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As shown in the figure, two particle each of mass $m$ tied at the ends of a light string of length $2 a$ 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 $2 x$ is
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Verified Answer
The correct answer is:
$\frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}$
Given length $=a$
Equating the vertical components of force,
$$
2 \tan \theta=\frac{F}{m a^{\prime}}
$$
or
$$
\begin{array}{r}
a^{\prime}=\frac{F}{2 m \tan \theta}=\frac{F}{2 m\left(\frac{\sqrt{a^2-x^2}}{x}\right)}=\frac{F x}{2 m \sqrt{a^2-x^2}} \\
\left(\because \tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{\sqrt{a^2-x^2}}{x}\right)
\end{array}
$$
Equating the vertical components of force,
$$
2 \tan \theta=\frac{F}{m a^{\prime}}
$$
or
$$
\begin{array}{r}
a^{\prime}=\frac{F}{2 m \tan \theta}=\frac{F}{2 m\left(\frac{\sqrt{a^2-x^2}}{x}\right)}=\frac{F x}{2 m \sqrt{a^2-x^2}} \\
\left(\because \tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{\sqrt{a^2-x^2}}{x}\right)
\end{array}
$$
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