Mix Examples - Surface Areas and Volumes Questions in English

(A) Given: $R = 28 \, cm$,$r = 7 \, cm$,$h = 72 \, cm$.
First,calculate the slant height $l = \sqrt{h^2 + (R - r)^2} = \sqrt{72^2 + (28 - 7)^2} = \sqrt{5184 + 441} = \sqrt{5625} = 75 \, cm$.
Curved Surface Area $(CSA)$ $= \pi(R + r)l = \frac{22}{7} \times (28 + 7) \times 75 = 22 \times 5 \times 75 = 8250 \, cm^2$.
Total Surface Area $(TSA)$ $= \text{CSA} + \pi R^2 + \pi r^2 = 8250 + \frac{22}{7} \times (28^2 + 7^2) = 8250 + \frac{22}{7} \times (784 + 49) = 8250 + 22 \times 119 = 8250 + 2618 = 10868 \, cm^2$.
Volume $= \frac{1}{3} \pi h (R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 72 \times (28^2 + 7^2 + 28 \times 7) = \frac{22 \times 24}{7} \times (784 + 49 + 196) = \frac{528}{7} \times 1029 = 528 \times 147 = 77616 \, cm^3$.

(D) The $CSA$ (Curved Surface Area) of a cylinder is the area of the curved surface excluding the two circular bases.
For a cylinder with radius $r$ and height $h$,the $CSA$ is calculated by multiplying the circumference of the base $(2 \pi r)$ by the height $(h)$.
Therefore,$CSA = 2 \pi r h$.

(A) hemisphere is half of a sphere.
Since the total surface area of a sphere is $4 \pi r^{2}$,the curved surface area $(CSA)$ of a hemisphere is half of that value.
$CSA \text{ of a hemisphere} = \frac{1}{2} \times (4 \pi r^{2}) = 2 \pi r^{2}$.

(B) The volume of a cone with base radius $r$ and height $h$ is given by the formula $V = \frac{1}{3} \pi r^{2} h$.

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

(D) The curved surface area $(CSA)$ of a cone with radius $r$ and slant height $l$ is given by the formula $\pi r l$.
Here,$r$ represents the radius of the circular base and $l$ represents the slant height of the cone.

(A) sphere is a three-dimensional object where every point on the surface is at an equal distance $r$ from the center. The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^{3}$.
$A$ hemisphere is exactly half of a sphere. Therefore,the volume of a hemisphere is half the volume of a sphere.
Volume of a hemisphere $= \frac{1}{2} \times (\text{Volume of a sphere})$
Volume of a hemisphere $= \frac{1}{2} \times \frac{4}{3} \pi r^{3} = \frac{2}{3} \pi r^{3}$.

(B) hemisphere is half of a sphere.
The curved surface area $(CSA)$ of a hemisphere is $2 \pi r^{2}$.
The base of a hemisphere is a circle with an area of $\pi r^{2}$.
The total surface area $(TSA)$ is the sum of the curved surface area and the base area.
$TSA = CSA + \text{Base Area} = 2 \pi r^{2} + \pi r^{2} = 3 \pi r^{2}$.

(C) The volume $V$ of a sphere with radius $r$ is given by the formula:
$V = \frac{4}{3} \pi r^{3}$
Therefore,the correct option is $C$.

(D) The Total Surface Area $(TSA)$ of a cylinder consists of the curved surface area plus the area of the two circular bases.
Curved Surface Area $(CSA)$ $= 2 \pi r h$
Area of two circular bases $= 2 \times (\pi r^{2}) = 2 \pi r^{2}$
Total Surface Area $(TSA)$ $= 2 \pi r h + 2 \pi r^{2}$
Factoring out $2 \pi r$,we get $TSA = 2 \pi r(r+h)$.

(A) The volume of a cylinder with base radius $r$ and height $h$ is calculated by multiplying the area of the circular base by the height of the cylinder.
Area of the base $= \pi r^{2}$.
Therefore,Volume of the cylinder $= \text{Area of base} \times \text{height} = \pi r^{2} h$.

