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NEET PHYSICSWAVEEasy

Question

An air column, closed at one end and open at the other, resonates with a tuning fork when the smallest length of the column is 50 cm50 \text{ cm}. The next larger length of the column resonating with the same tuning fork is

A

100 cm100 \text{ cm}

B

150 cm150 \text{ cm}

C

200 cm200 \text{ cm}

D

66.7 cm66.7 \text{ cm}

Step-by-Step Solution

  1. Identify the Type of Air Column: An air column closed at one end and open at the other acts as a closed organ pipe.
  2. Resonance Conditions for Closed Pipe: For a closed organ pipe, resonance occurs at odd multiples of λ/4\lambda/4. The resonating lengths are given by L=(2n1)λ4L = (2n-1)\frac{\lambda}{4}, where n=1,2,3,n = 1, 2, 3, \dots
  3. Find the Smallest Length: The smallest length (fundamental mode, n=1n=1) is given as L1=λ4=50 cmL_1 = \frac{\lambda}{4} = 50 \text{ cm}.
  4. Calculate the Next Larger Length: The next larger length (first overtone, n=2n=2) will be L2=3λ4L_2 = \frac{3\lambda}{4}. Since λ4=50 cm\frac{\lambda}{4} = 50 \text{ cm}, we substitute this value into the equation for L2L_2: L2=3×50 cm=150 cmL_2 = 3 \times 50 \text{ cm} = 150 \text{ cm}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE. 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.

PHYSICSWAVEcolumnclosedresonatestuningsmallest

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