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

Solution

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/kg 

= 1864.5


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