Mass of a substance is changing with respect to its volume as m = -v^3 + 3v^2 -2v, what is the maximum specific gravity of the substance between v = 1 m^3 and v = 2 m^3? (Density of reference substance = 0.25 kg/m^3)(a) 1(b) 2(c) 3(d) 4I have been asked this question during a job interview.The query is from Specific Gravity in chapter Moles, Density & Concentration of Basic Chemical Engineering

Mass of a substance is changing with respect to its volume as m = -v^3

1.	Mass of a substance is changing with respect to its volume as m = -v^3 + 3v^2 -2v, what is the maximum specific gravity of the substance between v = 1 m^3 and v = 2 m^3? (Density of reference substance = 0.25 kg/m^3)(a) 1(b) 2(c) 3(d) 4I have been asked this question during a job interview.The query is from Specific Gravity in chapter Moles, Density & Concentration of Basic Chemical Engineering
Answer» Right option is (a) 1 The explanation: Density of substance D = m/V = -v^2 + 3v – 2, => Specific GRAVITY of substance S = D/0.25 = 4(-v^2 + 3v – 2), => dS/dv = 4(-2v + 3) = 0, => v = 1.5, => S is maximum at v = 1.5, => maximum specific gravity = 4(-1.5^2 + 3*1.5 – 2) = 1.

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