Volume of a Sphere Questions in English

(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:
$V = \frac{4}{3} \pi (2r)^{3}$
$V = \frac{4}{3} \pi (8r^{3})$
$V = \frac{32}{3} \pi r^{3}$

(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.

(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 $488 \, cm^3$.

(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 $38.808 \, cm^3$.

(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 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.

(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) The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Given the radius $r = 11.2 \, cm$.
Substituting the values,we get:
$V = \frac{4}{3} \times \frac{22}{7} \times (11.2)^3 \, cm^3$
$V = \frac{4}{3} \times \frac{22}{7} \times 1404.928 \, cm^3$
$V = \frac{123633.664}{21} \, cm^3$
$V \approx 5887.32 \, cm^3$.

(B) The shot-put is a solid sphere. The mass of the sphere is calculated by multiplying its volume by its density.
Step $1$: Calculate the volume of the sphere.
The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Given $r = 4.9 \, cm$,we have:
$V = \frac{4}{3} \times \frac{22}{7} \times (4.9)^3$
$V = \frac{4}{3} \times \frac{22}{7} \times 4.9 \times 4.9 \times 4.9$
$V = \frac{4}{3} \times 22 \times 0.7 \times 4.9 \times 4.9$
$V \approx 493.0 \, cm^3$.
Step $2$: Calculate the mass of the shot-put.
Mass = Volume $\times$ Density
Mass = $493.0 \, cm^3 \times 7.8 \, g/cm^3$
Mass = $3845.4 \, g$.
Step $3$: Convert the mass into kilograms.
Since $1 \, kg = 1000 \, g$,
Mass = $\frac{3845.4}{1000} \, kg \approx 3.85 \, kg$.

(A) The volume of a hemispherical bowl is given by the formula $V = \frac{2}{3} \pi r^{3}$.
Given the radius $r = 3.5\, cm = \frac{7}{2}\, cm$.
Substituting the values:
$V = \frac{2}{3} \times \frac{22}{7} \times (\frac{7}{2})^{3}$
$V = \frac{2}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2}$
$V = \frac{1}{3} \times 11 \times \frac{49}{2} = \frac{539}{6} \approx 89.83\, cm^{3}$.

(D) Given,radius $(r) = 7 \, cm$.
The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Substituting the given values:
$V = \frac{4}{3} \times \frac{22}{7} \times (7)^3 \, cm^3$
$V = \frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7 \, cm^3$
$V = \frac{4 \times 22 \times 7 \times 7}{3} \, cm^3$
$V = \frac{4312}{3} \, cm^3$
Converting the improper fraction to a mixed fraction:
$V = 1437 \frac{1}{3} \, cm^3$.
Thus,the required volume is $1437 \frac{1}{3} \, cm^3$.

(A) Given,radius $(r) = 0.63 \, m$.
The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Substituting the values:
$V = \frac{4}{3} \times \frac{22}{7} \times (0.63)^3 \, m^3$
$V = \frac{4}{3} \times \frac{22}{7} \times \frac{63}{100} \times \frac{63}{100} \times \frac{63}{100} \, m^3$
$V = 4 \times 22 \times \frac{1}{3} \times \frac{1}{7} \times \frac{63 \times 63 \times 63}{1000000} \, m^3$
$V = 4 \times 22 \times \frac{1}{21} \times \frac{250047}{1000000} \, m^3$
$V = 88 \times \frac{11907}{1000000} \, m^3$
$V = \frac{1047816}{1000000} \, m^3 = 1.047816 \, m^3$.
Rounding to two decimal places,the volume is approximately $1.05 \, m^3$.

(B) The amount of water displaced by a solid object is equal to the volume of the object.
Diameter of the ball $= 28 \, cm$.
Radius of the ball $(r) = \frac{28}{2} \, cm = 14 \, cm$.
Volume of the spherical ball $= \frac{4}{3} \pi r^3$.
Volume $= \frac{4}{3} \times \frac{22}{7} \times (14)^3 \, cm^3$.
Volume $= \frac{4}{3} \times \frac{22}{7} \times 14 \times 14 \times 14 \, cm^3$.
Volume $= \frac{4 \times 22 \times 2 \times 14 \times 14}{3} \, cm^3 = \frac{34496}{3} \, cm^3$.
Converting to mixed fraction: $34496 \div 3 = 11498$ with a remainder of $2$.
So,Volume $= 11498 \frac{2}{3} \, cm^3$.
Thus,the amount of water displaced is $11498 \frac{2}{3} \, cm^3$.

