Mix Examples - Surface Areas and Volumes Questions in English

(C) The volume of the metal sphere is given by the formula $V_s = \frac{4}{3} \pi r^3$,where $r = 12\, cm$.
$V_s = \frac{4}{3} \pi (12)^3 = \frac{4}{3} \pi (1728) = 2304 \pi\, cm^3$.
The volume of one cone is given by $V_c = \frac{1}{3} \pi r^2 h$,where $r = 3\, cm$ and $h = 4\, cm$.
$V_c = \frac{1}{3} \pi (3)^2 (4) = \frac{1}{3} \pi (9)(4) = 12 \pi\, cm^3$.
The number of cones obtained is the ratio of the volume of the sphere to the volume of one cone:
$N = \frac{V_s}{V_c} = \frac{2304 \pi}{12 \pi} = 192$.
Thus,$192$ cones are obtained.

(A) $1$. Volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
$2$. Total volume of the five spheres = $\frac{4}{3}\pi (1^3 + 3^3 + 4^3 + 5^3 + 8^3) = \frac{4}{3}\pi (1 + 27 + 64 + 125 + 512) = \frac{4}{3}\pi (729)$.
$3$. Let the radius of the new large sphere be $R$. Then $\frac{4}{3}\pi R^3 = \frac{4}{3}\pi (729)$,which implies $R^3 = 729$,so $R = 9\, cm$.
$4$. Volume of one cylinder = $\pi r^2 h = \pi (3^2)(4) = 36\pi\, cm^3$.
$5$. Number of cylinders = $\frac{\text{Total Volume}}{\text{Volume of one cylinder}} = \frac{\frac{4}{3}\pi (729)}{36\pi} = \frac{972\pi}{36\pi} = 27$ cylinders.

(A) The surface area of a sphere is given by the formula $S = 4\pi r^{2}$.
Given $S = 1256 \, cm^{2}$ and $\pi = 3.14$.
$1256 = 4 \times 3.14 \times r^{2}$
$1256 = 12.56 \times r^{2}$
$r^{2} = \frac{1256}{12.56} = 100$
$r = 10 \, cm$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^{3}$.
$V = \frac{4}{3} \times 3.14 \times (10)^{3}$
$V = \frac{4}{3} \times 3.14 \times 1000$
$V = \frac{12560}{3} = 4186 \frac{2}{3} \, cm^{3}$.

(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 $4\pi r_1^2$ and the curved surface area of a hemisphere is $2\pi r_2^2$.
Given the ratio: $\frac{4\pi r_1^2}{2\pi r_2^2} = \frac{9}{2}$.
Simplifying this,we get $\frac{2r_1^2}{r_2^2} = \frac{9}{2}$,which implies $\frac{r_1^2}{r_2^2} = \frac{9}{4}$.
Taking the square root,we find $\frac{r_1}{r_2} = \frac{3}{2}$.
The volume of a sphere is $V_1 = \frac{4}{3}\pi r_1^3$ and the volume of a hemisphere is $V_2 = \frac{2}{3}\pi r_2^3$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{\frac{4}{3}\pi r_1^3}{\frac{2}{3}\pi r_2^3} = 2 \times \left(\frac{r_1}{r_2}\right)^3$.
Substituting the ratio $\frac{r_1}{r_2} = \frac{3}{2}$,we get $\frac{V_1}{V_2} = 2 \times \left(\frac{3}{2}\right)^3 = 2 \times \frac{27}{8} = \frac{27}{4}$.
Thus,the ratio of their volumes is $27:4$.

(C) The volume of the cubic tank is $V_{total} = 15.625 \, m^3$.
Since the tank is a cube, the side length $s$ is given by $s = \sqrt[3]{15.625} = 2.5 \, m$.
The current depth of water is $h = 1.3 \, m$.
The volume of water currently in the tank is $V_{current} = \text{side} \times \text{side} \times \text{depth} = 2.5 \times 2.5 \times 1.3 = 6.25 \times 1.3 = 8.125 \, m^3$.
The volume of water already used is $V_{used} = V_{total} - V_{current} = 15.625 - 8.125 = 7.5 \, m^3$.

