In the given circuit, the voltage across the base emitter junction is
 

 

Answer (B)

Since R₁ and R₂ are in series, the equivalent resistance is R_eq= R₁+R₂=16+2=18$\Omega$

Let us consider V₂ be the voltage at point Q.

Then the current through the series resistance is 

$i=\frac{9-0}{18}=\frac{1}{2}=0.5 mA$

At point Q

$\left(V_2-0\right)=0.5\times 2=1 mA$