A point source is emitting sound waves of intensity 16 times 10^-8 mathrm~W m^-2$16 \times 10^{-8} \mathrm{~W m}^{-2}$ at the origin. The difference in intensity (magnitude only) at two points located at a distance of 2 mathrm~m$2 \mathrm{~m}$ and 4 mathrm~m$4 \mathrm{~m}$ from the origin respectively will be ________ times 10^-8 mathrm~W m^-2$\times 10^{-8} \mathrm{~W m}^{-2}$.
Numerical Answer Type:
Enter a numerical valueAnswer: 3 to 3
Solution & Explanation
### Related Formula
I = fracP4pi r^2$$I = \frac{P}{4\pi r^2}$$I propto frac1r^2$$I \propto \frac{1}{r^2}$$
### Core Logic
For a point source, the intensity I$I$ is inversely proportional to the square of the distance from the source.
If the given intensity is at r_0 = 1mathrmm$r_0 = 1\mathrm{m}$ (usually implied if "at origin" is stated alongside a base value, though the phrasing "at the origin" is ambiguous), we can assume I_0 = 16 times 10^-8$I_0 = 16 \times 10^{-8}$.
The official NTA answer assumes the initial given intensity was at r=2mathrmm$r=2\mathrm{m}$. Wait, if I = 16 times 10^-8$I = 16 \times 10^{-8}$ was the source power factor or the intensity at r=2$r=2$, let's reverse-engineer the answer `3`.
If I propto 1/r^2$I \propto 1/r^2$, and I_1$I_1$ at r=2$r=2$ and I_2$I_2$ at r=4$r=4$:
I_2 = I_1 (r_1/r_2)^2 = I_1 (2/4)^2 = I_1 / 4$I_2 = I_1 (r_1/r_2)^2 = I_1 (2/4)^2 = I_1 / 4$.
Difference Delta I = I_1 - I_2 = I_1 - I_1/4 = 3 I_1 / 4$\Delta I = I_1 - I_2 = I_1 - I_1/4 = 3 I_1 / 4$.
If this difference equals 3, then I_1$I_1$ must be 4$4$. But 16 times 10^-8$16 \times 10^{-8}$ is given.
Wait, if I$I$ at 1mathrmm$1\mathrm{m}$ is 16 times 10^-8$16 \times 10^{-8}$:
I(2mathrmm) = frac162^2 = 4 times 10^-8$I(2\mathrm{m}) = \frac{16}{2^2} = 4 \times 10^{-8}$.
I(4mathrmm) = frac164^2 = 1 times 10^-8$I(4\mathrm{m}) = \frac{16}{4^2} = 1 \times 10^{-8}$.
Delta I = 4 - 1 = 3 times 10^-8$\Delta I = 4 - 1 = 3 \times 10^{-8}$.
This perfectly matches.
### Step 1: Calculate Intensity at r = 2m and r = 4m
Assume the intensity I_0 = 16 times 10^-8 mathrm~W m^-2$I_0 = 16 \times 10^{-8} \mathrm{~W m}^{-2}$ represents the reference intensity at r=1mathrmm$r=1\mathrm{m}$.
Intensity at r=2 mathrm~m$r=2 \mathrm{~m}$:
I_1 = fracI_02^2 = frac16 times 10^-84 = 4 times 10^-8 mathrm~W m^-2$$I_1 = \frac{I_0}{2^2} = \frac{16 \times 10^{-8}}{4} = 4 \times 10^{-8} \mathrm{~W m}^{-2}$$
Intensity at r=4 mathrm~m$r=4 \mathrm{~m}$:
I_2 = fracI_04^2 = frac16 times 10^-816 = 1 times 10^-8 mathrm~W m^-2$$I_2 = \frac{I_0}{4^2} = \frac{16 \times 10^{-8}}{16} = 1 \times 10^{-8} \mathrm{~W m}^{-2}$$
### Step 2: Calculate the Difference
Magnitude of intensity difference:
Delta I = |I_1 - I_2| = (4 - 1) times 10^-8 = 3 times 10^-8 mathrm~W m^-2$$\Delta I = |I_1 - I_2| = (4 - 1) \times 10^{-8} = 3 \times 10^{-8} \mathrm{~W m}^{-2}$$
### Pattern Recognition
Although the question text ("at the origin") was poorly drafted (intensity at r=0$r=0$ would be infinite), the numerical setup expects you to treat 16 times 10^-8$16 \times 10^{-8}$ as the P/4pi$P/4\pi$ constant multiplier for the 1/r^2$1/r^2$ dropoff.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Keywords:#difference in intensity#JEE Main 2024 Evening Q59#Waves JEE Main 2024#Sound Intensity and Distance JEE Main 2024
More Waves Previous-Year Questions — Page 4
Q45jee_main_2024_31_jan_morningOrgan Pipes
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60mathrm~cm$60\mathrm{~cm}$, the length of the closed pipe will be:
A.60mathrm~cm$60\mathrm{~cm}$
B.45mathrm~cm$45\mathrm{~cm}$
C.30mathrm~cm$30\mathrm{~cm}$
D.15mathrm~cm$15\mathrm{~cm}$
Solution
### Related Formula
f_textclosed, fundamental = fracv4L_c$$f_{\text{closed, fundamental}} = \frac{v}{4L_c}$$f_textopen, 1st overtone = frac2v2L_o$$f_{\text{open, 1st overtone}} = \frac{2v}{2L_o}$$
### Core Logic
Organ Pipes diagram for Q45 - JEE Main 2024 MorningOrgan Pipes diagram for Q45 - JEE Main 2024 Morning
For a closed organ pipe, the fundamental frequency (1st harmonic) is:
f_1 = fracvlambda = fracv4L_1$$f_1 = \frac{v}{\lambda} = \frac{v}{4L_1}$$
where L_1$L_1$ is the length of the closed pipe.
For an open organ pipe, the first overtone (2nd harmonic) is:
f_2 = frac2v2L_2 = fracvL_2$$f_2 = \frac{2v}{2L_2} = \frac{v}{L_2}$$
where L_2$L_2$ is the length of the open pipe (L_2 = 60mathrm\,cm$L_2 = 60\mathrm{\,cm}$).
### Step 2: Equating Frequencies
Given f_1 = f_2$f_1 = f_2$:
fracv4L_1 = fracvL_2$$\frac{v}{4L_1} = \frac{v}{L_2}$$L_2 = 4L_1$L_2 = 4L_1$60 = 4 times L_1$$60 = 4 \times L_1$$L_1 = 15mathrm\,cm$$L_1 = 15\mathrm{\,cm}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
More Waves Questions — jee_main_2024_30_january_evening
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