$A$ transistor has a current amplification factor of $60$. In a $CE$ amplifier, input resistance is $1 \, k\Omega$ and output voltage is $0.01 \, V$. The transconductance is (in $SI$ units)

The output characteristics of an $n-p-n$ transistor represent,($I_C =$ collector current,$V_{CE} =$ potential difference between collector and emitter,$I_B =$ base current,$V_{BB} =$ voltage given to base,$V_{BE} =$ the potential difference between base and emitter)

The thinnest part of a transistor is . . . . . .

Assertion: In a transistor,the base is made thin.
Reason: $A$ thin base makes the transistor stable.

$A$ common emitter amplifier circuit,built using an $npn$ transistor,is shown in the figure. Its $dc$ current gain is $250$,$R_C = 1\,k\Omega$ and $V_{CC} = 10\,V$. What is the minimum base current for $V_{CE}$ to reach saturation? (in $\mu A$)

For a common base configuration of a $PNP$ transistor,$\frac{I_C}{I_E} = 0.96$. Then,the maximum current gain in common emitter configuration will be: