Mix Examples - Surface Areas and Volumes Questions in English

(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.

(B) The formula for the volume of a sphere with radius $R$ is $V = \frac{4}{3} \pi R^{3}$.
Given that the radius of the sphere is $R = 2r$.
Substituting this value into the formula:
$V = \frac{4}{3} \pi (2r)^{3}$
$V = \frac{4}{3} \pi (8r^{3})$
$V = \frac{32}{3} \pi r^{3}$

(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.

(B) The total surface area of a cone is given by the formula: $\text{Total Surface Area} = \text{Area of the base} + \text{Curved surface area}$.
Given,radius $R = \frac{r}{2}$ and slant height $L = 2l$.
$\text{Total Surface Area} = \pi R^2 + \pi RL$
Substituting the values:
$= \pi \left(\frac{r}{2}\right)^2 + \pi \left(\frac{r}{2}\right) \times (2l)$
$= \pi \left(\frac{r^2}{4}\right) + \pi rl$
$= \pi r \left(\frac{r}{4} + l\right)$
$= \pi r \left(l + \frac{r}{4}\right)$.

(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:
$\text{Length} = \sqrt{10^2 + 10^2 + 5^2}$
$= \sqrt{100 + 100 + 25}$
$= \sqrt{225}$
$= 15 \text{ m}$.

(C) The balloon is hemispherical in shape.
The surface area of a hemispherical balloon with radius $r$ is given by the formula $S = 3\pi r^2$.
Let $r_1 = 6\,cm$ and $r_2 = 12\,cm$ be the radii in the two cases.
The ratio of the surface areas is $\frac{S_1}{S_2} = \frac{3\pi r_1^2}{3\pi r_2^2} = \frac{r_1^2}{r_2^2}$.
Substituting the values,we get $\frac{S_1}{S_2} = \frac{6^2}{12^2} = \frac{36}{144} = \frac{1}{4}$.
Therefore,the ratio of the surface areas of the two balloons is $1:4$.

(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) False.
For the largest possible right circular cone cut out of a cube of edge $r$,the height of the cone $(h)$ must be equal to the edge of the cube,so $h = r \, cm$.
The diameter of the base of the cone must be equal to the edge of the cube,so the radius of the base $(R)$ is $\frac{r}{2} \, cm$.
The volume of a cone is given by the formula $V = \frac{1}{3} \pi R^2 h$.
Substituting the values: $V = \frac{1}{3} \pi \left(\frac{r}{2}\right)^2 \cdot r = \frac{1}{3} \pi \left(\frac{r^2}{4}\right) \cdot r = \frac{1}{12} \pi r^3 \, cm^3$.
Since $\frac{1}{12} \pi r^3 \neq \frac{1}{6} \pi r^3$,the given statement is False.

(A) Let the radius of the sphere be $r$.
Given that the height and diameter of the cylinder are equal to the diameter of the sphere.
Since the diameter of the sphere is $2r$,the height of the cylinder $h = 2r$ and the diameter of the cylinder is $2r$.
Therefore,the radius of the cylinder is $r$.
Volume of the sphere $= \frac{4}{3} \pi r^3$.
Volume of the cylinder $= \pi r^2 h = \pi r^2 (2r) = 2 \pi r^3$.
Two-thirds of the volume of the cylinder $= \frac{2}{3} \times (2 \pi r^3) = \frac{4}{3} \pi r^3$.
Since the volume of the sphere is equal to $\frac{4}{3} \pi r^3$,the statement is True.

(FALSE) Let the original radius of the cone be $r$ and the original height be $h$.
The volume of the original cone is $V_1 = \frac{1}{3} \pi r^{2} h$.
Now,the new radius $r' = \frac{r}{2}$ and the new height $h' = 2h$.
The new volume $V_2 = \frac{1}{3} \pi (r')^{2} h' = \frac{1}{3} \pi \left(\frac{r}{2}\right)^{2} (2h)$.
Simplifying this,$V_2 = \frac{1}{3} \pi \left(\frac{r^2}{4}\right) (2h) = \frac{1}{2} \left(\frac{1}{3} \pi r^{2} h\right) = \frac{1}{2} V_1$.
Since the new volume is half of the original volume,the statement is False.

(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 $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 4^2 \times 3$
$= \frac{22}{7} \times 16 = \frac{352}{7} \approx 50.29 \, cm^3$.