(C) The amount of water displaced by a solid object is equal to its volume.
Diameter of the solid spherical ball $= 0.21 \, m$.
Radius $(r) = \frac{0.21}{2} \, m = \frac{21}{200} \, m$.
Volume of a sphere $= \frac{4}{3} \pi r^3$.
Volume $= \frac{4}{3} \times \frac{22}{7} \times \left( \frac{21}{200} \right)^3 \, m^3$.
Volume $= \frac{4}{3} \times \frac{22}{7} \times \frac{21}{200} \times \frac{21}{200} \times \frac{21}{200} \, m^3$.
Volume $= \frac{4 \times 22 \times 21 \times 21 \times 21}{3 \times 7 \times 200 \times 200 \times 200} \, m^3$.
Volume $= \frac{4 \times 22 \times 3 \times 21 \times 21}{3 \times 8000000} \, m^3 = \frac{88 \times 441}{8000000} \, m^3 = \frac{38808}{8000000} \, m^3 = 0.004851 \, m^3$.

(D) Given: Diameter of the metallic ball $= 4.2\, cm$.
Radius $(r) = \frac{4.2}{2}\, cm = 2.1\, cm$.
Volume of the spherical ball $= \frac{4}{3} \pi r^3$.
Volume $= \frac{4}{3} \times \frac{22}{7} \times (2.1)^3\, cm^3 = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1\, cm^3$.
Volume $= 4 \times 22 \times 0.1 \times 2.1 \times 2.1\, cm^3 = 38.808\, cm^3$.
Density of the metal $= 8.9\, g/cm^3$.
Mass of the ball $= \text{Density} \times \text{Volume} = 8.9\, g/cm^3 \times 38.808\, cm^3$.
Mass $= 345.3912\, g \approx 345.39\, g$.
Thus,the mass of the ball is $345.39\, g$ (approx.).

(A) Let the radius of the earth be $R$ and the radius of the moon be $r$.
Given that the diameter of the moon is $\frac{1}{4}$ of the diameter of the earth,it follows that the radius of the moon is $\frac{1}{4}$ of the radius of the earth,i.e.,$r = \frac{R}{4}$.
The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Volume of the earth $(V_E)$ $= \frac{4}{3} \pi R^3$.
Volume of the moon $(V_M)$ $= \frac{4}{3} \pi r^3 = \frac{4}{3} \pi \left(\frac{R}{4}\right)^3 = \frac{4}{3} \pi \left(\frac{R^3}{64}\right) = \frac{1}{64} \left(\frac{4}{3} \pi R^3\right)$.
Therefore,$V_M = \frac{1}{64} V_E$.
The fraction of the volume of the earth that is the volume of the moon is $\frac{V_M}{V_E} = \frac{1}{64}$.

(B) Diameter of the hemisphere $= 10.5\, cm$.
Radius of the hemisphere $(r) = \frac{10.5}{2}\, cm = 5.25\, cm$.
Volume of the hemispherical bowl $= \frac{2}{3} \pi r^3$.
Volume $= \frac{2}{3} \times \frac{22}{7} \times (5.25)^3\, cm^3$.
Volume $= \frac{2}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 5.25\, cm^3 = 303.1875\, cm^3$.
Since $1000\, cm^3 = 1\, l$,the capacity in litres is $\frac{303.1875}{1000}\, l = 0.3031875\, l$.
Rounding to three decimal places,the capacity is approximately $0.303\, l$.

(C) Let the radius of each small sphere be $r$ and the radius of the new large sphere be $r^{\prime}$.
The volume of one small sphere is given by $V = \frac{4}{3} \pi r^{3}$.
The total volume of $27$ small spheres is $27 \times \frac{4}{3} \pi r^{3}$.
When these are melted to form a new sphere of radius $r^{\prime}$,the volume of the new sphere is $\frac{4}{3} \pi (r^{\prime})^{3}$.
Since the volume remains constant during melting and recasting,we have:
$\frac{4}{3} \pi (r^{\prime})^{3} = 27 \times \frac{4}{3} \pi r^{3}$
Dividing both sides by $\frac{4}{3} \pi$,we get:
$(r^{\prime})^{3} = 27 r^{3}$
Taking the cube root on both sides:
$r^{\prime} = \sqrt[3]{27 r^{3}} = 3r$.
Thus,the radius of the new sphere is $3r$.

(D) Let the radius of each small sphere be $r$ and the radius of the new large sphere be $R$.
Given that $27$ small spheres are melted to form one large sphere,the volume remains conserved.
Volume of $27$ small spheres $= 27 \times (\frac{4}{3} \pi r^3)$.
Volume of the large sphere $= \frac{4}{3} \pi R^3$.
Equating the volumes: $27 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$.
$27 r^3 = R^3$,which implies $R = 3r$.
The surface area of a small sphere is $S = 4 \pi r^2$.
The surface area of the large sphere is $S' = 4 \pi R^2 = 4 \pi (3r)^2 = 4 \pi (9r^2) = 36 \pi r^2$.
Now,the ratio $S : S' = \frac{4 \pi r^2}{36 \pi r^2} = \frac{1}{9}$.
Thus,the ratio is $1: 9$.

