A triode has a plate resistance of 10 k and an amplification factor...

For a triode, the change in plate current $\Delta I_p$ is given by:

$\Delta I_p = \dfrac{\Delta V_p}{r_p} + g_m \Delta V_g$

where $g_m = \dfrac{\mu}{r_p}$.

To bring the plate current back to its initial value of $10 \text{ mA}$, the net change in plate current from the initial state must be zero ($\Delta I_p = 0$).

Substituting $\Delta I_p = 0$ into the equation:

$0 = \dfrac{\Delta V_p}{r_p} + \dfrac{\mu}{r_p} \Delta V_g$

$\Delta V_p = -\mu \Delta V_g$

The change in grid voltage from the initial state is:

$\Delta V_g = -5.0 - (-7.5) = 2.5 \text{ V}$

Using the given amplification factor $\mu = 20$:

$\Delta V_p = -20 \times 2.5 = -50 \text{ V}$

The new plate voltage is:

$V_p = 250 - 50 = 200 \text{ V}$

Alternatively, from the intermediate state ($I_p = 15 \text{ mA}$, $V_p = 250 \text{ V}$), to reduce the current to $10 \text{ mA}$ ($\Delta I_p = -5 \text{ mA}$) at a constant grid voltage:

$\Delta V_p = \Delta I_p \times r_p = (-5 \times 10^{-3}) \times (10 \times 10^3) = -50 \text{ V}$

The required plate voltage is $250 - 50 = 200 \text{ V}$.

Answer: $200 \text{ V}$