Measurement of Volume and Surface Area Questions in English

(A) Radius of the coin $r = \frac{1.5}{2} = 0.75 \, cm = \frac{3}{4} \, cm$.
Thickness of the coin $h = 2 \, mm = 0.2 \, cm = \frac{1}{5} \, cm$.
Volume of one coin $= \pi r^2 h = \pi \times (\frac{3}{4})^2 \times \frac{1}{5} = \pi \times \frac{9}{16} \times \frac{1}{5} = \frac{9\pi}{80} \, cm^3$.
Radius of the cylinder $R = \frac{6}{2} = 3 \, cm$.
Height of the cylinder $H = 8 \, cm$.
Volume of the cylinder $= \pi R^2 H = \pi \times (3)^2 \times 8 = 72\pi \, cm^3$.
Number of coins $= \frac{\text{Volume of the cylinder}}{\text{Volume of one coin}} = \frac{72\pi}{\frac{9\pi}{80}} = 72 \times \frac{80}{9} = 8 \times 80 = 640$.

(B) Let $r \, cm$ be the radius of the base of the cone.
Since the volume of the material remains constant during casting,the volume of the hemisphere equals the volume of the cone.
Volume of hemisphere = $\frac{2}{3} \pi R^3$,where $R = 6 \, cm$.
Volume of cone = $\frac{1}{3} \pi r^2 h$,where $h = 75 \, cm$.
Equating the volumes: $\frac{2}{3} \pi (6)^3 = \frac{1}{3} \pi r^2 (75)$.
Canceling $\frac{1}{3} \pi$ from both sides: $2 \times 216 = r^2 \times 75$.
$r^2 = \frac{432}{75} = 5.76$.
Taking the square root: $r = \sqrt{5.76} = 2.4 \, cm$.

(B) The total surface area $(TSA)$ of a cylinder is given by $2 \pi r h + 2 \pi r^2 = 231 \, cm^2$.
Given that the curved surface area $(CSA)$ is $(2/3)$ of the total surface area:
$CSA = 2 \pi r h = (2/3) \times 231 = 154 \, cm^2$.
Since $TSA = CSA + 2 \pi r^2$,we have $231 = 154 + 2 \pi r^2$,which implies $2 \pi r^2 = 77 \, cm^2$.
Thus,$\pi r^2 = 77/2 = 38.5 \, cm^2$.
Using $\pi = 22/7$,we get $(22/7) \times r^2 = 38.5$,so $r^2 = (38.5 \times 7) / 22 = 12.25$.
Therefore,$r = \sqrt{12.25} = 3.5 \, cm$ (or $7/2 \, cm$).
Now,substitute $r$ into the $CSA$ formula: $2 \times (22/7) \times (7/2) \times h = 154$.
$22 \times h = 154$,so $h = 154 / 22 = 7 \, cm$.
The volume of the cylinder is $V = \pi r^2 h = (22/7) \times (7/2) \times (7/2) \times 7 = 269.5 \, cm^3$.

(B) The volume of water required to fill the tank is $440 \ m^3$ in $10 \ min$.
Therefore,the volume of water flowing through the pipe per minute is $\frac{440}{10} = 44 \ m^3/min$.
The volume of water flowing through a circular pipe per minute is given by the formula $V = \pi r^2 h$,where $h$ is the speed of water per minute and $r$ is the inner radius.
Given $h = 7 \ m/min$ and $V = 44 \ m^3/min$,we have:
$44 = \frac{22}{7} \times r^2 \times 7$
Simplifying the equation:
$44 = 22 \times r^2$
$r^2 = \frac{44}{22} = 2$
$r = \sqrt{2} \ m$.

(B) The volume of the remaining solid is calculated by subtracting the volume of the cone from the volume of the cylinder.
Volume of cylinder = $\pi r^2 h = \pi \times 6^2 \times 10 = 360\pi \ cm^3$.
Volume of cone = $\frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times 6^2 \times 10 = 120\pi \ cm^3$.
Remaining volume = $360\pi - 120\pi = 240\pi \ cm^3$.
Using $\pi \approx 3.14159$,the volume is $240 \times 3.14159 \approx 753.98 \ cm^3$,which rounds to $754.3 \ cm^3$ when using $\pi \approx \frac{22}{7}$ $(240 \times \frac{22}{7} \approx 754.28 \ cm^3)$.

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

(A) The radius of the base of the cylinder is $r = \frac{28}{2} = 14 \, m$.
Let $h$ be the depth (height) of the tank.
The capacity (volume) of the tank is given by $V = \pi r^2 h = 6160 \, m^3$.
Substituting the values: $\frac{22}{7} \times 14 \times 14 \times h = 6160$.
$22 \times 2 \times 14 \times h = 6160 \implies 616 \times h = 6160 \implies h = 10 \, m$.
The inner curved surface area is $A = 2 \pi r h = 2 \times \frac{22}{7} \times 14 \times 10 = 880 \, m^2$.
The cost of painting the inner curved surface at the rate of ₹ $2.8$ per $m^2$ is $880 \times 2.8 = ₹ 2464$.

