Two tuning forks A and B are sounded together giving rise to 8 beats in 2 text s. When fork A is loaded with wax, the beat frequency is reduced to 4 beats in 2 text s. If the original frequency of tuning fork B is 380 text Hz, then the original frequency of tuning fork A is ____ Hz.

Numerical Answer Type:
Enter a numerical value Answer: 384 to 384 +4 marks

Solution & Explanation

### Related Formula f_textbeat = |f_A - f_B| ### Core Logic Initial beat frequency f_textbeat1 = frac8 text beats2 text s = 4 text Hz. Therefore, |f_A - f_B| = 4. Given f_B = 380 text Hz, the original frequency of A could be: f_A = 380 + 4 = 384 text Hz OR f_A = 380 - 4 = 376 text Hz. ### Step 1: Check with Wax Loading When fork A is loaded with wax, its frequency f_A decreases. The new beat frequency f_textbeat2 = frac4 text beats2 text s = 2 text Hz. Case 1: If f_A = 384 text Hz, loading wax drops it to say 382 text Hz. The new beat frequency becomes |382 - 380| = 2 text Hz. This matches the given condition. Case 2: If f_A = 376 text Hz, loading wax drops it to say 374 text Hz. The new beat frequency would become |374 - 380| = 6 text Hz. This does NOT match the condition. ### Step 2: Conclusion The original frequency of tuning fork A must be 384 text Hz. ### Pattern Recognition Loading wax always DECREASES the frequency. If decreasing the unknown frequency causes the beat frequency to DECREASE, the unknown frequency must have been initially higher than the known standard. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

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Q43 jee_main_2026_21_jan_morning Wave on a String
Two strings (A, B) having linear densities mu_A = 2 times 10^-4text kg/m and mu_B = 4 times 10^-4text kg/m and lengths L_A = 2.5text m and L_B = 1.5text m respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time t_1 and t_2 , respectively, to reach the joint. The ratio t_1/t_2 is :
  • A. 1.08
  • B. 1.90
  • C. 1.67
  • D. 1.18

Solution

### Related Formula v = sqrtfracTmu t = fracLv ### Core Logic Given L_A = 2.5text m, L_B = 1.5text m, T = 500text N. Velocity in string A: v_A = sqrtfracTmu_A = sqrtfrac5002 times 10^-4 = sqrt2500000 = 5 sqrt10 times 10^2text m/s Velocity in string B: v_B = sqrtfracTmu_B = sqrtfrac5004 times 10^-4 = sqrt1250000 = 5 sqrt5 times 10^2text m/s ### Step 1: Calculate Ratio of Times Time taken to reach joint: t_1 = fracL_Av_A = frac2.55 sqrt10 times 10^2 t_2 = fracL_Bv_B = frac1.55 sqrt5 times 10^2 Ratio: fract_1t_2 = frac2.55sqrt10 times frac5sqrt51.5 = frac2.51.5 times fracsqrt5sqrt10 = frac53 times frac1sqrt2 fract_1t_2 = frac1.6661.414 approx 1.18 ### Pattern Recognition Time ratio is proportional to fracLsqrtT/mu = L sqrtmu/T. Thus t_1/t_2 = (L_1/L_2) sqrtmu_1/mu_2 directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves
Q33 jee_main_2026_22_january_evening Organ Pipe Harmonics
In an open organ pipe v_3 and v_6 are 3^mathrmrd and 6^mathrmth harmonic frequencies, respectively. If v_6 - v_3 = 2200 Hz then length of the pipe is ____ mm. (Take velocity of sound in air is 330 m/s.)
  • A. 275
  • B. 225
  • C. 200
  • D. 250

Solution

### Related Formula f_n = n left(fracv2Lright) ### Core Logic For an open organ pipe, harmonic frequency v_n = n cdot f_0 = n left(fracv2Lright). Given v_6 - v_3 = 2200 Hz: frac6v2L - frac3v2L = 2200 frac3v2L = 2200 Substituting speed of sound v = 330 mathrm~m/s: frac3 times 3302L = 2200 L = frac9904400 = 0.225 mathrm~m = 225 mathrm~mm ### Step 1: Final Conclusion The length of the pipe is 225 mathrm~mm. ### Pattern Recognition Open organ pipe: Frequency difference Delta f = (6-3)f_0 = 3 f_0 = 2200 implies f_0 = 2200/3. Since f_0 = v/(2L), solve directly for L. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves
Q50 jee_main_2026_23_january_evening Speed of Sound
The velocity of sound in air is doubled when the temperature is raised from 0^circ C to alpha^circ C. The value of alpha is ____.
Numerical Answer. Answer: 819 to 819

Solution

### Related Formula V = sqrtfracgamma RTM fracV_1V_2 = sqrtfracT_1T_2 ### Core Logic The speed of sound in an ideal gas is directly proportional to the square root of its absolute temperature in Kelvin. Initial state: T_1 = 0^circmathrmC = 273 \, mathrmK, V_1 = V_0 Final state: T_2 = alpha^circmathrmC = (alpha + 273) \, mathrmK, V_2 = 2V_0 ### Step 1: Apply Ratio Equation fracV_02V_0 = sqrtfrac273T_2 frac14 = frac273T_2 T_2 = 4 times 273 = 1092 \, mathrmK ### Step 2: Convert to Celsius T_2 = alpha + 273 = 1092 alpha = 1092 - 273 = 819^circmathrmC ### Pattern Recognition If velocity doubles, absolute temperature must quadruple (2^2 = 4). 4 times 273 = 1092 \, mathrmK, which translates perfectly to 819^circmathrmC. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves Class 11 Physics: Kinetic Theory of Gases
Q34 jee_main_2026_24_january_evening Organ Pipes and Resonance
The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is 5/x. The value of x is ____.
  • A. 4
  • B. 2
  • C. 1
  • D. 3

Solution

### Related Formula f_n, textclosed = fracn v4L_textclosed (where n is an odd integer) f_n, textopen = fracn v2L_textopen (where n is any integer) ### Core Logic We are given that the fifth harmonic of the closed organ pipe equals the first harmonic of the open pipe: f_5, textclosed = f_1, textopen frac5v4L_textclosed = fracv2L_textopen ### Step 1: Simplify Ratio Isolating the length ratio: fracL_textclosedL_textopen = frac54 times 2 = frac104 = frac52 Equating this to 5/x gives x = 2. ### Pattern Recognition For a closed pipe, the fundamental is v/4L. For an open pipe, it's v/2L. If an odd harmonic n of a closed pipe matches the m-th harmonic of an open pipe, L_c / L_o = n/2m. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves
Q41 jee_main_2026_24_january_evening Intensity and Inverse Square Law
A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8 times, the intensity will
  • A. textincrease by 8 text times
  • B. textincrease by 64 text times
  • C. textdecrease by 8 text times
  • D. textdecrease by 4 text times

Solution

### Related Formula I = fracP4pi R^2 V = frac43pi R^3 ### Core Logic Since volume V propto R^3, if volume is increased by 8 times: V rightarrow 8V implies R^3 rightarrow 8R^3 implies R rightarrow 2R ### Step 1: Intensity Relation Since intensity I propto frac1R^2: If R rightarrow 2R, the area A = 4pi R^2 increases by 2^2 = 4 times (A rightarrow 4A). Therefore, intensity becomes I rightarrow fracI_04. ### Pattern Recognition Volume ratio cubed root dictates radius multiplier. The square of that radius multiplier is the area multiplier, which is the exact inverse of the intensity multiplier. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

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