When the load resistance of an amplifier is modified from 4 k to 8...

The voltage gain $A_v$ of an amplifier is given by the relation:

$A_v = \dfrac{\mu R_L}{r_p + R_L}$

where $\mu$ is the amplification factor, $R_L$ is the load resistance, and $r_p$ is the internal resistance.

For the first case, $R_L = 4 \text{ k}\Omega$ and $A_v = 10$:

$10 = \dfrac{4\mu}{r_p + 4}$

$10r_p + 40 = 4\mu \Rightarrow 5r_p + 20 = 2\mu$

For the second case, $R_L = 8 \text{ k}\Omega$ and $A_v = 12$:

$12 = \dfrac{8\mu}{r_p + 8}$

$12r_p + 96 = 8\mu \Rightarrow 3r_p + 24 = 2\mu$

Equating the expressions for $2\mu$:

$5r_p + 20 = 3r_p + 24$

$2r_p = 4 \Rightarrow r_p = 2 \text{ k}\Omega$

Substituting the value of $r_p$ into the equation for $2\mu$:

$2\mu = 5(2) + 20 = 30$

$\mu = 15$