Volume of a Combination of Solids Questions in English

(D) The formula for the volume of a cylinder is $V = \pi r^{2} h$.
Given that the volume $V = 550 \, cm^{3}$ and the radius $r = 5 \, cm$.
Substituting the values into the formula:
$550 = \frac{22}{7} \times (5)^{2} \times h$
$550 = \frac{22}{7} \times 25 \times h$
$h = \frac{550 \times 7}{22 \times 25}$
$h = \frac{3850}{550}$
$h = 7 \, cm$.
Therefore,the height of the cylinder is $7 \, cm$.

(B) The formula for the volume of a cone is $V = \frac{1}{3} \pi r^{2} h$.
Given radius $r = 5 \, cm$ and height $h = 21 \, cm$.
Substituting the values into the formula:
$V = \frac{1}{3} \times \frac{22}{7} \times (5)^{2} \times 21$
$V = \frac{1}{3} \times \frac{22}{7} \times 25 \times 21$
$V = 22 \times 25 \times \frac{21}{21}$
$V = 22 \times 25 = 550 \, cm^{3}$.

(B) The volume of a cone is given by the formula: $V = \frac{1}{3} \times \text{Area of base} \times \text{height}$.
Given,Area of base $= 60 \, cm^{2}$ and height $= 15 \, cm$.
Substituting these values into the formula:
$V = \frac{1}{3} \times 60 \times 15$
$V = 20 \times 15 = 300 \, cm^{3}$.
Therefore,the volume of the cone is $300 \, cm^{3}$.

(A) Let the radius of both the cone and the cylinder be $r$ and their height be $h$.
The volume of a cone is given by $V_{cone} = \frac{1}{3} \pi r^2 h$.
The volume of a cylinder is given by $V_{cylinder} = \pi r^2 h$.
The ratio of the volume of the cone to the volume of the cylinder is $\frac{V_{cone}}{V_{cylinder}} = \frac{\frac{1}{3} \pi r^2 h}{\pi r^2 h} = \frac{1}{3}$.
Therefore,the ratio is $1: 3$.

(C) The volume of a cylinder is given by the formula $V_{cylinder} = \pi r^{2} h$.
The volume of a cone is given by the formula $V_{cone} = \frac{1}{3} \pi r^{2} h$.
Given that the radii $(r)$ and heights $(h)$ are equal for both shapes,we can compare their volumes:
$V_{cylinder} = \pi r^{2} h$
Since $V_{cone} = \frac{1}{3} \pi r^{2} h$,we can write $\pi r^{2} h = 3 \times (\frac{1}{3} \pi r^{2} h)$.
Therefore,$V_{cylinder} = 3 \times V_{cone}$.

(C) The formula for the volume of a cone is $V = \frac{1}{3} \pi r^{2} h$.
Given that the radius $r = 3 \, cm$ and the height $h = 3 \, cm$.
Substituting these values into the formula:
$V = \frac{1}{3} \times \pi \times (3)^{2} \times 3$
$V = \frac{1}{3} \times \pi \times 9 \times 3$
$V = 9 \pi \, cm^{3}$.

(B) The volume $V$ of a sphere with radius $r$ is given by the formula $V = \frac{4}{3} \pi r^3$.
Let the radii of the two spheres be $r_1$ and $r_2$. Given that $\frac{r_1}{r_2} = \frac{2}{3}$.
The ratio of their volumes 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: $\left(\frac{2}{3}\right)^3 = \frac{2^3}{3^3} = \frac{8}{27}$.
Thus,the ratio of their volumes is $8:27$.

(B) The volume of a cylinder is given by the formula $V = \pi r^2 h$.
Given radius $r = 1.4 \, m$ and height $h = 4 \, m$.
Substituting the values,$V = \frac{22}{7} \times (1.4)^2 \times 4$.
$V = \frac{22}{7} \times 1.96 \times 4 = 22 \times 0.28 \times 4 = 24.64 \, m^3$.
Since $1 \, m^3 = 1000 \, \text{litres}$,the capacity in litres is $24.64 \times 1000 = 24640 \, \text{litres}$.

