Surface Area of a Sphere Questions in English

(C) The balloon is hemispherical in shape.
The surface area of a hemispherical balloon with radius $r$ is given by the formula $S = 3\pi r^2$.
Let $r_1 = 6\,cm$ and $r_2 = 12\,cm$ be the radii in the two cases.
The ratio of the surface areas is $\frac{S_1}{S_2} = \frac{3\pi r_1^2}{3\pi r_2^2} = \frac{r_1^2}{r_2^2}$.
Substituting the values,we get $\frac{S_1}{S_2} = \frac{6^2}{12^2} = \frac{36}{144} = \frac{1}{4}$.
Therefore,the ratio of the surface areas of the two balloons is $1:4$.

(A) The surface area of a sphere is given by the formula $4 \pi r^{2}$.
Given the radius $r = 7 \, cm$.
Substituting the value of $r$ in the formula:
Surface Area $= 4 \times \frac{22}{7} \times (7 \, cm)^{2}$
$= 4 \times \frac{22}{7} \times 49 \, cm^{2}$
$= 4 \times 22 \times 7 \, cm^{2}$
$= 88 \times 7 \, cm^{2}$
$= 616 \, cm^{2}$.

(C) The diameter of the hollow sphere is $d = 7 \,m$.
The radius $r$ of the sphere is given by $r = \frac{d}{2} = \frac{7}{2} = 3.5 \,m$.
The area available to the motorcyclist for riding is the surface area of the sphere,which is calculated using the formula $A = 4 \pi r^2$.
Substituting the values:
$A = 4 \times \frac{22}{7} \times 3.5 \times 3.5$
$A = 4 \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$A = 4 \times 22 \times \frac{7}{4}$
$A = 22 \times 7 = 154 \,m^2$.
Thus,the area available for riding is $154 \,m^2$.

(D) Since only the rounded surface of the dome is to be painted,we need to find the curved surface area of the hemisphere.
Circumference of the base of the dome $= 17.6 \, m$.
Let the radius be $r$. Then,$2\pi r = 17.6$.
$r = \frac{17.6}{2 \times \pi} = \frac{17.6 \times 7}{2 \times 22} = 0.8 \times 3.5 = 2.8 \, m$.
Curved surface area of the hemisphere $= 2\pi r^2 = 2 \times \frac{22}{7} \times 2.8 \times 2.8 = 49.28 \, m^2$.
Cost of painting $100 \, cm^2 = Rs. 5$.
Since $1 \, m^2 = 10000 \, cm^2$,the cost of painting $1 \, m^2 = \frac{5}{100} \times 10000 = Rs. 500$.
Total cost of painting $= 49.28 \times 500 = Rs. 24640$.

(A) The radius $(r)$ of the sphere is $10.5 \, cm$.
The formula for the surface area of a sphere is $4 \pi r^{2}$.
Substituting the given values:
Surface area $= 4 \times \frac{22}{7} \times (10.5)^{2} \, cm^{2}$
$= 4 \times \frac{22}{7} \times 10.5 \times 10.5 \, cm^{2}$
$= 4 \times 22 \times 1.5 \times 10.5 \, cm^{2}$
$= 88 \times 15.75 \, cm^{2}$
$= 1386 \, cm^{2}$
Therefore,the surface area of the sphere is $1386 \, cm^{2}$.

(B) The radius $(r)$ of the sphere is $5.6 \, cm$.
The formula for the surface area of a sphere is $4 \pi r^{2}$.
Substituting the given values:
Surface area $= 4 \times \frac{22}{7} \times (5.6)^{2} \, cm^{2}$
$= 4 \times \frac{22}{7} \times 5.6 \times 5.6 \, cm^{2}$
$= 4 \times 22 \times 0.8 \times 5.6 \, cm^{2}$
$= 88 \times 4.48 \, cm^{2}$
$= 394.24 \, cm^{2}$.
Therefore,the surface area of the sphere is $394.24 \, cm^{2}$.

