A cylindrical milk steel storage tank has 4.2 m as diameter and is 4.5 m high

(a) Find quantity of it can contain

(b) How much steel sheet was actually used to make this closed tank, if 1/7 of steel sheet was wasted in making the tank

Dear Student,
Formula used-
​​​​​Diameter=2 radius

Please find below the solution to the asked query:

Given :  Cylindrical milk steel storage tank has 4.2 m as diameter and is 4.5 m high .

a ) quantity of milk it can contain =  Volume of cylindrical milk steel storage tank 

We know volume of cylinder = $\mathrm{\pi }$r² h  , So

Volume of cylindrical milk steel storage tank = $\frac{22}{7} \times 2.1^2 \times 4.5 = 22 \times 0.3 \times 2.1 \times 4.5$ = 62.37 m³

We know 1 m³ =  1000 liter  , So

Quantity of milk it can contain = 62.37 m³ = 62370 liter                                      ( Ans )

b ) Steel sheet was actually used to make this closed tank , if $\frac{1}{7}$ of steel sheet was wasted in making the tank .

So,

Total steel sheet  = Total surface area of tank ( As we have closed tank ) + $\frac{1}{7}$Total surface area of tank

We know Total surface area of cylinder  = 2 $\mathrm{\pi }$ r ( r  + h ) , So

Total steel sheet  = $2 \times \frac{22}{7} \times 2.1 \left( 2.1 + 4.5 \right) + \frac{1}{7}\left[2 \times \frac{22}{7} \times 2.1 \left( 2.1 + 4.5 \right)\right] = 2 \times 22 \times 0.3 \times 6.6 + \frac{1}{7}\left[2 \times 22 \times 0.3 \times 6.6\right] = 87.12+\frac{1}{7}\times 87.12 = 87.12+12.4457 = 99.5657\approx \mathbf{99}\mathbf{.}\mathbf{57}\mathbf{ }{\mathbf{m}}^{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{(}\mathbf{ }\mathbf{Ans}\mathbf{ }\mathbf{)}$
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