(B) The total surface area of a cone is the sum of its curved surface area and the area of its circular base.
Curved Surface Area $= \pi rl$
Base Area $= \pi r^{2}$
Total Surface Area $= \pi rl + \pi r^{2} = \pi r(l + r)$
Where $r$ is the radius of the base and $l$ is the slant height of the cone.

(C) The diameter of the sphere is $d = 10 \, cm$.
Therefore,the radius $r$ is $r = \frac{d}{2} = \frac{10}{2} = 5 \, cm$.
The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Substituting the value of $r$ into the formula:
$V = \frac{4}{3} \pi (5)^3$
$V = \frac{4}{3} \pi (125)$
$V = \frac{500}{3} \pi \, cm^3$.

(B) $5$-rupee coin is shaped like a cylinder with a very small height $h$.
The total surface area of a cylinder is given by the sum of the areas of its two circular bases and its curved surface area.
Total Surface Area $= 2 \times (\text{Area of base}) + \text{Curved Surface Area}$.
Total Surface Area $= 2(\pi r^2) + 2 \pi r h$.
Factoring out $2 \pi r$,we get $2 \pi r(r + h)$.
Thus,the correct formula is $2 \pi r(r + h)$.

(A) The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Given the radius $r = 1.5 \text{ cm} = \frac{3}{2} \text{ cm}$.
Substituting the value of $r$ into the formula:
$V = \frac{4}{3} \pi \left( \frac{3}{2} \right)^3$
$V = \frac{4}{3} \pi \left( \frac{27}{8} \right)$
$V = \left( \frac{4}{3} \times \frac{27}{8} \right) \pi$
$V = \left( \frac{1}{1} \times \frac{9}{2} \right) \pi$
$V = 4.5 \pi \text{ cm}^3$.

(B) Let the radii of the two spheres be $r_1$ and $r_2$. Given that the ratio of their radii is $r_1 : r_2 = 3 : 5$.
The surface area of a sphere is given by the formula $S = 4\pi r^2$.
The ratio of the surface areas of the two spheres is $\frac{S_1}{S_2} = \frac{4\pi r_1^2}{4\pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2$.
Substituting the given ratio: $\frac{S_1}{S_2} = \left(\frac{3}{5}\right)^2 = \frac{9}{25}$.
Thus,the ratio of their surface areas is $9: 25$.

(C) Given: Radii of the frustum are $R = 10 \, cm$ and $r = 7 \, cm$. The height $h = 4 \, cm$.
First,we calculate the slant height $l$ of the frustum using the formula $l = \sqrt{h^2 + (R - r)^2}$.
Substituting the values: $l = \sqrt{4^2 + (10 - 7)^2} = \sqrt{16 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, cm$.
The curved surface area of a frustum of a cone is given by the formula $CSA = \pi (R + r) l$.
Substituting the values: $CSA = \pi (10 + 7) \times 5 = \pi (17) \times 5 = 85\pi \, cm^2$.
Thus,the curved surface area is $85\pi \, cm^2$.

(D) The formula for the volume of a frustum of a cone is $V = \frac{1}{3} \pi h (r_1^2 + r_2^2 + r_1 r_2)$.
Given: $r_1 = 6 \, cm$,$r_2 = 4 \, cm$,and $h = 9 \, cm$.
Substituting the values into the formula:
$V = \frac{1}{3} \pi (9) (6^2 + 4^2 + 6 \times 4)$
$V = 3 \pi (36 + 16 + 24)$
$V = 3 \pi (76)$
$V = 228 \pi \, cm^3$.
Thus,the volume is $228 \pi \, cm^3$.

