If ${R_p} = 7 \text{ k}\Omega$ and ${g_m} = 2.5 \text{ millimho}$, then on increasing the plate voltage by $50 \text{ V}$, how much must the grid voltage be changed so that the plate current remains the same? (in $V$)

The grid in a triode valve is used

The relation between $I_p$ and $V_p$ for a triode is $I_p = (0.125V_p - 7.5) \text{ mA}$. Keeping the grid potential constant at $1 \text{ V}$,the value of $r_p$ will be . . . . . . $\text{k}\Omega$.

In a diode,when there is a saturation current,the plate resistance will be

On applying a potential of $-1 \ V$ at the grid of a triode,the following relation between plate voltage $V_p$ $(V)$ and plate current $I_p$ (in $mA$) is found:
$I_p = 0.125 V_p - 7.5$
If on applying $-3 \ V$ potential at the grid and $300 \ V$ potential at the plate,the plate current is found to be $5 \ mA$,then the amplification factor of the triode is:

In the mutual characteristic curve of a triode,which point represents the cut-off voltage?