AP EAMCET · PHYSICS · Laws of Motion
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
- 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
Given length \(=a\) Equating the vertical components of force, \(2 T \sin \theta=F\) ...(i) Equating the horizontal forces, \(T \cos \theta=m a^{\prime}\) \( 2 \tan \theta=\frac{F}{m a^{\prime}} \) or…
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