(C) The radius $(r)$ of the sphere is $14 \, cm$.
The formula for the surface area of a sphere is $4 \pi r^2$.
Substituting the given values:
Surface area $= 4 \times \frac{22}{7} \times (14)^2 \, cm^2$
$= 4 \times \frac{22}{7} \times 14 \times 14 \, cm^2$
$= 4 \times 22 \times 2 \times 14 \, cm^2$
$= 88 \times 28 \, cm^2$
$= 2464 \, cm^2$.
Therefore,the surface area of the sphere is $2464 \, cm^2$.

(D) The radius $(r)$ of the sphere is given by:
$r = \frac{\text{Diameter}}{2} = \frac{14 \, cm}{2} = 7 \, cm$.
The formula for the surface area of a sphere is $4 \pi r^2$.
Substituting the values:
Surface Area $= 4 \times \frac{22}{7} \times (7 \, cm)^2$
$= 4 \times \frac{22}{7} \times 49 \, cm^2$
$= 4 \times 22 \times 7 \, cm^2$
$= 88 \times 7 \, cm^2$
$= 616 \, cm^2$.
Therefore,the surface area of the sphere is $616 \, cm^2$.

(A) Given,diameter of the sphere $d = 21 \, cm$.
Radius $(r) = \frac{d}{2} = \frac{21}{2} \, cm = 10.5 \, cm$.
The surface area of a sphere is given by the formula $4 \pi r^{2}$.
Substituting the values: $\text{Surface Area} = 4 \times \frac{22}{7} \times (10.5)^{2} \, cm^{2}$.
$= 4 \times \frac{22}{7} \times 10.5 \times 10.5 \, cm^{2}$.
$= 4 \times \frac{22}{7} \times \frac{21}{2} \times \frac{21}{2} \, cm^{2}$.
$= 22 \times 3 \times 21 \, cm^{2} = 1386 \, cm^{2}$.
Thus,the surface area of the sphere is $1386 \, cm^{2}$.

(B) Given: Diameter $(d) = 3.5 \, cm$.
Radius $(r) = \frac{d}{2} = \frac{3.5}{2} = 1.75 \, cm$.
The surface area of a sphere is given by the formula: $A = 4 \pi r^2$.
Substituting the values: $A = 4 \times \frac{22}{7} \times (1.75)^2$.
$A = 4 \times \frac{22}{7} \times 1.75 \times 1.75$.
$A = 4 \times \frac{22}{7} \times \frac{7}{4} \times 1.75$.
$A = 22 \times 1.75 = 38.5 \, cm^2$.
Thus,the surface area of the sphere is $38.5 \, cm^2$.

(D) The initial radius $(r_1)$ of the spherical balloon is $7 \, cm$.
The final radius $(r_2)$ of the spherical balloon after pumping air is $14 \, cm$.
The surface area of a sphere is given by the formula $A = 4 \pi r^2$.
The ratio of the surface areas is given by $\frac{A_1}{A_2} = \frac{4 \pi r_1^2}{4 \pi r_2^2} = \left( \frac{r_1}{r_2} \right)^2$.
Substituting the values,we get $\left( \frac{7}{14} \right)^2 = \left( \frac{1}{2} \right)^2 = \frac{1}{4}$.
Therefore,the ratio of the surface areas is $1:4$.

(B) Let the radius of the sphere be $r \, cm$.
The surface area of a sphere is given by the formula $4 \pi r^2$.
Given that the surface area is $154 \, cm^2$,we have:
$4 \pi r^2 = 154$
Substituting $\pi = \frac{22}{7}$:
$4 \times \frac{22}{7} \times r^2 = 154$
Solving for $r^2$:
$r^2 = \frac{154 \times 7}{4 \times 22}$
$r^2 = \frac{7 \times 7}{4} = \left(\frac{7}{2}\right)^2$
Taking the square root of both sides:
$r = \frac{7}{2} = 3.5 \, cm$
Thus,the radius of the sphere is $3.5 \, cm$.

