A) 200
B) 210
C) 205
D) 215
Correct Answer: C
Solution :
Let the frequency of tunning fork be N As the frequency of vibration string \[\propto \frac{1}{\text{length}\ \text{of}\text{string}}\] For sonometer wire of length 20 cm, frequency must be (N + 5) and that for the sonometer wire of length 21cm, the frequency must be (N ? 5) as in each case the tunning fork produces 5 beats/sec with sonometer wire Hence \[{{n}_{1}}{{l}_{1}}={{n}_{2}}{{l}_{2}}\] Þ \[(N+5)\times 20=(N-5)\times 21\] Þ \[N=205Hz.\]
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