$
Total surface area of a cylinder $= 2 \pi r(h + r)$
$= 2 \times \frac{22}{7} \times 28 \times (90 + 28) \, cm^2$
$= 176 \times 118 \, cm^2$
$= 20,768 \, cm^2$

(B) For the given cylinder,height $(h) = 30 \, cm$ and curved surface area $= 1320 \, cm^2$.
The formula for the curved surface area of a cylinder is $A = \pi d h$,where $d$ is the diameter.
Substituting the given values: $1320 = \frac{22}{7} \times d \times 30$.
Solving for $d$: $d = \frac{1320 \times 7}{22 \times 30}$.
$d = \frac{1320 \times 7}{660} = 2 \times 7 = 14 \, cm$.
Thus,the diameter of the cylinder is $14 \, cm$.

$(440 \ cm^2)$ For a metal hollow cylinder: height $(h) = 70\, cm$.
Inner radius $(r) = 10\, cm$.
Outer radius $(R) = \text{Inner radius} + \text{Thickness of metal} = 10 + 1 = 11\, cm$.
The difference between the areas of the outer and inner curved surfaces is given by:
$\text{Difference} = 2\pi Rh - 2\pi rh = 2\pi h(R - r)$.
Substituting the values:
$\text{Difference} = 2 \times \frac{22}{7} \times 70 \times (11 - 10) = 2 \times 22 \times 10 \times 1 = 440\, cm^2$.

(D) The formula for the curved surface area $(CSA)$ of a cylinder is given by $CSA = 2 \pi rh$.
Given:
Height $(h)$ = $35 \, cm$
Radius $(r)$ = $15 \, cm$
Using $\pi = \frac{22}{7}$:
$CSA = 2 \times \frac{22}{7} \times 15 \times 35$
$CSA = 2 \times 22 \times 15 \times 5$
$CSA = 44 \times 75$
$CSA = 3300 \, cm^2$
Therefore,the correct option is $D$.

(A) The diameter of the cylinder is $d = 28 \text{ cm}$,so the radius is $r = d/2 = 14 \text{ cm}$.
The height of the cylinder is $h = 50 \text{ cm}$.
The formula for the total surface area of a cylinder is $A = 2\pi r(r + h)$.
Substituting the values: $A = 2 \times (22/7) \times 14 \times (14 + 50)$.
$A = 2 \times 22 \times 2 \times 64$.
$A = 88 \times 64 = 5632 \text{ cm}^2$.

(B) The formula for the curved surface area $(CSA)$ of a cylinder is $CSA = 2\pi rh$,where $r$ is the radius and $h$ is the height.
Given: $CSA = 3960\, cm^{2}$ and $r = 21\, cm$.
Using $\pi = \frac{22}{7}$,we have:
$3960 = 2 \times \frac{22}{7} \times 21 \times h$
$3960 = 2 \times 22 \times 3 \times h$
$3960 = 132 \times h$
$h = \frac{3960}{132}$
$h = 30\, cm$.
Therefore,the height of the cylinder is $30\, cm$.

(C) The curved surface area $(CSA)$ of a cylinder is given by the formula: $CSA = 2 \pi r h$,where $r$ is the radius and $h$ is the height.
Given: $CSA = 550 \, cm^2$ and $h = 50 \, cm$.
Substituting the values: $550 = 2 \times \frac{22}{7} \times r \times 50$.
$550 = \frac{2200}{7} \times r$.
$r = \frac{550 \times 7}{2200} = \frac{7}{4} = 1.75 \, cm$.
The diameter $(d)$ is $2 \times r = 2 \times 1.75 = 3.5 \, cm$.

(A) Given: Radius $(r)$ = $10\,cm$,Height $(h)$ = $30\,cm$,Rate = ₹ $3/cm^2$,$\pi = 3.14$.
The formula for the curved surface area $(CSA)$ of a cylinder is $2\pi rh$.
$CSA$ = $2 \times 3.14 \times 10 \times 30 = 1884\,cm^2$.
Total cost = $CSA$ $\times$ Rate = $1884 \times 3 = 5652$.
Thus,the cost of painting the outer curved surface is ₹ $5652$.

