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 :

Solution(By Examveda Team)

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 :
$$\eqalign{ & = \left( {27 \times \frac{{1243}}{{18}}} \right)kg \cr & = \left( {\frac{{3729}}{2}} \right)kg \cr & = 1864.5\,kg \cr} $$