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Waves question

2024 · 30 Jan · Shift 2 · Q85
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Waves question

2024 · 30 Jan · Shift 2 · Q85

JEE MainPhysicsWavesNumerical+4 / −1
A point source is emitting sound waves of intensity 16×10−8 Wm−216 \times 10^{-8} \mathrm{~Wm}^{-2}16×10−8 Wm−2 at the origin. The difference in intensity (magnitude only) at two points located at a distances of 2m2 m2m and 4m4 m4m from the origin respectively will be ‾\underline{\hspace{2cm}}​×10−8 Wm−2\times 10^{-8} \mathrm{~Wm}^{-2}×10−8 Wm−2.
Numerical answer
View written solutionFree

Correct answer: 3

  1. For a point source, sound intensity varies inversely as the square of distance:

I∝1r2I \propto \frac{1}{r^2}I∝r21​

So,

I=kr2I = \frac{k}{r^2}I=r2k​

where kkk is a constant.

  1. Given that the source emits sound waves of intensity

16×10−8 Wm−216 \times 10^{-8}\ \text{Wm}^{-2}16×10−8 Wm−2

This is the intensity at unit distance, so

k=16×10−8k = 16 \times 10^{-8}k=16×10−8

Hence,

I(r)=16×10−8r2I(r) = \frac{16 \times 10^{-8}}{r^2}I(r)=r216×10−8​

  1. Intensity at r=2 mr = 2\,\text{m}r=2m:

I1=16×10−822=16×10−84=4×10−8 Wm−2I_1 = \frac{16 \times 10^{-8}}{2^2} = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8}\ \text{Wm}^{-2}I1​=2216×10−8​=416×10−8​=4×10−8 Wm−2

  1. Intensity at r=4 mr = 4\,\text{m}r=4m:

I2=16×10−842=16×10−816=1×10−8 Wm−2I_2 = \frac{16 \times 10^{-8}}{4^2} = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8}\ \text{Wm}^{-2}I2​=4216×10−8​=1616×10−8​=1×10−8 Wm−2

  1. Difference in intensity:

∣I1−I2∣=∣4−1∣×10−8=3×10−8 Wm−2|I_1 - I_2| = |4 - 1| \times 10^{-8} = 3 \times 10^{-8}\ \text{Wm}^{-2}∣I1​−I2​∣=∣4−1∣×10−8=3×10−8 Wm−2

Therefore, the required integer is:

3\boxed{3}3​

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