2. $p, q$ are natural numbers, and there are $11 q-1 \geqslant 15 \psi, 10 p \geqslant 7 q$ $+1, \frac{7}{10}<\frac{q}{p}<\frac{11}{15}$. Then the smallest $q=$ $\qquad$ .
2. $q=7$.

Solve $\left.\begin{array}{l}11 q-1 \geqslant 15 p, \\ 10 p \geqslant 7 q+1\end{array}\right\} \Rightarrow 11 q-1 \geqslant \frac{3}{2}(7 q+1) \Rightarrow q$
$\geqslant 5$. When $q=5,6$, check and find it does not meet the requirements, $q=7$ when, $p=5$.