(D) Since the spheres are made of the same metal,their density $\rho$ is constant. The weight $W$ of a sphere is given by $W = \text{Volume} \times \text{Density} = \frac{4}{3} \pi r^3 \rho$.
Thus,the ratio of weights is equal to the ratio of the cubes of their radii: $\frac{W_1}{W_2} = \frac{r_1^3}{r_2^3}$.
Given $W_1 = 5920\, g$ and $W_2 = 740\, g$,we have $\frac{5920}{740} = \frac{r_1^3}{r_2^3}$.
Simplifying the ratio: $8 = \frac{r_1^3}{r_2^3}$,which implies $\frac{r_1}{r_2} = \sqrt[3]{8} = 2$.
The diameter of the smaller sphere is $5\, cm$,so its radius $r_2 = \frac{5}{2} = 2.5\, cm$.
Therefore,the radius of the larger sphere is $r_1 = 2 \times r_2 = 2 \times 2.5 = 5\, cm$.

(A) The diameter of the cylindrical glass is $d = 7\, cm$, so the radius is $r = 3.5\, cm$.
The height of the milk in the glass is $h = 12\, cm$.
The volume of milk in one glass is given by the formula $V = \pi r^2 h$.
$V = \frac{22}{7} \times 3.5 \times 3.5 \times 12 = 22 \times 0.5 \times 3.5 \times 12 = 462\, cm^3$.
Since $1000\, cm^3 = 1\, litre$, the volume of milk in one glass is $0.462\, litres$.
For $1600$ students, the total milk required is $1600 \times 0.462 = 739.2\, litres$.

(A) When a right triangle with sides $6 \, cm, 8 \, cm$ and $10 \, cm$ is revolved about the side $8 \, cm$,the side $8 \, cm$ becomes the height $(h)$ of the cone,and the side $6 \, cm$ becomes the radius $(r)$ of the base. The hypotenuse $10 \, cm$ becomes the slant height $(l)$ of the cone.
Given: $r = 6 \, cm, h = 8 \, cm, l = 10 \, cm$.
Volume of the cone $V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times (6)^2 \times 8 = \frac{1}{3} \times 3.14 \times 36 \times 8 = 3.14 \times 12 \times 8 = 301.44 \, cm^3$.
Curved surface area of the cone $CSA = \pi r l = 3.14 \times 6 \times 10 = 188.4 \, cm^2$.

(C) $1$. The internal diameter of the hemispherical tank is $14\, m$,so the radius $r = 7\, m$.
$2$. The volume of a hemispherical tank is given by $V = \frac{2}{3} \pi r^3$.
$3$. $V = \frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 = \frac{2}{3} \times 22 \times 49 = \frac{2156}{3} = 718 \frac{2}{3}\, m^3$.
$4$. We know that $1\, m^3 = 1\, kilolitre$. Therefore,the tank contains $50\, m^3$ of water.
$5$. The volume of water to be pumped = (Total capacity) - (Volume already present).
$6$. Volume to be pumped = $718 \frac{2}{3} - 50 = 668 \frac{2}{3}\, m^3$.

(D) Let the radius of the sphere and the cylinder be $r$. Let the height of the cylinder be $h$.
Given that the volumes are equal:
Volume of sphere = $\frac{4}{3} \pi r^3$
Volume of cylinder = $\pi r^2 h$
Equating the two: $\frac{4}{3} \pi r^3 = \pi r^2 h$
$h = \frac{4}{3} r$
The diameter of the cylinder is $d = 2r$.
We need to find the percentage by which the diameter exceeds the height:
Difference = $d - h = 2r - \frac{4}{3} r = \frac{2}{3} r$
Percentage = $\frac{d - h}{h} \times 100 = \frac{(2/3)r}{(4/3)r} \times 100 = \frac{2}{4} \times 100 = 50\%$.

(B) The statement is False.
In a right circular cone,the radius $(r)$,height $(h)$,and slant height $(l)$ form a right-angled triangle where the slant height is the hypotenuse.
According to the Pythagorean theorem,the relationship is $l^2 = r^2 + h^2$.
Given: $r = 3 \, cm$,$h = 5 \, cm$,and $l = 4 \, cm$.
Calculating $r^2 + h^2 = 3^2 + 5^2 = 9 + 25 = 34$.
Calculating $l^2 = 4^2 = 16$.
Since $16 \neq 34$,the given dimensions cannot form a right circular cone because the slant height must be the longest side of the right-angled triangle formed by the radius and the height.