(B) The formula for the volume of a cone is $V = \frac{1}{3} \pi r^{2} h$.
Given radius $r = 6 \, cm$ and height $h = 21 \, cm$.
Substituting the values,we get $V = \frac{1}{3} \times \frac{22}{7} \times (6)^{2} \times 21$.
$V = \frac{1}{3} \times \frac{22}{7} \times 36 \times 21$.
$V = 22 \times 36 = 792 \, cm^{3}$.

(C) The formula for the volume of a cone is $V = \frac{1}{3} \pi r^{2} h$.
Given: $V = 9504\,cm^{3}$,$r = 18\,cm$,and $\pi = \frac{22}{7}$.
Substituting the values into the formula:
$9504 = \frac{1}{3} \times \frac{22}{7} \times (18)^{2} \times h$
$9504 = \frac{1}{3} \times \frac{22}{7} \times 324 \times h$
$9504 = 22 \times 108 \times \frac{h}{7}$
$9504 = 2376 \times \frac{h}{7}$
$h = \frac{9504 \times 7}{2376}$
$h = 4 \times 7 = 28\,cm$.
Thus,the height of the cone is $28\,cm$.

(B) The volume of a frustum of a cone is given by the formula: $V = \frac{1}{3} \pi h (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2})$.
Given: $r_{1} = 3\,cm$,$r_{2} = 2\,cm$,and $h = 21\,cm$.
Substituting the values into the formula:
$V = \frac{1}{3} \times \frac{22}{7} \times 21 \times (3^{2} + 2^{2} + 3 \times 2)$
$V = 22 \times (9 + 4 + 6)$
$V = 22 \times 19$
$V = 418\,cm^{3}$.
Thus,the capacity of the glass is $418\,cm^{3}$.

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

(C) The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^{3}$.
Given the radius $r = 1.5\,cm$.
Substituting the value of $r$ into the formula:
$V = \frac{4}{3} \times \pi \times (1.5)^{3}$
$V = \frac{4}{3} \times \pi \times 1.5 \times 1.5 \times 1.5$
$V = 4 \times \pi \times 0.5 \times 1.5 \times 1.5$
$V = 2 \times \pi \times 2.25$
$V = 4.5 \pi\,cm^{3}$.

(A) The volume $V$ of a sphere with radius $r$ is given by the formula $V = \frac{4}{3} \pi r^3$.
If the radius is doubled,the new radius $r' = 2r$.
The new volume $V'$ is given by $V' = \frac{4}{3} \pi (r')^3$.
Substituting $r' = 2r$ into the formula,we get $V' = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi (8r^3) = 8 \times (\frac{4}{3} \pi r^3)$.
Therefore,$V' = 8V$.
Thus,the volume becomes $8$ times the original volume.

(A) The volume of a cone with radius $r$ and height $h$ is given by the formula $V = \frac{1}{3} \pi r^{2} h$.
This formula represents one-third of the volume of a cylinder with the same base radius and height.

(D) Let the radii of the two spheres be $r_1$ and $r_2$. Given that $r_1 : r_2 = 3 : 4$.
The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Therefore,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 $\frac{V_1}{V_2} = \left( \frac{3}{4} \right)^3 = \frac{3^3}{4^3} = \frac{27}{64}$.
Thus,the ratio of their volumes is $27:64$.

(C) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Given that $V = 4.5 \pi \, cm^3$,we have:
$4.5 \pi = \frac{4}{3} \pi r^3$
Dividing both sides by $\pi$:
$4.5 = \frac{4}{3} r^3$
$r^3 = 4.5 \times \frac{3}{4} = \frac{45}{10} \times \frac{3}{4} = \frac{9}{2} \times \frac{3}{4} = \frac{27}{8}$
Taking the cube root of both sides:
$r = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} = 1.5 \, cm$
The diameter $d$ is $2r$:
$d = 2 \times 1.5 = 3 \, cm$.

