JEE Main & Advanced Sample Paper JEE Main Sample Paper-26

  • question_answer
    A mass of M kg is suspended by a weightless string. The minimum force that is required to displace it until the string makes an angle of \[{{45}^{o}}\] with the initial vertical direction is:

    A)  \[Mg(\sqrt{2}+1)\]        

    B)  \[Mg\sqrt{2}\]

    C)  \[\frac{Mg}{\sqrt{2}}\]                         

    D)  \[Mg(\sqrt{2}-1)\]

    Correct Answer: D

    Solution :

    Here, the constant horizontal force required to take the body from position 1 to position 2 can be calculate by using work-energy theorem. Let us assume that body is taken slowly so, that its speed does not change, then \[\Delta K=0\] \[{{W}_{F}}+{{W}_{Mg}}\,+{{W}_{tenstion}}=\Delta KE\] (Symbols have their usual meaning) \[{{W}_{F}}\,=F\times l\sin {{45}^{0}}\] \[{{W}_{Mg}}\,=Mg(l-\cos {{45}^{0}}),\,\,{{W}_{tension}}\,=0\] \[\therefore \,\,F=Mg\,(\sqrt{2}-1)\]

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