(A) Let the radii of the two spheres be $r_1$ and $r_2$.
The surface area of a sphere is given by $A = 4\pi r^2$.
The ratio of the surface areas is $\frac{4\pi r_1^2}{4\pi r_2^2} = \frac{r_1^2}{r_2^2} = \frac{81}{49}$.
Taking the square root of both sides,we get $\frac{r_1}{r_2} = \sqrt{\frac{81}{49}} = \frac{9}{7}$.
The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$.
The ratio of the volumes is $\frac{\frac{4}{3}\pi r_1^3}{\frac{4}{3}\pi r_2^3} = \frac{r_1^3}{r_2^3} = \left(\frac{r_1}{r_2}\right)^3$.
Substituting the ratio of radii,we get $\left(\frac{9}{7}\right)^3 = \frac{9^3}{7^3} = \frac{729}{343}$.
Thus,the statement is True.

(B) The lateral surface area $(LSA)$ of a cube is given by the formula $LSA = 4a^2$,where $a$ is the length of the edge.
Given,edge $a = 5 \, cm$.
Substituting the value in the formula:
$LSA = 4 \times (5)^2$
$LSA = 4 \times 25$
$LSA = 100 \, cm^2$.
The statement claims the lateral surface area is $150 \, cm^2$,which is incorrect.
Therefore,the statement is False.

(TRUE) The volume of a cylinder is given by the formula: $V = \text{Base Area} \times \text{Height}$.
Given:
Base Area = $154 \, cm^{2}$
Height = $20 \, cm$
Calculating the volume:
$V = 154 \, cm^{2} \times 20 \, cm = 3080 \, cm^{3}$.
Since the calculated volume matches the given value, the statement is True.

(B) The length of the longest rod that can be placed in a rectangular box of dimensions $l \times b \times h$ is given by the space diagonal formula: $d = \sqrt{l^2 + b^2 + h^2}$.
Given dimensions are $l = 12\, cm$,$b = 4\, cm$,and $h = 3\, cm$.
Calculating the diagonal: $d = \sqrt{12^2 + 4^2 + 3^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13\, cm$.
The maximum length of a bar that can fit inside the box is $13\, cm$.
Since the bar is $15\, cm$ long and $15\, cm > 13\, cm$,the bar cannot fit in the box.
Therefore,the statement is False.

(B) The formula for the curved surface area of a cylinder is $A = 2\pi rh$,where $r$ is the radius and $h$ is the height.
Given: $r = 10\, cm$,$A = 440\, cm^{2}$,and taking $\pi = 22/7$.
Substituting the values: $440 = 2 \times (22/7) \times 10 \times h$.
$440 = (440/7) \times h$.
$h = (440 \times 7) / 440$.
$h = 7\, cm$.
Therefore,the height of the cylinder is $7\, cm$.

(C) The formula for the total surface area $(TSA)$ of a cube with edge length '$a$' is given by:
$TSA = 6a^2$
Given that $TSA = 864 \, cm^2$,we can set up the equation:
$6a^2 = 864$
Divide both sides by $6$:
$a^2 = \frac{864}{6}$
$a^2 = 144$
Taking the square root of both sides:
$a = \sqrt{144}$
$a = 12 \, cm$
Therefore,the edge of the cube is $12 \, cm$.

(D) The volume of a cuboid is given by the formula: $V = \text{length} \times \text{breadth} \times \text{height}$.
Given:
Length $(l)$ = $30 \, cm$
Breadth $(b)$ = $20 \, cm$
Height $(h)$ = $15 \, cm$
Substituting these values into the formula:
$V = 30 \, cm \times 20 \, cm \times 15 \, cm$
$V = 600 \, cm^2 \times 15 \, cm$
$V = 9000 \, cm^3$.
Therefore, the correct option is $D$.

(A) The curved surface area $(CSA)$ of a hemisphere is given by the formula $CSA = 2\pi r^{2}$.
Given,$2\pi r^{2} = 306 \, cm^{2}$.
The total surface area $(TSA)$ of a hemisphere is given by the formula $TSA = 3\pi r^{2}$.
We can express $TSA$ in terms of $CSA$ as: $TSA = \frac{3}{2} \times (2\pi r^{2})$.
Substituting the given value: $TSA = \frac{3}{2} \times 306$.
$TSA = 3 \times 153 = 459 \, cm^{2}$.
Therefore,the total surface area is $459 \, cm^{2}$.

