Right answer is (b) 51 kg
To explain: Given N = 1200 rpm, ΔE = 2kJ = 2000 J, D = 1m, Cs = 0.02
Mean angular speed of engine,
ω = 2πN/60 = 125.66 rad/ sec
Fluctuation of energy of the flywheel is given by,
ΔE = Iω^2Cs = 1/2mR^2ω^2Cs For solid disc I = mR^2/2
m = 51 kg