(C) Given: $r + h = 37\, m$ and Total Surface Area $(TSA) = 2\pi r(r + h) = 1628\, m^2$.
Substituting the value of $(r + h)$ in the $TSA$ formula:
$2 \times \frac{22}{7} \times r \times 37 = 1628$
$r = \frac{1628 \times 7}{2 \times 22 \times 37} = \frac{1628 \times 7}{1628} = 7\, m$.
Since $r + h = 37$,then $h = 37 - 7 = 30\, m$.
Circumference of the base $= 2\pi r = 2 \times \frac{22}{7} \times 7 = 44\, m$.
Volume of the cylinder $= \pi r^2 h = \frac{22}{7} \times 7^2 \times 30 = 22 \times 7 \times 30 = 4620\, m^3$.
Thus,the circumference is $44\, m$ and the volume is $4620\, m^3$.

(B) When the paper is rolled along its length,the length of the paper becomes the circumference of the base of the cylinder,and the width of the paper becomes the height of the cylinder.
Given: Circumference $(2 \pi r) = 22 \text{ cm}$ and height $(h) = 10 \text{ cm}$.
From $2 \pi r = 22$,we have $2 \times \frac{22}{7} \times r = 22$,which gives $r = \frac{7}{2} \text{ cm}$.
The volume of the cylinder $(V)$ is given by the formula $V = \pi r^2 h$.
Substituting the values: $V = \frac{22}{7} \times (\frac{7}{2})^2 \times 10$.
$V = \frac{22}{7} \times \frac{49}{4} \times 10 = 22 \times \frac{7}{4} \times 10 = 11 \times 7 \times 5 = 385 \text{ cm}^3$.

(D) The formula for the total surface area of a cylinder is $2 \pi r(h + r)$.
The formula for the lateral surface area of a cylinder is $2 \pi rh$.
The ratio is given by $\frac{2 \pi r(h + r)}{2 \pi rh}$.
Simplifying the expression,we get $\frac{h + r}{h}$.
Given $r = 80 \text{ cm}$ and $h = 20 \text{ cm}$.
Substituting the values: $\frac{20 + 80}{20} = \frac{100}{20} = 5$.
Thus,the ratio is $5:1$.

(B) Let the radius of the cylinder be $r$ and its height be $h$. The volume of the cylinder is $V_{cylinder} = \pi r^{2} h$.
The volume of a right cone with the same radius $r$ and height $h$ is $V_{cone} = \frac{1}{3} \pi r^{2} h$.
Let $x$ be the number of cones required to store the water from the cylinder.
Therefore,$x \times V_{cone} = V_{cylinder}$.
Substituting the values,$x \times (\frac{1}{3} \pi r^{2} h) = \pi r^{2} h$.
Solving for $x$,we get $x = 3$.

(C) The radius of the cylindrical bucket is $r = \frac{28}{2} = 14\, cm$.
The height of the bucket is $h = 72\, cm$.
The volume of water in the bucket is given by $V = \pi r^2 h$.
$V = \frac{22}{7} \times 14 \times 14 \times 72 = 22 \times 2 \times 14 \times 72 = 44,352\, cm^3$.
Let the height of the water level in the rectangular tank be $H$.
The volume of water in the tank is $V = \text{length} \times \text{breadth} \times H = 66 \times 28 \times H$.
Equating the volumes: $66 \times 28 \times H = 44,352$.
$H = \frac{44,352}{66 \times 28} = \frac{44,352}{1,848} = 24\, cm$.

(C) The formula for the volume of a cylinder is $V = \pi r^2 h$.
Given: length $(h)$ = $70\, cm$,diameter = $2\, cm$,so radius $(r)$ = $1\, cm$.
Volume = $\frac{22}{7} \times (1)^2 \times 70 = 220\, cm^3$.
Weight of $1\, cm^3$ of iron = $10\, grams$.
Total weight = $220 \times 10 = 2200\, grams$.
To convert into $kg$,divide by $1000$: $\frac{2200}{1000} = 2.2\, kg$.

(A) Let the initial radius and height of the cylinder be $r$ and $h$,respectively.
The curved surface area of the original cylinder is given by $A_1 = 2 \pi r h$.
For the new cylinder,the radius $r' = 2r$ and the height $h' = \frac{h}{2}$.
The curved surface area of the new cylinder is $A_2 = 2 \pi r' h' = 2 \pi (2r) \left( \frac{h}{2} \right) = 2 \pi r h$.
Therefore,the ratio between the new curved surface area and the previous curved surface area is $\frac{A_2}{A_1} = \frac{2 \pi r h}{2 \pi r h} = 1:1$.