(B) Let the radius $r = 4x$ and the slant height $l = 7x$.
The curved surface area $(CSA)$ of a cone is given by $\pi rl$.
Substituting the values,$CSA = \pi(4x)(7x) = 28\pi x^2$.
The total surface area $(TSA)$ of a cone is given by $\pi r(r + l)$.
Substituting the values,$TSA = \pi(4x)(4x + 7x) = \pi(4x)(11x) = 44\pi x^2$.
The ratio of the total surface area to the curved surface area is $\frac{TSA}{CSA} = \frac{44\pi x^2}{28\pi x^2} = \frac{44}{28} = \frac{11}{7}$.
Thus,the ratio is $11:7$.

(C) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Given $V = 4500 \pi \text{ cm}^3$,we have:
$\frac{4}{3} \pi r^3 = 4500 \pi$
$r^3 = 4500 \times \frac{3}{4}$
$r^3 = 1125 \times 3 = 3375$
$r = \sqrt[3]{3375} = 15 \text{ cm}$.
The diameter $d$ is $2r = 2 \times 15 = 30 \text{ cm}$.

(D) Let the radii of the two cylinders be $r_1$ and $r_2$,and their heights be $h_1$ and $h_2$. Given that $h_1 = h_2 = h$ and $r_1:r_2 = 4:5$.
The volume of a cylinder is given by the formula $V = \pi r^2 h$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{\pi r_1^2 h}{\pi r_2^2 h} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^2$.
Substituting the given ratio: $\frac{V_1}{V_2} = \left(\frac{4}{5}\right)^2 = \frac{16}{25}$.
Therefore,the ratio of their volumes is $16:25$.

(A) We know that $1 \, m = 100 \, cm$.
Therefore,$1 \, m^3 = (100 \, cm) \times (100 \, cm) \times (100 \, cm) = 1,000,000 \, cm^3$.
Since $1000 \, cm^3 = 1 \, \text{litre}$,we can write $1 \, cm^3 = \frac{1}{1000} \, \text{litre}$.
Thus,$1,000,000 \, cm^3 = \frac{1,000,000}{1000} \, \text{litres} = 1000 \, \text{litres}$.
Hence,$1 \, m^3 = 1000 \, \text{litres}$.

(B) We know that $1 \text{ litre}$ is defined as the volume of a cube with side length $10 \text{ cm}$.
Since the volume of a cube is given by $(\text{side})^3$,we have:
$1 \text{ litre} = (10 \text{ cm})^3 = 10 \text{ cm} \times 10 \text{ cm} \times 10 \text{ cm} = 1000 \text{ cm}^3$.
Therefore,$1 \text{ litre} = 1000 \text{ cm}^3$.

(C) Let the radii of the two cylinders be $r_1$ and $r_2$,and their heights be $h_1$ and $h_2$.
Given that $r_1:r_2 = 3:2$ and $h_1:h_2 = 8:11$.
The volume of a cylinder is given by the formula $V = \pi r^2 h$.
The ratio of the volumes of the two cylinders is $\frac{V_1}{V_2} = \frac{\pi r_1^2 h_1}{\pi r_2^2 h_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right)$.
Substituting the given values: $\frac{V_1}{V_2} = \left(\frac{3}{2}\right)^2 \times \left(\frac{8}{11}\right) = \frac{9}{4} \times \frac{8}{11}$.
Calculating the result: $\frac{9 \times 8}{4 \times 11} = \frac{72}{44} = \frac{18}{11}$.
Thus,the ratio of their volumes is $18:11$.

(D) The total surface area of a cone is given by the formula: $TSA = \pi r(r + l)$,where $r$ is the radius and $l$ is the slant height.
Given: $r = 5 \, cm$ and $l = 9 \, cm$.
Substituting the values into the formula:
$TSA = \pi \times 5 \times (5 + 9)$
$TSA = \pi \times 5 \times 14$
$TSA = 70 \pi \, cm^2$.
Using $\pi \approx \frac{22}{7}$:
$TSA = 70 \times \frac{22}{7} = 10 \times 22 = 220 \, cm^2$.
Therefore,the total surface area is $220 \, cm^2$.

