JEE Advanced · Physics · 5. Laws of Motion
Two particles of mass \(m\) each are tied at the ends of a light string of length \(2 a\). 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 centre \(P\) (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 \(2 x\), is

- A \(\frac{F}{2 m} \frac{a}{\sqrt{a^2-x^2}}\)
- B \(\frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}\)
- C \(\frac{F}{2 m} \frac{x}{a}\)
- D \(\frac{F}{2 m} \frac{\sqrt{a^2-x^2}}{x}\)
Answer & Solution
Correct Answer
(B) \(\frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}\)
Step-by-step Solution
Detailed explanation
6.
\(2 T \cos \theta=F\)
\(\therefore T=\frac{F}{2} \sec \theta\)
Acceleration of particle \(=\frac{T \sin \theta}{m}=\frac{F \tan \theta}{2 m}\) \(=\frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}\)
\(\therefore\) Option (b) is correct.

\(2 T \cos \theta=F\)
\(\therefore T=\frac{F}{2} \sec \theta\)
Acceleration of particle \(=\frac{T \sin \theta}{m}=\frac{F \tan \theta}{2 m}\) \(=\frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}\)
\(\therefore\) Option (b) is correct.

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