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 $4.5 \times 3 \times h$ = 50

So, h = $50/ 13.5$

h = 100/27

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

= lb + 2h(l + b)

= $[4.5 \times 3 + 2 \times 100/ 27 (4.5 + 3)]$ sq m

= $(13.5 + 200/ 27 \times 7.5)$

= $(27/ 2 + 500/ 9)$

= $1243/ 18$

Therefore weight of lead = $(27 \times 1243/ 18)$

= $3729/ 2$ kg

= 1864.5

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