Is barium sulfate soluble in dilute mineral acid?
The key point is to have an acid with such a high $\ce{H+}$ activity that in large extent shifts the below reaction to the left :
$$\ce{  HSO4-  + H2O &lt;=&gt; SO4^2- + H3O+ } \\ K_\mathrm{a2}=1.0 \cdot  10^{-2}$$

Then activity of sulphate ions from dissolution of $\ce{BaSO4}$ is decreased by the equilibrium reaction and the solubility increases.

$$\begin{align}
K_\mathrm{s}&amp;=a_{\ce{Ba^2+}}\cdot a_{\ce{SO4^2-}}=1.1\times 10^{-10}\\
\frac{K_\mathrm{a2}}{a_{\ce{H+}}}&amp;=\frac{a_{\ce{SO4^2-}}}{a_{\ce{HSO4^-}}}\\
\end{align}$$

The  effective solubility product 
$$K_\mathrm{s,ef}=K_\mathrm{s} \cdot    \frac{a_{\ce{H+}}}{K_\mathrm{a2}}$$

Note that despite activity coefficients {gamma} decrease with increasing ionic strength, for concentrated acids this tendency reverses and raises very rapidly

[HCl]  Gamma
0.0005 0.975
0.01 0.904
0.1 0.796
1 0.809
2 1.01
5 2.38
10 10.44
12 17.25


Therefore, $a_{\ce{H+}}$ of 10 M HCl = 104.4.

$$K_\mathrm{s,ef}=K_\mathrm{s} \cdot    \frac{104.4}{K_\mathrm{a2}}\\=1.1\cdot 10^{-10}\cdot 1.044\cdot 10^4=1.148\cdot 10^{-6} \mathrm{~ mol^2 L^{-2}}$$

Solubility would be then $ M \cdot \sqrt{K_\mathrm{s,ef}}=233.4 \cdot 0.001072=0.25\mathrm{~g/L}$

Conclusion: The $\ce{BaSO4}$ solubility in concentrated $\ce{HCl}$ is limited, in diluted one is very low. The solubility in concentrated sulphuric acid would be much higher, as $a_{\ce{H+}}$ would be very high.