The frequency of a tuning fork is 512 Hz. It produces resonance in a closed organ pipe of length 16 cm (speed of sound 340 m/s). The pipe resonates in MCQ with Answer and Explanation
The frequency of a tuning fork is 512 Hz. It produces resonance in a closed organ pipe of length 16 cm (speed of sound 340 m/s). The pipe resonates in
A. Fundamental mode
B. Second overtone
C. First overtone
D. End correction needed
Answer: Option A
Solution (By JKSSB Mock Tests)
For closed pipe, fundamental f = v/(4L) = 340/(4×0.16) = 340/0.64 ≈ 531 Hz, close to 512 Hz, slight difference due to end correction. So fundamental. Approx match, SSC type.
Explanation:
By conservation of angular momentum: I1w1 = I2w2. Moment of inertia of a solid sphere is 2/5 MR^2. So, (R1^2) * (2pi/T1) = (R2^2) * (2pi/T2). If radius halves, R2 = R1/2. R1^2 / T1 = (R1/2)^2 / T2. T2 = T1 / 4. A day is 24 hours, so 24/4 = exactly 6 hours.
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