Solution

(i) Radius of sphere $=7 cm$

Volume of sphere $=^{\frac{4}{3} \pi r^{3}}$ $=[\frac{4}{3} \times \frac{22}{7} \times(7)^{3}] cm^{3}$ $=(\frac{4312}{3}) cm^{3}$ $=1437 \frac{1}{3} cm^{3}$

Therefore, the volume of the sphere is $1437 \frac{1}{3} cm^{3}$.

(ii) Radius of sphere $=0.63 m$

$\frac{4}{3} \pi r^{3}$

Volume of sphere $=$

$=[\frac{4}{3} \times \frac{22}{7} \times(0.63)^{3}] m^{3}$

$=1.0478 m^{3}$

Therefore, the volume of the sphere is $1.05 m^{3}$ (approximately).

Solution

(i) Radius ( $r$ ) of ball $=$

$(\frac{28}{2}) cm=14 cm$

Volume of ball $=\frac{4}{3} \pi r^{3}$

$=[\frac{4}{3} \times \frac{22}{7} \times(14)^{3}] cm^{3}$

$=11498 \frac{2}{3} cm^{3}$

Therefore, the volume of the sphere is $11498 \frac{2}{3} cm^{3}$.

$ \begin{aligned} & \text{ (ii)Radius ( } r \text{ ) of ball }=\frac{(\frac{0.21}{2}) m}{\frac{4}{3} \pi r^{3}}=0.105 m \\ & \text{ Volume of ball }= \\ & =[\frac{4}{3} \times \frac{22}{7} \times(0.105)^{3}] m^{3} \\ & =0.004851 m^{3} \end{aligned} $

Therefore, the volume of the sphere is $0.004851 m^{3}$.

Solution

$ (\frac{4.2}{2}) cm=2.1 cm $

Radius ( $r$ ) of metallic ball

Volume of metallic ball $=$

$ \frac{4}{3} \pi r^{3} $

$=[\frac{4}{3} \times \frac{22}{7} \times(2.1)^{3}] cm^{3}$

$=38.808 cm^{3}$

Density $=\frac{\text{ Mass }}{\text{ Volume }}$

Mass $=$ Density $\times$ Volume

$=(8.9 \times 38.808) g$

$=345.3912 g$

Hence, the mass of the ball is $345.39 g$ (approximately).

Solution

$ \frac{d}{2} $

Let the diameter of earth be $d$. Therefore, the radius of earth will be

Diameter of moon will be $\frac{\frac{d}{4}}{}$ and the radius of moon will be $\frac{d}{8}$.

Volume of moon $=$

$ \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi(\frac{d}{8})^{3}=\frac{1}{512} \times \frac{4}{3} \pi d^{3} $

$ \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi(\frac{d}{2})^{3}=\frac{1}{8} \times \frac{4}{3} \pi d^{3} $

$\frac{\text{ Volume of moon }}{\text{ Volume of earth }}=\frac{\frac{1}{512} \times \frac{4}{3} \pi d^{3}}{\frac{1}{8} \times \frac{4}{3} \pi d^{3}}$

$ =\frac{1}{64} $

$\Rightarrow$ Volume of moon $=\frac{1}{64}$ Volume of earth

Therefore, the volume of moon is $\frac{1}{64}$ of the volume of earth.

Solution

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

Volume of hemispherical bowl $=\frac{2}{3} \pi r^{3}$

$=[\frac{2}{3} \times \frac{22}{7} \times(5.25)^{3}] cm^{3}$

$=303.1875 cm^{3}$

$(\frac{303.1875}{1000})$ litre

Capacity of the bowl $=$

$=0.3031875$ litre $=0.303$ litre (approximately)

$=5.25 cm$

Radius ( $r$ ) of hemispherical bowl $=$

Therefore, the volume of the hemispherical bowl is 0.303 litre.

