JEE Main 2012 Physics question
A uniform tube of length 60.5 cm is held vertically with its lower end dipped in water. A sound source of frequency 500 Hz sends sound waves into the tube. When the length of tube above water is 16 cm and again when it is 50 cm, the tube resonates with the source of sound. Two lowest frequencies (in Hz ), to which tube will resonate when it is taken out of water, are (approximately).
Answer: (4)
Solution
Successive resonances differ by : λ = 2(50 − 16) = 68 cm, so v = 340 m/s, and end correction e: cm.
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