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Q.

Two masses each equal to m are lying on X-axis at (- a,0) and (+ a, 0) respectively as shown in fig.They are connected by a light string. A force F is applied at the origin and along the Y-axis. As a result, the masses move towards each other What is the acceleration of each mass ? Assume the instantaneous position of the masses as (- x,0) and {x, 0) respectively

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a

Fm⋅xa2−x2

b

F2m⋅xa2−x2

c

2Fm⋅xa2−x2

d

2Fm⋅a2−x2x

answer is B.

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Detailed Solution

See figure.From figure F=2Tcos⁡θ or T=F/(2cos⁡θ)The force responsible for motion of masses on X-axis is Tsin⁡θ∴ ma=Tsin⁡θ=F2cos⁡θ×sin⁡θ =F2tan⁡θ=F2×OBOA=F2×x(a2−x2)So, a=F2m×x(a2−x2).
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