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NEET Sample Paper NEET Sample Test Paper-84

  • question_answer
    The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate?

    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 \[{{v}_{3}}=\frac{1000}{2}Hz\]


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