(A) Given,the diameter of the solid hemisphere $d = 21 \, cm$.
Therefore,the radius $r = \frac{d}{2} = \frac{21}{2} \, cm = 10.5 \, cm$.
The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^3$.
Substituting the values,we get $V = \frac{2}{3} \times \frac{22}{7} \times (10.5)^3$.
$V = \frac{2}{3} \times \frac{22}{7} \times 10.5 \times 10.5 \times 10.5$.
$V = \frac{44}{21} \times 1157.625$.
$V = 2425.5 \, cm^3$.

(B) Let the edge length of the cube be $x \ cm$.
The surface area of a cube is given by the formula $S = 6x^2 \ cm^2$.
The volume of a cube is given by the formula $V = x^3 \ cm^3$.
According to the problem,the numerical values of the surface area and volume are equal:
$6x^2 = x^3$
Dividing both sides by $x^2$ (since $x \neq 0$):
$6 = x$
Therefore,the edge length of the cube is $6 \ cm$.

(C) The volume $V$ 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.
Given: $V = 616 \, cm^3$ and $h = 12 \, cm$.
Using $\pi \approx \frac{22}{7}$,we have:
$616 = \frac{1}{3} \times \frac{22}{7} \times r^2 \times 12$
$616 = 4 \times \frac{22}{7} \times r^2$
$616 = \frac{88}{7} \times r^2$
$r^2 = \frac{616 \times 7}{88}$
$r^2 = 7 \times 7 = 49$
$r = 7 \, cm$.
The diameter $d$ is $2r = 2 \times 7 = 14 \, cm$.

(D) Let the diameters of the two spheres be $d_1$ and $d_2$. Given that $d_1 : d_2 = 3 : 5$.
Since the radius $r = d/2$,the ratio of their radii $r_1 : r_2$ is also $3 : 5$.
The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.
The ratio of the volumes of the two spheres is $\frac{V_1}{V_2} = \frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3} = \left( \frac{r_1}{r_2} \right)^3$.
Substituting the given ratio,we get $\left( \frac{3}{5} \right)^3 = \frac{27}{125}$.
Therefore,the ratio of their volumes is $27 : 125$.

(A) cube has $6$ identical square faces.
If the length of each edge of the cube is $a$,then the area of one face is $a \times a = a^{2}$.
Since there are $6$ such faces,the total surface area of the cube is $6 \times a^{2} = 6 a^{2}$.

(B) cuboid has $6$ rectangular faces.
Let the length be $l$,breadth be $b$,and height be $h$.
The area of the opposite faces are $(l \times b)$,$(b \times h)$,and $(h \times l)$.
Since there are two of each face,the total surface area is $2(lb + bh + hl)$.

(C) The total surface area of a cone consists of the area of the circular base and the curved surface area of the cone.
$1$. The area of the circular base is $\pi r^2$.
$2$. The curved surface area of the cone is $\pi r l$,where $r$ is the radius and $l$ is the slant height.
$3$. Total Surface Area = $\text{Base Area} + \text{Curved Surface Area} = \pi r^2 + \pi r l$.
$4$. Factoring out $\pi r$,we get $\pi r(r + l)$ or $\pi r(l + r)$.

(D) cube has $6$ faces in total. The lateral surface area refers to the area of the four side faces,excluding the top and bottom faces. If the side length of the cube is $a$,the area of one face is $a^{2}$. Therefore,the lateral surface area is $4 \times a^{2} = 4 a^{2}$.

(A) The total surface area of a cylinder consists of the area of the two circular bases and the curved surface area.
Area of two circular bases $= 2 \times (\pi r^2) = 2 \pi r^2$.
Curved surface area of the cylinder $= 2 \pi rh$.
Total surface area $= 2 \pi r^2 + 2 \pi rh$.
Factoring out $2 \pi r$,we get $2 \pi r(r + h)$ or $2 \pi r(h + r)$.

(B) hemisphere is half of a sphere. The total surface area of a sphere is $4 \pi r^{2}$.
Therefore,the curved surface area of a hemisphere is half of the surface area of a sphere.
Curved surface area of a hemisphere $= \frac{1}{2} \times 4 \pi r^{2} = 2 \pi r^{2}$.

(C) The surface area of a sphere with radius $r$ is given by the formula $A = 4 \pi r^{2}$.
Thus,the correct option is $C$.

(D) The lateral surface area of a cuboid is defined as the sum of the areas of its four vertical faces (excluding the top and bottom faces).
For a cuboid with length $l$,breadth $b$,and height $h$,the area of the four vertical faces is given by the perimeter of the base multiplied by the height.
Lateral Surface Area $= (2l + 2b) \times h = 2h(l + b)$.

