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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2001

  • question_answer
    The scale of a spring balance reading from 0 to 10 kg is 0.25 m long. A body suspended from the balance oscillates vertically with a period of\[\pi /10\,s\]s. The mass suspended is: (neglect the mass of the spring)

    A)  10 kg                                    

    B)  0.98 kg

    C)         5kg                                       

    D)  20 kg

    E)  4kg

    Correct Answer: B

    Solution :

    \[k=\frac{F}{l}=\frac{10\times g}{0.25}=40\,g\,N/m\] \[T=2\pi \sqrt{\left( \frac{m}{k} \right)}\Rightarrow {{T}^{2}}=\frac{4{{\pi }^{2}}m}{k}\] \[\Rightarrow \]               \[{{\left( \frac{\pi }{10} \right)}^{2}}=\frac{4{{\pi }^{2}}m}{40g}\] \[\therefore \]  \[m=0.98kg\]


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