Question

As shown in the figure, two particles, each of mass ${}^{\prime }{m}^{\prime }$tied at the ends of a light string of length $2a$ are kept on a frictionless horizontal surface. When the mid point (P) of the string is pulled vertically upwards with a small but constant force $F$. the particles move towards each other on the surface. Magnitude of acceleration of each particle, when the separation between them becomes $2x$ is

• A. $\frac{F}{2m}\frac{a}{\sqrt{a^2-x^2}}$
• B. $\frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}}$
• C. $\frac{F}{2m}\frac{x}{a}$
• D. $\frac{F}{2m}\frac{\sqrt{a^2-x^2}}{x}$

Answer 1

Answer: (B)

Given length $=a$
Equating the vertical components of force,
$2T\sin \theta =F$ ... (i)
Equating the horizontal forces,
$T\cos \theta =m{a}^{\prime }$ ...(ii)

Dividing Eq. (i) by Eq. (ii), we get
$2\tan \theta =\frac{F}{m{a}^{\prime }}$
or $a^{\prime }=\frac{F}{2m\tan \theta }=\frac{F}{2m\left(\frac{\sqrt{a^2-x^2}}{x}\right)}=\frac{Fx}{2m\sqrt{a^2-x^2}}$
$\left(\because \tan \theta =\frac{\text{ perpendicular }}{\text{ base }}=\frac{\sqrt{a^2-x^2}}{x}\right)$