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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

A mass M kg is suspended by a weightless string. The horizontal force that is required to displace it untill the string makes an angle $45^{\circ}$ with initial vertical direction is

\[(a)\;Mg(\sqrt 2 +1)\quad (b)\;Mg \sqrt 2 \quad (c)\;\frac{mg}{\sqrt 2}\quad (d)\;Mg(\sqrt 2-1)\]

1 Answer

Work done by trension -Work done by force =Work done by gravitational force
$0+F \times AB =Mg \times AC$
$F=Mg \bigg(\large\frac{AC}{AB}\bigg)$
$AB= l\sin 45$
$\qquad=\large\frac{l}{\sqrt 2}$
$AC=OC-OA$
$\qquad=l-l \cos 45$
$\qquad=Mg \bigg(\large\frac{1-\frac{1}{2}}{\Large\frac{1}{\sqrt 2}}\bigg)$
$AC=OC-OA$
$\qquad=l-l \cos 45$ (l-length of string)
Therefore $F=Mg(\sqrt 2-1)$
Hence d is the correct answer. 

 

answered Jul 12, 2013 by meena.p
edited Feb 10, 2014 by meena.p
 

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