(N/A) The volume of a sphere $V = \frac{4}{3} \pi r^{3}$.
Let the initial radius be $r$. If the radius is increased by $10 \%$,the new radius $r'$ becomes $r + 0.1r = 1.1r$.
The new volume $V'$ is given by $V' = \frac{4}{3} \pi (r')^{3} = \frac{4}{3} \pi (1.1r)^{3}$.
Calculating $(1.1)^{3} = 1.331$,so $V' = 1.331 \times (\frac{4}{3} \pi r^{3}) = 1.331V$.
The increase in volume is $\Delta V = V' - V = 1.331V - V = 0.331V$.
The percentage increase in volume is $\frac{\Delta V}{V} \times 100 = \frac{0.331V}{V} \times 100 = 33.1 \%$.
Thus,the volume increases by $33.1 \%$.

(B) The internal volume of the rectangular box is calculated as: $V_{\text{box}} = 16 \, cm \times 8 \, cm \times 8 \, cm = 1024 \, cm^3$.
The volume of a single sphere with radius $r = 2 \, cm$ is given by the formula $V_{\text{sphere}} = \frac{4}{3} \pi r^3$.
Substituting the values: $V_{\text{sphere}} = \frac{4}{3} \times 3.14 \times (2)^3 = \frac{4}{3} \times 3.14 \times 8 = \frac{100.48}{3} \approx 33.4933 \, cm^3$.
The total volume occupied by $16$ such spheres is: $V_{\text{total spheres}} = 16 \times 33.4933 = 535.8933 \, cm^3$.
The volume of the preservative liquid is the remaining space in the box: $V_{\text{liquid}} = V_{\text{box}} - V_{\text{total spheres}} = 1024 - 535.8933 = 488.1067 \, cm^3$.
Rounding to the nearest integer,we get $488 \, cm^3$.

(C) Let the edge of the cube be $x \, m$.
Since the volume of the cube is $x^3$,we have $x^3 = 15.625 \, m^3$.
Taking the cube root,$x = \sqrt[3]{15.625} = 2.5 \, m$.
The total capacity of the tank is $15.625 \, m^3$ with a height of $2.5 \, m$.
The volume of water currently in the tank with depth $1.3 \, m$ is calculated as: $\text{Volume} = \text{length} \times \text{width} \times \text{depth} = 2.5 \times 2.5 \times 1.3 = 8.125 \, m^3$.
The volume of water already used is the difference between the total capacity and the current volume: $15.625 \, m^3 - 8.125 \, m^3 = 7.5 \, m^3$.

(D) The amount of water displaced by a solid spherical ball is equal to the volume of the solid spherical ball.
The diameter of the ball is $d = 4.2 \, cm$,so the radius is $r = \frac{d}{2} = \frac{4.2}{2} = 2.1 \, cm$.
The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Substituting the values: $V = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1$.
$V = 4 \times \frac{22}{7} \times 0.7 \times 2.1 \times 2.1$.
$V = 4 \times 22 \times 0.1 \times 2.1 \times 2.1 = 88 \times 0.1 \times 4.41 = 8.8 \times 4.41 = 38.808 \, cm^3$.
Thus,the amount of water displaced is $38.808 \, cm^3$.

(A) The area of canvas required for a conical tent is equal to its curved surface area,which is given by the formula $\pi r l$.
First,we calculate the slant height $(l)$ using the formula $l = \sqrt{r^{2} + h^{2}}$,where $r = 12 \, m$ and $h = 3.5 \, m$.
$l = \sqrt{(12)^{2} + (3.5)^{2}} = \sqrt{144 + 12.25} = \sqrt{156.25} = 12.5 \, m$.
Now,calculate the curved surface area:
Area $= \pi r l = \frac{22}{7} \times 12 \times 12.5$.
Area $= \frac{22 \times 150}{7} \approx 471.428 \, m^{2}$.
Rounding to two decimal places,the area required is $471.42 \, m^{2}$.

