Solution

(i) Radius ( $r$ ) of sphere $=10.5 cm$

Surface area of sphere $=4 n r^{2}$

$=[4 \times \frac{22}{7} \times(10.5)^{2}] cm^{2}$

$=(4 \times \frac{22}{7} \times 10.5 \times 10.5) cm^{2}$

$=(88 \times 1.5 \times 10.5) cm^{2}$

$=1386 cm^{2}$

Therefore, the surface area of a sphere having radius $10.5 cm$ is $1386 cm^{2}$.

(ii) Radius( $r$ ) of sphere $=5.6 cm$

Surface area of sphere $=4 \pi r^{2}$ $=[4 \times \frac{22}{7} \times(5.6)^{2}] cm^{2}$

$=(88 \times 0.8 \times 5.6) cm^{2}$

$=394.24 cm^{2}$

Therefore, the surface area of a sphere having radius $5.6 cm$ is $394.24 cm^{2}$.

(iii) Radius ( $r$ ) of sphere $=14 cm$

Surface area of sphere $=4 \pi r^{2}$

$=[4 \times \frac{22}{7} \times(14)^{2}] cm^{2}$

$=(4 \times 44 \times 14) cm^{2}$

$=2464 cm^{2}$

Therefore, the surface area of a sphere having radius $14 cm$ is $2464 cm^{2}$.

Solution

(i) $Radius(r)$ of sphere $=\frac{\text{ Diameter }}{2}=(\frac{14}{2}) cm=7 cm$

Surface area of sphere $=4 n r^{2}$

$=(4 \times \frac{22}{7} \times(7)^{2}) cm^{2}$ $=(88 \times 7) cm^{2}$ $=616 cm^{2}$

Therefore, the surface area of a sphere having diameter $14 cm$ is $616 cm^{2}$.

(ii) Radius ( $r$ ) of sphere $=\frac{21}{2}=10.5 cm$

Surface area of sphere $=4 n r^{2}$

$=[4 \times \frac{22}{7} \times(10.5)^{2}] cm^{2}$

$=1386 cm^{2}$

Therefore, the surface area of a sphere having diameter $21 cm$ is $1386 cm^{2}$.

(iii) Radius ( $r$ ) of sphere $=\frac{\frac{3.5}{2}=1.75}{} m$

Surface area of sphere $=4 n r^{2}$

$=[4 \times \frac{22}{7} \times(1.75)^{2}] m^{2}$

$=38.5 m^{2}$

Therefore, the surface area of the sphere having diameter $3.5 m$ is $38.5 m^{2}$.

Solution

Radius ( $r$ ) of hemisphere $=10 cm$

Total surface area of hemisphere $=$ CSA of hemisphere + Area of circular end of hemisphere

$ \begin{aligned} & =2 \pi r^{2}+\pi r^{2} \\ & =3 \pi r^{2} \\ & =[3 \times 3.14 \times(10)^{2}] cm^{2} \\ & =942 cm^{2} \end{aligned} $

Therefore, the total surface area of such a hemisphere is $942 cm^{2}$.

Solution

Radius ( $r_1$ ) of spherical balloon $=7 cm$

Radius $(r_2.$ ) of spherical balloon, when air is pumped into it $=14 cm$

$ \begin{aligned} \text{ Required ratio } & =\frac{\text{ Ini }}{\text{ Surface area af }} \\ & =\frac{4 \pi r_1^{2}}{4 \pi r_2^{2}}=(\frac{r_1}{r_2})^{2} \\ & =(\frac{7}{14})^{2}=\frac{1}{4} \end{aligned} $

Therefore, the ratio between the surface areas in these two cases is $1: 4$.

Solution

$ =(\frac{10.5}{2}) cm=5.25 cm $

Inner radius ( $r$ ) of hemispherical bowl

Surface area of hemispherical bowl $=2 \pi r^{2}$

$=[2 \times \frac{22}{7} \times(5.25)^{2}] cm^{2}$

$=173.25 cm^{2}$

Cost of tin-plating $100 cm^{2}$ area $=$ Rs 16

Cost of tin-plating $173.25 cm^{2}$ area $=Rs(\frac{16 \times 173.25}{100})=Rs 27.72$

Therefore, the cost of tin-plating the inner side of the hemispherical bowl is Rs 27.72 .

Solution

Let the radius of the sphere be $r$.

Surface area of sphere $=154$

$ \begin{aligned} & \therefore 4 \pi r^{2}=154 cm^{2} \\ & r^{2}=(\frac{154 \times 7}{4 \times 22}) cm^{2}=(\frac{7 \times 7}{2 \times 2}) cm^{2} \\ & r=(\frac{7}{2}) cm=3.5 cm \end{aligned} $

Therefore, the radius of the sphere whose surface area is $154 cm^{2}$ is $3.5 cm$.

Solution

Let the diameter of earth be $d$. Therefore, the diameter of moon will be $\frac{d}{4}$.

$ \begin{aligned} & \text{ Radius of earth }=\frac{\frac{d}{2}}{\frac{1}{2} \times \frac{d}{4}=\frac{d}{8}} \\ & \text{ Radius of moon }=\frac{4 \pi(\frac{d}{8})^{2}}{4 \pi(\frac{d}{2})^{2}} \\ & \text{ Surface area of moon }=\frac{4 \pi(\frac{d}{8})^{2}}{4 \pi(\frac{d}{2})^{2}} \\ & \text{ Surface area of earth }=\frac{4}{\text{ Required ratio }} \\ & =\frac{4}{64}=\frac{1}{16} \end{aligned} $

Therefore, the ratio between their surface areas will be $1: 16$.

Solution

Inner radius of hemispherical bowl $=5 cm$

Thickness of the bowl $=0.25 cm$

$\therefore$ Outer radius $(r)$ of hemispherical bowl $=(5+0.25) cm$

$=5.25 cm$

Outer CSA of hemispherical bowl $=2 \pi r^{2}$

$ =2 \times \frac{22}{7} \times(5.25 cm)^{2}=173.25 cm^{2} $

Therefore, the outer curved surface area of the bowl is $173.25 cm^{2}$.

Solution

(i) Surface area of sphere $=4 \pi r^{2}$

(ii) Height of cylinder $=r+r=2 r$

Radius of cylinder $=r$

CSA of cylinder $=2 nrh$

$=2 \pi r(2 r)$

$=4 \Gamma r^{2}$

(iii)

Required ratio $=\frac{\text{ Surface area of sphere }}{\text{ CSA of cylinder }}$

$=\frac{4 \pi r^{2}}{4 \pi r^{2}}$

$=\frac{1}{1}$

Therefore, the ratio between these two surface areas is $1: 1$.