(D) The radius of the cylindrical jar $R = 24/2 = 12\, cm$.
The rise in the water level $h = 67.5\, mm = 6.75\, cm$.
The volume of the spherical ball is equal to the volume of the water displaced by it.
Volume of displaced water $= \pi R^2 h = \pi \times (12)^2 \times 6.75\, cm^3$.
Let $r$ be the radius of the spherical ball. The volume of the ball is given by $\frac{4}{3} \pi r^3$.
Equating the volumes: $\frac{4}{3} \pi r^3 = \pi \times 144 \times 6.75$.
$r^3 = \frac{144 \times 6.75 \times 3}{4} = 36 \times 6.75 \times 3 = 108 \times 6.75 = 729$.
$r^3 = 9^3$,which implies $r = 9\, cm$.
The diameter of the ball $= 2r = 2 \times 9 = 18\, cm$.

(B) The volume of a solid cone is given by $V_{cone} = \frac{1}{3} \pi r^2 h_{cone}$.
The volume of a solid cylinder is given by $V_{cylinder} = \pi r^2 h_{cylinder}$.
Since the material is converted from the cone to the cylinder,their volumes are equal: $V_{cone} = V_{cylinder}$.
Therefore,$\frac{1}{3} \pi r^2 h_{cone} = \pi r^2 h_{cylinder}$.
Given that the radii are equal,we can cancel $\pi r^2$ from both sides:
$\frac{1}{3} h_{cone} = h_{cylinder}$.
Given $h_{cylinder} = 5\,cm$,we have $\frac{1}{3} h_{cone} = 5\,cm$.
Thus,$h_{cone} = 3 \times 5\,cm = 15\,cm$.

(B) The volume of a cylinder is given by $V = \pi r^2 h$.
Let the initial radius be $r_1$ and height be $h_1$. Then $V_1 = \pi r_1^2 h_1 = 3850\, mm^3$.
For the new cylinder,the length is doubled $(h_2 = 2h_1)$ and the diameter is halved,which means the radius is also halved $(r_2 = r_1 / 2)$.
The new volume $V_2$ is given by $V_2 = \pi r_2^2 h_2$.
Substituting the new values: $V_2 = \pi (r_1 / 2)^2 (2h_1) = \pi (r_1^2 / 4) (2h_1) = \frac{1}{2} \pi r_1^2 h_1$.
Since $V_1 = \pi r_1^2 h_1 = 3850$,we have $V_2 = \frac{1}{2} \times 3850 = 1925\, mm^3$.

(C) The radius $(r)$ of each cylindrical pillar is half of the diameter: $r = \frac{50\, cm}{2} = 25\, cm = 0.25\, m = \frac{1}{4}\, m.$
The height $(h)$ of each pillar is $4\, m.$
The curved surface area of one cylindrical pillar is given by the formula $2\pi rh.$
Curved surface area of one pillar $= 2 \times 3.14 \times \frac{1}{4} \times 4 = 6.28\, m^{2}.$
The total curved surface area for $50$ such pillars $= 50 \times 6.28\, m^{2} = 314\, m^{2}.$
The cleaning rate is $50$ $paise$ per $m^{2},$ which is equal to $₹ 0.50$ per $m^{2}.$
Total labour charges $= 314\, m^{2} \times ₹ 0.50/m^{2} = ₹ 157.$

(A) Let the original radius be $r$ and the original height be $h$. The original volume $V$ is given by $V = \pi r^2 h$.
If the new radius is $R = 2r$ and the new height is $H$,the new volume $V'$ must be equal to $V$.
So,$V' = \pi R^2 H = \pi (2r)^2 H = \pi (4r^2) H = 4 \pi r^2 H$.
Since $V' = V$,we have $4 \pi r^2 H = \pi r^2 h$.
Dividing both sides by $\pi r^2$,we get $4H = h$,which implies $H = \frac{1}{4} h$.
Therefore,the height should be $\frac{1}{4}$ of the original height.

(B) The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Let the radius of the large ball be $R = \frac{3}{2} = 1.5 \, cm$.
Let the radii of the three smaller balls be $r_1, r_2,$ and $r_3$.
Given $r_1 = \frac{1.5}{2} = 0.75 \, cm$ and $r_2 = \frac{2}{2} = 1 \, cm$.
Since the volume remains constant during melting and recasting:
$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi r_1^3 + \frac{4}{3} \pi r_2^3 + \frac{4}{3} \pi r_3^3$
$R^3 = r_1^3 + r_2^3 + r_3^3$
$(1.5)^3 = (0.75)^3 + (1)^3 + r_3^3$
$3.375 = 0.421875 + 1 + r_3^3$
$3.375 = 1.421875 + r_3^3$
$r_3^3 = 3.375 - 1.421875 = 1.953125$
$r_3 = \sqrt[3]{1.953125} = 1.25 \, cm$.
The diameter of the third ball is $2 \times r_3 = 2 \times 1.25 = 2.5 \, cm$.

