Mix Examples - Surface Areas and Volumes Questions in English

(D) The surface area of a sphere is given by the formula $S = 4 \pi r^2$.
Let the radii of the three spheres be $r_1 = 33 \, cm$,$r_2 = 44 \, cm$,and $r_3 = 48 \, cm$.
Let the radius of the fourth sphere be $R$.
According to the problem,the surface area of the fourth sphere is equal to the sum of the surface areas of the first three spheres:
$4 \pi R^2 = 4 \pi r_1^2 + 4 \pi r_2^2 + 4 \pi r_3^2$
Dividing both sides by $4 \pi$,we get:
$R^2 = r_1^2 + r_2^2 + r_3^2$
$R^2 = (33)^2 + (44)^2 + (48)^2$
$R^2 = 1089 + 1936 + 2304$
$R^2 = 5329$
Taking the square root of both sides:
$R = \sqrt{5329} = 73 \, cm$.
Thus,the radius of the fourth sphere is $73 \, cm$.

(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 $( ho)$ = $7.8\, g/cm^3$
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 $30 \, cm$.

(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 $cm$: $0.7 \times 100 = 70\, cm$.

(A) For a given cone:
Radius $(r) = 10 \, cm$ and slant height $(l) = 26 \, cm$.
Using the relation $l^2 = r^2 + h^2$,we find the height $(h)$:
$h^2 = l^2 - r^2$
$h^2 = (26)^2 - (10)^2$
$h^2 = 676 - 100 = 576$
$h = \sqrt{576} = 24 \, cm$.
The volume of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times 3.14 \times (10)^2 \times 24$
$V = 3.14 \times 100 \times 8$
$V = 314 \times 8 = 2512 \, cm^3$.

(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) Given: Radius of the base $(r)$ = $14 \, cm$,Slant height $(l)$ = $50 \, cm$.
First,we find the vertical height $(h)$ of the cone using the relation $r^2 + h^2 = l^2$.
$14^2 + h^2 = 50^2$
$196 + h^2 = 2500$
$h^2 = 2500 - 196 = 2304$
$h = \sqrt{2304} = 48 \, cm$.
The volume of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 14 \times 14 \times 48$
$V = \frac{1}{3} \times 22 \times 2 \times 14 \times 48$
$V = 22 \times 2 \times 14 \times 16$
$V = 9856 \, cm^3$.
Therefore,the volume of the cone is $9856 \, cm^3$.

(A) The formula for the volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$,where $\pi r^2$ is the area of the base.
Given,the area of the base $(A)$ = $1386 \, cm^2$ and the height $(h)$ = $9 \, cm$.
Substituting these values into the formula:
$V = \frac{1}{3} \times A \times h$
$V = \frac{1}{3} \times 1386 \times 9$
$V = 1386 \times 3$
$V = 4158 \, cm^3$.
Therefore,the volume of the cone is $4158 \, cm^3$.

(B) Given: Radius $(r) = 7 \, cm$, Slant height $(l) = 25 \, cm$.
First, find the vertical height $(h)$ using the relation $l^2 = r^2 + h^2$.
$h^2 = l^2 - r^2 = 25^2 - 7^2 = 625 - 49 = 576$.
$h = \sqrt{576} = 24 \, cm$.
The volume $(V)$ of the cone is given by $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 24 = 22 \times 7 \times 8 = 1232 \, cm^3$.
Since $1000 \, cm^3 = 1 \, litre$, the capacity is $\frac{1232}{1000} = 1.232 \, litres$.

(C) Given: Radius $(r)$ = $21 \,cm$,Slant height $(l)$ = $29 \,cm$.
First,find the height $(h)$ of the cone using the relation $l^2 = r^2 + h^2$.
$h^2 = l^2 - r^2 = 29^2 - 21^2 = 841 - 441 = 400$.
$h = \sqrt{400} = 20 \,cm$.
The volume $(V)$ of a cone is given by $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 21 \times 21 \times 20 = 22 \times 3 \times 7 \times 20 = 9240 \,cm^3$.
Since $1000 \,cm^3 = 1 \,litre$,the capacity is $9240 / 1000 = 9.240 \,litres$.