(D) Since both spheres are made of the same metal,their density $\rho$ is the same.
Weight $W = \text{Volume} \times \text{Density} \times g = \frac{4}{3} \pi r^3 \rho g$.
Therefore,the ratio of the weights is equal to the ratio of the volumes: $\frac{W_1}{W_2} = \frac{r_1^3}{r_2^3}$.
Given $W_1 = 5920 \, g$,$W_2 = 740 \, g$,and diameter of the smaller sphere $d_2 = 5 \, cm$,so $r_2 = 2.5 \, cm$.
$\frac{5920}{740} = \frac{r_1^3}{(2.5)^3}$.
$8 = \frac{r_1^3}{15.625}$.
$r_1^3 = 8 \times 15.625 = 125$.
$r_1 = \sqrt[3]{125} = 5 \, cm$.

$(C)$ The radius of the cylindrical glass $r = \text{diameter} \div 2 = 7 \div 2 = 3.5 \, cm$.
The height of the milk in the glass $h = 12 \, cm$.
The volume of milk in one glass $V = \pi r^2 h = \frac{22}{7} \times (3.5)^2 \times 12$.
$V = \frac{22}{7} \times 12.25 \times 12 = 22 \times 1.75 \times 12 = 462 \, cm^3$.
Since $1000 \, cm^3 = 1 \, \text{litre}$, the volume of milk in one glass is $462 \div 1000 = 0.462 \, \text{litres}$.
Total milk needed for $1600$ students $= 1600 \times 0.462 \, \text{litres} = 739.2 \, \text{litres}$.

(A) The length (height) of the cylindrical roller is $h = 2.5\, m$ and the radius is $r = 1.75\, m$.
The area covered by the roller in one revolution is equal to its lateral surface area (curved surface area).
Lateral Surface Area $= 2\pi rh = 2 \times \frac{22}{7} \times 1.75 \times 2.5$.
$= 2 \times \frac{22}{7} \times \frac{175}{100} \times 2.5 = 2 \times 22 \times 0.25 \times 2.5 = 44 \times 0.625 = 27.5\, m^{2}$.
The total area covered on the road is $5500\, m^{2}$.
Number of revolutions $= \frac{\text{Total area covered}}{\text{Area covered in one revolution}}$.
Number of revolutions $= \frac{5500}{27.5} = 200$ revolutions.

(B) The volume of the overhead tank is $V = 40 \, m \times 25 \, m \times 15 \, m = 15,000 \, m^3$.
Since $1 \, m^3 = 1,000$ litres,the total capacity of the tank in litres is $15,000 \times 1,000 = 15,000,000$ litres.
The daily water requirement for the village is $5,000 \times 75 = 375,000$ litres.
The number of days the water will last is given by the total capacity divided by the daily requirement:
$\text{Number of days} = \frac{15,000,000}{375,000} = \frac{15,000}{375} = 40$ days.

(C) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Let the radius of the large laddoo be $R = 5 \, cm$ and the radius of the smaller laddoos be $r = 2.5 \, cm$.
Since the total amount of material remains the same,the total volume of the smaller laddoos must equal the volume of the large laddoo.
Let $n$ be the number of smaller laddoos.
$n \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$n = \frac{R^3}{r^3} = (\frac{R}{r})^3$
$n = (\frac{5}{2.5})^3 = (2)^3 = 8$.
Therefore,$8$ laddoos of radius $2.5 \, cm$ can be made.

(D) When a right triangle is revolved about one of its sides (containing the right angle),a cone is formed.
Here,the side about which the triangle is revolved becomes the height $(h)$ of the cone,and the other side becomes the radius $(r)$ of the base.
Given:
Height $(h) = 8 \, cm$
Radius $(r) = 6 \, cm$
Slant height $(l) = 10 \, cm$
The curved surface area of a cone is given by the formula: $\text{CSA} = \pi r l$
Substituting the values:
$\text{CSA} = \frac{22}{7} \times 6 \times 10$
$\text{CSA} = \frac{1320}{7} \approx 188.57 \, cm^2$
Thus,the curved surface area of the solid formed is $188.57 \, cm^2$.

(A) Length of the rectangular surface $= 6\, m = 600\, cm$.
Breadth of the rectangular surface $= 4\, m = 400\, cm$.
Depth (rainfall) $= 1\, cm$.
Volume of rain water collected $= \text{Length} \times \text{Breadth} \times \text{Depth} = 600 \times 400 \times 1 = 240,000\, cm^3$.
Let the height of the water in the cylindrical vessel be $h\, cm$.
The internal radius of the cylindrical vessel $r = 20\, cm$.
Volume of water in the cylindrical vessel $= \pi r^2 h = 3.14 \times (20)^2 \times h = 3.14 \times 400 \times h = 1256h\, cm^3$.
Equating the volumes: $1256h = 240,000$.
$h = \frac{240,000}{1256} \approx 191.08\, cm$.
Rounding to the nearest integer,the height is $191\, cm$.

