The problem states that $Y$ is inversely proportional to the square of $x$, and it provides the condition $Y = 6$ when $x = 4$. It asks for the value of $Y$ when $x = 8$.

Step-by-step Solution:

1. Since $Y$ is inversely proportional to $x^2$, we can write the relationship as: $Y \propto \frac{1}{x^2}$ This means: $Y = \frac{k}{x^2}$ where $k$ is a constant.

2. To find $k$, we use the condition that $Y = 6$ when $x = 4$: $6 = \frac{k}{4^2}$ $6 = \frac{k}{16}$ Solving for $k$: $k = 6 \times 16 = 96$

3. Now that we know $k = 96$, we can find $Y$ when $x = 8$. Using the equation: $Y = \frac{96}{8^2} = \frac{96}{64}$ Simplifying: $Y = \frac{3}{2}$

Thus, the value of $Y$ when $x = 8$ is $\frac{3}{2}$, which is $1 \frac{1}{2}$.

Answer:

The correct option is: $\boxed{1 \frac{1}{2}}$

Would you like further details or have any questions?

Here are some related questions to explore:

1. How do you solve proportionality problems in general?
2. What happens when $Y$ is directly proportional to $x^2$ instead?
3. How do you find the constant $k$ in an inverse proportionality problem?
4. What is the difference between inverse and direct proportionality?
5. How does the behavior of $Y$ change as $x$ increases in inverse square proportionality?

Tip: When dealing with proportionality, always identify the relationship and find the constant before applying it to new values.