(A) The volume of a cylinder is given by the formula $V = \pi r^2 h$,where $V$ is the volume,$r$ is the radius,and $h$ is the height.
Given: $V = 550 \, cm^3$,$r = 5 \, cm$,and $\pi \approx \frac{22}{7}$.
Substituting the values into the formula: $550 = \frac{22}{7} \times (5)^2 \times h$.
$550 = \frac{22}{7} \times 25 \times h$.
$550 = \frac{550}{7} \times h$.
$h = 550 \times \frac{7}{550}$.
$h = 7 \, cm$.

(B) The formula for the volume of a cone is $V = \frac{1}{3} \pi r^{2} h$.
Given that the radius $r = 5 \, cm$ and the height $h = 12 \, cm$.
Substituting these values into the formula:
$V = \frac{1}{3} \times \pi \times (5)^{2} \times 12$
$V = \frac{1}{3} \times \pi \times 25 \times 12$
$V = \pi \times 25 \times 4$
$V = 100 \pi \, cm^{3}$.
Therefore,the volume is $100 \, \pi \, cm^{3}$.

(C) Let the diameters of the two cones be $d_1$ and $d_2$,and their heights be $h_1$ and $h_2$.
The ratio of diameters is $d_1:d_2 = 2:3$,so the ratio of their radii is $r_1:r_2 = 2:3$.
The ratio of their heights is $h_1:h_2 = 9:4$.
The volume of a cone is given by $V = \frac{1}{3}\pi r^2 h$.
The ratio of their volumes is $\frac{V_1}{V_2} = \frac{\frac{1}{3}\pi r_1^2 h_1}{\frac{1}{3}\pi r_2^2 h_2} = \left(\frac{r_1}{r_2}\right)^2 \times \left(\frac{h_1}{h_2}\right)$.
Substituting the given ratios: $\frac{V_1}{V_2} = \left(\frac{2}{3}\right)^2 \times \left(\frac{9}{4}\right) = \frac{4}{9} \times \frac{9}{4} = 1$.
Thus,the ratio of their volumes is $1:1$.

(D) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$,where $r$ is the radius.
Let the original radius be $r_1 = r$.
Then the original volume is $V_1 = \frac{4}{3} \pi r^3$.
If the radius is doubled,the new radius becomes $r_2 = 2r$.
The new volume is $V_2 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = 8 \times (\frac{4}{3} \pi r^3)$.
Therefore,$V_2 = 8 \times V_1$.
Thus,the volume becomes $8$ times the volume of the original sphere.

(A) The formula for the curved surface area $(CSA)$ of a cone is given by $CSA = \pi r l$,where $r$ is the radius and $l$ is the slant height.
Given: $r = 12 \, cm$ and $l = 20 \, cm$.
Substituting the values into the formula:
$CSA = \pi \times 12 \times 20$
$CSA = 240 \pi \, cm^2$.
Therefore,the curved surface area is $240 \pi \, cm^2$.

(B) The formula for the volume of a cylinder is $V = \pi r^{2} h$.
Given that the radius $r = 5 \, cm$ and the height $h = 5 \, cm$.
Substituting these values into the formula:
$V = \pi \times (5)^{2} \times 5$
$V = \pi \times 25 \times 5$
$V = 125\pi \, cm^{3}$.
Therefore,the volume is $125\pi \, cm^{3}$.

(C) The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^{3}$.
Given the radius $r = 3 \, cm$.
Substituting the value of $r$ into the formula:
$V = \frac{2}{3} \times \pi \times (3)^{3}$
$V = \frac{2}{3} \times \pi \times 27$
$V = 2 \times \pi \times 9$
$V = 18 \pi \, cm^{3}$.
Therefore,the volume is $18 \pi \, cm^{3}$.

