Right choice is (a) 0.133 Ω
To elaborate: Inverse Hybrid parameter g11 is given by, g11 = \(\frac{I_1}{V_1}\), when I2 = 0.
Therefore short circuiting the terminal Y-Y’, we get,
V1 = I1 ((5||5) + 5)
= I1 \(\left(\left(\frac{5×5}{5+5}\right) + 5\right)\)
= 7.5I1
∴ \(\frac{I_1}{V_1} = \frac{1}{7.5}\) = 0.133 Ω
Hence g11 = 0.133 Ω.