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Two particles of mass 'm' each are tied at the ends of a light string of length '2a'. The whole system is kept on frictionless horizontal surface with the string held tight so
that each mass is at a distance 'a' from the center P (as shown in figure). Now, the midpoint 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:

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$\frac{F}{2 m} \sqrt{\frac{a^{2} - x^{2}}{x}}$

$\frac{F}{2 m} \frac{a}{\sqrt{a^{2} - x^{2}}}$

$\frac{F}{2 m} \frac{x}{\sqrt{a^{2} - x^{2}}}$

$\frac{F}{2 m} \frac{x}{a}$

answer is B.

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Detailed Solution

$2 T_{x} \cos θ = F \Rightarrow T_{x} = \frac{Fa}{2 \sqrt{a^{2} - x^{2}}}$
$Acceleration of particle = \frac{T_{x} \sin θ}{m} = \frac{F}{2 m} \frac{x}{\sqrt{a^{2} - x^{2}}}$

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