(D) Let the radius of the base of both the cone and the cylinder be $r$. Since they have the same base area,their radii are equal.
Let the slant height of the cone be $l$ and the height of the cylinder be $h = 2 \, m$.
The curved surface area of the cone is given by $A_{cone} = \pi r l$.
The curved surface area of the cylinder is given by $A_{cylinder} = 2 \pi r h$.
According to the problem,$A_{cone} = A_{cylinder}$.
Therefore,$\pi r l = 2 \pi r h$.
Dividing both sides by $\pi r$ (since $r \neq 0$),we get $l = 2h$.
Substituting $h = 2 \, m$,we get $l = 2 \times 2 = 4 \, m$.
Thus,the slant height of the cone is $4 \, m$.

(B) Let the radius of both the cylinder and the cone be $r$.
Let the height of the cylinder be $h$ and the slant height of the cone be $l$.
Given that the height of the cylinder is equal to the slant height of the cone,so $h = l$.
The curved surface area of the cylinder is $CSA_{cylinder} = 2 \pi r h$.
The curved surface area of the cone is $CSA_{cone} = \pi r l$.
Since $h = l$,we can substitute $l$ with $h$ in the formula for the cone's curved surface area: $CSA_{cone} = \pi r h$.
The ratio of the curved surface area of the cylinder to the cone is $\frac{2 \pi r h}{\pi r h} = \frac{2}{1}$.
Therefore,the required ratio is $2:1$.

(A) The volume of the cubical block of side $1 \, m$ is $V_{cube} = (1)^3 = 1 \, m^3$.
For a cylinder of the largest possible volume carved from this cube,the diameter of the base must be equal to the side of the cube ($d = 1 \, m$,so radius $r = 0.5 \, m$) and the height must be equal to the side of the cube $(h = 1 \, m)$.
The volume of the cylinder is $V_{cylinder} = \pi r^2 h = \pi \times (0.5)^2 \times 1 = \frac{\pi}{4} \, m^3$.
Taking $\pi \approx \frac{22}{7}$,the volume of the cylinder is $V_{cylinder} = \frac{22}{7 \times 4} = \frac{11}{14} \, m^3$.
The volume of the remaining wood is $V_{remaining} = V_{cube} - V_{cylinder} = 1 - \frac{11}{14} = \frac{3}{14} \, m^3$.

(C) The volume of a cylinder is given by the formula $V_{\text{cylinder}} = \pi r^2 h$,where $r$ is the radius and $h$ is the height.
Given,$h = 3 \text{ cm}$,so $V_{\text{cylinder}} = \pi r^2 (3) = 3 \pi r^2 \text{ cm}^3$.
The volume of a cone is given by the formula $V_{\text{cone}} = \frac{1}{3} \pi r^2 H$,where $H$ is the height of the cone.
Since the cylinder is recast into a cone of the same radius,their volumes must be equal.
Therefore,$3 \pi r^2 = \frac{1}{3} \pi r^2 H$.
Dividing both sides by $\pi r^2$,we get $3 = \frac{1}{3} H$.
Multiplying by $3$,we find $H = 9 \text{ cm}$.

(C) Total volume of rain water = Area $\times$ height.
Given,Area = $1\, km^2 = (1000\, m)^2 = 1,000,000\, m^2$.
Height = $2\, cm = 0.02\, m$.
Total volume = $1,000,000\, m^2 \times 0.02\, m = 20,000\, m^3$.
Volume of collected water = $50\%$ of $20,000\, m^3 = 0.5 \times 20,000 = 10,000\, m^3$.
Base area of the pool = $100\, m \times 10\, m = 1,000\, m^2$.
Increase in water level = $\frac{\text{Volume collected}}{\text{Base area of pool}} = \frac{10,000\, m^3}{1,000\, m^2} = 10\, m$.
Therefore,the water level in the pool would have increased by $10\, m$.

(B) Given: Perpendicular height $h = 4 \frac{2}{3} \text{ m} = \frac{14}{3} \text{ m}$.
Base diameter $d = 6 \text{ m}$,so radius $r = \frac{d}{2} = 3 \text{ m}$.
The volume of a conical tent is given by $V = \frac{1}{3} \pi r^2 h$.
Substituting the values: $V = \frac{1}{3} \times \frac{22}{7} \times (3)^2 \times \frac{14}{3}$.
$V = \frac{1}{3} \times \frac{22}{7} \times 9 \times \frac{14}{3} = 22 \times 2 = 44 \text{ m}^3$.
Since $11$ persons sleep in the tent,the average volume of air per person is $\frac{\text{Total Volume}}{\text{Number of persons}} = \frac{44}{11} = 4 \text{ m}^3$.

