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Manipal Medical Manipal Medical Solved Paper-2004

  • question_answer
    Three blocks of masses\[{{m}_{1}},{{m}_{2}}\]and\[{{m}_{3}}\]are connected by massless strings as shown on a frictionless table. They are pulled with   a   force   of 40 N. If\[{{m}_{1}}\text{=}10\text{ }kg,\text{ }{{m}_{2}}=6\]kg and\[{{m}_{3}}=4kg,\]then tension\[{{T}_{2}}\]will be:

    A)  10 N        

    B)  20 N

    C)  32 N        

    D)  40 N

    Correct Answer: C

    Solution :

     Here: \[{{m}_{1}}=10kg,{{m}_{2}}=6kg,{{m}_{3}}=4kg,\] \[F=40N\] Since, the table is frictionless i.e., it is smooth therefore, force on the blocks is given by \[F=({{m}_{1}}+{{m}_{2}}+{{m}_{3}})a\] \[\Rightarrow \] \[a=\frac{F}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}\] \[=\frac{40}{10+6+4}=\frac{40}{20}\] \[=2\text{ }m/{{s}^{2}}\] Now the tension between 10 kg and 6 kg masses is given by \[{{T}_{2}}=({{m}_{1}}+{{m}_{2}})a\] \[=(10+6)2=16\times 2\] \[=32N\]


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