(B) Volume is defined as the amount of space occupied by a three-dimensional object.
In the International System of Units $(SI)$,the base unit of length is the meter $(m)$.
Since volume is calculated as the product of three dimensions (length $\times$ width $\times$ height),its unit is the cube of the unit of length.
Therefore,the unit of volume is $m \times m \times m = m^3$.
- $m$ is a unit of length.
- $m^2$ is a unit of area.
- $cm^2$ is also a unit of area.
Thus,the correct option is $B$.

(B) The volume of a cubical tank is given by the formula $V = \text{side}^3$.
Given the side length is $1\, m$,the volume $V = (1\, m)^3 = 1\, m^3$.
We know that $1\, m^3 = 1000\, \text{litres}$.
Therefore,the maximum amount of water that can be stored in the tank is $1000\, \text{litres}$.

(D) The volume $V$ of a cubical tank is given by the formula $V = \text{length} \times \text{breadth} \times \text{height}$.
Given dimensions are $2 \,m \times 2 \,m \times 2 \,m = 8 \,m^{3}$.
To convert $m^{3}$ to $cm^{3}$, we know that $1 \,m = 100 \,cm$, so $1 \,m^{3} = (100 \,cm)^{3} = 1,000,000 \,cm^{3} = 10^{6} \,cm^{3}$.
Thus, $8 \,m^{3} = 8 \times 10^{6} \,cm^{3}$. (Option $A$ is correct).
We also know that $1 \,m^{3} = 1000 \text{ litres}$.
Therefore, $8 \,m^{3} = 8000 \text{ litres}$. (Option $C$ is correct).
Since $1000 \text{ litres} = 1 \text{ kilolitre}$, $8000 \text{ litres} = 8 \text{ kilolitres}$. (Option $B$ is correct).
Since all options $A, B,$ and $C$ are correct, the correct choice is $D$.

(D) We know that $1\,m = 100\,cm = 10^2\,cm$.
To find the value of $1\,m^3$ in $cm^3$,we cube both sides of the equation:
$1\,m^3 = (10^2\,cm)^3$
$1\,m^3 = (10^2)^3\,cm^3$
Using the power rule $(a^m)^n = a^{m \times n}$,we get:
$1\,m^3 = 10^{2 \times 3}\,cm^3 = 10^6\,cm^3$.
Therefore,$1\,m^3 = 10^6\,cm^3$.

(B) The diameter of the hemisphere is $d = 1 \, cm$.
Therefore,the radius $r$ is $r = \frac{d}{2} = \frac{1}{2} \, cm$.
The formula for the volume of a hemisphere is $V = \frac{2}{3} \pi r^3$.
Substituting the value of $r$ into the formula:
$V = \frac{2}{3} \pi \left( \frac{1}{2} \right)^3$
$V = \frac{2}{3} \pi \left( \frac{1}{8} \right)$
$V = \frac{2 \pi}{24} = \frac{\pi}{12} \, cm^3$.

(D) To match the given parts,we identify the standard volume formulas:
$1$. The volume of a cuboid with length $l$,breadth $b$,and height $h$ is given by $V = lbh$. Thus,$1$ matches with $d$.
$2$. The volume of a cone with radius $r$ and height $h$ is given by $V = \frac{1}{3} \pi r^{2} h$. Thus,$2$ matches with $a$.
$3$. The volume of a cylinder with radius $r$ and height $h$ is given by $V = \pi r^{2} h$. Thus,$3$ matches with $b$.
$4$. The volume of a sphere with radius $r$ is given by $V = \frac{4}{3} \pi r^{3}$. Thus,$4$ matches with $c$.
Therefore,the correct matching is $(1-d), (2-a), (3-b), (4-c)$.