What is the new steady state frequency, of a 10 MW load increase occurs in an area with frequency 50 Hz, governor regulation R = 0.1 pu., and frequency
dependent component of load change D = 0.8 p.u. on 100 MVA base?

Correct option is B

​$$\Delta P_D = 10 \, \text{MW (positive because load increases)} \\\text{Base MVA} = 100 \, \text{MVA} \\\text{Frequency } f = f_i = 50 \, \text{Hz} \\\\\Rightarrow D = 0.8 \, \text{pu} \\D = 0.8 \times \frac{\text{Base MVA}}{\text{Frequency}} \\D = 0.8 \times \frac{100}{50} \\\therefore D = 1.6 \, \text{MW/Hz} \\R = 0.1 \, \text{pu} \\\frac{1}{R} = \frac{1}{0.1} = 10 \, \text{pu} \\\frac{1}{R} = 10 \times \frac{\text{Base MVA}}{\text{Frequency}} \\\frac{1}{R} = 10 \times \frac{100}{50} \\\therefore \frac{1}{R} = 20 \, \text{MW/Hz} \\\\\text{Applying all in the below equation:} \\\Rightarrow \Delta f = f_n - f_i = \frac{-\Delta P_D}{\left(D + \frac{1}{R}\right)} \\\Rightarrow f_n - 50 = \frac{-10}{1.6 + 20} \\\Rightarrow f_n = -0.463 + 50 \\\\\therefore \text{New steady-state frequency} = f_n = 49.537 \, \text{Hz}$$​​