2. Solved Papers
  3. WB JEE Medical

WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    A smooth massless string passes over a smooth fixed pulley. Two masses \[{{m}_{1}}\]and \[{{m}_{2}},({{m}_{1}}>{{m}_{2}})\] are tied at the two ends of the string. The masses are allowed to move under gravity starting from rest. The total external force acting on the two masses is

    A) \[{{m}_{2}},({{m}_{1}}>{{m}_{2}})\]

    B)  \[\frac{{{({{m}_{1}}-{{m}_{2}})}^{2}}}{{{m}_{1}}+{{m}_{2}}}g\]

    C)  \[({{m}_{1}}-{{m}_{2}})g\]

    D)  \[\frac{{{({{m}_{1}}+{{m}_{2}})}^{2}}}{{{m}_{1}}-{{m}_{2}}}g\]

    Correct Answer: B

    Solution :

     We know that Acceleration, \[{{a}_{CM}}={{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}\times g\,\] \[(\because \,{{m}_{_{1}}}>{{m}_{2}})\] So, resultant external force, \[F=({{m}_{1}}+{{m}_{2}}){{a}_{CM}}\] \[=({{m}_{1}}+{{m}_{2}})\times {{\left( \frac{{{m}_{1}}-{{m}_{2}}}{{{m}_{1}}+{{m}_{2}}} \right)}^{2}}\times g\] \[=\frac{{{({{m}_{1}}-{{m}_{2}})}^{2}}}{({{m}_{1}}+{{m}_{2}})}\times g\]


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