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 making an angle of \[45{}^\circ \] with the initial vertical direction is

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

B) \[Mg(\sqrt{2}+1)\]     

C) \[Mg\sqrt{2}\] 

D)        \[\frac{Mg}{\sqrt{2}}\]

Correct Answer: A

Solution :

Work done in displacement is equal to gain in potential energy of mass Work done \[=F\times l\sin 45{}^\circ =\frac{Fl}{\sqrt{2}}\] Gain in potential energy\[=Mg(l-l\cos 45{}^\circ )\] \[=Mgl\left( 1-\frac{1}{\sqrt{2}} \right)\therefore \frac{Fl}{r2}=\frac{Mgl(\sqrt{2}-1)}{\sqrt{2}}\] Or \[F=Mg(\sqrt{2}-1)\]