SSC JE Civil Engineering 30-Oct-2020 Shift 1 Paper

'X' % of Moisture content is a sludge means that 'X; kg of water + (100 - x) kg of solids leads to formation of 100 Kg wet sludge.
Total solids = 210 kg/day (Given)
Moisture Content of sludge = 95% (Given)
95 kg of water + 5 kg of solids = 100 kg of wet sludge
5 kg of solids → 100 kg of wet sludge
1 kg of solids → $\frac{100}{5}$ = 20 kg of wet sludge
210 kg of solids → (20 × 210) = 4200 kg/day of wet sludge
Weight of wet sludge in one day = 4200 kg/day
Density of wet sludge = ρws = 1.05 × 1000 = 1050 kg/m${}^3$
Volume of wet sludge, ${\mathrm{V}}_{\mathrm{ws}}=\frac{\mathrm{m}}{{\mathrm{\rho }}_{\mathrm{ws}}}=\frac{4200}{1050}=4{\mathrm{m}}^3$