Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: A sound wave has higher speed in solids than gases. Reason R: Gases have higher value of Bulk modulus than solids. In the light of the above statements, choose the correct answer from the options given below.

Solution & Explanation

### Related Formula v = sqrtfracmathrmBrho ### Core Logic Assertion A: Sound velocity relies on structural elasticity bounds. Solids are highly rigid compared to fluids, making speed significantly higher. (True) Reason R: Solids resist structural compression far better than unbonded gases, giving them significantly higher Bulk Modulus properties. Thus, statement R is completely false. ### Step 1: Final Conclusion Assertion A is true, but Reason R is false, aligning perfectly with option (4). ### Pattern Recognition Even though density rho is higher for solids, the corresponding elastic modulus parameter increases by several orders of magnitude, dominating the structural velocity index. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

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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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