Mix Examples - Surface Areas and Volumes Questions in English

(B) By observing the figure,we can see that the upper part of the funnel is in the shape of a cone (specifically,a conical part) and the lower part is in the shape of a cylinder.
Therefore,a funnel is a combination of a cone and a cylinder.

(B) When an object is submerged in a container completely full of water,the volume of water that overflows is equal to the volume of the submerged object.
Here,the object is a spherical marble with radius $r = 2.1 \, cm$.
The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Substituting the values: $V = \frac{4}{3} \times \frac{22}{7} \times (2.1)^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1$.
$V = 4 \times 22 \times 0.1 \times 2.1 \times 2.1$.
$V = 88 \times 0.1 \times 4.41$.
$V = 8.8 \times 4.41 = 38.808 \, cm^3$.
Rounding to one decimal place,the volume of water that flows out is $38.8 \, cm^3$.

(C) The volume of the cubical ice cream brick is given by $V_{cube} = a^3$,where $a = 22 \, cm$.
$V_{cube} = 22 \times 22 \times 22 = 10648 \, cm^3$.
The volume of one ice cream cone is given by $V_{cone} = \frac{1}{3} \pi r^2 h$,where $r = 2 \, cm$ and $h = 7 \, cm$.
Using $\pi = \frac{22}{7}$,we get $V_{cone} = \frac{1}{3} \times \frac{22}{7} \times 2^2 \times 7 = \frac{1}{3} \times 22 \times 4 = \frac{88}{3} \, cm^3$.
The number of children who will get the ice cream is the total volume of the brick divided by the volume of one cone.
Number of children $= \frac{V_{cube}}{V_{cone}} = \frac{10648}{88/3} = \frac{10648 \times 3}{88}$.
$10648 \div 88 = 121$.
Number of children $= 121 \times 3 = 363$.

(D) frustum of a cone is formed by cutting a cone with a plane parallel to its base.
Let the height of the frustum be $h$,and the radii of the two circular ends be $r_{1}$ and $r_{2}$.
The formula for the volume $V$ of a frustum of a cone is derived from the difference between the volumes of the original large cone and the smaller cone removed from the top.
The standard formula for the volume of a frustum of a cone is given by:
$V = \frac{1}{3} \pi h (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2})$
Thus,the correct option is $D$.

(A) To find the volume of the largest right circular cone that can be cut from a cube,the diameter of the base of the cone must be equal to the edge of the cube,and the height of the cone must be equal to the edge of the cube.
Given edge of the cube,$a = 4.2 \,cm$.
Therefore,the radius of the cone,$r = \frac{a}{2} = \frac{4.2}{2} = 2.1 \,cm$.
The height of the cone,$h = a = 4.2 \,cm$.
The volume of a cone is given by the formula $V = \frac{1}{3} \pi r^2 h$.
Substituting the values: $V = \frac{1}{3} \times \frac{22}{7} \times (2.1)^2 \times 4.2$.
$V = \frac{1}{3} \times \frac{22}{7} \times 4.41 \times 4.2$.
$V = \frac{1}{3} \times 22 \times 0.63 \times 4.2$.
$V = 22 \times 0.21 \times 4.2$.
$V = 4.62 \times 4.2 = 19.404 \,cm^3$.
Rounding to one decimal place,we get $19.4 \,cm^3$.

(B) sharpened pencil consists of a long cylindrical body and a conical tip at one end.
Therefore,the shape of a sharpened pencil is the combination of a cylinder and a cone.
Thus,the correct option is $B$.

(C) By observing the given figure,a surahi is formed by joining a spherical body with a cylindrical neck.
Therefore,a surahi is the combination of a sphere and a cylinder.

(D) By observing the given figure,it is clear that the plumbline is formed by joining the base of a hemisphere to the base of a cone.
Therefore,the plumbline is a combination of a hemisphere and a cone.

(A) glass (tumbler) is typically wider at the top and narrower at the bottom,with two circular bases of different radii.
This geometric shape,which is obtained by cutting a cone with a plane parallel to its base,is known as the frustum of a cone.
Therefore,the shape of a glass is in the form of a frustum of a cone.

