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AMU Medical AMU Solved Paper-2012

  • question_answer
    Three blocks, of masses \[{{m}_{1}}=2.0,\,\,{{m}_{2}}=4.0\] and \[{{m}_{3}}=6.0\,\,kg\] are connected by strings on a frictionless inclined plane of \[{{60}^{o}}\], as shown J the figure. A force F = 120 N is applied upward along the incline to the uppermost block, causing an upward movement of the blocks, if connecting cords are light. The values of tensions \[{{T}_{1}}\] and  in the cords are

    A)  \[{{T}_{1}}=20\,N,\,{{T}_{2}}=60\,N\]

    B)  \[{{T}_{1}}=60\,N,\,{{T}_{2}}=60\,N\]

    C)  \[{{T}_{1}}=30\,N,\,{{T}_{2}}=50\,N\]  

    D)  \[{{T}_{1}}=20\,N,\,{{T}_{2}}=100\,N\]

    Correct Answer: A

    Solution :

                    \[{{T}_{2}}=\frac{({{m}_{1}}+{{m}_{2}})F}{({{m}_{1}}+{{m}_{2}}+{{m}_{3}})}\]                 \[=\frac{(2+4)\times 120}{(2+4+6)}\]                 \[=\frac{6\times 120}{12}\]                 \[{{T}_{2}}=60\,N\]                 \[{{T}_{1}}=\frac{{{m}_{1}}F}{({{m}_{1}}+{{m}_{2}}+{{m}_{3}})}\]                 \[=\frac{2\times 120}{12}\]                 = 20 N


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