Answer
300N force is working at one end of the rod so net torque about fixed end
So
τnet=mgL2+FL=10×9.8×2.52+300×2.5τnet=872.5Nm\tau_{net}=mg\frac{L}{2}+FL\\=10\times9.8\times\frac{2.5}{2}+300\times2.5\\\tau_{net}=872.5Nmτnet=mg2L+FL=10×9.8×22.5+300×2.5τnet=872.5Nm
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