The value of $2\underset{0}{\overset{\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}{\int <!--\; \int \; -->}}f(x)g(x)dx-<!--\; -\; -->\underset{0}{\overset{\frac{\mathrm{\pi <!--\; \pi \; -->}}{2}}{\int <!--\; \int \; -->}}g(x)dx$ is ______.

The correct answer is 0.

To find the value of the given expression, we consider the standard functions from this JEE Advanced problem defined on the interval $[0,\frac{\pi }{2}]$:
$f\left(x\right)={\sin }^{2}x$
$g\left(x\right)=\sqrt{\frac{\pi x}{2}-x^2}$

First, let us analyze the behavior of the function $g\left(x\right)$ under the reflection transformation $x\to \frac{\pi }{2}-x$:
$g(\frac{\pi }{2}-x)=\sqrt{\frac{\pi }{2}(\frac{\pi }{2}-x)-{(\frac{\pi }{2}-x)}^2}$
$=\sqrt{\frac{{\pi }^2}{4}-\frac{\pi x}{2}-(\frac{{\pi }^2}{4}-\pi x+x^2)}$
$=\sqrt{\frac{\pi x}{2}-x^2}=g\left(x\right)$

Thus, we have:
$g(\frac{\pi }{2}-x)=g\left(x\right)$

Now, let us define the first integral as $I$:
$I={\int }_0^{\frac{\pi }{2}}f\left(x\right)g\left(x\right)dx={\int }_0^{\frac{\pi }{2}}{\sin }^{2}xg\left(x\right)dx$

Using the property of definite integrals ${\int }_a^{b}h\left(x\right)dx={\int }_a^{b}h(a+b-x)dx$, we substitute $x\to \frac{\pi }{2}-x$:
$I={\int }_0^{\frac{\pi }{2}}{\sin }^2(\frac{\pi }{2}-x)g(\frac{\pi }{2}-x)dx$
Since ${\sin }^2(\frac{\pi }{2}-x)={\cos }^{2}x$ and $g(\frac{\pi }{2}-x)=g\left(x\right)$, we get:
$I={\int }_0^{\frac{\pi }{2}}{\cos }^{2}xg\left(x\right)dx$

Adding the two equations for $I$ together:
$2I={\int }_0^{\frac{\pi }{2}}{\sin }^{2}xg\left(x\right)dx+{\int }_0^{\frac{\pi }{2}}{\cos }^{2}xg\left(x\right)dx$
$2I={\int }_0^{\frac{\pi }{2}}({\sin }^{2}x+{\cos }^{2}x)g\left(x\right)dx$

Using the fundamental trigonometric identity ${\sin }^{2}x+{\cos }^{2}x=1$, the equation simplifies to:
$2I={\int }_0^{\frac{\pi }{2}}g\left(x\right)dx$

Substituting $I={\int }_0^{\frac{\pi }{2}}f\left(x\right)g\left(x\right)dx$ back, we find:
$2{\int }_0^{\frac{\pi }{2}}f\left(x\right)g\left(x\right)dx={\int }_0^{\frac{\pi }{2}}g\left(x\right)dx$

Subtracting ${\int }_0^{\frac{\pi }{2}}g\left(x\right)dx$ from both sides, we get the value of the target expression:
$2{\int }_0^{\frac{\pi }{2}}f\left(x\right)g\left(x\right)dx-{\int }_0^{\frac{\pi }{2}}g\left(x\right)dx=0$