(B) By observing the figure,the gilli consists of a central cylindrical part with two conical ends attached to its circular faces.
Therefore,the shape is a combination of:
$=$ Cone $+$ Cylinder $+$ Cone
$=$ Two cones and a cylinder

(C) The shape of a shuttlecock consists of two parts:
$1$. The cork part,which is in the shape of a hemisphere.
$2$. The feathered part,which is in the shape of a frustum of a cone.
Therefore,the shuttlecock is a combination of a frustum of a cone and a hemisphere.

(D) When a cone is cut by a plane parallel to its base,it is divided into two parts: a smaller cone at the top and a remaining part at the bottom. The remaining part,which has two circular bases of different radii,is known as the frustum of a cone.

(A) Given,edge of the cube $= 22 \, cm$.
Volume of the cube $= (22)^3 = 10648 \, cm^3$.
Since $\frac{1}{8}$ of the space remains unfilled,the space filled by marbles $= 10648 \times (1 - \frac{1}{8}) = 10648 \times \frac{7}{8} = 9317 \, cm^3$.
Diameter of a marble $= 0.5 \, cm$,so radius $r = 0.25 \, cm$.
Volume of one spherical marble $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (0.25)^3 = \frac{4}{3} \times \frac{22}{7} \times \frac{1}{64} = \frac{88}{1344} = \frac{11}{168} \approx 0.065476 \, cm^3$.
Number of marbles $= \frac{\text{Filled volume}}{\text{Volume of one marble}} = \frac{9317}{11/168} = \frac{9317 \times 168}{11} = 847 \times 168 = 142296$.
Rounding to the nearest provided option,the correct answer is $142244$.

(B) Given,internal diameter of the spherical shell $= 4 \, cm$ and external diameter $= 8 \, cm$.
Internal radius of the spherical shell,$r_1 = \frac{4}{2} = 2 \, cm$.
External radius of the spherical shell,$r_2 = \frac{8}{2} = 4 \, cm$.
Volume of the spherical shell $= \frac{4}{3} \pi (r_2^3 - r_1^3) = \frac{4}{3} \pi (4^3 - 2^3) = \frac{4}{3} \pi (64 - 8) = \frac{4}{3} \pi (56) = \frac{224}{3} \pi \, cm^3$.
Let the height of the cone be $h \, cm$ and the radius of the base be $R = \frac{8}{2} = 4 \, cm$.
Since the shell is recast into a cone,their volumes are equal:
Volume of cone $= \frac{1}{3} \pi R^2 h = \frac{1}{3} \pi (4)^2 h = \frac{16}{3} \pi h$.
Equating the volumes: $\frac{16}{3} \pi h = \frac{224}{3} \pi$.
$16h = 224$.
$h = \frac{224}{16} = 14 \, cm$.
Thus,the height of the cone is $14 \, cm$.

(C) Given,dimensions of the cuboid $= 49 \, cm \times 33 \, cm \times 24 \, cm$.
Volume of the cuboid $= \text{length} \times \text{breadth} \times \text{height} = 49 \times 33 \times 24 = 38808 \, cm^3$.
Let the radius of the sphere be $r$.
Volume of the sphere $= \frac{4}{3} \pi r^3$.
Since the cuboid is moulded into a sphere,their volumes are equal:
$\frac{4}{3} \pi r^3 = 38808$
$\frac{4}{3} \times \frac{22}{7} \times r^3 = 38808$
$r^3 = \frac{38808 \times 3 \times 7}{4 \times 22}$
$r^3 = 441 \times 21 = 9261$
$r = \sqrt[3]{9261} = 21 \, cm$.
Thus,the radius of the sphere is $21 \, cm$.

(D) Volume of the wall $= 270 \times 300 \times 350 = 28,350,000 \,cm^3$.
Since $\frac{1}{8}$ of the wall's volume is occupied by mortar,the volume occupied by the bricks is $\frac{7}{8}$ of the total volume.
Volume occupied by bricks $= 28,350,000 \times \frac{7}{8} = 24,806,250 \,cm^3$.
Volume of one brick $= 22.5 \times 11.25 \times 8.75 = 2,214.84375 \,cm^3$.
Number of bricks $= \frac{\text{Volume occupied by bricks}}{\text{Volume of one brick}} = \frac{24,806,250}{2,214.84375} = 11,200$.
Therefore,the number of bricks used is $11,200$.

