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

The equation of the ellipse is $\dfrac{x^2}{16} + \dfrac{y^2}{b^2} = 1$. For this ellipse, $a^2 = 16$. The foci are at $(\pm ae, 0)$, where $e = \sqrt{1 - \dfrac{b^2}{16}}$. Thus, $ae = \sqrt{16 - b^2}$. The foci are at $(\pm \sqrt{16 - b^2}, 0)$.

The equation of the hyperbola is $\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}$. Dividing by $\dfrac{1}{25}$, we get $\dfrac{x^2}{144/25} - \dfrac{y^2}{81/25} = 1$. Here, $a^2 = \dfrac{144}{25}$ and $b^2 = \dfrac{81}{25}$.

For the hyperbola, the eccentricity $e_h$ is given by $e_h = \sqrt{1 + \dfrac{b^2}{a^2}} = \sqrt{1 + \dfrac{81/25}{144/25}} = \sqrt{1 + \dfrac{81}{144}} = \sqrt{\dfrac{144 + 81}{144}} = \sqrt{\dfrac{225}{144}} = \dfrac{15}{12} = \dfrac{5}{4}$.

The foci of the hyperbola are at $(\pm a_h e_h, 0)$, where $a_h = \sqrt{\dfrac{144}{25}} = \dfrac{12}{5}$. Thus, $a_h e_h = \dfrac{12}{5} \times \dfrac{5}{4} = 3$. The foci are at $(\pm 3, 0)$.

Equating the foci of the ellipse and the hyperbola, we have $\sqrt{16 - b^2} = 3$. Squaring both sides, $16 - b^2 = 9$, which gives $b^2 = 16 - 9 = 7$.

Answer: $7$