A uniform rod of length l and mass m is suspended on two vertical inextensible strings as shown in the fig. Then tension T in the left string at the instant, when right string snaps it?



a) T = mg                b) T = $\frac{mg}{2}$               c) T = $\frac{mg}{4}$                  d) T = zero 

Dear Student,

Please find below the solution to the asked query:

When the right string snaps, then the rod starts rotating about the left string. Then the angular acceleration of the rod is,

$$\begin{gathered} Before snaping the string, the tension in the strings are, \\ 2T=mg\Rightarrow T=\frac{mg}{2} \\ Torque on the rod is, \\ \tau =mg\left(\frac{l}{2}\right)\Rightarrow I\alpha =\frac{mgl}{2}\Rightarrow \frac{1}{3}m{l}^2\alpha =\frac{mgl}{2}\Rightarrow \alpha =\frac{3g}{2l} \\ So, the acceleration with which the centre of mass falls down is, \\ a=\frac{l\alpha }{2}=\frac{3g}{4} \\ So, \\ ma=F_{down}-F_{up}\Rightarrow m\frac{3g}{4}=mg-T\text{'}\Rightarrow T\text{'}=mg-\frac{3mg}{4} \\ \Rightarrow T\text{'}=\frac{mg}{4} \end{gathered}$$

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