Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2008

  • question_answer A mass of M kg is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of \[45{}^\circ \] 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 calculated by using  work-energy theorem. Let us assume that body is taken slowly so that its speed doesn't change,  then                     \[={{W}_{F}}+{{W}_{Mg}}+{{W}_{tension}}\] [symbols have their usual meanings] \[={{W}_{F}}=F\times l\,\sin \,{{45}^{o}},\] \[{{W}_{ & Mg}}={{M}_{g}}(l-l\,\cos \,{{45}^{o}}),\]\[{{W}_{tension}}=0\] \[\therefore \]  \[F=Mg(\sqrt{2}-1)\]


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