Subject: General Aptitude · Chapter: Volume and Surface Area · Exam: 2009 · Marks: · Difficulty:
A cistern, open at the top, is to be lined with sheet of lead which weights $27 \mathrm{kg} / \mathrm{m}^{2}$. The cistern is 4.5 m long and 3 m wide and holds $50 \mathrm{m}^{3}$. The weight of lead required is
(a)1660.5 kg
(b)1764.5 kg
(c)1860.5 kg
(d)1864.5 kg
Answer
Answer (as printed): D
Explanation
Let the depth of the cistern be $h$ metres. Then, $4.5 \times 3 \times h=50 \Rightarrow h=\frac{50}{13.5}=\frac{100}{27}$. Area of sheet required $=l b+2(b h+l h)=l b+2 h(l+b)$ $$\begin{aligned} & =\left[4.5 \times 3+2 \times \frac{100}{27}(4.5+3)\right] \mathrm{m}^{2} \\ & =\left(13.5+\frac{200}{27} \times 7.5\right) \mathrm{m}^{2}=\left(\frac{27}{2}+\frac{500}{9}\right) \mathrm{m}^{2}=\frac{1243}{18} \mathrm{m}^{2} . \\ & \therefore \text { Weight of lead }=\left(27 \times \frac{1243}{18}\right) \mathrm{kg}=\left(\frac{3729}{2}\right) \mathrm{kg}=1864.5 \mathrm{kg} . \end{aligned}$$