Fraction of uncertainty resolved
An aside, but I don't really see the value in this given the posterior variance already measures the "remaining uncertainty" for the assumed model.
Consider a conjugate normal model, where, the prior variance is $$\sigma^2_0$$ and the posterior variance is
$$ \left(\frac{1}{\sigma_0^2} + \frac{n}{\sigma^2}\right)^{-1} $$
Therefore, in this example, the fraction resolved is
$$ \frac{n\sigma_0^2}{n\sigma_0^2 + \sigma^2} $$
For simplicity, suppose that $$\sigma^2=1$$ then this is just
$$ \frac{n\sigma_0^2}{n\sigma_0^2 + 1} $$
Now, suppose that $$n=1, \sigma_0^2=1$$, then
$$ \begin{aligned} \mathbb{V}[\theta|x] &= 1/2\ \text{FR} &= 1/2 \end{aligned} $$
Or $$\sigma^2_0 = 1/2$$
$$ \begin{aligned} \mathbb{V}[\theta|x] &= 1/3\ \text{FR} &= 1/3 \end{aligned} $$
Alternatively, suppose that $$n=1,\sigma_0^2=10$$ then
$$ \begin{aligned} \mathbb{V}[\theta|x] &= 10/11\ \text{FR} &= 10/11 \end{aligned} $$
So, a less informative prior results in a much higher fraction resolved for the same number of observations. The posterior variance is higher in this case, but we have resolved a higher fraction of uncertainty. Why consider the fraction resolved instead of the actual uncertainty from the posterior variance?
Suppose instead $$n=2,\sigma_0^2=10$$
$$ \begin{aligned} \mathbb{V}[\theta|x] &= 10/21\ \text{FR} &= 20/21 \end{aligned} $$
Here, the posterior variance is about 1/2 the same as for $$\sigma_0^2=1,n=1$$, but in this case we have resolved >95% of the uncertainty. Compared to only 1/2 resolved for the stronger prior.
To resolve lots of uncertainty, we just need to specify a diffuse prior, which then results in higher actual uncertainty. Parameter combinations with more diffuse priors will have more uncertainty resolved with fewer observations but may have higher overall variance.