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The problem states that the intensity \( I \) of the sound is inversely proportional to the square of the distance \( d \). Mathematically, this is written as:

\[
I \propto \frac{1}{d^2}
\]

This can be expressed as:

\[
I = \frac{k}{d^2}
\]

where \( k \) is a constant of proportionality. We are given that when \( d = 2 \), \( I = 30 \). Using this information, we can find \( k \).

\[
30 = \frac{k}{2^2} = \frac{k}{4}
\]

Solving for \( k \):

\[
k = 30 \times 4 = 120
\]

Now that we know \( k = 120 \), we can find \( I \) when \( d = 10 \):

\[
I = \frac{120}{10^2} = \frac{120}{100} = 1.2
\]

Thus, the intensity \( I \) when \( d = 10 \) is **1.2 watts/m²**.

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Would you like further clarification or details on any step? Here are some related questions to explore this concept further:

1. What is the general formula for inverse proportionality?
2. How would the intensity change if the distance was halved?
3. What happens to the intensity if the distance triples?
4. How do you derive the constant of proportionality in problems like this?
5. How is sound intensity affected by obstacles or mediums other than air?

**Tip:** In problems involving inverse proportionality, doubling the distance reduces the intensity by a factor of 4, as intensity decreases with the square of the distance.

The intensity of the sound, I watts/m², received from a loudspeaker is inversely proportional to the square of the distance, d meters, from the loudspeaker. When d = 2, I = 30. Work out the value of I when d = 10.

Learn how to calculate the intensity of sound based on inverse proportionality to distance squared. The problem involves finding the constant of proportionality and applying it to a new distance. Suitable for students in Grades 9-11.

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