Surface Areas and Volumes Test - 27

Given, radius $$r=5.6$$ cm.

We know, the surface area of a sphere of radius $$ r= 4\pi { r}^{ 2 } $$

$$= 4 \times \dfrac {22}{7} \times 5.6 \times 5.6 $$

$$= 394.24  {cm}^{2} $$.

Therefore, option $$A$$ is correct.

The surface area of the cube of side '$$a$$' is given by,
$$S=6a^2$$
Here, $$a=0.5 $$ cm
$$\therefore$$ the total surface area of a cube $$=6(0.5)^2=1.5 cm^2$$

Diameter of the bowl $$ = \cfrac {Diameter}{2} = \cfrac {10.5}{2} = \cfrac {21}{4}$$

Volume of a hemisphere $$ = \cfrac { 2 }{ 3 } \pi { r }^{ 3 } $$

$$= \cfrac {2}{3} \times \cfrac {22}{7} \times \cfrac {21}{4} \times \cfrac {21}{4} \times \cfrac {21}{4}$$

$$ = 303.2  {cm}^{3} $$

One litre of milk requires $$ 1000{cm}^{3} $$ of volume. 

Hence in $$ 303.2 {cm}^{3}, \cfrac {303.2}{1000} = 0.303 $$ litres of milk can be filled.

Total surface area of a cylinder of Radius "$$R$$" and height "$$h$$" $$ = 2\pi R(R + h)$$
Radius of the base of the cylinder $$ = \dfrac {28}{2}  = 14  $$ cm

Hence, total surface area of the cylinder $$ =2\times \dfrac { 22 }{ 7 } \times 14(14 + 20) $$

$$= 2992 cm^2 $$

Surface area of a sphere $$ = 4\pi { r }^{ 2 } = 154  $$ 

$$ => 4 \times \cfrac {22}{7}\times { r }^{ 2 }  = 154  {cm}^{2} $$

$$ => { r }^{ 2 } = \cfrac {49}{4}  $$$$ =\cfrac {7}{2}  cm $$
Volume of a sphere $$ = \cfrac { 4 }{ 3 } \pi { r }^{ 3 } $$

$$= \cfrac {4}{3} \times \cfrac {22}{7} \times \cfrac {7}{2} \times \cfrac {7}{2} \times \cfrac {7}{2} = 179 \cfrac {2}{3}  {cm}^{3}$$

Given that, the inner diameter of a hemispherical dome is $$28\ m$$.

To find out: The volume of the dome.

Radius of the hemispherical dome, $$ r= \dfrac {28}{2} = 14  \ m$$

We know that, the volume of a hemisphere of radius $$r$$ is $$\dfrac { 2 }{ 3 }

\pi { r }^{ 3 }$$

$$\therefore \ Volume = \dfrac {2}{3} \times \dfrac {22}{7} \times 14 \times 14 \times 14\quad \quad \left[\because \ \pi=\dfrac{22}{7}\right]$$

$$\Rightarrow \dfrac{120736}{21}$$

$$\therefore \ Volume=5749.33  \ {m}^{3} $$

Hence, the volume of the given hemispherical dome is $$5749.33  \ {m}^{3}$$.

Let $$r$$ cm be the radius of the sphere. Its surface area $$=55.44{cm}^{2}$$
$$\Rightarrow$$ $$4\pi {r}^{2}=55.44$$
$$\Rightarrow$$ $$4\times \cfrac{22}{7}\times {r}^{2}=55.44$$
$$\Rightarrow$$ $${r}^{2}=\cfrac{55.44\times 7}{4\times 22}=4.41={(2.1)}^{2}$$
$$\Rightarrow$$ $$r=2.1cm$$
now, volume of the sphere$$=\cfrac{4}{3}\pi {r}^{3}=\cfrac{1}{3}(4\pi {r}^{2})\times r=\cfrac{1}{3}\times 55.44\times 2.1{cm}^{3}$$
$$=55.55\times 0.7{cm}^{3}$$
Hence the volume of the sphere $$=38.808{cm}^{3}$$

$$r=\cfrac{3.5}{2}mm$$
Capacity of the  capsule $$=\cfrac{4}{3}\pi {r}^{3}$$
$$=\cfrac{4}{3}\times \cfrac{22}{7}\times \cfrac{3.5}{2}\times \cfrac{3.5}{2}\times \cfrac{3.5}{2}{mm}^{3}$$
$$=\cfrac{4}{3}\times \cfrac{22}{7}\times \cfrac{7}{4}\times \cfrac{7}{4}\times \cfrac{7}{4}{mm}^{3}=\cfrac{11}{24}\times 49{mm}^{3}$$
$$=\cfrac{539}{24}{mm}^{3}=22.346{mm}^{3}$$

Radius of cylinder $$=3 \text{ cm}$$