An unknown transistor needs to be identified as a npn or pnp type. A multimeter, with +ve and -ve terminals, is used to measure resistance between different terminals of transistor. If terminal 2 is the base of the transistor then which of the following is correct for a pnp transistor?

First, recall the basic structure of a bipolar junction transistor. In a $$pnp$$ device the sequence of semiconductor types from emitter to collector is $$p \; - \; n \; - \; p$$. Therefore

$$\text{Emitter (E)} : p\text{-type}, \qquad \text{Base (B)} : n\text{-type}, \qquad \text{Collector (C)} : p\text{-type}. $$

A multimeter in the resistance (or diode-test) mode applies a small d.c. voltage internally: its red lead is at a higher potential than its black lead. So

$$\text{Red lead} = +\text{ve}, \qquad \text{Black lead} = -\text{ve}. $$

For a single $$p$$-$$n$$ junction we have the standard diode rule:

When $$p$$-side is connected to the positive lead and $$n$$-side to the negative lead, the junction is forward biased and the meter shows a low resistance.
When the polarities are reversed, the junction is reverse biased and the meter shows a high resistance.

Now, in a transistor there are two separate $$p$$-$$n$$ junctions: the emitter-base junction and the collector-base junction. Because the base of a $$pnp$$ transistor is $$n$$-type, both of those junctions have the $$n$$-type side at the base.

We are told that terminal 2 is the base. Let us examine the four possible measurements one by one, always applying the diode rule stated above.

Case A: $$\text{(+ve lead on 2, -ve lead on 3)}$$
Base (n) is at positive potential, terminal 3 (p) is at negative potential, so the junction is reverse biased. Hence the resistance must be high, not low. So statement A is incorrect.

Case B: $$\text{(+ve lead on 2, -ve lead on 1)}$$
Again the base (n) is positive and terminal 1 (p) is negative, giving reverse bias and therefore a high resistance. This agrees with what is written in option B, so option B is consistent.

Case C: $$\text{(+ve lead on 1, -ve lead on 2)}$$
Now terminal 1 (p) is positive and the base (n) is negative, so the emitter-base junction is forward biased. The meter should show a low resistance, but option C claims it is high; hence option C is wrong.

Case D: $$\text{(+ve lead on 3, -ve lead on 2)}$$
Here terminal 3 (p) is positive and the base (n) is negative, producing forward bias on the collector-base junction. The resistance should be low, yet option D says it is high, so option D is also wrong.

Among the four possibilities, only option B matches the correct behaviour of the two $$p$$-$$n$$ junctions in a $$pnp$$ transistor when the meter leads are connected as stated.

Hence, the correct answer is Option B.