Surface Areas and Volumes Test - 22

A coin is in the shape of a cylinder.

Volume of earth dug out

Curved Surface area of a frustum of a cone  $$=\pi l({ r }_{ 1 }+{ r }_{ 2 })$$ where $$l$$ is the slant height $$ = \sqrt { { h }^{ 2 }+{ ({ r }_{ 1 }-{ r }_{ 2 }) }^{ 2 } } $$ $$h$$ is the height. 
$$ { r }_{ 1 } $$ and $${ r }_{ 2 }$$ are the radii of the lower and upper ends of a frustum of a cone.

So, $$l = \sqrt { { 16  }^{ 2 }+{ (20 - 8) }^{ 2 } } $$

$$\therefore l = 20  cm $$
Hence, Curved surface area of this frustum of a cone  $$=\pi \times 20 \times (20 + 8) = 560 \pi  {cm}^{2} $$ 

We know that, the volume of the cuboid  =$$lbh$$ and the volume of the cylinder = $$\pi r^2h$$

Volume of iron block = volume of a cuboid  $$=(440\, \times\, 260\, \times\, 100)\, cm^{3}$$          {$$1m=100cm$$}

Internal radius of the pipe $$= 30 cm$$

External radius of the pipe = internal radius + thickness $$ = (30 + 5) cm = 35 cm.$$

Let the length of the pipe be $$h$$ cm.

Volume of hollow cylindrical pipe = (external volume) - (Internal volume)

$$=[\pi\, \times\, (35)^{2}\, \times\, h\, -\, \pi\,

\times\, (30)^{2}\, \times\, h]\, cm^{3}$$

= $$\pi h {(35)^{2}\, -\, \pi h (30)^{2}}\, cm^{3}$$

= $$(35^2-30^2)\, \pi h\, cm^{3} = (325\, \pi h)\,

cm^{3}$$

As per question,

Volume of cylindrical pipe = Volume of iron bar

$$\therefore\, 325\, \pi h\, =\, 440\, \times\, 260\, \times\, 100$$

$$\Rightarrow\, h\, =\, \displaystyle  \left(\frac{440\,

\times\, 260\, \times\, 100}{325}\, \times\, \frac{7}{22} \right)\, cm$$

$$\Rightarrow\, h\, =\, \displaystyle \left(\frac {11200}{100} \right)\, m=$$ 1$$12 m$$.

Hence, the length of the pipe is 112 m.

$$\because$$ Diameter of the base $$=10.5cm$$
$$\therefore$$ Radius of the base $$(r)=\cfrac{10.5}{2}cm=5.25cm$$
Slant height $$(l)=10cm$$
$$\therefore$$ Curved surface area of the cone
$$=\pi r l=\cfrac{22}{7}\times 5.25\times 10=165{cm}^{2}$$

Volume of the toy $$ = $$ Volume of the hemispherical part $$ + $$ Volume of conical part
Volume of a hemisphere of radius 'r' $$ = \frac { 2 }{ 3 } \pi { r }^{ 3 } $$

Volume of a cone $$ = \frac { 1 }{ 3 } \pi { r }^{ 2 }h $$  where r

is the radius of the base of the cone and h is the height.

The hemisphere and the cone will have the same radius r $$ = 4.2  cm $$

Since total length of the toy is $$ 10.2 $$ cm, the length of the conical part will be $$ 10.2 - 4.2 = 6  cm $$
Hence, Volume of the solid $$ =(\frac { 2 }{ 3 } \times \frac {22}{7} \times { 4.2 }^{ 3 }) +  (\frac { 1 }{ 3 } \times \frac {22}{7} \times { 4.2 }^{ 2 }\times 6 )

=  266.112  {cm}^{3} $$

Volume of the solid $$ = $$ Volume of the cylindrical part $$ + $$ Volume of conical part

Volume of Cylinder of Radius "R"(7cm) and height "h"(40cm) $$ = \pi { R }^{2 }h $$

Volume of a frustum of a cone $$ = \dfrac { 1 }{ 3 } \pi h({ R }^{ 2 }+{ r }^{ 2 }+Rr) $$, where $$h$$ is the height and $$R$$ and $$r$$ are the radii of the lower and upper ends of a frustum of a cone.

The amount of canvas used to make the tent $$ = $$ Curved surface area of cylindrical part $$ + $$ Curved surface area of the conical part.

Curved Surface Area of a cylinder of radius $$R$$ and height $$h$$ $$= 2\pi Rh$$

Total surface area of the cone $$=\pi r(l+r)$$
$$=\cfrac{22}{7}\times 7\times (14+7)=\cfrac{22}{7}\times 7\times 21=462{cm}^{2}$$
Hence the total surface area of the cone is $$462{cm}^{2}$$