Question

As shown in the figure, two particles, each of mass $'m'$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,
$2 T \sin \theta=F$ ... (i)
Equating the horizontal forces,
$T \cos \theta=m a'$ ...(ii)

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