Question

What is the tension in the string at $ A $ immediately after the string breaks at $ B $ ?
$ \left( A \right)\dfrac{{mg}}{5} \\
  \left( B \right)\dfrac{{2mg}}{{31}} \\
  \left( C \right)\dfrac{{mg}}{{37}} \\
  \left( D \right)\dfrac{{5mg}}{{37}} \\ $

Answer 1

Hint :In order to solve his question, we are going to first compute the centre of mass position and as we know that the torque about the centre of mass is zero, so the torque due to tension at $ A $ is equal to that of string at $ B $ , the tension at $ A $ does not change just after the string breaks.
The torque at any point is given by
 $ \tau = F \times r $
Where, $ F $ is forced at the point and $ r $ is the radial distance of the point from the centre of mass.
The torque about centre of mass is zero
 $ {\tau _{CM}} = 0 $

Complete Step By Step Answer:
The centre of mass of the system is given by
 $ {x_{CM}} = \dfrac{{{\rho _0}\int\limits_0^L {{x^4}dx} }}{{{\rho _0}\int\limits_0^L {{x^3}dx} }} = \dfrac{{\dfrac{{{L^5}}}{5}}}{{\dfrac{{{L^4}}}{4}}} = \dfrac{{4L}}{5} $
The torque about the centre of mass is zero, so the torque due to tension at $ A $ , $ {T_A} $ is equal to that of string at $ B $ , $ {T_B} $
Thus,
$ {T_A} \times \dfrac{{4L}}{5} = {T_B} \times \dfrac{L}{5} \\
   \Rightarrow {T_A} = \dfrac{{{T_B}}}{4} - - - \left( 1 \right) \\ $
The sum of the tensions at $ A $ and $ B $ is equal and opposite to the weight
 $ {T_A} + {T_B} = mg - - - \left( 2 \right) $
Solving $ \left( 1 \right) $ and $ \left( 2 \right) $ , we get
 $ {T_A} = \dfrac{{mg}}{5} $ , $ {T_B} = \dfrac{{4mg}}{5} $ ;
Now just after the string breaks, the tension of the string at $ A $ remains the same, i.e. there is no change in the value of the tension at the string
So, $ {T_A} = \dfrac{{mg}}{5} $
Hence, the option $ \left( A \right)\dfrac{{mg}}{5} $ is the correct answer.

Note :
It is to be noted that the torque of a string rotating about a point depends upon its distance with the centre of mass and the torque at the centre of mass is always zero because the radial distance there is zero, here the driving force is the tension of the string and the gravity is also present at the string.