Question 50

A cistern, open at the top. is to be lined with sheet lead which weights $$27 kg/m^3$$". The cistern is 4.5 m long and 3 m wide and holds 50 $$m^3$$'. The weight of lead required is

Let the depth of the cistern be h metersΒ 

ThenΒ 4.5Γ—3Γ—hΒ = 50Β 

So, h =Β 50/13.5

h = 100/27

Area of sheet required = lb + 2 (bh + lh)Β 

= lb + 2h(l + b)

Β =Β [4.5Γ—3+2Γ—100/27(4.5+3)]Β sq mΒ 

Β =Β (13.5+200/27Γ—7.5)

Β =Β (27/2+500/9)

Β =Β 1243/18

Therefore weight of lead =Β (27Γ—1243/18)

=Β Β 3729/2Β kgΒ 

= 1864.5

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