(A) The curved surface area of a cone is given by the formula $\pi r l$,where $r$ is the radius of the base and $l$ is the slant height of the cone.

(B) The curved surface area of a cylinder is calculated by multiplying the circumference of its base $(2 \pi r)$ by its height $(h)$.
Therefore,the formula for the curved surface area of a cylinder is $2 \pi r h$.

(C) hemisphere is half of a sphere. The curved surface area of a hemisphere is $2 \pi r^{2}$.
The base of the hemisphere is a circle with an area of $\pi r^{2}$.
Therefore,the total surface area of a hemisphere is the sum of the curved surface area and the base area:
$\text{Total Surface Area} = 2 \pi r^{2} + \pi r^{2} = 3 \pi r^{2}$.

(D) We know that $1 \, m = 100 \, cm$.
Therefore,$1 \, m^{2} = 1 \, m \times 1 \, m$.
Substituting the value of $1 \, m$ in centimeters,we get:
$1 \, m^{2} = (100 \, cm) \times (100 \, cm) = 10000 \, cm^{2}$.
Thus,$1 \, m^{2} = 10000 \, cm^{2}$.

(A) By definition,$1$ hectare is the area of a square with a side length of $100$ meters.
Therefore,$1$ hectare $= 100 \text{ } m \times 100 \text{ } m = 10000 \text{ } m^{2}$.

(B) We know that $1\, m = 100\, cm$.
Therefore,$1\, m^{3} = (1\, m) \times (1\, m) \times (1\, m)$.
Substituting the value of $1\, m$ in centimeters: $1\, m^{3} = (100\, cm) \times (100\, cm) \times (100\, cm)$.
$1\, m^{3} = 100 \times 100 \times 100\, cm^{3}$.
$1\, m^{3} = 1000000\, cm^{3}$.

(C) In a right circular cone,the slant height $(l)$,the radius of the base $(r)$,and the vertical height $(h)$ form a right-angled triangle.
According to the Pythagorean theorem,the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Therefore,$l^{2} = r^{2} + h^{2}$.

(D) The formula for the total surface area of a cube is $6a^2$,where $a$ is the length of the edge.
Given,$a = 7\, cm$.
Total Surface Area $= 6 \times (7)^2$
$= 6 \times 49$
$= 294\, cm^2$.

(A) The volume of a cube is given by the formula $V = a^3$,where $a$ is the length of the edge of the cube.
Given,the edge of the cube $a = 12 \, cm$.
Therefore,the volume $V = (12 \, cm)^3 = 12 \times 12 \times 12 \, cm^3$.
$V = 144 \times 12 \, cm^3 = 1728 \, cm^3$.
Thus,the volume of the cube is $1728 \, cm^3$.

(B) The dimensions of the cuboidal box are length $(l)$ = $25 \, cm$,breadth $(b)$ = $15 \, cm$,and height $(h)$ = $8 \, cm$.
The formula for the total surface area of a cuboid is $2(lb + bh + lh)$.
Substituting the given values:
Total Surface Area = $2(25 \times 15 + 15 \times 8 + 25 \times 8)$
Total Surface Area = $2(375 + 120 + 200)$
Total Surface Area = $2(695)$
Total Surface Area = $1390 \, cm^2$.

(C) Given: Diameter of the sphere $(d)$ = $14 \, cm$.
Radius of the sphere $(r)$ = $d / 2 = 14 / 2 = 7 \, cm$.
The formula for the surface area of a sphere is $A = 4 \pi r^2$.
Substituting the values: $A = 4 \times (22 / 7) \times (7)^2$.
$A = 4 \times (22 / 7) \times 49$.
$A = 4 \times 22 \times 7$.
$A = 88 \times 7 = 616 \, cm^2$.

(D) The formula for the volume of a cone is $V = \frac{1}{3} \pi r^2 h$.
Given:
Radius $(r)$ = $7 \, cm$
Height $(h)$ = $30 \, cm$
Using $\pi = \frac{22}{7}$:
$V = \frac{1}{3} \times \frac{22}{7} \times (7)^2 \times 30$
$V = \frac{1}{3} \times \frac{22}{7} \times 49 \times 30$
$V = 22 \times 7 \times 10$
$V = 1540 \, cm^3$.