(A) Given,height of the conical tent,$h = 9\, m$.
Circumference of the base $= 2\pi r = 44\, m$.
$2 \times \frac{22}{7} \times r = 44$.
$r = \frac{44 \times 7}{2 \times 22} = 7\, m$.
Volume of the air in the tent is equal to the volume of the cone,$V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 9$.
$V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 9$.
$V = 22 \times 7 \times 3 = 462\, m^3$.

(C) The volume of the conical vessel is given by the formula $V = \frac{1}{3} \pi r^2 h$.
Substituting the given values,$r = 2 \ cm$ and $h = 3 \ cm$:
$V = \frac{1}{3} \times \pi \times (2)^2 \times 3 = 4\pi \ cm^3$.
When this liquid is transferred to a cylindrical jar of radius $R = 2 \ cm$,let the height of the kerosene level be $h_{jar}$.
The volume of the liquid remains the same,so the volume in the cylinder is $V = \pi R^2 h_{jar}$.
$4\pi = \pi \times (2)^2 \times h_{jar}$.
$4\pi = 4\pi \times h_{jar}$.
Therefore,$h_{jar} = 1 \ cm$.

(B) Let the radius of the solid cylinder be $r\,cm$ and its height be $H = 9\,cm$ (assuming the height of the solid cylinder is $9\,cm$ based on the volume calculation provided).
The volume of the hollow cylinder is given by $V = \pi(R^2 - r_{in}^2)h$,where $R = 4.3\,cm$,$r_{in} = 1.1\,cm$,and $h = 3\,cm$.
$V = \pi(4.3^2 - 1.1^2) \times 3 = \pi(18.49 - 1.21) \times 3 = \pi(17.28) \times 3 = 51.84\pi\,cm^3$.
The volume of the solid cylinder is $V = \pi r^2 H$. Given $H = 9\,cm$:
$51.84\pi = \pi r^2 \times 9$.
$r^2 = \frac{51.84}{9} = 5.76$.
$r = \sqrt{5.76} = 2.4\,cm$.

(C) Let the height of the cylinder be $H$ and its radius be $r$. Since the cone fits exactly,its radius is also $r$.
The volume of the cylinder is $V_{cyl} = \pi r^2 H$.
The volume of the cone is $V_{cone} = \frac{1}{3} \pi r^2 h$.
The total volume of the solid is $V_{total} = V_{cyl} + V_{cone} = \pi r^2 H + \frac{1}{3} \pi r^2 h$.
According to the problem,$V_{total} = 3 \times V_{cone}$.
Therefore,$\pi r^2 H + \frac{1}{3} \pi r^2 h = 3 \times (\frac{1}{3} \pi r^2 h)$.
Simplifying the equation:
$\pi r^2 H + \frac{1}{3} \pi r^2 h = \pi r^2 h$.
Subtract $\frac{1}{3} \pi r^2 h$ from both sides:
$\pi r^2 H = \pi r^2 h - \frac{1}{3} \pi r^2 h$.
$\pi r^2 H = \frac{2}{3} \pi r^2 h$.
Dividing both sides by $\pi r^2$,we get $H = \frac{2}{3} h$.

(D) $1$. Calculate the volume of the earth dug out from the well:
Radius of the well $(r)$ = $11.2 / 2 = 5.6 \, m$.
Depth of the well $(h)$ = $8 \, m$.
Volume $(V)$ = $\pi r^2 h = (22/7) \times (5.6)^2 \times 8 = (22/7) \times 31.36 \times 8 = 788.48 \, m^3$.
$2$. Calculate the area of the circular embankment:
The width of the embankment is $7 \, m$. The inner radius is $r = 5.6 \, m$, and the outer radius $(R)$ is $5.6 + 7 = 12.6 \, m$.
Area of the embankment = $\pi (R^2 - r^2) = \pi (12.6^2 - 5.6^2) = \pi (12.6 - 5.6)(12.6 + 5.6) = \pi (7)(18.2) = (22/7) \times 7 \times 18.2 = 22 \times 18.2 = 400.4 \, m^2$.
$3$. Calculate the height of the embankment $(H)$:
Volume of earth = Area of embankment $\times$ Height of embankment $(H)$.
$788.48 = 400.4 \times H$.
$H = 788.48 / 400.4 = 1.969... \approx 1.97 \, m$. (Note: Given the options provided in the prompt were inconsistent with the calculation, the correct mathematical approach yields $H \approx 1.97 \, m$. Based on standard problem structures, the logic above is the correct procedure).

