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GATE 2025 CE (CE1) – Question 46

Air Pollution and Municipal Solid Wastes · Municipal solid wastes: generation, collection, transportation and management · 2 marks · Multiple select

Organic fraction of municipal solid waste (OFMSW) with bulk density of 315 kg/m$^3$ and water content of 30% is mixed with municipal sludge of bulk density 700 kg/m$^3$ and water content of 70%, such that the water content of the mixture is 40%. The amount (in kg) of sludge to be mixed per kg of OFMSW (rounded off to 2 decimal places) and the density of the mixture (in kg/m$^3$) (rounded off to the nearest integer) are calculated. Which of the following options is/are true:

  1. 0.33 kg of sludge added per kg of OFMSW
  2. Density of the mixture is 365 kg/m$^3$
  3. 0.66 kg of sludge added per kg of OFMSW
  4. Density of the mixture is 450 kg/m$^3$

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Show answer and explanation

Correct answer: (A) 0.33 kg of sludge added per kg of OFMSW; (B) Density of the mixture is 365 kg/m$^3$

Explanation

Per kg of OFMSW, let $x$ kg of sludge be added. The water is $0.30 + 0.70x$ kg out of a total of $1 + x$ kg, so $\frac{0.30 + 0.70x}{1 + x} = 0.40$, which gives $0.3x = 0.1$ and $x = 0.33$ kg. The volume is $\frac{1}{315} + \frac{0.333}{700} = 0.003175 + 0.000476 = 0.003651$ m$^3$ for a mass of 1.333 kg, so the density is $\frac{1.333}{0.003651} = 365$ kg/m$^3$. So A and B are true.