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NEET Sample Paper NEET Sample Test Paper-18

  • question_answer
    A mass is suspended separately by two springs of spring constant\[{{k}_{1}}\]and \[{{k}_{2}}\]in successive order. The time periods of oscillations in the two cases are \[{{T}_{1}}\] and \[{{T}_{2}},\]respectively. If the same mass be suspended by connecting two springs in parallel (as shown in the figure) then the time period of the oscillation is T. The correct relation is:

    A) \[{{T}^{2}}=T_{1}^{2}+T_{2}^{2}\]            

    B) \[{{T}^{-2}}=T_{1}^{-2}+T_{2}^{-2}\]

    C) \[{{T}^{-1}}={{T}_{1}}^{-1}+{{T}_{2}}^{-1}\]    

    D) \[T={{T}_{1}}+{{T}_{2}}\]   

    Correct Answer: B

    Solution :

    \[{{T}_{1}}=2\pi \sqrt{\frac{m}{{{k}_{1}}}},{{T}_{2}}=2\pi \sqrt{\frac{m}{{{k}_{2}}}}\] \[T=2\pi \sqrt{\frac{m}{({{k}_{1}}+{{k}_{2}})}}\] \[{{T}_{1}}^{-2}+{{T}_{2}}^{-2}=\frac{{{k}_{1}}}{4{{\pi }^{2}}m}+\frac{{{k}_{2}}}{4{{\pi }^{2}}m}\] \[=\frac{{{k}_{1}}+{{k}_{2}}}{4{{\pi }^{2}}m}\] \[={{T}^{-2}}\] Hence, the correction option is [b].


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