(C) Let the radius of the earth be $R$.
Since the diameter of the moon is one-fourth of the diameter of the earth,the radius of the moon will also be one-fourth of the radius of the earth.
Therefore,the radius of the moon $= \frac{R}{4}$.
The surface area of a sphere is given by the formula $A = 4 \pi r^2$.
Surface area of the earth $= 4 \pi R^2$.
Surface area of the moon $= 4 \pi \left( \frac{R}{4} \right)^2 = 4 \pi \left( \frac{R^2}{16} \right) = \frac{\pi R^2}{4}$.
The ratio of the surface area of the moon to the surface area of the earth is:
$\frac{\text{Surface area of the moon}}{\text{Surface area of the earth}} = \frac{\frac{\pi R^2}{4}}{4 \pi R^2} = \frac{1}{4 \times 4} = \frac{1}{16}$.
Thus,the ratio of their surface areas is $1:16$.

(D) Inner radius $(r) = 5 \, cm$
Thickness $= 0.25 \, cm$
Outer radius $(R) = r + \text{thickness} = 5 + 0.25 = 5.25 \, cm$
The outer curved surface area of a hemispherical bowl is given by the formula $2 \pi R^2$.
Outer curved surface area $= 2 \times \frac{22}{7} \times (5.25)^2 \, cm^2$
$= 2 \times \frac{22}{7} \times 5.25 \times 5.25 \, cm^2$
$= 2 \times \frac{22}{7} \times \frac{525}{100} \times \frac{525}{100} \, cm^2$
$= 2 \times 22 \times \frac{75}{100} \times 5.25 \, cm^2$
$= 44 \times 0.75 \times 5.25 \, cm^2$
$= 173.25 \, cm^2$

(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 $((A_1 - A_2) / A_1) \times 100\% = 0.4375 \times 100\% = 43.75\%$.

(D) Given,the radius of the sphere $(r) = 14 \, cm$.
The formula for the surface area of a sphere is $A = 4 \pi r^2$.
Substituting the value of $r = 14 \, cm$ and $\pi = \frac{22}{7}$:
$A = 4 \times \frac{22}{7} \times 14 \times 14 \, cm^2$
$A = 4 \times 22 \times 2 \times 14 \, cm^2$
$A = 88 \times 28 \, cm^2$
$A = 2464 \, cm^2$.
Thus,the surface area of the sphere is $2464 \, cm^2$.

(B) Given,the radius of the sphere $(r) = 3.5\, cm = \frac{7}{2}\, cm$.
The surface area of a sphere is given by the formula $A = 4\pi r^{2}$.
Substituting the values:
$A = 4 \times \frac{22}{7} \times \left(\frac{7}{2}\right)^{2}$
$A = 4 \times \frac{22}{7} \times \frac{49}{4} = 22 \times 7 = 154\, cm^{2}$.
The cost of plating gold is ₹ $150$ per $cm^{2}$.
Total cost = $\text{Surface area} \times \text{Rate}$
Total cost = $154 \times 150 = 23,100$.
Therefore,the cost of plating gold on its surface is ₹ $23,100$.

(C) Let the radius of the sphere be $r_1$ and the radius of the hemisphere be $r_2$.
Given that the ratio of their radii is $r_1 : r_2 = 3 : 2$,so $\frac{r_1}{r_2} = \frac{3}{2}$.
The surface area of a sphere is $4 \pi r_1^2$.
The total surface area of a hemisphere is $3 \pi r_2^2$.
Now,the ratio of the surface area of the sphere to the total surface area of the hemisphere is:
$\frac{4 \pi r_1^2}{3 \pi r_2^2} = \frac{4}{3} \times \left(\frac{r_1}{r_2}\right)^2$
$= \frac{4}{3} \times \left(\frac{3}{2}\right)^2$
$= \frac{4}{3} \times \frac{9}{4}$
$= \frac{3}{1}$
Therefore,the required ratio is $3:1$.