(A) Given,diameter of the cylinder $= 2 \, cm$.
Therefore,radius of the cylinder $(R) = 1 \, cm$ and height $(h) = 16 \, cm$.
Volume of the cylinder $= \pi R^2 h = \pi \times (1)^2 \times 16 = 16 \pi \, cm^3$.
Let the radius of each solid sphere be $r \, cm$.
Volume of one sphere $= \frac{4}{3} \pi r^3$.
Since $12$ spheres are made from the cylinder,the total volume of $12$ spheres equals the volume of the cylinder.
$12 \times (\frac{4}{3} \pi r^3) = 16 \pi$.
$16 \pi r^3 = 16 \pi$.
$r^3 = 1$,which implies $r = 1 \, cm$.
Diameter of each sphere $= 2r = 2 \times 1 = 2 \, cm$.

(B) Given,the radius of the top of the bucket,$R = 28 \, cm$.
The radius of the bottom of the bucket,$r = 7 \, cm$.
The slant height of the bucket,$l = 45 \, cm$.
Since the bucket is in the form of a frustum of a cone,the curved surface area is given by the formula:
$\text{Curved Surface Area} = \pi l (R + r)$
Substituting the given values:
$\text{Curved Surface Area} = \frac{22}{7} \times 45 \times (28 + 7)$
$= \frac{22}{7} \times 45 \times 35$
$= 22 \times 45 \times 5$
$= 22 \times 225 = 4950 \, cm^2$.
Thus,the curved surface area of the bucket is $4950 \, cm^2$.

(C) Given,diameter of cylinder $=$ Diameter of hemisphere $= 0.5\, cm$ (since both hemispheres are attached to the cylinder).
$\therefore$ Radius of cylinder $(r) =$ radius of hemisphere $(r) = \frac{0.5}{2} = 0.25\, cm$ (since diameter $= 2 \times$ radius).
Total length of capsule $= 2\, cm$.
Length of the cylindrical part of the capsule $(h) = \text{Total length} - (\text{Radius of left hemisphere} + \text{Radius of right hemisphere})$.
$h = 2 - (0.25 + 0.25) = 2 - 0.5 = 1.5\, cm$.
Now,capacity of the capsule $=$ Volume of cylindrical part $+ 2 \times$ Volume of hemisphere.
Capacity $= \pi r^2 h + 2 \times (\frac{2}{3} \pi r^3) = \pi r^2 (h + \frac{4}{3} r)$.
Substituting the values:
Capacity $= \frac{22}{7} \times (0.25)^2 \times (1.5 + \frac{4}{3} \times 0.25)$.
Capacity $= \frac{22}{7} \times 0.0625 \times (1.5 + 0.3333) = \frac{22}{7} \times 0.0625 \times 1.8333 \approx 0.36\, cm^3$.
Hence,the capacity of the capsule is $0.36\, cm^3$.

(D) The curved surface area of a single solid hemisphere is given by $2 \pi r^{2}$.
When two solid hemispheres of the same base radius $r$ are joined together along their circular bases, they form a complete solid sphere.
The curved surface area of the resulting solid sphere is the sum of the curved surface areas of the two individual hemispheres.
Therefore, the curved surface area of the new solid $= 2 \pi r^{2} + 2 \pi r^{2} = 4 \pi r^{2}$.

(A) sphere is enclosed in a right circular cylinder such that it touches the top,bottom,and the curved surface of the cylinder.
Since the sphere touches the curved surface of the cylinder,the diameter of the sphere must be equal to the diameter of the cylinder.
The radius of the cylinder is given as $r \, cm$,so the diameter of the cylinder is $2r \, cm$.
Therefore,the diameter of the sphere is $2r \, cm$.

(B) When a solid object is reshaped into another form,the total amount of material used remains the same. Therefore,the volume of the new shape remains unaltered.