(D) Let the diameters of the two cylinders be $d_1$ and $d_2$,and their heights be $h_1$ and $h_2$.
Given,$d_1:d_2 = 3:4$,so the radii $r_1:r_2 = 3:4$.
Given,$h_1:h_2 = 4:5$.
The volume of a cylinder is given by $V = \pi r^2 h$.
The ratio of the volumes 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 ratios: $\frac{V_1}{V_2} = \left(\frac{3}{4}\right)^2 \times \left(\frac{4}{5}\right) = \frac{9}{16} \times \frac{4}{5} = \frac{9 \times 1}{4 \times 5} = \frac{9}{20}$.
Thus,the ratio of their volumes is $9:20$.

(C) cube has $6$ faces in total. Each face of a cube with side length $l$ has an area of $l^2$.
If the cube is open at one end,it means one face is missing.
Therefore,the total surface area of the cube open at one end is the sum of the areas of the remaining $5$ faces.
Total surface area $= 5 \times l^2 = 5l^2$.

(A) cube is a three-dimensional shape having $6$ identical square faces.
If the side length of the cube is $l$,then the area of one square face is $l \times l = l^{2}$.
Since there are $6$ such faces in a closed cube,the total surface area is the sum of the areas of all $6$ faces.
Therefore,Total Surface Area $= 6 \times l^{2} = 6 l^{2}$.

(A) cylinder has two circular bases.
Since the base of a cylinder is a circle with radius $r$,the area of the base is given by the formula for the area of a circle.
Area of a circle $= \pi r^{2}$.
Therefore,the area of the base of a cylinder is $\pi r^{2}$.

(A) The curved surface area of a cylinder is the area of the side surface excluding the top and bottom circular bases.
For a cylinder with radius $r$ and height $h$,the curved surface area is calculated by multiplying the circumference of the base $(2 \pi r)$ by the height $(h)$.
Therefore,the formula is $2 \pi r h$.

(B) 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 $= 2 \pi r h$.
Total surface area $= 2 \pi r^2 + 2 \pi r h$.
Factoring out $2 \pi r$,we get $2 \pi r(r + h)$.

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

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

(B) The surface area of a sphere with radius $r$ is given by the formula $A = 4 \pi r^{2}$.
This represents the total area covered by the outer boundary of the sphere.

(C) sphere has a total surface area of $4 \pi r^{2}$.
Since a hemisphere is exactly half of a sphere,its curved surface area is half of the total surface area of the sphere.
Curved surface area of a hemisphere $= \frac{1}{2} \times (4 \pi r^{2}) = 2 \pi r^{2}$.

(C) In a right circular cone,the slant height $(l)$,the vertical height $(h)$,and the radius of the base $(r)$ 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,$l^{2} = h^{2} + r^{2}$.
Taking the square root on both sides,we get $l = \sqrt{h^{2} + r^{2}}$.

(C) hemisphere is half of a sphere. The curved surface area of a sphere is $4\pi r^{2}$,so the curved surface area of a hemisphere is $2\pi r^{2}$.
Additionally,the base of a 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: $2\pi r^{2} + \pi r^{2} = 3\pi r^{2}$.

(D) cone is a three-dimensional geometric shape that tapers smoothly from a flat base to a point called the apex or vertex.
By definition,the base of a right circular cone is a circle.
Therefore,the correct option is $D$.

(B) The volume $V$ of a sphere with radius $r$ is given by the formula $V = \frac{4}{3} \pi r^{3}$.
Therefore,the correct option is $B$.

(A) sphere has a volume of $V = \frac{4}{3} \pi r^{3}$.
Since a hemisphere is exactly half of a sphere,its volume is half of the volume of the sphere.
Therefore,the volume of a hemisphere $= \frac{1}{2} \times (\frac{4}{3} \pi r^{3}) = \frac{2}{3} \pi r^{3}$.

(B) $1-$ rupee coin is cylindrical in shape.
The volume of a cylinder is given by the formula $V = \pi r^{2} h$,where $r$ is the radius of the base and $h$ is the thickness (height) of the coin.
Therefore,the volume of a $1-$ rupee coin is $V = \pi r^{2} h$.