(C) The surface area of a sphere is given by the formula $A = 4 \pi r^2$.
Given radius $r = 30 \, cm$ and $\pi = 3.14$.
Substituting the values into the formula:
$A = 4 \times 3.14 \times (30)^2$
$A = 4 \times 3.14 \times 900$
$A = 12.56 \times 900$
$A = 11304 \, cm^2$.

(A) The formula for the surface area of a sphere is $A = 4\pi r^2$.
Given,radius $r = 50\, cm$ and $\pi = 3.14$.
Substituting the values into the formula:
$A = 4 \times 3.14 \times (50)^2$
$A = 4 \times 3.14 \times 2500$
$A = 12.56 \times 2500$
$A = 31400\, cm^2$.

(B) The surface area of a sphere is given by the formula $A = 4 \pi r^2$.
Given radius $r = 1.25 \, m$ and $\pi = 3.14$.
Substituting the values into the formula:
$A = 4 \times 3.14 \times (1.25)^2$
$A = 4 \times 3.14 \times 1.5625$
$A = 12.56 \times 1.5625$
$A = 19.625 \, m^2$.
Therefore,the correct option is $B$.

(C) Given,the diameter of the sphere $d = 21\, cm$.
Therefore,the radius $r = \frac{d}{2} = \frac{21}{2} = 10.5\, cm$.
The surface area of a sphere is given by the formula $A = 4\pi r^2$.
Substituting the values,we get $A = 4 \times \frac{22}{7} \times (10.5)^2$.
$A = 4 \times \frac{22}{7} \times 10.5 \times 10.5$.
$A = 4 \times 22 \times 1.5 \times 10.5$.
$A = 88 \times 15.75 = 1386\, cm^2$.

(D) The diameter of the sphere is $d = 49\, cm$.
Therefore,the radius $r = d / 2 = 49 / 2 = 24.5\, cm$.
The formula for the surface area of a sphere is $A = 4 \pi r^2$.
Substituting the values,we get $A = 4 \times (22 / 7) \times (49 / 2) \times (49 / 2)$.
$A = 4 \times (22 / 7) \times (2401 / 4)$.
$A = (22 / 7) \times 2401$.
$A = 22 \times 343 = 7546\, cm^2$.

(A) Given,diameter of the sphere $d = 3.5\,m$.
Radius $r = \frac{d}{2} = \frac{3.5}{2} = 1.75\,m$.
The surface area of a sphere is given by the formula $A = 4\pi r^2$.
Substituting the values,$A = 4 \times \frac{22}{7} \times (1.75)^2$.
$A = 4 \times \frac{22}{7} \times 1.75 \times 1.75$.
$A = 4 \times 22 \times 0.25 \times 1.75$.
$A = 88 \times 0.4375 = 38.5\,m^2$.

(A) Given radius $r = 14\, cm$.
$1$. The curved surface area $(CSA)$ of a hemisphere is given by the formula $2\pi r^{2}$.
$CSA = 2 \times \frac{22}{7} \times 14 \times 14 = 2 \times 22 \times 2 \times 14 = 1232\, cm^{2}$.
$2$. The total surface area $(TSA)$ of a hemisphere is given by the formula $3\pi r^{2}$.
$TSA = 3 \times \frac{22}{7} \times 14 \times 14 = 3 \times 22 \times 2 \times 14 = 1848\, cm^{2}$.
Thus,the curved surface area is $1232\, cm^{2}$ and the total surface area is $1848\, cm^{2}$.

(N/A) Given,radius $r = 17.5 \, cm = \frac{35}{2} \, cm$.
Curved Surface Area of a hemisphere = $2\pi r^2$.
$= 2 \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = 11 \times 5 \times 35 = 1925 \, cm^2$.
Total Surface Area of a hemisphere = $3\pi r^2$.
$= 3 \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2} = 3 \times 11 \times 5 \times \frac{35}{2} = 165 \times 17.5 = 2887.5 \, cm^2$.

