Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    A spring of length L and uniform cross-section is spread on a smooth plane. One of its ends is pulled by a force F. Find the tension in it at a distance I from this end.

    A)  \[\frac{l}{2}F\]                                 

    B) \[\frac{L}{l}F\]

    C) \[\left( 1-\frac{l}{L} \right)F\]                    

    D) \[\left( 1+\frac{l}{L} \right)F\]

    Correct Answer: C

    Solution :

    Mass per unit length of string  \[=\frac{M}{L}\]acceleration in string due to force F                 \[a=\frac{F}{M}\]                 Tension at a distance ( from end (Q) is T = force on part PO                       = mass of part \[PO\times \]acceleration or        \[T=\frac{M}{L}(L-l)\times \frac{F}{M}\] \[\Rightarrow \]               \[T=\left( 1-\frac{l}{L} \right)F\]


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