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

$\begin{array}{l}\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}\\ a=\sqrt{\frac{144}{25}},b=\sqrt{\frac{81}{25}},e=\sqrt{1+\frac{81}{144}}=\frac{15}{12}=\frac{5}{4}\end{array}$

$\therefore \text{ Foci }=(\pm 3,0)$

$\therefore$   foci of ellipse= foci of hyperbola 

$\therefore$   for ellipse $ae=3$ but $a=4$,

$\therefore e=\frac{3}{4}$

Then $b^2=a^2\left(1-e^2\right)$

$\Rightarrow b^2=16\left(1-\frac{9}{16}\right)=7$