QJEEM24Ph07a9: JEE Main 2024 Physics - Class 11 JEEM24Ph07a9 Solution: A point source is emitting sound waves of intensity at the…

To solve this problem, we need to understand the relationship between the intensity of sound waves and the distance from the source. The intensity $I$ of sound waves from a point source decreases with the square of the distance $r$ from the source, according to the inverse square law, which can be expressed as:

$I \propto \frac{1}{r^2}$

This means that if the distance is doubled, the intensity becomes one-fourth of its initial value because $(2r)^2 = 4r^2$ .

The initial intensity given at the origin (source) is:

$I_0 = 16 \times 10^{-8} \mathrm{~Wm}^{-2}$

Let's call $I_1$ the intensity at $r = 2 \text{ m}$ and $I_2$ the intensity at $r = 4 \text{ m}$ . Using the inverse square law, we can write:

$I_1 = \frac{I_0}{(2)^2} = \frac{I_0}{4}$

and

$I_2 = \frac{I_0}{(4)^2} = \frac{I_0}{16}$

Now substitute the given value for $I_0$ to find $I_1$ and $I_2$ :

$I_1 = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \mathrm{~Wm}^{-2}$

$I_2 = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \mathrm{~Wm}^{-2}$

Now to find the difference in intensity (magnitude only) between the two points, we subtract $I_2$ from $I_1$ :

$\Delta I = | I_1 - I_2 |$

$\Delta I = | 4 \times 10^{-8} - 1 \times 10^{-8} |$

$\Delta I = 3 \times 10^{-8} \mathrm{~Wm}^{-2}$

So the difference in intensity (magnitude only) at the two points is:

$3 \times 10^{-8} \mathrm{~Wm}^{-2}$