Q. 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) $Mg\sqrt{2}$
B) $\frac{Mg}{\sqrt{2}}$
C) $Mg\left( \sqrt{2} - 1 \right)$
D) $Mg\left( \sqrt{2} + 1 \right)$
Explanation
(c) Work done by horizontal force
$W = F \times S = F \times l\sin\theta$ ….(i)
Increment in potential energy of mass M is
$U = Mgh = Mg\left( l - l\cos\theta \right) = Mgl(1 - \cos{\theta)}$ …..(ii)
[image:6_image73.emf]
From equation (i) and (ii)
$Fl\sin\theta = Mgl(1 - \cos\theta)$
$\Rightarrow Fl\frac{1}{\sqrt{2}} = Mgl\left( 1 - \frac{1}{\sqrt{2}} \right)$ [As $\theta = 45{^\circ}$]
$\therefore F = Mg(\sqrt{2} - 1)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.