BVP Medical BVP Medical Solved Paper-2010

  • question_answer
    A partly hanging uniform chain of length L is resting on a rough horizontal table. If \[{{E}_{e}}\] is the maximum possible length that can hang in equilibrium. The coefficient of friction between the chain and table is

    A) \[{{E}_{ph}}\]                                   

    B) \[\frac{{{E}_{e}}}{{{E}_{ph}}}\]

    C)  \[\frac{v}{c}\]                                  

    D) \[\frac{v}{2c}\]

    Correct Answer: A

    Solution :

                    If \[{{y}_{2}}=b\sin \frac{2\pi }{\lambda }[(vt-x)+{{x}_{0}}]\] is the mass/length, then weight of hanging length \[{{x}_{0}}=(\lambda /2)\] Weight of chain on the table \[\left| a-b \right|\]                 \[a+b\]                 \[\sqrt{{{a}^{2}}+{{b}^{2}}}\]                 \[\sqrt{{{a}^{2}}+{{b}^{2}}+2ab\,\cos \,x}\] Equating  \[\frac{{{\varepsilon }_{0}}A}{d}\left[ \frac{{{k}_{1}}}{{{k}_{2}}}+\frac{{{k}_{2}}{{k}_{3}}}{{{k}_{2}}+{{k}_{3}}} \right]\]                 \[\frac{{{\varepsilon }_{0}}A}{d}\left[ \frac{{{k}_{1}}}{{{k}_{2}}}+\frac{({{k}_{2}}+{{k}_{3}})}{{{k}_{2}}{{k}_{3}}} \right]\]


You need to login to perform this action.
You will be redirected in 3 sec spinner