Solution

Inner radius $(r_1)$ of hemispherical tank $=1 m$ Thickness of

$ \begin{aligned} & \Rightarrow 4 n r^{2}=154 cm^{2} \\ & \Rightarrow r^{2}=(\frac{154 \times 7}{4 \times 22}) cm^{2} \\ & \Rightarrow r=(\frac{7}{2}) cm=3.5 cm \\ & \text{ Volume of sphere }= \\ & =[\frac{4}{3} \times \frac{22}{7} \times(3.5)^{3}] cm^{3} \\ & =179 \frac{2}{3} cm^{3} \\ & \text{ hemispherical tank }=1 cm=0.01 m \\ & \text{ Outer radius }(r_2) \text{ of hemispherical tank }=(1+0.01) m=1.01 m \\ & \text{ Volume of iron used to make such a tank } \pi=\frac{2}{3}(r_2^{3}-r_1^{3}) \\ & =[\frac{2}{3} \times \frac{22}{7} \times{(1.01)^{3}-(1)^{3}}] m^{3} \\ & =[\frac{44}{21} \times(1.030301-1)] m^{3} \\ & =0.06348 m^{3} \quad \text{ (approximately) } \end{aligned} $

Solution

Let radius of sphere be $r$.

Surface area of sphere $=154 cm^{2}$

Therefore, the volume of the sphere is $179 \frac{2}{3} cm^{3}$.

Solution

(i) Cost of white-washing the dome from inside $=$ Rs 498.96

Cost of white-washing $1 m^{2}$ area $=$ Rs 2

Therefore, CSA of the inner side of dome $=(\frac{498.96}{2}) m^{2}$

$=249.48 m^{2}$

(ii) Let the inner radius of the hemispherical dome be $r$.

CSA of inner side of dome $=249.48 m^{2}$

$2 \pi r^{2}=249.48 m^{2}$

$ \begin{aligned} & \Rightarrow 2 \times \frac{22}{7} \times r^{2}=249.48 m^{2} \\ & \Rightarrow r^{2}=(\frac{249.48 \times 7}{2 \times 22}) m^{2}=39.69 m^{2} \end{aligned} $

$\Rightarrow r=6.3 m$

Volume of air inside the dome $=$ Volume of hemispherical dome

$=\frac{2}{3} \pi r^{3}$

$=[\frac{2}{3} \times \frac{22}{7} \times(6.3)^{3}] m^{3}$

$=523.908 m^{3}$

$=523.9 m^{3}$ (approximately)

Therefore, the volume of air inside the dome is $523.9 m^{3}$.

Solution

(i)Radius of 1 solid iron sphere $=r$

Volume of 1 solid iron sphere

$ \begin{aligned} = & \frac{4}{3} \pi r^{3} \\ & =27 \times \frac{4}{3} \pi r^{3} \end{aligned} $

Volume of 27 solid iron spheres

27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be $r^{\prime}$.

Volume of new solid iron sphere $=\frac{4}{3} \pi r^{3}$

$ \begin{aligned} & \frac{4}{3} \pi r^{\prime 3}=27 \times \frac{4}{3} \pi r^{3} \\ & r^{\prime 3}=27 r^{3} \\ & r^{\prime}=3 r \end{aligned} $

(ii) Surface area of 1 solid iron sphere of radius $r=4 \pi r^{2}$

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

$ \begin{aligned} & =4 n(3 r)^{2}=36 n r^{2} \\ & \frac{S}{S^{\prime}}=\frac{4 \pi r^{2}}{36 \pi r^{2}}=\frac{1}{9}=1: 9 \end{aligned} $

Solution

$ =(\frac{3.5}{2}) mm=1.75 mm $

Radius ( $r$ ) of capsule

Volume of spherical capsule $=\frac{4}{3} \pi r^{3}$

$ \begin{aligned} & =[\frac{4}{3} \times \frac{22}{7} \times(1.75)^{3}] mm^{3} \\ & =22.458 mm^{3} \\ & =22.46 mm^{3} \text{ (approximately) } \end{aligned} $

Therefore, the volume of the spherical capsule is $22.46 mm^{3}$.