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 2006]

A) \[Mg(\sqrt{2}+1)\]

B) \[Mg\sqrt{2}\]

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

D) \[Mg(\sqrt{2}-1)\]

Correct Answer: D

Solution :

[d] 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

\[F=Mg(\sqrt{2}-1)\]