Correct choice is (b) 7.5 Ω
To explain I would say: Hybrid parameter h11 is given by, h11 = \(\frac{V_1}{I_1}\), when V2 = 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
∴ \(\frac{V_1}{I_1}\) = 7.5
Hence h11 = 7.5 Ω.