(B) The tube is a hollow cylinder.
Outer diameter $= 16 \, cm$, so outer radius $(R) = 16 / 2 = 8 \, cm$.
Thickness of the iron sheet $= 2 \, cm$.
Inner radius $(r) = R - \text{thickness} = 8 \, cm - 2 \, cm = 6 \, cm$.
Length (height) of the cylinder $(h) = 100 \, cm$.
The volume of iron used is the difference between the outer volume and the inner volume of the cylinder.
Volume $= \pi (R^2 - r^2) h$
Volume $= \frac{22}{7} \times (8^2 - 6^2) \times 100$
Volume $= \frac{22}{7} \times (64 - 36) \times 100$
Volume $= \frac{22}{7} \times 28 \times 100$
Volume $= 22 \times 4 \times 100 = 8800 \, cm^3$.

(A) The diameter of the semi-circular sheet is $28 \, cm$,so its radius $R = 14 \, cm$. When this sheet is bent to form an open conical cup,the radius of the sheet becomes the slant height $(l)$ of the cone.
Therefore,$l = 14 \, cm$.
The circumference of the semi-circular arc becomes the circumference of the base of the cone. The length of the semi-circular arc is $\pi R = \pi \times 14$.
Let $r$ be the radius of the base of the cone. Then,$2 \pi r = 14 \pi$,which gives $r = 7 \, cm$.
Now,we find the height $(h)$ of the cone using the relation $l^2 = r^2 + h^2$:
$h = \sqrt{l^2 - r^2} = \sqrt{14^2 - 7^2} = \sqrt{196 - 49} = \sqrt{147} \approx 12.124 \, cm$.
The capacity (volume) of the conical cup is $V = \frac{1}{3} \pi r^2 h$.
$V = \frac{1}{3} \times \frac{22}{7} \times 7 \times 7 \times 12.124 \approx 622.38 \, cm^3$. Given the options,the closest value is $622.16 \, cm^3$.

(A) $(i)$ The area of the ground covered by the tent is the area of the circular base:
$\text{Area} = \pi r^{2} = \frac{22}{7} \times 5^{2} = \frac{550}{7} \, m^{2}$.
Number of students $= \frac{\text{Total ground area}}{\text{Area per student}} = \frac{550/7}{5/7} = \frac{550}{5} = 110$.
$(ii)$ The curved surface area of the cone is $165 \, m^{2}$.
$\pi r l = 165 \Rightarrow \frac{22}{7} \times 5 \times l = 165$.
$l = \frac{165 \times 7}{110} = 10.5 \, m$.
Using $l^{2} = r^{2} + h^{2}$,we find the height $h$:
$h = \sqrt{l^{2} - r^{2}} = \sqrt{(10.5)^{2} - 5^{2}} = \sqrt{110.25 - 25} = \sqrt{85.25} \approx 9.233 \, m$.
Volume of the cone $= \frac{1}{3} \pi r^{2} h = \frac{1}{3} \times \frac{22}{7} \times 25 \times 9.233 \approx 242.1 \, m^{3}$.

(C) Internal diameter of the hemispherical tank $= 14\, m$.
Internal radius $(r)$ of the hemispherical tank $= 14\, m / 2 = 7\, m$.
Volume of the hemispherical tank $= \frac{2}{3} \pi r^{3} = \frac{2}{3} \times \frac{22}{7} \times (7)^{3} = \frac{2}{3} \times 22 \times 49 = \frac{2156}{3} \approx 718.66\, m^{3}$.
The tank already contains $50$ kilolitres of water. Since $1\, m^{3} = 1000$ litres,$50$ kilolitres $= 50,000$ litres $= 50\, m^{3}$.
Volume of water to be pumped into the tank $= \text{Total capacity} - \text{Existing volume} = 718.66\, m^{3} - 50\, m^{3} = 668.66\, m^{3}$.