(A) The formula for the surface area of a sphere is $A = 4 \pi r^2$.
Given that $A = 7850 \, cm^2$ and $\pi = 3.14$,we substitute these values into the formula:
$7850 = 4 \times 3.14 \times r^2$
$7850 = 12.56 \times r^2$
$r^2 = \frac{7850}{12.56}$
$r^2 = 625$
$r = \sqrt{625} = 25 \, cm$.
Therefore,the radius of the sphere is $25 \, cm$.

(B) Given: Inner radius $(r)$ = $4 \, cm$,Thickness $(t)$ = $0.2 \, cm$.
The outer radius $(R)$ = $r + t = 4 + 0.2 = 4.2 \, cm$.
The formula for the curved surface area of a hemisphere is $2 \pi R^2$.
Outer curved surface area = $2 \times \frac{22}{7} \times (4.2)^2$.
$= 2 \times \frac{22}{7} \times 4.2 \times 4.2$.
$= 2 \times 22 \times 0.6 \times 4.2$.
$= 44 \times 2.52 = 110.88 \, cm^2$.

(C) Let the initial radius of the sphere be $r_1 = r$.
The initial surface area is $A_1 = 4 \pi r^2$.
The radius is decreased by $20 \%$,so the new radius $r_2 = r - 0.20r = 0.8r$.
The new surface area is $A_2 = 4 \pi (0.8r)^2 = 4 \pi (0.64r^2) = 0.64(4 \pi r^2) = 0.64 A_1$.
The decrease in surface area is $A_1 - A_2 = A_1 - 0.64 A_1 = 0.36 A_1$.
The percentage decrease is $\frac{0.36 A_1}{A_1} \times 100 \% = 36 \%$.

(D) The surface area of a sphere is given by the formula $S = 4 \pi r^2$.
Let the radii of the three spheres be $r_1 = 33 \, cm$,$r_2 = 44 \, cm$,and $r_3 = 48 \, cm$.
Let the radius of the fourth sphere be $R$.
According to the problem,the surface area of the fourth sphere is equal to the sum of the surface areas of the first three spheres:
$4 \pi R^2 = 4 \pi r_1^2 + 4 \pi r_2^2 + 4 \pi r_3^2$
Dividing both sides by $4 \pi$,we get:
$R^2 = r_1^2 + r_2^2 + r_3^2$
$R^2 = (33)^2 + (44)^2 + (48)^2$
$R^2 = 1089 + 1936 + 2304$
$R^2 = 5329$
Taking the square root of both sides:
$R = \sqrt{5329} = 73 \, cm$.
Thus,the radius of the fourth sphere is $73 \, cm$.

(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 given by $4\pi r_1^2$.
The curved surface area of a hemisphere is given by $2\pi r_2^2$.
According to the problem,the surface areas are equal:
$4\pi r_1^2 = 2\pi r_2^2$
Dividing both sides by $2\pi$,we get:
$2r_1^2 = r_2^2$
Taking the square root on both sides:
$\sqrt{2}r_1 = r_2$
Therefore,the ratio of their radii is:
$\frac{r_1}{r_2} = \frac{1}{\sqrt{2}}$
Thus,the ratio is $1: \sqrt{2}$.

(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) 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 $C$.

(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:
$\text{Total Surface Area} = 2 \pi r^{2} + \pi r^{2} = 3 \pi r^{2}$.

(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: $A = 4 \times (22 / 7) \times (7)^2$.
$A = 4 \times (22 / 7) \times 49$.
$A = 4 \times 22 \times 7$.
$A = 88 \times 7 = 616 \, cm^2$.

(C) The surface area of a sphere is given by the formula $A = 4 \pi r^2$,where $r$ is the radius.
Given $A = 2464 \, cm^2$ and taking $\pi = \frac{22}{7}$:
$4 \times \frac{22}{7} \times r^2 = 2464$
$r^2 = \frac{2464 \times 7}{4 \times 22}$
$r^2 = \frac{2464 \times 7}{88}$
$r^2 = 28 \times 7 = 196$
$r = \sqrt{196} = 14 \, cm$
The diameter $d = 2r = 2 \times 14 = 28 \, cm$.