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NEET PHYSICSWAVE OPTICSMedium

Question

The number of possible natural oscillations of the air column in a pipe closed at one end of a length of 85 cm85 \text{ cm} whose frequencies lie below 1250 Hz1250 \text{ Hz} is: (velocity of sound 340 ms1340 \text{ ms}^{-1})

A

4

B

5

C

7

D

6

Step-by-Step Solution

  1. Identify the formula for a closed organ pipe: The natural frequencies for a pipe closed at one end are odd harmonics, given by f=(2n1)v4Lf = (2n - 1) \frac{v}{4L}, where vv is the speed of sound, LL is the length of the pipe, and n=1,2,3,n = 1, 2, 3, \dots .
  2. Calculate the fundamental frequency (f1f_1): Given v=340 m/sv = 340 \text{ m/s} and L=85 cm=0.85 mL = 85 \text{ cm} = 0.85 \text{ m}. f1=v4L=3404×0.85=3403.4=100 Hzf_1 = \frac{v}{4L} = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \text{ Hz}
  3. Determine the possible frequencies: A closed pipe produces only odd multiples of the fundamental frequency. The frequencies are: f1=1×100=100 Hzf_1 = 1 \times 100 = 100 \text{ Hz} f2=3×100=300 Hzf_2 = 3 \times 100 = 300 \text{ Hz} f3=5×100=500 Hzf_3 = 5 \times 100 = 500 \text{ Hz} f4=7×100=700 Hzf_4 = 7 \times 100 = 700 \text{ Hz} f5=9×100=900 Hzf_5 = 9 \times 100 = 900 \text{ Hz} f6=11×100=1100 Hzf_6 = 11 \times 100 = 1100 \text{ Hz} f7=13×100=1300 Hzf_7 = 13 \times 100 = 1300 \text{ Hz}
  4. Count frequencies below 1250 Hz1250 \text{ Hz}: The possible natural frequencies below 1250 Hz1250 \text{ Hz} are 100, 300, 500, 700, 900, and 1100 Hz. There are a total of 6 such frequencies.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE OPTICS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVE OPTICSnumberpossiblenaturaloscillationscolumn

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