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

Given:
Plate resistance, $r_p = 10 \text{ k}\Omega$
Amplification factor, $\mu = 20$
Initial plate current, $I_{p1} = 10 \text{ mA}$
Final plate current, $I_{p2} = 15 \text{ mA}$
Initial grid voltage, $V_{g1} = -7.5 \text{ V}$

The mutual conductance $g_m$ is given by:
$g_m = \dfrac{\mu}{r_p} = \dfrac{20}{10 \text{ k}\Omega} = 2 \text{ mA/V}$

The change in plate current is:
$\Delta I_p = I_{p2} - I_{p1} = 15 \text{ mA} - 10 \text{ mA} = 5 \text{ mA}$

For a constant plate voltage ($\Delta V_p = 0$), the change in plate current is related to the change in grid voltage by:
$\Delta I_p = g_m \Delta V_g$

Substituting the values:
$5 \text{ mA} = 2 \text{ mA/V} \times \Delta V_g$
$\Delta V_g = 2.5 \text{ V}$

The new grid voltage $V_{g2}$ is:
$V_{g2} = V_{g1} + \Delta V_g = -7.5 \text{ V} + 2.5 \text{ V} = -5.0 \text{ V}$