BHU PMT BHU PMT (Mains) Solved Paper-2009

  • question_answer
    System shown in the figure is released from rest with mass 2 kg in contact with ground. Pulley and spring are massless and the friction is absent everywhere. The speed of 5 kg block when 2 kg block leaves the contact with the ground is (force   constant   of   spring k = 40 \[N{{m}^{-1}}\]and\[g=10\text{ }m{{s}^{-2}}\])

    A)  \[\sqrt{2}\,m{{s}^{-1}}\]                             

    B) \[2\sqrt{2}\,m{{s}^{-1}}\]

    C)  \[2\,m{{s}^{-1}}\]                          

    D) \[4\sqrt{2}\,m{{s}^{-1}}\]

    Correct Answer: B

    Solution :

                     Let\[x\]be the extension in the string when 2 kg block leaves the contact with ground. Then tension in the                 spring should be equal to weight of 2 kg block.                 \[kx=2g\]or \[x=\frac{2g}{k}=\frac{2\times 10}{40}=\frac{1}{2}m\]                 Now, from conservation of mechanical energy                 \[mgx=\frac{1}{2}k{{x}^{2}}+\frac{1}{2}m{{v}^{2}}\]                 Or           \[v=\sqrt{2gx-\frac{k{{x}^{2}}}{m}}\]                 \[=\sqrt{2\times 10\times \frac{1}{2}-\frac{40}{4\times 5}}=2\sqrt{2}m{{s}^{-1}}\]


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