(A) The curved surface area of a cylinder is given by the formula $A_{cylinder} = 2 \pi r h$,where $r$ is the radius and $h$ is the height.
The slant surface area (lateral surface area) of a cone is given by the formula $A_{cone} = \pi r l$,where $r$ is the radius and $l$ is the slant height.
Given that the cone has a slant height $l = 2h$ and the same base radius $r$ as the cylinder,the area of the cone is:
$A_{cone} = \pi \times r \times (2h) = 2 \pi r h$.
Comparing the two areas:
Ratio $= A_{cylinder} : A_{cone} = 2 \pi r h : 2 \pi r h = 1 : 1$.

(B) The volume of a cone is given by $V_{cone} = \frac{1}{3} \pi r^2 h_1$,where $r$ is the radius and $h_1$ is the height of the cone.
The volume of a cylinder is given by $V_{cylinder} = \pi r^2 h_2$,where $r$ is the radius and $h_2$ is the height of the cylinder.
Since the material is converted from one shape to another,their volumes remain equal: $V_{cone} = V_{cylinder}$.
$\frac{1}{3} \pi r^2 h_1 = \pi r^2 h_2$.
Given that the radii are equal and the height of the cylinder $h_2 = 5\, cm$,we can cancel $\pi r^2$ from both sides:
$\frac{1}{3} h_1 = 5$.
$h_1 = 5 \times 3 = 15\, cm$.
Therefore,the height of the cone is $15\, cm$.

(A) The volume of the cube is $V_{cube} = a^3 = 343 \, c.c.$
Therefore,the edge length of the cube is $a = \sqrt[3]{343} = 7 \, cm$.
Since the cone is fitted inside the cube such that its base touches the edges of one face,the diameter of the base of the cone is equal to the side of the cube,$d = 7 \, cm$. Thus,the radius $r = 3.5 \, cm$.
The height of the cone is equal to the side of the cube,$h = 7 \, cm$.
The volume of the cone is given by $V_{cone} = \frac{1}{3} \pi r^2 h$.
$V_{cone} = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 7$.
$V_{cone} = \frac{1}{3} \times 22 \times 12.25 = \frac{269.5}{3} \approx 89.83 \, c.c.$
Rounding to the nearest integer,the volume is approximately $90 \, c.c.$

(C) The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$.
Given $r = 50 \ cm$ and $h = 25 \ cm$,the volume is $V = \frac{1}{3} \pi (50)^2 (25) = \frac{62500 \pi}{3} \ cm^3$.
When the cone is recast into a sphere,the volume remains constant. Let $R$ be the radius of the sphere.
Volume of the sphere $= \frac{4}{3} \pi R^3$.
Equating the volumes: $\frac{4}{3} \pi R^3 = \frac{62500 \pi}{3}$.
$4 R^3 = 62500 \Rightarrow R^3 = 15625$.
Taking the cube root,$R = \sqrt[3]{15625} = 25 \ cm$.
The surface area of the sphere is $A = 4 \pi R^2$.
$A = 4 \times \frac{22}{7} \times (25)^2 = 4 \times \frac{22}{7} \times 625 = \frac{55000}{7} \approx 7857.14 \ cm^2$.

(A) Let the radius $r = 5x$ and height $h = 12x$.
The volume of a cone is given by $V = \frac{1}{3} \pi r^2 h$.
Given $V = 314 \frac{3}{7} = \frac{2200}{7} \text{ m}^3$.
Substituting the values: $\frac{1}{3} \times \frac{22}{7} \times (5x)^2 \times (12x) = \frac{2200}{7}$.
$\frac{1}{3} \times \frac{22}{7} \times 25x^2 \times 12x = \frac{2200}{7}$.
$\frac{22}{7} \times 100x^3 = \frac{2200}{7}$.
$100x^3 = 100$.
$x^3 = 1$,so $x = 1$.
Therefore,the radius $r = 5x = 5 \times 1 = 5 \text{ m}$.

(A) Let the radius of the base be $r$. Since they have the same height,the height of the cone and the cylinder is $h = r$.
$1$. Volume of the cone: $V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (r) = \frac{1}{3} \pi r^3$.
$2$. Volume of the hemisphere: $V_2 = \frac{2}{3} \pi r^3$.
$3$. Volume of the cylinder: $V_3 = \pi r^2 h = \pi r^2 (r) = \pi r^3$.
Ratio $V_1 : V_2 : V_3 = \frac{1}{3} \pi r^3 : \frac{2}{3} \pi r^3 : \pi r^3$.
Dividing by $\frac{1}{3} \pi r^3$,we get the ratio $1 : 2 : 3$.

