Pls give solution of this question

Solution:
Initially,


$$\begin{gathered} T_1+T_1=mg \\ \Rightarrow 2T_1=mg \\ \Rightarrow T_1=\frac{mg}{2}...\left(1\right) \end{gathered}$$
Now, after just cutting one of the string the rod will start rotating about one of its end (P).
So, torque about end P,



$$\begin{gathered} \tau =T_2\times 0+mg\times \frac{L}{2}=\frac{mgL}{2} \\ \Rightarrow I_P\times \alpha =\frac{mgL}{2} \\ \Rightarrow \frac{m{L}^2}{3}\times \alpha =\frac{mgL}{2} \left[\mathrm{Since}, I_P=\frac{m{L}^2}{12}+m\times {\left(\frac{L}{2}\right)}^2=\frac{m{L}^2}{3}\right] \\ \Rightarrow \alpha =\frac{3g}{2L} ..... \left(2\right) \\ \mathrm{But}, \mathrm{acceleration} \mathrm{of} \mathrm{the} \mathrm{rod}, \\ a=\alpha \times \frac{L}{2}=\frac{3g}{2L}\times \frac{L}{2}=\frac{3g}{4} .....\left(3\right) \\ \mathrm{According} \mathrm{to} \mathrm{newton}\text{'}\mathrm{s} \mathrm{second} \mathrm{law}: \\ mg-T_2=ma \\ \Rightarrow mg-T_2=m\times \frac{3g}{4} \\ \Rightarrow T_2=mg-\frac{3mg}{4} \\ \Rightarrow T_2=\frac{mg}{4} ......\left(4\right) \\ \mathrm{Hence}, \mathrm{the} \mathrm{ratio}: \\ \frac{T_2}{T_1}=\frac{\left({\displaystyle \frac{mg}{4}}\right)}{\left({\displaystyle \frac{mg}{2}}\right)}=\frac{1}{2} \\ \Rightarrow T_2 : T_1=1 : 2 \end{gathered}$$

Hence, the correct answer is option (A).