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JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • 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?     JEE Main Online Paper (Held On 19 April 2016)

    A) 1100 Hz                                

    B) 1000 Hz

    C) 166 Hz                  

    D) 100 Hz

    Correct Answer: B

    Solution :

                    Total length of sonometer wire, \[l=110cm=1.1m\] Length of wire is in ratio, 6 : 3 : 2 i.e. 60 cm, 30 cm, 20 cm. Tension in the wire, T = 400 N Mass per unit length, m = 0.01 kg Minimum common frequency = ? As we know, Frequency, \[v=\frac{1}{2}\sqrt{\frac{T}{m}}=\frac{1000}{11}Hz\] Similarly, \[{{v}_{1}}=\frac{1000}{6}Hz\] \[{{v}_{2}}=\frac{1000}{3}Hz\] \[{{v}_{3}}=\frac{1000}{2}Hz\] Hence common frequency = 1000 Hz


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