Correct Answer: $\sqrt{1-\frac{w^1}{w}}$

Solution : Given: The area of the circle with radius $y$ is $w$. The difference between the areas of the bigger circle (with radius $y$) and that of the smaller circle (with radius $x$) is $w^1$.
We know that the area of the circle is $\pi r^2$.
According to the question,
$\pi y^2-\pi x^2=w^1$
⇒ $\pi (y^2-x^2)=w^1$
⇒ $(y^2-x^2)=\frac{w^1}{\pi}$
Dividing the above equation by $y^2$, we get,
⇒ $(1-\frac{x^2}{y^2})=\frac{w^1}{\pi y^2}$
Since, $w=\pi y^2$
⇒ $(1-\frac{x^2}{y^2})=\frac{w^1}{w}$
⇒ $\frac{x^2}{y^2}=1-\frac{w^1}{w}$
⇒ $\frac{x}{y}=\sqrt{1-\frac{w^1}{w}}$
Hence, the correct answer is $\sqrt{1-\frac{w^1}{w}}$.