In a silicon transistor, the base current is changed by $ 20 $ change of $ 0.02V $ in base to emitter voltage and a change of $ 2mA $ in the collector current.
Find the input resistance, $ {\beta _{ac}} $ .
If the transistor is used as an amplifier. Find the voltage gain of the amplifier with the load resistance $ 5k $ amplifier.
A. $ a)100 \\
       b) - 500 \\ $
B. $ a)1000 \\
              b) - 500 \\ $
C. $ a)2000 \\
       b) - 5000 \\ $
D.$ a)10 \\
       b) - 5 \\ $

Hint :Here first we first convert the units into proper form, then use the formula to find the input resistance first then find out the current gain for ac alters by collector and base currents. Then voltage gain can be calculated by the additional load that is given as $ 5k $ .
For input resistance, $ {R_i} = \dfrac{{\vartriangle {V_i}}}{{\vartriangle {I_b}}} $ .
For alter beta current gain; $ {\beta _{ac}} = \dfrac{{\vartriangle {I_c}}}{{\vartriangle {I_b}}} $ , and voltage gain,
 $ {A_v} = - {\beta _{ac}}\dfrac{{{R_0}}}{{{R_i}}} $ .

Complete Step By Step Answer:
In order to solve this question first see what we have,
We have, change in collector current that is $ 2mA $ and base current which is $ 20\mu A $ .
We have also changed the input voltage as $ 0.02V $ .
Now,
 $ {R_i} = \dfrac{{\vartriangle {V_i}}}{{\vartriangle {I_b}}} = \dfrac{{0.02V}}{{20\mu A}} = 1000\Omega $
To find the alter ac current gain we have;
 $ {\beta _{ac}} = \dfrac{{\vartriangle {I_c}}}{{\vartriangle {I_b}}} = \dfrac{{2mA}}{{20\mu A}} = 100 $
Also,
$ {g_m} = \dfrac{{\vartriangle {I_c}}}{{\vartriangle {V_i}}} = \dfrac{{2mA}}{{0.02V}} = \dfrac{{2 \times {{10}^{ - 3}}}}{{0.02}} = 0.1{\Omega ^{ - 1}} \\
    \\ $
In the third part of the question, when it is used as an amplifier;
Load gain is given as $ 5k $ .
So, voltage gain;
 $ {A_v} = - {\beta _{ac}}\dfrac{{{R_0}}}{{{R_i}}} = - 100 \times \dfrac{{5k\Omega }}{{1k\Omega }} = - 500 $
Since $ {g_m} $ is equal to $ \dfrac{{{\beta _{ac}}}}{n} $ .
Option A is the correct answer.

Note :
Here note that before applying the formula it is required to convert the units. The solution above in the current gain reveals that for an ac input at the base, the collector current will be above hundred times the magnitude of the base current. Generally AC and DC betas tend to have similar magnitudes at lower levels.