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

Given ellipse: $\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1$ ...(i)
Given hyperbola: $\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}$ ...(ii)
**For Hyperbola:**
Rewriting (ii) in standard form:
$\dfrac{x^2}{\dfrac{144}{25}} - \dfrac{y^2}{\dfrac{81}{25}} = 1$
Here, $a^2 = \dfrac{144}{25}$ and $b^2 = \dfrac{81}{25}$
For hyperbola, $c^2 = a^2 + b^2$
$\Rightarrow c^2 = \dfrac{144}{25} + \dfrac{81}{25} = \dfrac{225}{25} = 9$
$\therefore c = 3$
So, foci of hyperbola $= (\pm 3,\ 0)$
**For Ellipse:**
Here, $a^2 = 16$
For ellipse, $c^2 = a^2 - b^2$
$\Rightarrow c^2 = 16 - b^2$
Since foci of ellipse and hyperbola coincide:
$\Rightarrow 16 - b^2 = 9$
$\Rightarrow b^2 = 16 - 9$
$\therefore \boxed{b^2 = 7}$