The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, then the value of  $b^2$  is 

hyperbola :

$\begin{array}{l}\frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1\\ a^2=\frac{144}{25},b^2=\frac{81}{25}\\ e=\sqrt{\frac{a^2+b^2}{a^2}}\\ =\sqrt{1+\frac{b^2}{a^2}}\end{array}$

$\begin{array}{l}\Rightarrow \sqrt{1+\frac{81}{144}}=\frac{15}{12}\\ foci=(\pm ae,0)\\ =(\pm 3,0)\\ fociofellipse=(\pm ae,0)\end{array}$

$\begin{array}{l}=(\pm 4e,0)\\ 4e=3\\ e=\frac{3}{4}\\ e^2=\frac{9}{16}\\ 1-\frac{b^2}{9^2}=\frac{9}{16}\\ \Rightarrow b^2=7\end{array}$