(C) Given,diameter of the upper end of the bucket $D = 44 \, cm$,so radius $R = 22 \, cm$.
Diameter of the lower end of the bucket $d = 24 \, cm$,so radius $r = 12 \, cm$.
Height of the bucket $h = 35 \, cm$.
The shape of the bucket is a frustum of a cone.
The capacity (volume) of the bucket is given by the formula: $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr)$.
Substituting the values: $V = \frac{1}{3} \times \frac{22}{7} \times 35 \times (22^2 + 12^2 + 22 \times 12)$.
$V = \frac{1}{3} \times 22 \times 5 \times (484 + 144 + 264)$.
$V = \frac{110}{3} \times 892$.
$V = \frac{98120}{3} \approx 32706.67 \, cm^3$.
Since $1000 \, cm^3 = 1 \, L$,the capacity is $32706.67 / 1000 \approx 32.7 \, L$.

(A) right circular cone is a solid generated by the revolution of a right-angled triangle about one of its sides containing the right angle.
When a plane cuts a right circular cone parallel to its base,the resulting cross-section is always a circle.
Note that the portion of the cone between the base and the plane is called the frustum of the cone,but the cross-section itself is a circle.

(A) Let the radii of the two spheres be $r_1$ and $r_2$,respectively.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
Given the ratio of volumes $V_1 : V_2 = 64 : 27$,we have:
$\frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \frac{64}{27}$
$\frac{r_1^3}{r_2^3} = \frac{64}{27} = \left(\frac{4}{3}\right)^3$
Taking the cube root on both sides,we get:
$\frac{r_1}{r_2} = \frac{4}{3}$
The surface area of a sphere is given by $A = 4 \pi r^2$.
The ratio of their surface areas is:
$\frac{A_1}{A_2} = \frac{4 \pi r_1^2}{4 \pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2$
Substituting the value of $\frac{r_1}{r_2}$:
$\frac{A_1}{A_2} = \left(\frac{4}{3}\right)^2 = \frac{16}{9}$
Thus,the ratio of their surface areas is $16:9$.

(A) True.
The curved surface area of the combined shape is the sum of the curved surface area of the cone and the curved surface area of the cylinder.
The curved surface area of a cone is given by $\pi r l$,where $l = \sqrt{h^{2} + r^{2}}$ is the slant height.
Thus,the curved surface area of the cone is $\pi r \sqrt{h^{2} + r^{2}}$.
The curved surface area of a cylinder is $2 \pi r h$.
Therefore,the total curved surface area of the combined shape is $\pi r \sqrt{h^{2} + r^{2}} + 2 \pi r h$.

(B) False.
Let $r$ be the radius of the original steel ball and $r_{1}$ be the radius of each new ball formed after melting.
The volume of the original ball is $V = \frac{4}{3} \pi r^{3}$.
The volume of eight new balls is $8 \times V_{1} = 8 \times \left( \frac{4}{3} \pi r_{1}^{3} \right)$.
Since the volume remains constant during melting,we have $\frac{4}{3} \pi r^{3} = 8 \times \frac{4}{3} \pi r_{1}^{3}$.
Dividing both sides by $\frac{4}{3} \pi$,we get $r^{3} = 8 r_{1}^{3}$.
Taking the cube root on both sides,we get $r = 2 r_{1}$,which implies $r_{1} = \frac{r}{2}$.
Therefore,the radius of each new ball is $\frac{1}{2}$ of the original radius,not $\frac{1}{8}$.

(B) False.
The total surface area of a single cube with side $a$ is $6 a^{2}$.
When two such identical cubes are joined end to end,two faces (each of area $a^{2}$) are hidden inside the resulting cuboid.
The dimensions of the resulting cuboid are length $l = 2a$,breadth $b = a$,and height $h = a$.
The total surface area of a cuboid is given by the formula $2(lb + bh + hl)$.
Substituting the values: $2((2a \times a) + (a \times a) + (a \times 2a)) = 2(2a^{2} + a^{2} + 2a^{2}) = 2(5a^{2}) = 10 a^{2}$.
Therefore,the statement is false.

(FALSE) False.
The total surface area of the lattu (top) is the sum of the curved surface area of the hemisphere and the curved surface area of the cone.
This is because the circular base of the hemisphere and the circular base of the cone are joined together and are not exposed to the outside. Therefore,they are not included in the total surface area of the combined solid.

