Correct Answer - Option 3 : 0.1

**Concept:**

As the speed of the flywheel changes from maximum and minimum speeds i.e. ω1 to ω2, the maximum fluctuation of energy:

\({\rm{Δ }}E = I{ω ^2}{C_s}\)

Mean Speed, \(ω = \frac{{{ω _1}\ +\ {ω _2}}}{2} \)

Coefficient of fluctuation of speed, \({C_s} = \frac{{{ω _1}\ -\ {ω _2}}}{ω } \)

**Calculation:**

**Given:**

ω_{1} = 110 rad/s, ω_{2} = 90 rad/s, ΔE = 200 N-m

\(ω = \frac{{{ω _1}\ +\ {ω _2}}}{2} =\frac{110\;+\;90}{2}=100\) rad/s

\({C_s} = \frac{{{ω _1}\; - \;{ω _2}}}{ω } = \frac{{110\ -\ 90}}{{100}} = \frac{{20}}{{100}}=0.2\)

\(I= \frac{{{\rm{Δ }}E}}{{{ω ^2}{C_s}}} = \frac{{200}}{{{{\left( {100} \right)}^2}\ \times\ 0.2}} = 0.1\)