Textbook - Surface Areas and Volumes Questions in English

(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) Inner radius $(r) = 1 \,m$.
Thickness $= 1 \,cm = 0.01 \,m$.
Outer radius $(R) = 1 \,m + 0.01 \,m = 1.01 \,m$.
The volume of iron used is the difference between the outer volume and the inner volume of the hemispherical shell.
Volume of iron $= \frac{2}{3} \pi R^3 - \frac{2}{3} \pi r^3 = \frac{2}{3} \pi (R^3 - r^3)$.
Volume $= \frac{2}{3} \times \frac{22}{7} \times ((1.01)^3 - (1)^3) \,m^3$.
Volume $= \frac{44}{21} \times (1.030301 - 1) \,m^3$.
Volume $= \frac{44}{21} \times 0.030301 \,m^3 \approx 0.06348 \,m^3$.

(D) Let $r$ be the radius of the sphere.
The surface area of a sphere is given by $4 \pi r^2$.
Given,$4 \pi r^2 = 154$.
$r^2 = \frac{154}{4 \pi} = \frac{154 \times 7}{4 \times 22} = \frac{7 \times 7}{4} = \left(\frac{7}{2}\right)^2$.
Therefore,$r = \frac{7}{2} \, cm$.
Now,the volume of the sphere is given by $V = \frac{4}{3} \pi r^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^3 \, cm^3$.
$V = \frac{4}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times \frac{7}{2} \, cm^3$.
$V = \frac{11 \times 7 \times 7}{3} \, cm^3 = \frac{539}{3} \, cm^3 = 179 \frac{2}{3} \, cm^3$.
Thus,the required volume of the sphere is $179 \frac{2}{3} \, cm^3$.

(A) Total cost of white-washing $= Rs. 4989.60$
Rate of white-washing $= Rs. 20$ per $m^2$
We know that,$\text{Area} = \frac{\text{Total cost}}{\text{Rate}}$
$\text{Inside surface area} = \frac{4989.60}{20} = 249.48 \, m^2$
Thus,the inside surface area of the dome is $249.48 \, m^2$.

(B) Total cost of white-washing $= Rs. 4989.60$; Rate of white-washing $= Rs. 20$ per $m^2$.
$\therefore \text{Inner Surface Area} = \frac{\text{Total cost}}{\text{Rate}} = \frac{4989.60}{20} = 249.48 \, m^2$.
Let $r$ be the radius of the hemispherical dome.
$\therefore$ Inner Surface Area $= 2 \pi r^2 = 249.48$.
$2 \times \frac{22}{7} \times r^2 = 249.48 \Rightarrow r^2 = \frac{249.48 \times 7}{2 \times 22} = \frac{1746.36}{44} = 39.69$.
$r = \sqrt{39.69} = 6.3 \, m$.
Volume of air inside the dome $= \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times (6.3)^3$.
$= \frac{2}{3} \times \frac{22}{7} \times 6.3 \times 6.3 \times 6.3 = 2 \times 22 \times 0.3 \times 6.3 \times 6.3 = 523.908 \, m^3$.
Thus,the volume of air inside the dome is approximately $523.9 \, m^3$.

(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 $22.46 \, mm^3$.

(A) $1$. External Surface Area to be polished:
External dimensions: $H = 110\, cm, B = 85\, cm, D = 25\, cm$.
External surface area (excluding the front face) $= 2(H \times D) + (B \times D) + 2(H \times B) = 2(110 \times 25) + (85 \times 25) + 2(110 \times 85) = 5500 + 2125 + 18700 = 26325\, cm^2$.
Area of the front face (excluding the openings) $= (110 \times 85) - 3(100 \times 75) = 9350 - 22500$ (This is not correct,let's re-evaluate).
Correct approach: External surface area to be polished $= 2(110 \times 25) + (85 \times 25) + 2(110 \times 85) - 3(100 \times 75) = 5500 + 2125 + 18700 - 22500 = 3825\, cm^2$.
$2$. Internal Surface Area to be painted:
Internal dimensions of each shelf: $H' = 30\, cm, B' = 75\, cm, D' = 20\, cm$.
Area of $3$ shelves $= 3 \times [2(H' \times D') + (B' \times D') + 2(H' \times B')] = 3 \times [2(30 \times 20) + (75 \times 20) + 2(30 \times 75)] = 3 \times [1200 + 1500 + 4500] = 3 \times 7200 = 21600\, cm^2$.
$3$. Total Expenses:
Cost of polishing $= 3825 \times 0.20 = ₹ 765$.
Cost of painting $= 21600 \times 0.10 = ₹ 2160$.
Total cost $= 765 + 2160 = ₹ 2925$.

(N/A) $1$. For the spheres:
Diameter of each sphere = $21 \, cm$,so radius $r = 10.5 \, cm$.
The surface area of one sphere is $4 \pi r^2 = 4 \times \frac{22}{7} \times 10.5 \times 10.5 = 1386 \, cm^2$.
However,a small part of the sphere is covered by the support. The support is a cylinder of radius $1.5 \, cm$. The area of the circle covered on the sphere is $\pi \times (1.5)^2 = \frac{22}{7} \times 2.25 \approx 7.07 \, cm^2$.
Surface area to be painted silver per sphere = $1386 - 7.07 = 1378.93 \, cm^2$.
For $8$ spheres,total silver area = $8 \times 1378.93 = 11031.44 \, cm^2$.
Cost of silver paint = $11031.44 \times 0.25 = ₹ 2757.86$.
$2$. For the supports:
Each support is a cylinder with radius $r = 1.5 \, cm$ and height $h = 7 \, cm$.
Curved surface area of one cylinder = $2 \pi rh = 2 \times \frac{22}{7} \times 1.5 \times 7 = 66 \, cm^2$.
Total area to be painted black for $8$ supports = $8 \times 66 = 528 \, cm^2$.
Cost of black paint = $528 \times 0.05 = ₹ 26.40$.
Total cost = $2757.86 + 26.40 = ₹ 2784.26$.

(A) Let the initial diameter of the sphere be $D_1$. Then the initial radius $r_1 = D_1 / 2$.
The initial curved surface area $A_1 = 4 \pi r_1^2$.
When the diameter is decreased by $25\%$,the new diameter $D_2 = D_1 - 0.25 D_1 = 0.75 D_1$.
The new radius $r_2 = D_2 / 2 = 0.75 (D_1 / 2) = 0.75 r_1$.
The new curved surface area $A_2 = 4 \pi r_2^2 = 4 \pi (0.75 r_1)^2 = 4 \pi (0.5625 r_1^2) = 0.5625 A_1$.
The decrease in surface area is $A_1 - A_2 = A_1 - 0.5625 A_1 = 0.4375 A_1$.
The percentage decrease is $((A_1 - A_2) / A_1) \times 100\% = 0.4375 \times 100\% = 43.75\%$.