Two tuning forks P and Q sounded together and 6 beats per second are heard. P is in unison with a 30cm air column open at both ends and Q is in resonance when length of air column is increased by 2cm. The frequencies of forks P and Q are
A
90Hz and 84Hz
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B
100Hz and 106Hz
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C
96Hz and 90Hz
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D
206Hz and 200Hz
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Solution
The correct option is C96Hz and 90Hz We know that f=v2I (Fundamental frequency of an open organ pipe). Number of beats heard =6 ⇒fP−fQ=6 or v2(0.3)−v2(0.32)=6 or v=2×0.3×0.32×60.32−0.3 Now, fP=v2(0.3)=0.32×60.02=96Hz and fQ=v2(0.32)=0.3×60.02=90Hz.