Surface Areas and Volumes Test - 50

Outer Surface area of the building $$ = $$ Curved surface area of cylindrical part $$ + $$ Curved surface area if the conical part. Curved Surface Area of a Cylinder of Radius "R" and height "h" $$ = 2\pi Rh$$


Radius of the cylindrical part $$ = \frac {Diameter}{2} = \frac {4.2}{2}  = 2.1  m $$ Curved surface area of a cone $$= \pi rl$$  where r is the radius of the cone and l is the slant height.

Radius of the conical part $$ = \frac {Diameter}{2} = \frac {4.2}{2} = 2.1  m $$

For a cone, l $$ = \sqrt { {h }^{ 2 }+  {r}^{ 2 } } $$ where h is the height. 

Hence, l $$ = \sqrt { { 2.1 }^{2 }+  {2.1}^{ 2 } } $$

$$ l = 2.1 \sqrt{2}  m $$

Hence, surface area of building $$ = (2

\times \frac {22}{7} \times 2.1 \times 4) + (\frac {22}{7} \times 2.1\times 2.1 \sqrt {2} ) = 72.4 {m}^{2} $$
Volume of the building $$ = $$ Volume of the cylindrical part $ + $$ Volume of conical part

Volume of a Cylinder of Radius "R" and height "h" $$ = \pi{ R }^{ 2 }h $$

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.

Hence, Volume of the pillar $$ = (\frac { 22 }{ 7 } \times 2.1\times 2.1\times 4)

+  (\frac { 1 }{ 3 } \times \frac {22}{7} \times { 2.1 }^{ 2 }\times 2.1 )

=  65.142   {m}^{3} $$

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.
Hence, volume of the frustum of the cone $$ = \dfrac { 1 }{ 3 } \times \dfrac {22}{7} \times 21 ({ 8 }^{ 2 }+{ 18 }^{ 2 }+ 8 \times 18) =  11704  \ \text{cm}^{3} $$

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

Hence, in $$ 11704\ \text {cm}^{3} , \dfrac {11704}{1000} = 11.704 $$ litres of milk can be occupied.

Cost of milk which can completely fill the vessel at the rate of Rs. $$ 10 $$ per litre $$=$$ $$ 10 \times 11.704 = $$ Rs. $$  117.04 $$

Volume of the pillar $$ = $$ Volume of the cylindrical part $$ + $$ Volume of conical part
Volume of a Cylinder of Radius "R" and height "h" $$ = \pi{ R }^{ 2 }h $$

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.

Hence, Volume of the pillar $$ = (\frac { 22 }{ 7 } \times 8\times 8\times 240)

+  (\frac { 1 }{ 3 } \times \frac {22}{7} \times { 8 }^{ 2 }\times 36 ) = 50688  {cm}^{3} $$

If one cu cm wieighs $$ 7.8 $$ grams, then $$ 50688{cm}^{3} $$ weighs $$ 50688 \times 7.8 = 395366.4  $$ grams  or $$ 395.37  kg $$

Radius of first spherical ball $$ = \cfrac {2}{2} cm= 1  cm $$

Radius of second spherical ball $$ = \cfrac {2 \sqrt [3] {14.5}}{2} cm= \sqrt [3] {14.5}  cm $$

Let the radius of the third ball be $$ r $$ cm.

Radius of the main spherical ball $$ = \cfrac {5}{2}  cm  = 2.5  cm$$

Volume of main spherical ball $$ = $$ Volume of spherical ball  $$ 1 + $$Volume of spherical ball $$ 2 + $$ Volume of spherical ball $$ 3 $$ 

Volume of a sphere $$ = \cfrac { 4 }{ 3 } \pi { r }^{ 3 } $$

So, $$ \cfrac { 4 }{ 3 } \pi \times { 2.5 }^{ 3 } = \cfrac { 4 }{ 3 } \pi { 1 }^{ 3 } + \cfrac { 4 }{ 3 } \pi { (\sqrt [3] {14.5})  }^{ 3 } + \cfrac { 4 }{ 3 } \pi { r }^{ 3 } $$

$$ \Rightarrow  { 2.5 }^{ 3 } = { 1 }^{ 3 } + { (\sqrt [3] {14.5})  }^{ 3 } + { r }^{ 3 } $$

$$\Rightarrow { r }^{ 3 } = 15.625 - 1 - 14.5 $$

$$ \Rightarrow { r }^{ 3 } = 0.125 = 0.5 \times 0.5 \times 0.5 $$

$$ \Rightarrow r  = 0.5   $$

The radius of the third spherical ball is $$0.5\,\, cm$$
Hence, the diameter of the third spherical ball is $$ 2 \times 0.5\,cm = 1  cm $$.

The volume of the grain silo can be found by adding the volumes of all the solids of which it is composed. 

The silo is made up of a cylinder with height $$10$$ feet and base radius $$5$$ feet and two cones, each having height $$5$$ feet and base radius $$5$$ feet. 

The formulas volume of cylinder $$\pi { r }^{ 2 }{ h }$$ and volume of cone $$\dfrac { 1 }{ 3 } \pi { r }^{ 2 }{ h }$$ can be used to determine the total volume of the silo. 

Since the two cones have identical dimensions, the total volume, in cubic feet, of the silo is given by:  

$$V=\pi { (5) }^{ 2 }{ (10) }+(2)\left( \dfrac { 1 }{ 3 }  \right) \pi { (5) }^{ 2 }{ (5) }$$

$$=\left( \dfrac { 4 }{ 3 }  \right) (250)\pi$$

$$=1047.2$$ cubic feet.(approx)

Volume of block with dimensions