(A) True.
The actual capacity of a vessel is defined as the total volume of the empty space inside it that can be filled with a substance. In the given figure,a hemispherical depression is carved out from the base of the cylinder. Therefore,the volume of the space available inside the vessel is the volume of the cylinder minus the volume of the hemisphere.

(B) False.
The curved surface area of one hemisphere is $2 \pi r^{2}$.
When two identical solid hemispheres are joined together along their circular bases,the bases are hidden inside the new solid.
Therefore,the total surface area of the resulting solid consists only of the two curved surfaces.
Total surface area $= 2 \pi r^{2} + 2 \pi r^{2} = 4 \pi r^{2}$.
Since $4 \pi r^{2} \neq 6 \pi r^{2}$,the statement is False.

(B) False.
The total surface area of a single cylinder with radius $r$ and height $h$ is given by $2 \pi rh + 2 \pi r^2$.
When one cylinder is placed on top of another cylinder of the same radius and height,the two circular faces (one from each cylinder) are hidden at the point of contact.
The new shape is a single cylinder with radius $r$ and total height $H = 2h$.
The total surface area of this new cylinder is the sum of its curved surface area and the areas of its two circular bases.
Total Surface Area $= 2 \pi r H + 2 \pi r^2 = 2 \pi r(2h) + 2 \pi r^2 = 4 \pi rh + 2 \pi r^2$.
Since $4 \pi rh + 2 \pi r^2 \neq 4 \pi rh + 4 \pi r^2$,the given statement is False.

(B) False.
The total surface area of the combined solid is the sum of the curved surface area of the cone,the curved surface area of the cylinder,and the area of the bottom base of the cylinder.
$1$. Curved surface area of the cone = $\pi r l$,where slant height $l = \sqrt{r^2 + h^2}$.
$2$. Curved surface area of the cylinder = $2\pi rh$.
$3$. Area of the bottom base of the cylinder = $\pi r^2$.
Total surface area = $\pi r l + 2\pi rh + \pi r^2$
$= \pi r (l + 2h + r)$
$= \pi r [\sqrt{r^2 + h^2} + 2h + r]$.
Since the calculated expression $\pi r [\sqrt{r^2 + h^2} + 2h + r]$ is not equal to the given expression $\pi [\sqrt{r^2 + h^2} + 3r + 2h]$,the statement is False.

(B) False.
Since the solid ball is exactly fitted inside the cubical box of side $a$,the diameter of the ball must be equal to the side of the cube,which is $a$.
Therefore,the radius of the ball $r = \frac{a}{2}$.
The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^{3}$.
Substituting $r = \frac{a}{2}$ into the formula,we get:
$V = \frac{4}{3} \pi \left( \frac{a}{2} \right)^{3} = \frac{4}{3} \pi \left( \frac{a^{3}}{8} \right) = \frac{1}{6} \pi a^{3}$.
Since $\frac{1}{6} \pi a^{3} \neq \frac{4}{3} \pi a^{3}$,the given statement is False.

(B) False.
The correct formula for the volume of the frustum of a cone is $V = \frac{1}{3} \pi h(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2}),$ where $h$ is the vertical height of the frustum and $r_{1}, r_{2}$ are the radii of the circular ends. The given expression contains a negative sign instead of a positive sign for the term $r_{1}r_{2}$.

(TRUE) True.
We know that the capacity (volume) of a cylindrical vessel is given by $V_{cylinder} = \pi r^{2} h$.
The capacity (volume) of a hemisphere is given by $V_{hemisphere} = \frac{2}{3} \pi r^{3}$.
From the figure,the hemispherical portion is raised upward at the bottom,which means it occupies space inside the cylinder. Therefore,the capacity of the vessel is the volume of the cylinder minus the volume of the hemisphere.
Capacity of the vessel = $V_{cylinder} - V_{hemisphere}$
$= \pi r^{2} h - \frac{2}{3} \pi r^{3}$
$= \pi r^{2} (h - \frac{2}{3} r)$
$= \frac{\pi r^{2}}{3} (3h - 2r)$.