(B) The Total Surface Area $(TSA)$ of a cone is given by the formula: $TSA = \pi r(r + l) = \pi r^2 + \pi rl$.
The Curved Surface Area $(CSA)$ of a cone is given by the formula: $CSA = \pi rl$.
To find how much the $TSA$ exceeds the $CSA$,we subtract the $CSA$ from the $TSA$:
Difference $= TSA - CSA$
Difference $= (\pi r^2 + \pi rl) - (\pi rl)$
Difference $= \pi r^2 + \pi rl - \pi rl$
Difference $= \pi r^2$.
Thus,the total surface area of a cone exceeds its curved surface area by $\pi r^2$.

(B) $1-$ rupee coin is shaped like a thin cylinder.
For a cylinder with radius $r$ and height $h$,the curved surface area $(CSA)$ is given by the formula $A = 2 \pi r h$.
Here,$h$ represents the thickness of the coin.

(B) frustum of a cone is formed when a cone is cut by a plane parallel to its base.
Let $h$ be the height of the frustum,and $r_{1}$ and $r_{2}$ be the radii of the two circular bases.
The formula for the volume $V$ of a frustum of a cone is given by:
$V = \frac{1}{3} \pi h(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})$
Therefore,option $B$ is the correct formula.

(A) frustum of a cone is formed when a cone is cut by a plane parallel to its base.
Let $r_{1}$ and $r_{2}$ be the radii of the two circular ends of the frustum,and $l$ be the slant height of the frustum.
The formula for the Curved Surface Area $(CSA)$ of a frustum of a cone is derived from the difference between the lateral surface area of the original cone and the smaller cone removed from the top.
The formula is given by: $CSA = \pi l(r_{1} + r_{2})$.

(A) The total surface area $(TSA)$ of a frustum of a cone consists of the lateral surface area plus the areas of the two circular bases.
$1$. The lateral surface area of a frustum is given by $\pi l(r_{1} + r_{2})$,where $l$ is the slant height and $r_{1}, r_{2}$ are the radii of the two circular bases.
$2$. The area of the top circular base is $\pi r_{1}^{2}$.
$3$. The area of the bottom circular base is $\pi r_{2}^{2}$.
$4$. Therefore,the total surface area $(TSA)$ is the sum of these three parts: $TSA = \pi l(r_{1} + r_{2}) + \pi r_{1}^{2} + \pi r_{2}^{2}$.

(B) $5$-rupee coin is shaped like a thin cylinder.
The total surface area $(TSA)$ of a cylinder with radius $r$ and height $h$ is the sum of its curved surface area and the areas of its two circular bases.
$TSA = \text{Curved Surface Area} + 2 \times \text{Area of circular base}$
$TSA = 2 \pi r h + 2 \pi r^2$
By taking $2 \pi r$ as a common factor,we get:
$TSA = 2 \pi r(r + h)$
Therefore,the correct formula is $A = 2 \pi r(r + h)$.

(C) 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:
$1 \text{ litre} = (10 \text{ cm})^3 = 10 \times 10 \times 10 \text{ cm}^3 = 1000 \text{ cm}^3$.
Therefore,$1 \text{ litre} = 1000 \text{ cm}^3$.

(C) We know that $1 \, m = 100 \, cm$.
Therefore,$1 \, m^3 = (100 \, cm) \times (100 \, cm) \times (100 \, cm) = 1,000,000 \, cm^3$.
We also know that $1 \, \text{liter} = 1000 \, cm^3$.
To convert $cm^3$ to liters,we divide by $1000$.
Thus,$1,000,000 \, cm^3 / 1000 = 1000 \, \text{liters}$.
So,$1 \, m^3 = 1000 \, \text{liters}$.

(C) We know that $1 \ m^{3} = 1000 \ \text{litres}$.
Also,$1 \ \text{kilolitre} = 1000 \ \text{litres}$.
Therefore,$1 \ m^{3} = 1 \ \text{kilolitre}$.