If $-1<\sqrt{p}-\sqrt{100}<1$, then $-1<\sqrt{p}-10<1$ or $9<\sqrt{p}<11$. Since $\sqrt{p}$ is greater than 9, then $p$ is greater than $9^{2}=81$. Since $\sqrt{p}$ is less than 11, then $p$ is less than $11^{2}=121$. In other words, $81<p<121$. Since $p$ is a positive integer, then $82 \leq p \leq 120$. Therefore, there are $120-82+1=39$ such integers $p$.