Question

A uniform chain of length L and mass m is hanging vertically from its ends A and B which are close together. At a given instant the end B is released. What is the tension at A when B has fallen distance x (x < 1)?
A. $\dfrac{{mg}}{2}\left[ {1 + \dfrac{{3x}}{l}} \right]$
B. $mg\left[ {1 + \dfrac{{2x}}{l}} \right]$
C. $\dfrac{{mg}}{2}\left[ {1 + \dfrac{x}{l}} \right]$
D. $\dfrac{{mg}}{2}\left[ {1 + \dfrac{{4x}}{l}} \right]$

Answer 1

Hint:If we calculate the net weight experienced at A due to release of point B, we can get the value of tension. Because the tension balances this weight. So, if we calculate the weight added due to release of point B when it has fallen a distance x and add it to the initial weight due to half the length of the wire we can get the total effective weight. This value of force due to weight gives us the value of force of tension.

Complete step-by-step solution:
In order to solve this question, we need to calculate the net weight. Since this weight is balanced by the tension. let us suppose that the chain falls down to a distance $d\,x$ . Let the time taken be $\Delta t$
When the chain falls $d\,x$ distance downward then $\dfrac{{dx}}{2}$ length goes towards end A.
Now let us find the initial momentum of this part.
We know that momentum is mass times velocity. Let the weight of the chain be W. Then weight per unit length will be $\dfrac{W}{L}$ .So mass per unit length can be written as $\dfrac{W}{{Lg}}$ .
So mass of part $\dfrac{{dx}}{2}$ will be $\dfrac{W}{{Lg}} \times \dfrac{{dx}}{2}$
Therefore, initial momentum =$\left( {\dfrac{W}{{Lg}} \times \dfrac{{dx}}{2}} \right) \times v$
Velocity of this part falling freely can be found using the relation $v = \sqrt {2gh} $
Thus,
$v = \sqrt {2gdx} $
The final momentum will be zero.
Thus, the change in momentum will be $\Delta P = \left( {\dfrac{W}{{Lg}} \times \dfrac{{dx}}{2}} \right) \times v - 0 = \left( {\dfrac{W}{{Lg}} \times \dfrac{{dx}}{2}} \right) \times v$
Force is rate of change of momentum
Thus,
$F = \dfrac{{\Delta P}}{{\Delta t}} = \dfrac{{\left( {\dfrac{W}{{Lg}} \times \dfrac{{dx}}{2}} \right) \times v}}{{\Delta t}}$
$ \Rightarrow F = \dfrac{W}{{2Lg}} \times {v^2}$
Since velocity is the rate of change of displacement.
On substituting the value of velocity we get ,
$\therefore F = \dfrac{W}{L} \times dx$
Force of weight of $\dfrac{{dx}}{2}$ will be$\dfrac{W}{L}\dfrac{{dx}}{2}$
So, the net force can be written as
$F = \dfrac{W}{L}dx + \dfrac{W}{L}\dfrac{{dx}}{2}$
$\therefore F = \dfrac{3}{2}\dfrac{W}{L}dx$
In the beginning there is already a weight due to part $\dfrac{L}{2}$ .
That is weight of this part equal to $\dfrac{W}{L} \times \dfrac{L}{2}$ is already present
So Total weight can be written as
${F_T} = \dfrac{3}{2}\dfrac{W}{L}dx + \dfrac{W}{L} \times \dfrac{L}{2}$
So for x length we can write it as ${F_T} = \dfrac{3}{2}\dfrac{W}{L}x + \dfrac{W}{L} \times \dfrac{L}{2}$
The weight is balance by the tension so the force of tension will be
$T = \dfrac{{mg}}{2}\left( {1 + \dfrac{{3x}}{L}} \right)$
So, the correct answer is option A.

Note:- Remember that total tension will be due to the weight that is already present and the new weight due to release of point B. Don’t forget to take the initial weight of $\dfrac{L}{2}$ into consideration because total weight is balanced by the tension.