(A) The formula for the total surface area $(TSA)$ of a closed cylinder is given by $TSA = 2\pi r(r + h)$,where $r$ is the radius and $h$ is the height.
Given: $TSA = 748\, cm^2$ and $r = 7\, cm$.
Using $\pi = \frac{22}{7}$:
$748 = 2 \times \frac{22}{7} \times 7 \times (7 + h)$
$748 = 44 \times (7 + h)$
Divide both sides by $44$:
$7 + h = \frac{748}{44}$
$7 + h = 17$
$h = 17 - 7$
$h = 10\, cm$.

(B) The well is in the shape of a cylinder.
Given: Depth $(h)$ = $20 \, m$,Radius $(r)$ = $7 \, m$.
The inner curved surface area of the cylinder = $2 \pi r h$.
Area = $2 \times (22/7) \times 7 \times 20 = 2 \times 22 \times 20 = 880 \, m^2$.
The cost of cementing is ₹ $15$ per $m^2$.
Total cost = $880 \times 15 = ₹ 13,200$.

(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 $r_1 : r_2 = 3 : 7$ and $h_1 : h_2 = 28 : 15$.
The curved surface area $(CSA)$ of a cylinder is given by the formula $CSA = 2\pi rh$.
Therefore,the ratio of their curved surface areas is:
$\frac{CSA_1}{CSA_2} = \frac{2\pi r_1 h_1}{2\pi r_2 h_2} = \frac{r_1}{r_2} \times \frac{h_1}{h_2}$
Substituting the given ratios:
$\frac{CSA_1}{CSA_2} = \frac{3}{7} \times \frac{28}{15}$
$\frac{CSA_1}{CSA_2} = \frac{3}{15} \times \frac{28}{7} = \frac{1}{5} \times 4 = \frac{4}{5}$
Thus,the ratio of their curved surface areas is $4 : 5$.

(D) Let the radius of the cylinder be $r$ and the height be $h$.
The curved surface area $(CSA)$ of a cylinder is $2\pi rh$.
The total surface area $(TSA)$ of a cylinder is $2\pi rh + 2\pi r^2 = 2\pi r(h + r)$.
Given the ratio $CSA : TSA = 2 : 5$,we have:
$\frac{2\pi rh}{2\pi r(h + r)} = \frac{2}{5}$
$\frac{h}{h + r} = \frac{2}{5}$
Cross-multiplying gives $5h = 2(h + r)$.
$5h = 2h + 2r$
$3h = 2r$
Therefore,the ratio of the radius to the height is $\frac{r}{h} = \frac{3}{2}$,which is $3: 2$.

(A) $1$. Convert the radius from $cm$ to $m$: $r = 14 \, cm = 0.14 \, m$.
$2$. The formula for the curved surface area $(CSA)$ of one cylinder is $2 \pi rh$.
$3$. $CSA = 2 \times \frac{22}{7} \times 0.14 \times 5 = 4.4 \, m^2$.
$4$. Total $CSA$ for $20$ pillars = $20 \times 4.4 = 88 \, m^2$.
$5$. Total cost = $\text{Total Area} \times \text{Rate} = 88 \times 120 = ₹ 10,560$.

(N/A) Given: Slant height $(l) = 25 \,cm$,Curved Surface Area $(CSA) = 550 \,cm^{2}$.
$1$. Finding the radius $(r)$:
$CSA = \pi r l$
$550 = \frac{22}{7} \times r \times 25$
$r = \frac{550 \times 7}{22 \times 25} = 7 \,cm$.
$2$. Finding the height $(h)$:
Using the relation $l^{2} = h^{2} + r^{2}$:
$25^{2} = h^{2} + 7^{2}$
$625 = h^{2} + 49$
$h^{2} = 625 - 49 = 576$
$h = \sqrt{576} = 24 \,cm$.
$3$. Finding the total surface area $(TSA)$:
$TSA = \pi r(l + r)$
$TSA = \frac{22}{7} \times 7 \times (25 + 7)$
$TSA = 22 \times 32 = 704 \,cm^{2}$.

(A) For a given cone:
Radius $(r) = 7\, m$ and slant height $(l) = 12\, m$.
Area of tarpaulin used in the tent = Curved surface area of the cone.
Curved surface area $= \pi r l = \frac{22}{7} \times 7 \times 12 = 264\, m^2$.
Since the area of the tarpaulin is $l \times b$,where $b = 2\, m$:
$264 = l \times 2 \implies l = \frac{264}{2} = 132\, m$.
Thus,$132\, m$ of tarpaulin is required.
Cost of $1\, m$ tarpaulin = ₹ $48$.
Total cost = $132 \times 48 = ₹ 6336$.

