Two metal rods P and Q of equal original length have linear expansivities in the ratio 2:3. If they are heated through the same temperature difference, find the ratio of the increase in their lengths.

For metal P:

$$\Delta L_P = L_{0P} \alpha_P \Delta T$$

For metal Q:

$$\Delta L_Q = L_{0Q} \alpha_Q \Delta T$$

The ratio of linear expansivities of P to Q is $\frac{\alpha_P}{\alpha_Q} = \frac{2}{3}$.
The ratio of their original lengths (P to Q) is $\frac{L_{0P}}{L_{0Q}} = \frac{3}{4}$.

$$\frac{\Delta L_P}{\Delta L_Q} = \left(\frac{3}{4}\right) \times \left(\frac{2}{3}\right)$$

$$\frac{\Delta L_P}{\Delta L_Q} = \frac{3 \times 2}{4 \times 3} = \frac{6}{12} = \frac{1}{2}$$

The ratio of the increase in lengths of P to Q is 1:2.

Comparing this with the given options:

(a) 1:2

(b) 2:1

(c) 8:9

(d) 9:8

The correct option is (a).

The final answer is $\boxed{\text{1:2}}$.