(C) The diameter of the conical tent is $12 \ m$,so the radius $r = \frac{12}{2} = 6 \ m$.
The slant height $l = 6.3 \ m$.
The curved surface area of the cone is given by the formula $A = \pi r l$.
Substituting the values: $A = \frac{22}{7} \times 6 \times 6.3 = 22 \times 6 \times 0.9 = 118.8 \ m^2$.
The canvas is rectangular with a width $w = 2 \ m$. Let the length be $L$.
Since the area of the canvas must equal the curved surface area of the tent: $L \times w = A$.
$L \times 2 = 118.8$.
$L = \frac{118.8}{2} = 59.4 \ m$.

(C) Let the initial radius and height of the cylinder be $r$ and $h$,respectively.
The initial volume $V_{1}$ is given by the formula $V_{1} = \pi r^{2} h$.
According to the problem,the new radius $r' = 2r$ and the new height $h' = \frac{h}{2}$.
The new volume $V_{2}$ is given by $V_{2} = \pi (r')^{2} h' = \pi (2r)^{2} \left(\frac{h}{2}\right)$.
Simplifying the expression for $V_{2}$,we get $V_{2} = \pi (4r^{2}) \left(\frac{h}{2}\right) = 2 \pi r^{2} h$.
The ratio of the new volume to the previous volume is $\frac{V_{2}}{V_{1}} = \frac{2 \pi r^{2} h}{\pi r^{2} h} = \frac{2}{1}$.
Therefore,the ratio is $2:1$.

(C) The radius of the base of the tent is $r = 105/2 = 52.5\, m$.
The curved surface area of the cylindrical part is given by $2\pi rh = 2 \times (22/7) \times 52.5 \times 3 = 990\, m^2$.
The lateral surface area of the conical part is given by $\pi rl = (22/7) \times 52.5 \times 53 = 8745\, m^2$.
The total surface area of the canvas required is the sum of these two areas: $990 + 8745 = 9735\, m^2$.
Given the width of the canvas is $5\, m$,the length of the canvas is calculated as: $\text{Length} = \text{Total Area} / \text{Width} = 9735 / 5 = 1947\, m$.

(A) Let the initial radius of the cone be $r$ and the initial slant height be $l$.
The initial curved surface area of the cone is $A_1 = \pi rl$.
According to the problem,the new radius $r' = r + 0.20r = 1.2r$.
The new slant height $l' = 2l$.
The new curved surface area of the cone is $A_2 = \pi r' l' = \pi (1.2r)(2l) = 2.4 \pi rl$.
The increase in the curved surface area is $\Delta A = A_2 - A_1 = 2.4 \pi rl - \pi rl = 1.4 \pi rl$.
The percentage increase is $\frac{\Delta A}{A_1} \times 100 = \frac{1.4 \pi rl}{\pi rl} \times 100 = 140 \%$.
Therefore,the curved surface area of the cone will increase by $140 \%$.

(C) The radius of the base of the cone is $r = 5 \, m$ and the height is $h = 12 \, m$.
First,we calculate the slant height $l$ of the cone using the formula $l = \sqrt{r^2 + h^2}$.
$l = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \, m$.
The area of the canvas required for the conical tent is equal to the curved surface area of the cone,which is given by the formula $\pi r l$.
Area $= \pi \times 5 \times 13 = 65 \pi \, m^2$.
Thus,the required area is $65 \pi \, m^2$.

(A) $1$. Calculate the volume of the rectangular block: $V_{\text{block}} = \text{length} \times \text{width} \times \text{height} = 10 \text{ cm} \times 5 \text{ cm} \times 2 \text{ cm} = 100 \text{ cm}^3$.
$2$. Calculate the volume of the cone: $V_{\text{cone}} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3 \text{ cm})^2 \times 7 \text{ cm} = \frac{1}{3} \times \frac{22}{7} \times 9 \times 7 = 66 \text{ cm}^3$.
$3$. Calculate the volume of wood wasted: $V_{\text{wasted}} = V_{\text{block}} - V_{\text{cone}} = 100 \text{ cm}^3 - 66 \text{ cm}^3 = 34 \text{ cm}^3$.
$4$. Calculate the percentage of wood wasted: $\text{Percentage} = \left( \frac{V_{\text{wasted}}}{V_{\text{block}}} \right) \times 100 = \left( \frac{34}{100} \right) \times 100 = 34 \%$.