(A) Given,the diameter of the solid hemisphere is $7 \, cm$.
Radius $(r) = \frac{\text{diameter}}{2} = \frac{7}{2} \, cm = 3.5 \, cm$.
Curved surface area of a hemisphere $= 2 \pi r^2$.
$= 2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \, cm^2 = 77 \, cm^2$.
Total surface area of a hemisphere $= 3 \pi r^2$.
$= 3 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \, cm^2 = 115.5 \, cm^2$.

(A) Given radius $r = 2.1 \, m$.
Curved Surface Area of a hemisphere = $2 \pi r^2$.
$= 2 \times \frac{22}{7} \times 2.1 \times 2.1 = 2 \times 22 \times 0.3 \times 2.1 = 27.72 \, m^2$.
Total Surface Area of a hemisphere = $3 \pi r^2$.
$= 3 \times \frac{22}{7} \times 2.1 \times 2.1 = 3 \times 22 \times 0.3 \times 2.1 = 41.58 \, m^2$.

(A) Given: Diameter $(d)$ = $14 \, cm$.
Radius $(r)$ = $d / 2 = 14 / 2 = 7 \, cm$.
Curved Surface Area $(CSA)$ of a hemisphere = $2 \pi r^2$.
$CSA$ = $2 \times (22 / 7) \times 7 \times 7 = 2 \times 22 \times 7 = 308 \, cm^2$.
Total Surface Area $(TSA)$ of a hemisphere = $3 \pi r^2$.
$TSA$ = $3 \times (22 / 7) \times 7 \times 7 = 3 \times 22 \times 7 = 462 \, cm^2$.
Thus,the curved surface area is $308 \, cm^2$ and the total surface area is $462 \, cm^2$.

(N/A) Given diameter $d = 126 \, cm$,so radius $r = \frac{126}{2} = 63 \, cm$.
Curved Surface Area of a hemisphere $= 2 \pi r^2 = 2 \times \frac{22}{7} \times 63 \times 63 = 2 \times 22 \times 9 \times 63 = 24,948 \, cm^2$.
Total Surface Area of a hemisphere $= 3 \pi r^2 = 3 \times \frac{22}{7} \times 63 \times 63 = 3 \times 22 \times 9 \times 63 = 37,422 \, cm^2$.

(A) Given diameter $d = 2.8\,m$,so the radius $r = d/2 = 1.4\,m$.
The curved surface area $(CSA)$ of a hemisphere is given by $2\pi r^2$.
$CSA = 2 \times (22/7) \times (1.4)^2 = 2 \times (22/7) \times 1.96 = 2 \times 22 \times 0.28 = 12.32\,m^2$.
The total surface area $(TSA)$ of a hemisphere is given by $3\pi r^2$.
$TSA = 3 \times (22/7) \times (1.4)^2 = 3 \times (22/7) \times 1.96 = 3 \times 22 \times 0.28 = 18.48\,m^2$.
Thus,the curved surface area is $12.32\,m^2$ and the total surface area is $18.48\,m^2$.

(D) The dome is in the form of a hemisphere.
Radius $(r)$ = $4.2 \, m$.
Surface area of the hemispherical dome = Curved Surface Area $(CSA)$ = $2 \pi r^2$.
$CSA$ = $2 \times \frac{22}{7} \times 4.2 \times 4.2$.
$CSA$ = $2 \times 22 \times 0.6 \times 4.2 = 110.88 \, m^2$.
Cost of painting = Area $\times$ Rate.
Cost = $110.88 \times 18 = ₹ 1995.84$.

(A) The total surface area of a hemisphere is given by the formula $TSA = 3\pi r^2$,where $r$ is the radius.
Given $TSA = 4158 \, cm^2$.
So,$3 \times \frac{22}{7} \times r^2 = 4158$.
$r^2 = \frac{4158 \times 7}{3 \times 22}$.
$r^2 = \frac{4158 \times 7}{66} = 63 \times 7 = 441$.
$r = \sqrt{441} = 21 \, cm$.
The diameter $d = 2r = 2 \times 21 = 42 \, cm$.