(A) Let $V_{1}$ and $V_{2}$ be the volumes of two spheres.
$\therefore \frac{V_{1}}{V_{2}} = \frac{64}{27}$
$\Rightarrow \frac{\frac{4}{3} \pi r_{1}^{3}}{\frac{4}{3} \pi r_{2}^{3}} = \frac{64}{27}$
$\Rightarrow \frac{r_{1}^{3}}{r_{2}^{3}} = \frac{64}{27}$
$\Rightarrow \left(\frac{r_{1}}{r_{2}}\right)^{3} = \left(\frac{4}{3}\right)^{3}$
$\Rightarrow \frac{r_{1}}{r_{2}} = \frac{4}{3} \quad ...(1)$
Now,let $SA_{1}$ and $SA_{2}$ be the surface areas of the two spheres.
$\therefore \frac{SA_{1}}{SA_{2}} = \frac{4 \pi r_{1}^{2}}{4 \pi r_{2}^{2}}$
$\Rightarrow \frac{SA_{1}}{SA_{2}} = \left(\frac{r_{1}}{r_{2}}\right)^{2}$
$\Rightarrow \frac{SA_{1}}{SA_{2}} = \left(\frac{4}{3}\right)^{2} \quad [\text{Using } (1)]$
$\Rightarrow \frac{SA_{1}}{SA_{2}} = \frac{16}{9}$
$\therefore SA_{1} : SA_{2} = 16 : 9$
Thus,the ratio of the surface areas of the two spheres is $16 : 9$.

(B) Side of the cube $a = 4 \, cm$.
Volume of the cube $V_{cube} = a^3 = (4)^3 = 64 \, cm^3$.
Since the sphere touches the sides of the cube,the diameter of the sphere is equal to the side of the cube,$d = 4 \, cm$.
Radius of the sphere $r = d / 2 = 4 / 2 = 2 \, cm$.
Volume of the sphere $V_{sphere} = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (2)^3$.
$V_{sphere} = \frac{4}{3} \times \frac{22}{7} \times 8 = \frac{704}{21} \approx 33.52 \, cm^3$.
Volume of the gap = $V_{cube} - V_{sphere} = 64 - 33.52 = 30.48 \, cm^3$.

(C) Volume of a sphere is given by $V_s = \frac{4}{3} \pi r^3$.
Volume of a right circular cylinder is given by $V_c = \pi r^2 h$.
Given that the volumes are equal,we have $\frac{4}{3} \pi r^3 = \pi r^2 h$.
Simplifying this,we get $h = \frac{4}{3} r$.
The diameter of the cylinder is $d = 2r$.
The difference between the diameter and the height is $d - h = 2r - \frac{4}{3} r = \frac{2}{3} r$.
The percentage by which the diameter exceeds the height is $\frac{d - h}{h} \times 100 = \frac{\frac{2}{3} r}{\frac{4}{3} r} \times 100 = \frac{2}{4} \times 100 = 50 \%$.

(N/A) Radius of one circular plate $(r)$ $= 14\,cm$.
Thickness of one circular plate $= 3\,cm$.
As the plates are placed one above the other,the height $(h)$ of the cylinder formed by $30$ plates $= 30 \times 3 = 90\,cm$.
$(i)$ Total surface area of the cylinder $= 2\pi r(r + h) = 2 \times \frac{22}{7} \times 14 \times (14 + 90) = 44 \times 2 \times 104 = 88 \times 104 = 9152\,cm^2$.
$(ii)$ Volume of the cylinder $= \pi r^2 h = \frac{22}{7} \times 14 \times 14 \times 90 = 22 \times 2 \times 14 \times 90 = 44 \times 1260 = 55440\,cm^3$.

(A) For a given cuboidal box:
Length $(l) = 25 \, cm$
Breadth $(b) = 20 \, cm$
Height $(h) = 10 \, cm$
The formula for the total surface area of a cuboid is:
Total Surface Area $= 2(lb + bh + hl)$
Substituting the given values:
$= 2(25 \times 20 + 20 \times 10 + 10 \times 25) \, cm^2$
$= 2(500 + 200 + 250) \, cm^2$
$= 2(950) \, cm^2$
$= 1900 \, cm^2$
Thus,the total surface area of the cuboidal box is $1900 \, cm^2$.

(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 $B$.