(B) The statement is False.
The curved surface area of a frustum of a cone is given by the formula $A = \pi l(r_{1} + r_{2})$.
In this formula,the slant height $l$ is defined as $l = \sqrt{h^{2} + (r_{1} - r_{2})^{2}}$,where $h$ is the vertical height,and $r_{1}$ and $r_{2}$ are the radii of the two circular ends.
The given expression for the slant height $l = \sqrt{h^{2} + (r_{1} + r_{2})^{2}}$ is incorrect because it uses the sum of the radii instead of the difference of the radii.

(A) True.
The total surface area of the metallic sheet used is the sum of the areas of the individual parts that make up the object.
$1$. The bucket is a frustum of a cone $(ABCD)$,so we include its curved surface area.
$2$. The bucket has a circular base at the bottom of the frustum,which is also the top of the cylinder. However,since the bucket is open and mounted on a cylinder,the metallic sheet used includes the curved surface of the frustum,the curved surface of the cylinder,and the circular base at the bottom of the cylinder $(EF)$.
$3$. The area of the metallic sheet = (Curved surface area of frustum of a cone) + (Curved surface area of cylinder) + (Area of the circular base at the bottom).
As shown in the figure,$ABCD$ is a frustum of a cone and $CDEF$ is a hollow cylinder.

(N/A) The cone of maximum size carved out from a cube of edge $14 \, cm$ will have a base radius $r = 7 \, cm$ and height $h = 14 \, cm$.
The slant height $l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 14^2} = \sqrt{49 + 196} = \sqrt{245} = 7\sqrt{5} \, cm$.
Surface area of the cone $= \pi r l + \pi r^2 = \frac{22}{7} \times 7 \times 7\sqrt{5} + \frac{22}{7} \times 7^2 = 154\sqrt{5} + 154 = 154(\sqrt{5} + 1) \, cm^2$.
Surface area of the cube $= 6 \times (14)^2 = 6 \times 196 = 1176 \, cm^2$.
When the cone is carved out,the circular base of the cone is removed from one face of the cube,but the curved surface area of the cone is added to the total surface area.
Surface area of the remaining solid $= (\text{Total surface area of cube}) - (\text{Area of circular base of cone}) + (\text{Curved surface area of cone})$.
$= 1176 - \pi r^2 + \pi r l = 1176 - 154 + 154\sqrt{5} = (1022 + 154\sqrt{5}) \, cm^2$.

(D) The volume of the solid metallic sphere is given by the formula $V_s = \frac{4}{3} \pi r^3$,where $r = 10.5 \, cm$.
$V_s = \frac{4}{3} \pi (10.5)^3 \, cm^3$.
The volume of one cone is given by the formula $V_c = \frac{1}{3} \pi r^2 h$,where $r = 3.5 \, cm$ and $h = 3 \, cm$.
$V_c = \frac{1}{3} \pi (3.5)^2 \times 3 \, cm^3 = \pi (3.5)^2 \, cm^3$.
The number of cones formed is equal to the total volume of the sphere divided by the volume of one cone.
$\text{Number of cones} = \frac{V_s}{V_c} = \frac{\frac{4}{3} \pi (10.5)^3}{\pi (3.5)^2} = \frac{4 \times 10.5 \times 10.5 \times 10.5}{3 \times 3.5 \times 3.5} = 4 \times 3 \times 3 \times 3.5 = 126$.

(A) First,convert all dimensions to meters: Width $= 300 \,cm = 3 \,m$,Depth $= 120 \,cm = 1.2 \,m$,Speed $= 20 \,km/h = 20000 \,m/h$.
Volume of water flowing through the canal in $1 \,hour$ $= \text{Width} \times \text{Depth} \times \text{Speed} = 3 \times 1.2 \times 20000 = 72000 \,m^3$.
Volume of water flowing in $20 \,minutes$ $= \frac{72000 \times 20}{60} = 24000 \,m^3$.
Let the area to be irrigated be $A \,m^2$. The volume of water required for irrigation is $\text{Area} \times \text{Height of standing water}$.
Given height $= 8 \,cm = 0.08 \,m$.
So,$A \times 0.08 = 24000$.
$A = \frac{24000}{0.08} = 300000 \,m^2$.
Since $1 \,hectare = 10000 \,m^2$,the area in hectares $= \frac{300000}{10000} = 30 \,hectares$.