(A) For a single iron hollow box: length $(l) = 30 \, cm$,breadth $(b) = 15 \, cm$,and height $(h) = 10 \, cm$.
The volume of each cuboid $= l \times b \times h$.
$= 30 \times 15 \times 10 \, cm^3 = 4500 \, cm^3$.
Therefore,the volume of a carton containing $60$ such iron boxes $= 60 \times 4500 \, cm^3$.
$= 2,70,000 \, cm^3$.
Since $1 \, m^3 = 1,000,000 \, cm^3$,we convert the volume to cubic meters:
$= \frac{2,70,000}{1,000,000} \, m^3 = 0.27 \, m^3$.

(A) Given,the edge of the cube is $a = 6 \, cm$.
$1$. Total Surface Area of the cube:
Formula: $TSA = 6a^2$
$TSA = 6 \times (6)^2 = 6 \times 36 = 216 \, cm^2$.
$2$. Volume of the cube:
Formula: $V = a^3$
$V = (6)^3 = 6 \times 6 \times 6 = 216 \, cm^3$.

(C) For a given room,the dimensions are: length $(l) = 12 \, m$,width $(b) = 4 \, m$,and height $(h) = 3 \, m$.
The longest rod that can be placed in a rectangular room corresponds to the space diagonal of the cuboid.
The formula for the length of the diagonal is $d = \sqrt{l^2 + b^2 + h^2}$.
Substituting the given values:
$d = \sqrt{12^2 + 4^2 + 3^2} \, m$
$d = \sqrt{144 + 16 + 9} \, m$
$d = \sqrt{169} \, m$
$d = 13 \, m$.
Thus,the length of the longest rod is $13 \, m$.

(N/A) Let the length,breadth,and height of the cuboid be $l \, cm$,$b \, cm$,and $h \, cm$ respectively.
The areas of the three adjacent faces meeting at a vertex are given by $lb$,$bh$,and $hl$.
Given:
$lb = 700 \, cm^2$ --- $(i)$
$bh = 300 \, cm^2$ --- $(ii)$
$hl = 525 \, cm^2$ --- $(iii)$
Multiplying $(i)$,$(ii)$,and $(iii)$:
$(lb) \times (bh) \times (hl) = 700 \times 300 \times 525$
$l^2 b^2 h^2 = 110,250,000$
$(lbh)^2 = (10,500)^2$
$lbh = 10,500 \, cm^3$ (This is the volume of the cuboid).
To find the dimensions:
From $(i)$ and $(iii)$,$lb = 700$ and $hl = 525$. Dividing $(i)$ by $(iii)$ gives $b/h = 700/525 = 4/3$,so $b = 4h/3$.
Substitute $b$ into $(ii)$:
$(4h/3) \times h = 300$
$h^2 = 300 \times 3 / 4 = 225$
$h = 15 \, cm$.
Now,$b = 4(15)/3 = 20 \, cm$.
And $l = 700/20 = 35 \, cm$.
Thus,the dimensions are $35 \, cm$,$20 \, cm$,and $15 \, cm$,and the volume is $10,500 \, cm^3$.

(A) Given, capacity of the tank = $60,000$ litres.
Since $1$ m$^3 = 1,000$ litres, the volume of the tank in cubic meters is $60,000 / 1,000 = 60$ m$^3$.
The formula for the volume of a cuboid is $V = \text{length} \times \text{breadth} \times \text{height}$.
Substituting the given values: $60 = 5 \times 4 \times h$.
$60 = 20 \times h$.
$h = 60 / 20 = 3$ m.
Therefore, the height of the tank is $3$ m.

(B) $1$. Calculate the total volume of the tank: $V = \text{length} \times \text{breadth} \times \text{height} = 20\, \text{m} \times 13\, \text{m} \times 10\, \text{m} = 2600\, \text{m}^3$.
$2$. Convert the volume into litres: Since $1\, \text{m}^3 = 1000\, \text{litres}$, the total capacity is $2600 \times 1000 = 2,600,000\, \text{litres}$.
$3$. Calculate the daily water requirement for the village: $5000 \times 130\, \text{litres/day} = 650,000\, \text{litres/day}$.
$4$. Calculate the number of days the water will last: $\frac{2,600,000}{650,000} = 4\, \text{days}$.

