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JEE Main & Advanced JEE Main Paper (Held On 22 April 2013)

  • question_answer
    Two blocks of mass \[{{M}_{1}}=20\,kg\] and \[{{\operatorname{M}}_{2}}=12\operatorname{K}\operatorname{g},\] are connected by a metal rod of mass 8 kg. The system is pulled vertically up applying a force of 480 N as shown. The tension at mid-point of the rod is:   JEE Main  Online Paper (Held On 22 April 2013)

    A)  \[144\] N                             

    B)  \[96\] N

    C)  \[240\] N                             

    D)  \[192\] N

    Correct Answer: D

    Solution :

     Acceleration produced in upward direction \[\text{a =}\frac{\text{F}}{{{\text{M}}_{\text{1}}}\text{+}{{\text{M}}_{\text{2}}}\text{+ Mass of metal rod}}\] \[=\frac{480}{20+12+8}=12m{{s}^{-2}}\] Tension at the mid-point \[\text{T =}\left( {{\text{M}}_{\text{2}}}\text{+}\frac{\text{Mass of rod}}{\text{2}} \right)\text{a}\] \[=(12+4)\times 12=192N\]  


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