Question

Two particle of mass $m$ each are tied at the ends of a light string of length $2a$. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance ` $a$' from the center $P$ (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force $F$. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes $2x$ is :

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

Answer 1

The correct option is B $\frac{Fx}{2m\sqrt{a^2-x^2}}$

Assume the tension in string is $T$, the net force on particle is $T$.

The magnitude of acceleration $a=\frac{T}{m}$

From equilibrium $2T\sin \theta =F$

$\Rightarrow T=\frac{F}{2\sin \theta }$

$\sin \theta =\frac{\sqrt{a^2-x^2}}{x}$

substituting value of $\sin \theta$ in expression of $T$

$\Rightarrow T=\frac{xF}{2\sqrt{a^2-x^2}}$

acceleration = $\frac{T}{m}=$ $\frac{xF}{2m\sqrt{a^2-x^2}}$

Hence correct answer is option B.