(N/A) Let the edge of the cube be $a$. The volume of a cube is given by $V = a^{3}$.
Given $a^{3} = 3375\, cm^{3}$.
Taking the cube root of both sides,$a = \sqrt[3]{3375} = 15\, cm$.
The total surface area of a cube is given by $TSA = 6a^{2}$.
$TSA = 6 \times (15)^{2} = 6 \times 225 = 1350\, cm^{2}$.
Thus,the edge is $15\, cm$ and the total surface area is $1350\, cm^{2}$.

(N/A) Let the length $l = 30\, cm$,breadth $b = 25\, cm$,and height be $h\, cm$.
The lateral surface area $(LSA)$ of a cuboid is given by $2h(l + b) = 2h(30 + 25) = 110h\, cm^2$.
The total surface area $(TSA)$ of a cuboid is given by $2(lb + bh + lh) = 2(30 \times 25 + 25h + 30h) = 2(750 + 55h) = 1500 + 110h\, cm^2$.
According to the problem,$TSA = 2 \times LSA$.
$1500 + 110h = 2(110h)$.
$1500 + 110h = 220h$.
$110h = 1500$.
$h = \frac{1500}{110} = \frac{150}{11}\, cm$.
The volume $V = l \times b \times h = 30 \times 25 \times \frac{150}{11} = 750 \times \frac{150}{11} = \frac{112500}{11}\, cm^3$.

(N/A) The volume $V$ of a cuboid with dimensions $a, b, c$ is given by $V = abc$.
The total surface area $S$ of the cuboid is given by $S = 2(ab + bc + ca)$.
Now,consider the right-hand side $(RHS)$ of the expression:
$RHS = \frac{2}{S} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$
Substitute the expression for $S$:
$RHS = \frac{2}{2(ab + bc + ca)} \left( \frac{bc + ac + ab}{abc} \right)$
Simplify the expression:
$RHS = \frac{1}{ab + bc + ca} \cdot \frac{ab + bc + ca}{abc}$
Cancel the common term $(ab + bc + ca)$:
$RHS = \frac{1}{abc}$
Since $V = abc$,we have:
$RHS = \frac{1}{V}$
Thus,$\frac{1}{V} = \frac{2}{S} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)$ is proved.

(A) Let the dimensions of the cuboid be $l, b,$ and $h$. The areas of the three adjacent faces are given by $lb = 4000$,$bh = 2000$,and $hl = 3200$.
Multiplying these three equations: $(lb)(bh)(hl) = 4000 \times 2000 \times 3200$.
$(lbh)^2 = 25,600,000,000$.
$lbh = \sqrt{25,600,000,000} = 160,000 \, cm^3$.
Now,to find the dimensions:
$l = (lbh) / (bh) = 160,000 / 2000 = 80 \, cm$.
$b = (lbh) / (hl) = 160,000 / 3200 = 50 \, cm$.
$h = (lbh) / (lb) = 160,000 / 4000 = 40 \, cm$.
Thus,the dimensions are $80 \, cm, 50 \, cm, 40 \, cm$ and the volume is $160,000 \, cm^3$.

(C) Let the dimensions of the cuboid be $4x$,$3x$,and $1x$.
The total surface area of a cuboid is given by the formula: $TSA = 2(lb + bh + lh)$.
Substituting the given values: $3800 = 2((4x)(3x) + (3x)(x) + (4x)(x))$.
$3800 = 2(12x^{2} + 3x^{2} + 4x^{2})$.
$3800 = 2(19x^{2})$.
$3800 = 38x^{2}$.
$x^{2} = 100$,which implies $x = 10 \, cm$.
The dimensions are $4(10) = 40 \, cm$,$3(10) = 30 \, cm$,and $1(10) = 10 \, cm$.
The volume of the cuboid is $V = l \times b \times h = 40 \times 30 \times 10 = 12,000 \, cm^{3}$.

