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     AIEEE  Solved  Paper-2006

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             \[\Delta K=\]Change in kinetic energy\[=0\] \[={{W}_{F}}+{{W}_{Mg}}+{{W}_{tension}}\]                     ??.(i) (symbols have their usual meanings) \[{{W}_{F}}=F\times l\,\sin {{45}^{o}},\] \[{{W}_{Mg}}={{M}_{g}}(l-l\cos {{45}^{o}}),{{W}_{tension}}=0\] Putting these values in Eq. (i), we get \[F=Mg(\sqrt{2}-1)\]