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A point source is emitting sound waves of intensity 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 and 4 mathrm~m from the origin respectively will be ________ times 10^-8 mathrm~W m^-2.

Numerical Answer Type:
Enter a numerical value Answer: 3 to 3

Solution & Explanation

### Related Formula I = fracP4pi r^2 I propto frac1r^2 ### Core Logic For a point source, the intensity I is inversely proportional to the square of the distance from the source. If the given intensity is at r_0 = 1mathrmm (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. The official NTA answer assumes the initial given intensity was at r=2mathrmm. Wait, if I = 16 times 10^-8 was the source power factor or the intensity at r=2, let's reverse-engineer the answer `3`. If I propto 1/r^2, and I_1 at r=2 and I_2 at r=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. If this difference equals 3, then I_1 must be 4. But 16 times 10^-8 is given. Wait, if I at 1mathrmm is 16 times 10^-8: I(2mathrmm) = frac162^2 = 4 times 10^-8. I(4mathrmm) = frac164^2 = 1 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 represents the reference intensity at r=1mathrmm. Intensity at r=2 mathrm~m: I_1 = fracI_02^2 = frac16 times 10^-84 = 4 times 10^-8 mathrm~W m^-2 Intensity at r=4 mathrm~m: I_2 = fracI_04^2 = frac16 times 10^-816 = 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 ### Pattern Recognition Although the question text ("at the origin") was poorly drafted (intensity at r=0 would be infinite), the numerical setup expects you to treat 16 times 10^-8 as the P/4pi constant multiplier for the 1/r^2 dropoff. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

Reference Study Guides

Waves

More Waves Previous-Year Questions — Page 3

Q57 jee_main_2024_01_february_morning Beats
A tuning fork resonates with a sonometer wire of length 1mathrm~m stretched with a tension of 6mathrm~N. When the tension in the wire is changed to 54mathrm~N, the same tuning fork produces 12 beats per second with it. The frequency of the tuning fork is _______ mathrmHz.
Numerical Answer. Answer: 6 to 6

Solution

### Related Formula Fundamental frequency of a sonometer wire: f = frac12LsqrtfracTmu implies f propto sqrtT Beat frequency equation: f_textbeat = |f_textfork - f_textwire| ### Core Logic Let the tuning fork frequency be f. Initially, it resonates with the wire at 6mathrm~N: f = f_1 = frac12Lsqrtfrac6mu When the tension shifts to 54mathrm~N, the wire's new frequency becomes: f_2 = frac12Lsqrtfrac54mu Taking the ratio of frequencies: fracf_1f_2 = sqrtfrac654 = sqrtfrac19 = frac13 implies f_2 = 3f_1 = 3f ### Step 1: Apply Beat Conditions The new frequency f_2 creates 12 beats per second with the fork: f_2 - f = 12 implies 3f - f = 12 2f = 12 implies f = 6mathrm~Hz ### Pattern Recognition Resonance establishes a direct anchor equality (f_textfork = f_textinitial). Since tension changes by a factor of 9 (54/6), the frequency scales up by a factor of 3 (sqrt9). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves Class 11 Physics: Oscillations
Q59 jee_main_2024_30_jan_morning Resonance in Closed Organ Pipes
In a closed organ pipe, the frequency of fundamental note is 30 mathrm~Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 mathrm~Hz. If the organ pipe has a cross-sectional area of 2 mathrm~cm^2, the amount of water poured in the organ tube is ______ g. (Take speed of sound in air is 330 mathrm~m/s)
Numerical Answer. Answer: 400 to 400

Solution

### Related Formula f = fracv4L quad (textclosed organ pipe) ### Core Logic Adding water to a closed organ pipe reduces its effective air column length, thereby increasing the fundamental frequency. The change in length corresponds directly to the volume of water poured. ### Step 1: Calculate Initial Length fracv4ell_1 = 30 frac3304ell_1 = 30 ell_1 = frac330120 = frac114 mathrm~m ### Step 2: Calculate Final Length fracv4ell_2 = 110 frac3304ell_2 = 110 ell_2 = frac330440 = frac34 mathrm~m ### Step 3: Calculate Water Mass The change in length (water depth) is: Delta ell = ell_1 - ell_2 = frac114 - frac34 = frac84 = 2 mathrm~m = 200 mathrm~cm Change in volume (volume of water) = Area times Delta ell: V_textwater = 2 mathrm~cm^2 times 200 mathrm~cm = 400 mathrm~cm^3 Mass of water (M): M = textVolume times textDensity = 400 mathrm~cm^3 times 1 mathrm~g/cm^3 = 400 mathrm~g ### Pattern Recognition Always map frequency changes to length changes using f propto 1/L. Poured water height is exactly the length change Delta L. Watch out for CGS unit conversion (cm, grams). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves
Q37 jee_main_2024_31_jan_evening Speed of Sound in Gases
The speed of sound in oxygen at S.T.P. will be approximately: (textGiven, R = 8.3 text J K^-1 text, gamma = 1.4)
  • A. 310 text m/s
  • B. 333 text m/s
  • C. 341 text m/s
  • D. 325 text m/s

Solution

### Related Formula v = sqrtfracgamma RTM ### Core Logic For Oxygen (O_2) at standard temperature and pressure (S.T.P.): T = 273 text K M = 32 text g/mol = 32 times 10^-3 text kg/mol gamma = 1.4 R = 8.3 text J K^-1 text mol^-1 ### Step 1: Calculate Velocity v = sqrtfrac1.4 times 8.3 times 27332 times 10^-3 v = sqrtfrac3172.2632 times 10^-3 v = sqrt99.133 times 10^3 v = sqrt99133 approx 314.85 text m/s Approximating to the closest given option yields 310 text m/s. ### Pattern Recognition For diatomic gases around room temp or STP, velocities range roughly from 250 to 350 text m/s depending on molar mass (N_2 approx 334, O_2 approx 315). Recognize 315 is closest to option (1) due to standard approximations taken in exams. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves Class 11 Physics: Kinetic Theory of Gases
Q45 jee_main_2024_31_jan_morning Organ 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, the length of the closed pipe will be:
  • A. 60mathrm~cm
  • B. 45mathrm~cm
  • C. 30mathrm~cm
  • D. 15mathrm~cm

Solution

### Related Formula f_textclosed, fundamental = fracv4L_c f_textopen, 1st overtone = frac2v2L_o ### Core Logic
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
Organ Pipes diagram for Q45 - JEE Main 2024 Morning
For a closed organ pipe, the fundamental frequency (1st harmonic) is: f_1 = fracvlambda = fracv4L_1 where 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 where L_2 is the length of the open pipe (L_2 = 60mathrm\,cm). ### Step 2: Equating Frequencies Given f_1 = f_2: fracv4L_1 = fracvL_2 L_2 = 4L_1 60 = 4 times L_1 L_1 = 15mathrm\,cm ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

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