(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) Given: Radius $(r)$ = $35 \,cm$,Slant height $(l)$ = $91 \,cm$.
First,find the vertical height $(h)$ using the relation $l^2 = r^2 + h^2$.
$h = \sqrt{l^2 - r^2} = \sqrt{91^2 - 35^2} = \sqrt{(91 - 35)(91 + 35)} = \sqrt{56 \times 126} = \sqrt{7056} = 84 \,cm$.
The volume $(V)$ of the cone is given by $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 35 \times 35 \times 84 = 22 \times 35 \times 5 \times 84 / 3 = 22 \times 35 \times 5 \times 28 = 107800 \,cm^3$.
Since $1000 \,cm^3 = 1 \,litre$,the capacity is $107800 / 1000 = 107.8 \,litres$.

(N/A) When $\Delta ABC$ is revolved about side $AB$,the height of the cone $h = AB = 6.3 \, cm$ and the radius $r = BC = 8.4 \, cm$. The volume $V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (8.4)^2 \times 6.3 = 465.696 \, cm^3$. When $\Delta ABC$ is revolved about side $BC$,the height of the cone $h = BC = 8.4 \, cm$ and the radius $r = AB = 6.3 \, cm$. The volume $V_2 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (6.3)^2 \times 8.4 = 349.272 \, cm^3$.

(C) For a given sphere,the radius $(r) = 2.1 \, cm = \frac{21}{10} \, cm$.
The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Substituting the values:
$V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{21}{10}\right)^3 \, cm^3$
$V = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10} \, cm^3$
$V = 4 \times 22 \times \frac{1}{1} \times \frac{1}{1} \times \frac{0.1}{1} \times 2.1 \times 2.1 \, cm^3$ (simplifying $3 \times 7 = 21$ with one $21$ in the numerator)
$V = 88 \times 0.1 \times 4.41 \, cm^3$
$V = 8.8 \times 4.41 \, cm^3 = 38.808 \, cm^3$.

(D) For a given hemisphere:
Radius $(r) = \frac{\text{diameter}}{2} = \frac{12}{2} = 6 \, cm$.
Volume of the hemisphere $= \frac{2}{3} \pi r^3$.
Substituting the values:
Volume $= \frac{2}{3} \times 3.14 \times (6)^3 \, cm^3$.
Volume $= \frac{2}{3} \times 3.14 \times 216 \, cm^3$.
Volume $= 2 \times 3.14 \times 72 \, cm^3$.
Volume $= 452.16 \, cm^3$.

(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 $V = \frac{4}{3} \pi r^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times (21)^3 = \frac{4}{3} \times \frac{22}{7} \times 21 \times 21 \times 21$.
$V = 4 \times 22 \times 21 \times 21 = 88 \times 441 = 38808 \, cm^3$.

(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 given by $4\pi r_1^2$.
The curved surface area of a hemisphere is given by $2\pi r_2^2$.
According to the problem,the surface areas are equal:
$4\pi r_1^2 = 2\pi r_2^2$
Dividing both sides by $2\pi$,we get:
$2r_1^2 = r_2^2$
Taking the square root on both sides:
$\sqrt{2}r_1 = r_2$
Therefore,the ratio of their radii is:
$\frac{r_1}{r_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio is $1: \sqrt{2}$.

(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 $128$.

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

(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.

(B) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Let the radius of the original ball be $R = 9 \,cm$.
Let the radii of the three smaller balls be $r_1 = 1 \,cm$,$r_2 = 6 \,cm$,and $r_3$ (the unknown radius).
Since the volume remains conserved during melting and recasting,the volume of the original ball equals the sum of the volumes of the three smaller balls:
$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi r_1^3 + \frac{4}{3} \pi r_2^3 + \frac{4}{3} \pi r_3^3$
Dividing both sides by $\frac{4}{3} \pi$,we get:
$R^3 = r_1^3 + r_2^3 + r_3^3$
Substituting the given values:
$9^3 = 1^3 + 6^3 + r_3^3$
$729 = 1 + 216 + r_3^3$
$729 = 217 + r_3^3$
$r_3^3 = 729 - 217$
$r_3^3 = 512$
Taking the cube root on both sides:
$r_3 = \sqrt[3]{512} = 8 \,cm$.
Thus,the radius of the third ball is $8 \,cm$.