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Uttarakhand PMT Uttarakhand PMT 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 \[\text{10kg,}{{\text{m}}_{\text{2}}}\text{=6kg}\] and \[{{\text{m}}_{3}}\text{=6}\,\text{kg}\] then tension\[{{T}_{2}}\]will be:

    A)  10 N        

    B)  20 N

    C)  32 N        

    D)  40 N

    Correct Answer: C

    Solution :

     Here: \[{{m}_{1}}=10\text{ }kg,\text{ }{{\text{m}}_{2}}=6\text{ }kg,\text{ }{{\text{m}}_{3}}=4\text{ }kg,\] \[F=40N\] 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 \[{{T}_{2}}=({{m}_{1}}+{{m}_{2}})a\] \[=(10+6)2=16\times 2\] \[=32\text{ }N\]


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