A) 1100 Hz
B) 1000 Hz
C) 166 Hz
D) 100 Hz
Correct Answer: B
Solution :
[b] Total length of sonometer wire, \[l=110\,cm=1.1\,\,m\] Length of wire is in ratio, \[6:3:2\] i.e., \[60\,\,cm,\]\[30\,\,cm,\]\[20\,\,cm.\] Tension in the wire, \[T=400N\] Mass per unit length, \[m=0.01\,kg\] Minimum common frequency \[=?\] As we know, Frequency, \[v=\frac{1}{2l}\sqrt{\frac{T}{m}}=\frac{1000}{11}Hz\] Similarly, \[{{v}_{1}}=\frac{1000}{6}Hz\] \[{{v}_{2}}=\frac{1000}{3}Hz\] \[{{v}_{3}}=\frac{1000}{2}Hz\] Hence common frequency \[{{v}_{3}}=\frac{1000}{2}Hz\]You need to login to perform this action.
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