(B) Given:
Diameter of the conical tomb $(d)$ = $28\, m$.
Radius $(r)$ = $d / 2 = 28 / 2 = 14\, m$.
Slant height $(l)$ = $50\, m$.
The curved surface area $(CSA)$ of a cone is given by the formula:
$CSA = \pi r l$
$CSA = (22 / 7) \times 14 \times 50$
$CSA = 22 \times 2 \times 50 = 2200\, m^2$.
The rate of whitewashing is $80$ paise per $m^2$,which is equal to $Rs. 0.80$ per $m^2$.
Total cost = $CSA \times \text{Rate}$
Total cost = $2200 \times 0.80 = Rs. 1760$.

(A) Given: Area of the rectangular sheet $= 264 \text{ cm}^2$,Width $= 11 \text{ cm}$.
Length of the sheet $= \frac{\text{Area}}{\text{Width}} = \frac{264}{11} = 24 \text{ cm}$.
When the sheet is rolled along its breadth,the width of the sheet becomes the circumference of the base of the cylinder,and the length of the sheet becomes the height of the cylinder.
Circumference $= 2\pi r = 11 \text{ cm}$.
$r = \frac{11}{2\pi} = \frac{11 \times 7}{2 \times 22} = \frac{7}{4} \text{ cm}$.
Height $(h) = 24 \text{ cm}$.
Volume of the cylinder $= \pi r^2 h = \frac{22}{7} \times (\frac{7}{4})^2 \times 24$.
Volume $= \frac{22}{7} \times \frac{49}{16} \times 24 = 22 \times \frac{7}{16} \times 24 = 22 \times 7 \times 1.5 = 231 \text{ cm}^3$.

(C) Let the heights of the cylinder and cone be $h_1 = 2x$ and $h_2 = 3x$ respectively.
Let the radii of the cylinder and cone be $r_1 = 3y$ and $r_2 = 4y$ respectively.
The volume of a cylinder is given by $V_1 = \pi r_1^2 h_1$.
The volume of a cone is given by $V_2 = \frac{1}{3} \pi r_2^2 h_2$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\frac{1}{3} \pi r_2^2 h_2} = \frac{3 r_1^2 h_1}{r_2^2 h_2}$.
Substituting the given ratios: $\frac{V_1}{V_2} = \frac{3 \times (3y)^2 \times (2x)}{(4y)^2 \times (3x)} = \frac{3 \times 9y^2 \times 2x}{16y^2 \times 3x} = \frac{54}{48} = \frac{9}{8}$.
Thus,the ratio of their volumes is $9:8$.

(A) The volume of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$,where $r$ is the radius and $h$ is the height.
Let the original volume be $V_1 = \frac{1}{3} \pi r^2 h$.
If the height is doubled,the new height $h' = 2h$.
The new volume $V_2 = \frac{1}{3} \pi r^2 (2h) = 2 \times (\frac{1}{3} \pi r^2 h) = 2V_1$.
The increase in volume is $V_2 - V_1 = 2V_1 - V_1 = V_1$.
The percentage increase is given by $\frac{\text{Increase}}{\text{Original Volume}} \times 100 = \frac{V_1}{V_1} \times 100 = 100 \%$.

(B) The volume of a cube with side $a$ is given by $V = a^3$.
The volumes of the three cubes are $3^3 = 27 \text{ cm}^3$,$4^3 = 64 \text{ cm}^3$,and $5^3 = 125 \text{ cm}^3$.
When these cubes are melted to form a new cube,the total volume of the new cube is the sum of the volumes of the three individual cubes.
Total volume $= 27 + 64 + 125 = 216 \text{ cm}^3$.
Let the side of the new cube be $S$. Then $S^3 = 216$.
Taking the cube root on both sides,$S = \sqrt[3]{216} = 6 \text{ cm}$.

(D) Let the original radius of the cone be $r$ and the original height be $h$.
The original volume of the cone is $V_1 = \frac{1}{3} \pi r^2 h$.
Since both the radius and height are increased by $100 \%$,the new radius $r' = r + 100 \% \text{ of } r = 2r$ and the new height $h' = h + 100 \% \text{ of } h = 2h$.
The new volume of the cone is $V_2 = \frac{1}{3} \pi (r')^2 (h') = \frac{1}{3} \pi (2r)^2 (2h)$.
$V_2 = \frac{1}{3} \pi (4r^2) (2h) = 8 \times (\frac{1}{3} \pi r^2 h) = 8 V_1$.
Therefore,the volume of the cone becomes $8$ times the original volume.

(C) Let the original radius of the sphere be $r$.
Original volume $V_1 = \frac{4}{3} \pi r^3$.
If the radius is doubled,the new radius $R = 2r$.
New volume $V_2 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = 8 \times (\frac{4}{3} \pi r^3) = 8V_1$.
The increase in volume is $V_2 - V_1 = 8V_1 - V_1 = 7V_1$.
The percentage increase is $\frac{\text{Increase}}{\text{Original Volume}} \times 100 = \frac{7V_1}{V_1} \times 100 = 700 \%$.