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JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Motion of Connected Bodies

  • 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 \[{{T}_{3}}=40\,N\]. If \[{{m}_{1}}=10\,kg,\,{{m}_{2}}=6\,kg\] and \[{{m}_{3}}=4\,kg\], the tension \[{{T}_{2}}\] will be               [MP PMT/PET 1998]

    A)                         20 N               

    B)                         40 N            

    C)                         10 N               

    D)                         32 N            

    Correct Answer: D

    Solution :

                                \[{{T}_{2}}=({{m}_{1}}+{{m}_{2}})\times \frac{{{T}_{3}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}}=\frac{(10+6)\times 40}{20}\]\[=32\ N\]


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