(A) The diameter of the spherical capsule is $d = 3.5 \, mm$.
The radius $r$ of the capsule is given by $r = \frac{d}{2} = \frac{3.5}{2} = 1.75 \, mm$.
The volume $V$ of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Substituting the values,we get:
$V = \frac{4}{3} \times \frac{22}{7} \times (1.75)^3 \, mm^3$
$V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{4}\right)^3 \, mm^3$
$V = \frac{4}{3} \times \frac{22}{7} \times \frac{7 \times 7 \times 7}{4 \times 4 \times 4} \, mm^3$
$V = \frac{22 \times 7 \times 7}{3 \times 4 \times 4} \, mm^3 = \frac{1078}{48} \, mm^3 \approx 22.4583 \, mm^3$.
Rounding to two decimal places,the volume is approximately $22.46 \, mm^3$.

(C) For a given sphere,the radius $(r) = 2.1 \, cm = \frac{21}{10} \, cm$.
The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.
Substituting the values:
$V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{21}{10}\right)^3 \, cm^3$
$V = \frac{4}{3} \times \frac{22}{7} \times \frac{21}{10} \times \frac{21}{10} \times \frac{21}{10} \, cm^3$
$V = 4 \times 22 \times \frac{1}{1} \times \frac{1}{1} \times \frac{0.1}{1} \times 2.1 \times 2.1 \, cm^3$ (simplifying $3 \times 7 = 21$ with one $21$ in the numerator)
$V = 88 \times 0.1 \times 4.41 \, cm^3$
$V = 8.8 \times 4.41 \, cm^3 = 38.808 \, cm^3$.

(D) For a given hemisphere:
Radius $(r) = \frac{\text{diameter}}{2} = \frac{12}{2} = 6 \, cm$.
Volume of the hemisphere $= \frac{2}{3} \pi r^3$.
Substituting the values:
Volume $= \frac{2}{3} \times 3.14 \times (6)^3 \, cm^3$.
Volume $= \frac{2}{3} \times 3.14 \times 216 \, cm^3$.
Volume $= 2 \times 3.14 \times 72 \, cm^3$.
Volume $= 452.16 \, cm^3$.

(D) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Let the radii of the three spheres be $r_1 = 3 \,cm, r_2 = 4 \,cm$,and $r_3 = 5 \,cm$.
The total volume of the three spheres is $V_{total} = \frac{4}{3} \pi (r_1^3 + r_2^3 + r_3^3)$.
$V_{total} = \frac{4}{3} \pi (3^3 + 4^3 + 5^3) = \frac{4}{3} \pi (27 + 64 + 125) = \frac{4}{3} \pi (216)$.
Let the radius of the new sphere be $R$. Its volume is $V_{new} = \frac{4}{3} \pi R^3$.
Since the volume remains conserved,$V_{new} = V_{total}$.
$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi (216)$.
$R^3 = 216$.
$R = \sqrt[3]{216} = 6 \,cm$.

(B) The volume of a sphere is given by the formula $V = \frac{4}{3} \pi r^3$.
Let the radius of the original ball be $R = 9 \,cm$.
Let the radii of the three smaller balls be $r_1 = 1 \,cm$,$r_2 = 6 \,cm$,and $r_3$ (the unknown radius).
Since the volume remains conserved during melting and recasting,the volume of the original ball equals the sum of the volumes of the three smaller balls:
$\frac{4}{3} \pi R^3 = \frac{4}{3} \pi r_1^3 + \frac{4}{3} \pi r_2^3 + \frac{4}{3} \pi r_3^3$
Dividing both sides by $\frac{4}{3} \pi$,we get:
$R^3 = r_1^3 + r_2^3 + r_3^3$
Substituting the given values:
$9^3 = 1^3 + 6^3 + r_3^3$
$729 = 1 + 216 + r_3^3$
$729 = 217 + r_3^3$
$r_3^3 = 729 - 217$
$r_3^3 = 512$
Taking the cube root on both sides:
$r_3 = \sqrt[3]{512} = 8 \,cm$.
Thus,the radius of the third ball is $8 \,cm$.

(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) 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.

(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 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) The volume $V$ of a sphere is given by the formula $V = \frac{4}{3} \pi r^{3}$.
Given,radius $r = 2.1 \, m$.
Substituting the value of $r$ in the formula:
$V = \frac{4}{3} \times \frac{22}{7} \times (2.1)^{3}$
$V = \frac{4}{3} \times \frac{22}{7} \times 2.1 \times 2.1 \times 2.1$
$V = 4 \times 22 \times 0.1 \times 2.1 \times 2.1$
$V = 88 \times 0.1 \times 4.41$
$V = 8.8 \times 4.41 = 38.808 \, m^{3}$.