(B) Let $h$ be the height of the given cone. On dividing the cone through the midpoint of its axis and parallel to its base into two parts,we obtain a smaller cone at the top and a frustum at the bottom.
In two similar triangles,the ratio of corresponding sides is equal. Let $r$ be the radius of the smaller cone. By similarity of triangles,we have $\frac{r}{4} = \frac{h/2}{h} = \frac{1}{2}$.
Therefore,$r = 2 \, cm$.
Volume of the smaller cone $(V_1)$ = $\frac{1}{3} \pi r^2 (h/2) = \frac{1}{3} \pi (2)^2 (h/2) = \frac{2}{3} \pi h$.
Volume of the original cone $(V)$ = $\frac{1}{3} \pi (4)^2 h = \frac{16}{3} \pi h$.
Volume of the frustum $(V_2)$ = $V - V_1 = \frac{16}{3} \pi h - \frac{2}{3} \pi h = \frac{14}{3} \pi h$.
Ratio of the volume of the smaller cone to the volume of the frustum = $\frac{V_1}{V_2} = \frac{\frac{2}{3} \pi h}{\frac{14}{3} \pi h} = \frac{2}{14} = \frac{1}{7}$.
Thus,the ratio is $1:7$.

(C) Let the edges of the three cubes be $3x, 4x,$ and $5x \, \text{cm}$ respectively.
The volume of the three cubes is $V = (3x)^3 + (4x)^3 + (5x)^3 = 27x^3 + 64x^3 + 125x^3 = 216x^3 \, \text{cm}^3$.
Let the side of the new single cube be $a$. The volume of this new cube is $a^3 = 216x^3$,which implies $a = 6x$.
The diagonal of a cube with side $a$ is given by $d = a\sqrt{3}$.
Given that the diagonal is $12\sqrt{3} \, \text{cm}$,we have $a\sqrt{3} = 12\sqrt{3}$,so $a = 12 \, \text{cm}$.
Since $a = 6x$,we have $6x = 12$,which gives $x = 2$.
Therefore,the edges of the three cubes are $3(2) = 6 \, \text{cm}$,$4(2) = 8 \, \text{cm}$,and $5(2) = 10 \, \text{cm}$.

(D) Given,the edges of three solid cubes are $3 \, cm$,$4 \, cm$,and $5 \, cm$ respectively.
The volume of a cube is given by the formula $V = (\text{edge})^3$.
Volume of the first cube $= (3)^3 = 27 \, cm^3$.
Volume of the second cube $= (4)^3 = 64 \, cm^3$.
Volume of the third cube $= (5)^3 = 125 \, cm^3$.
Total volume of the three cubes $= 27 + 64 + 125 = 216 \, cm^3$.
Let the edge of the new single cube be $R \, cm$.
Then,the volume of the new cube $= R^3$.
Since the total volume remains constant,$R^3 = 216$.
Taking the cube root on both sides,$R = \sqrt[3]{216} = 6 \, cm$.
Thus,the edge of the resulting cube is $6 \, cm$.

(A) Given,dimensions of the cuboid $= 9 \, cm \times 11 \, cm \times 12 \, cm$.
Volume of the cuboid $= 9 \times 11 \times 12 = 1188 \, cm^3$.
Diameter of the spherical shot $= 3 \, cm$.
Radius of the shot,$r = \frac{3}{2} = 1.5 \, cm$.
Volume of one spherical shot $= \frac{4}{3} \pi r^3 = \frac{4}{3} \times \frac{22}{7} \times (1.5)^3$.
Volume of one shot $= \frac{4}{3} \times \frac{22}{7} \times 3.375 = \frac{99}{7} \approx 14.143 \, cm^3$.
Number of shots $= \frac{\text{Volume of cuboid}}{\text{Volume of one shot}} = \frac{1188}{99/7} = \frac{1188 \times 7}{99} = 12 \times 7 = 84$.
Thus,$84$ shots can be made.