(A) $1$. Volume of the cuboid = $length \times breadth \times height = 5 \,cm \times 4 \,cm \times 3 \,cm = 60 \,cm^3$.
$2$. Volume of one cube = $(side)^3 = (1 \,cm)^3 = 1 \,cm^3$.
$3$. Number of cubes = $\frac{\text{Volume of cuboid}}{\text{Volume of one cube}} = \frac{60 \,cm^3}{1 \,cm^3} = 60$.
$4$. Total surface area of the cuboid = $2(lb + bh + lh) = 2(5 \times 4 + 4 \times 3 + 3 \times 5) = 2(20 + 12 + 15) = 2(47) = 94 \,cm^2$.
$5$. Total surface area of one cube = $6(side)^2 = 6(1)^2 = 6 \,cm^2$.
$6$. Total surface area of $60$ cubes = $60 \times 6 \,cm^2 = 360 \,cm^2$.
$7$. Ratio = $\frac{94}{360} = \frac{47}{180}$ or $47:180$.

(A) Given:
Length $(l)$ = $10\, cm$
Breadth $(b)$ = $6\, cm$
Height $(h)$ = $4\, cm$
Density $(\rho)$ = $7.8\, g/cm^3$
Step $1$: Calculate the volume $(V)$ of the cuboid.
$V = l \times b \times h$
$V = 10\, cm \times 6\, cm \times 4\, cm = 240\, cm^3$
Step $2$: Calculate the mass $(m)$ of the cuboid.
Mass = Density $\times$ Volume
$m = 7.8\, g/cm^3 \times 240\, cm^3$
$m = 1872\, g$
Therefore,the mass of the cuboid is $1872\, g$.

(B) Let the edge of the fourth cube be $x\, cm$.
The volume of a cube with edge $a$ is given by $V = a^3$.
The volumes of the given cubes are $1^3 = 1\, cm^3$,$9^3 = 729\, cm^3$,and $10^3 = 1000\, cm^3$.
According to the problem,the sum of the volumes of the cubes with edges $9\, cm$ and $10\, cm$ is equal to the sum of the volumes of the cubes with edges $1\, cm$ and $x\, cm$.
So,$9^3 + 10^3 = 1^3 + x^3$.
$729 + 1000 = 1 + x^3$.
$1729 = 1 + x^3$.
$x^3 = 1728$.
Taking the cube root on both sides,$x = \sqrt[3]{1728} = 12\, cm$.
Thus,the edge of the fourth cube is $12\, cm$.

(C) Given for a solid cylinder:
Radius $(r) = 7 \, cm$
Volume $(V) = 4620 \, cm^{3}$
The formula for the volume of a cylinder is $V = \pi r^{2} h$.
Substituting the given values:
$4620 = \frac{22}{7} \times (7)^{2} \times h$
$4620 = \frac{22}{7} \times 49 \times h$
$4620 = 22 \times 7 \times h$
$4620 = 154 \times h$
Solving for height $(h)$:
$h = \frac{4620}{154}$
$h = 30 \, cm$
Thus,the height of the cylinder is $30 \, cm$.

(D) Given for a cylinder: radius $(r) = 14 \, cm$ and curved surface area $(CSA) = 4400 \, cm^2$.
The formula for the curved surface area of a cylinder is $CSA = 2 \pi r h$.
Substituting the given values: $4400 = 2 \times \frac{22}{7} \times 14 \times h$.
Simplifying the equation: $4400 = 2 \times 22 \times 2 \times h = 88 \times h$.
Therefore,$h = \frac{4400}{88} = 50 \, cm$.
The formula for the volume of a cylinder is $V = \pi r^2 h$.
Substituting the values: $V = \frac{22}{7} \times 14 \times 14 \times 50$.
$V = 22 \times 2 \times 14 \times 50 = 44 \times 700 = 30800 \, cm^3$.

(N/A) Given for a cylinder:
Curved surface area $(CSA)$ $= 3960 \, cm^{2}$
Volume $(V)$ $= 41580 \, cm^{3}$
We know that:
$CSA = 2 \pi r h = 3960 \, cm^{2}$ --- $(1)$
$V = \pi r^{2} h = 41580 \, cm^{3}$ --- $(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{V}{CSA} = \frac{\pi r^{2} h}{2 \pi r h} = \frac{41580}{3960}$
$\frac{r}{2} = 10.5$
$r = 21 \, cm$
Now,substitute $r = 21 \, cm$ in equation $(1)$:
$2 \times \frac{22}{7} \times 21 \times h = 3960$
$2 \times 22 \times 3 \times h = 3960$
$132 \times h = 3960$
$h = \frac{3960}{132} = 30 \, cm$
Thus,the radius is $21 \, cm$ and the height is $30 \, cm$.