Question

A cistern, open at the top. is to be lined with sheet lead which weights 27 𝑘𝑔 / 𝑚3. The cistern is 4.5 m long and 3 m wide and holds 50 𝑚3. The weight of lead required is

• A. 1764.60 kg
• B. 1864.62 kg
• C. 1660.62 kg
• D. 1860.62 kg

Answer 1

Answer: (B)

1864.62 kg

Answer 2

Let the depth of the cistern be h meters
Then 4.5$\times$3$\times$h = 50

So, h $=\frac{50}{13.5}$

h $=\frac{100}{27}$

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

$=[\frac{4.5\times3+2\times100}{27(4.5+3)}]$ sq m

$=(\frac{13.5+200}{27\times7.5})$

$=(\frac{27}{2}+\frac{500}{9})$

$=\frac{1243}{18}$

Therefore weight of lead $=(\frac{27\times1243}{18})=\frac{3729}{2}$ kg
= 1864.5.
The correct option is (B)