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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 meters
Then 4.5×3×h = 50
So, h = 50/13.5
h = 100/27
Area of sheet required = lb + 2 (bh + lh)
= lb + 2h(l + b)
= [4.5×3+2×100/27(4.5+3)] sq m
= (13.5+200/27×7.5)
= (27/2+500/9)
= 1243/18
Therefore weight of lead = (27×1243/18)
= 3729/2 kg
= 1864.5
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