A cistern, open at the top, is to be lined with sheet of lead which weights 27 kg/m2. The cistern is 4.5 m long and 3 m wide and holds 50 m3. The weight of lead required is :
Options:
A . 1660.5 kg
B . 1764.5 kg
C . 1860.5 kg
D . 1864.5 kg
Answer: Option D Let the depth of the cistern be h metres Then, $$\eqalign{ & 4.5 \times 3 \times h = 50 \cr & \Rightarrow h = \frac{{50}}{{13.5}} \cr & \Rightarrow h = \frac{{100}}{{27}} \cr} $$ Area of sheet required : $$\eqalign{ & = lb + 2\left( {bh + lh} \right) \cr & = lb + 2h\left( {l + b} \right) \cr & = \left[ {4.5 \times 3 + 2 \times \frac{{100}}{{27}} \times \left( {4.5 + 3} \right)} \right]{{\text{m}}^2} \cr & = \left( {13.5 + \frac{{200}}{{27}} \times 7.5} \right){\text{ }}{{\text{m}}^2} \cr & = \left( {\frac{{27}}{2} + \frac{{500}}{9}} \right){{\text{m}}^2} \cr & = \frac{{1243}}{{18}}{\text{ }}{{\text{m}}^2} \cr} $$ ∴ Weight of lead : $$\eqalign{ & = \left( {27 \times \frac{{1243}}{{18}}} \right)kg \cr & = \left( {\frac{{3729}}{2}} \right)kg \cr & = 1864.5\,kg \cr} $$
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