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 :

Answer: Option D

Let the depth of the cistern be h metres Then,
$$\begin{array}{rl}& 4.5\times 3\times h=50\\ & \Rightarrow h=\frac{50}{13.5}\\ & \Rightarrow h=\frac{100}{27}\end{array}$$
Area of sheet required :
$$\begin{array}{rl}& =lb+2\left(bh+lh\right)\\ & =lb+2h\left(l+b\right)\\ & =\left[4.5\times 3+2\times \frac{100}{27}\times \left(4.5+3\right)\right]{\text{m}}^2\\ & =\left(13.5+\frac{200}{27}\times 7.5\right){\text{m}}^2\\ & =\left(\frac{27}{2}+\frac{500}{9}\right){\text{m}}^2\\ & =\frac{1243}{18}{\text{m}}^2\end{array}$$
∴ Weight of lead :
$$\begin{array}{rl}& =\left(27\times \frac{1243}{18}\right)kg\\ & =\left(\frac{3729}{2}\right)kg\\ & =1864.5kg\end{array}$$