Question:

A cistern, open at the top. is to be lined with sheet lead which weights 27 π‘˜π‘” / π‘š3. The cistern is 4.5 m long and 3 m wide and holds 50 π‘š3. The weight of lead required is

Updated On: Sep 24, 2024
  • 1764.60 kg
  • 1864.62 kg
  • 1660.62 kg
  • 1860.62 kg
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The Correct Option is B

Solution and Explanation

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

So, h \(=\frac{50}{13.5}\) 

\(=\frac{100}{27}\) 

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)

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