Question

two particles of mass "m" each are tied at the ends of a light string of length "2a".The whole system is kept on a frictionless horizontal surface with the string held so that each mass is a distance a from the centre "P". Now the midpoint of the string is pulled vertically upwards with a small but constant force "F" as a result the particle moves towards each other on the surface the magnitude of acceleration when the separation between them becomes "2x" is :-

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Answer 1

Answer:

$\large \bold\red{(b) \frac{F}{2m} \frac{x}{ \sqrt{ {a}^{2} - {x}^{2} } }}$

Explanation:

Given,

Two particles each of mass 'm' are tied at the ends of a light string having length '2a'.

Now,

A constant force, F is applied at the centre vertically upwards.

So, Tension , T in the string will arise.

Therefore,

First of all draw the free body diagram of the given situation.

Note :- Refer to the attachment for FBD.

Now,

The acceleration of mass, m is due to the force $\bold{T \cos \alpha }$

Therefore,

we get,

$T \cos( \alpha ) = ma \\ \\ = > a = \frac{T \cos( \alpha ) }{m} \: \: \: \: ............(1)$

Also,

we have,

$F = 2T \sin( \alpha ) \\ \\ = > T = \frac{F}{2 \sin( \alpha ) } \: \: \: \: \: \: ..........(2)$

Therefore,

from equations (1) and (2),

we get,

$a = \frac{F \cos( \alpha ) }{2m \sin( \alpha ) } \\ \\ = > a = \frac{F}{2m \tan( \alpha ) }$

But,

we know that,

$\tan( \alpha ) = \frac{ \sqrt{ {a}^{2} - {x}^{2} } }{x}$

Therefore,

putting the values,

we get,

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

Hence,

the correct answer is,

$\large \bold{(b) \frac{F}{2m} \frac{x}{ \sqrt{ {a}^{2} - {x}^{2} } }}$

Answer 2

Answer:

FIRST READ HERE AND THEN LOOK AT THE ATTACHED PHOTO

Given : 2 masses each measuring "m" is tied by a string separated by a distance of 2a.

The midpoint of string is pulled by a constant force F. This leads to coming close of the masses.

To find : Net acceleration of any one of the mass when distance of separation becomes 2x.

Things to note:

1. Surface is Frictionless.

2. Gravitational force has to be neglected since no options has the term "G".

PLEASE DRAW THE DIAGRAM CAREFULLY.

CONCLUSION: The acceleration will be a component of the applied force F.

We have find that component, then our problem will be solved.

Now refer to the attached photo.

Tension develops in both of the strings.

It is related to the applied force F such that

2Tsinθ = F

=> T = F/(2sinθ) ...........(i)

Now the force when distance is 2x

= Tcosθ ..............(ii)

Putting value of (i) in (ii)

force = {F/(2sinθ)} × cosθ

force = Fcot(θ) /2

force = F/2 {x/√(a² + x²)}

Now acceleration = force/mass

acc. = F/2m {x/√(a² + x²)}

So the correct answer is 2nd option.