(B) Given, volume of the frustum $= 28.49 \, \text{L} = 28.49 \times 1000 \, \text{cm}^3 = 28490 \, \text{cm}^3$ (since $1 \, \text{L} = 1000 \, \text{cm}^3$).
Radius of the top $(r_1) = 28 \, \text{cm}$.
Radius of the bottom $(r_2) = 21 \, \text{cm}$.
Let the height of the bucket be $h \, \text{cm}$.
The volume of a frustum of a cone is given by $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.
Substituting the values: $28490 = \frac{1}{3} \times \frac{22}{7} \times h \times (28^2 + 21^2 + 28 \times 21)$.
$28490 = \frac{22}{21} \times h \times (784 + 441 + 588)$.
$28490 = \frac{22}{21} \times h \times 1813$.
$h = \frac{28490 \times 21}{22 \times 1813}$.
$h = \frac{598290}{39886} = 15 \, \text{cm}$.

(C) Let the cone be $ORN$ with base radius $r_1 = 8 \, cm$ and height $OM = 12 \, cm$.
Let $P$ be the mid-point of $OM$,then $OP = PM = \frac{12}{2} = 6 \, cm$.
Since the plane is parallel to the base,$\triangle OPD \sim \triangle OMN$.
By properties of similar triangles,$\frac{OP}{OM} = \frac{PD}{MN}$.
$\frac{6}{12} = \frac{PD}{8} \Rightarrow \frac{1}{2} = \frac{PD}{8} \Rightarrow PD = 4 \, cm$.
The plane divides the cone into a smaller cone (top) and a frustum (bottom).
Volume of smaller cone $V_1 = \frac{1}{3} \pi (PD)^2 (OP) = \frac{1}{3} \pi (4)^2 (6) = 32 \pi \, cm^3$.
Volume of the original cone $V = \frac{1}{3} \pi (MN)^2 (OM) = \frac{1}{3} \pi (8)^2 (12) = 256 \pi \, cm^3$.
Volume of the frustum $V_2 = V - V_1 = 256 \pi - 32 \pi = 224 \pi \, cm^3$.
The ratio of the volume of the smaller cone to the frustum is $V_1 : V_2 = 32 \pi : 224 \pi = 1 : 7$.

(D) Let the side length of a cube be $a \, cm$.
Given,the volume of the cube is $a^{3} = 64 \, cm^{3}$.
Therefore,$a = \sqrt[3]{64} = 4 \, cm$.
When two such cubes are joined end to end,they form a cuboid with dimensions:
Length $(l)$ = $2a = 2 \times 4 = 8 \, cm$
Breadth $(b)$ = $a = 4 \, cm$
Height $(h)$ = $a = 4 \, cm$
The surface area of the resulting cuboid is given by the formula $2(lb + bh + hl)$.
Surface Area = $2(8 \times 4 + 4 \times 4 + 4 \times 8)$
Surface Area = $2(32 + 16 + 32)$
Surface Area = $2(80) = 160 \, cm^{2}$.

(A) Given that,the side of the solid cube $(a) = 7 \, cm$.
The height of the conical cavity (cone),$h = 7 \, cm$.
The radius of the conical cavity (cone),$r = 3 \, cm$.
Volume of the cube $= a^3 = (7)^3 = 343 \, cm^3$.
Volume of the conical cavity (cone) $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3)^2 \times 7$.
$= \frac{1}{3} \times 22 \times 9 = 66 \, cm^3$.
Volume of the remaining solid $=$ Volume of the cube $-$ Volume of the conical cavity.
$= 343 - 66 = 277 \, cm^3$.
Thus,the volume of the remaining solid is $277 \, cm^3$.

(B) When two cones with the same base radius and height are joined together along their bases,the base area is hidden,and the total surface area of the resulting shape is the sum of the curved surface areas of the two cones.
Given:
Radius of the cone,$r = 8\, cm$
Height of the cone,$h = 15\, cm$
The slant height $l$ of each cone is given by $l = \sqrt{r^2 + h^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\, cm$.
The total surface area of the resulting shape is the sum of the curved surface areas of the two identical cones:
Surface Area $= 2 \times (\pi r l)$
$= 2 \times \pi \times 8 \times 17$
$= 272\pi\, cm^2$
Using $\pi \approx \frac{22}{7}$:
$= 272 \times \frac{22}{7} = \frac{5984}{7} \approx 854.85\, cm^2$.
Rounding to the nearest whole number,we get $855\, cm^2$.