Work Power Energy

Use conservation of energy to find the speeds.

$mg (l + 2l) = \frac{1}{2} m (l^2 + 4l^2) \omega^2 \\ 3gl = \frac{5}{2} l^2 \omega^2 \\ \omega = \sqrt{\frac{6g}{5l}} \\ \boxed{v_B = l \omega = \sqrt{\frac{6gl}{5}}}$

We therefore know that Option A is correct. Find the rod tensions using the following Newton's 2nd Law equations:

$T_{BC} - mg - T_{AB} = \frac{m v_B^2}{l} = \frac{6mg}{5} \\ T_{AB} - mg = \frac{m v_A^2}{2l} = \frac{12mg}{5} \implies \boxed{T_{AB} = \frac{17mg}{5}} \\ \implies \boxed{T_{BC} = \frac{28mg}{5}}$

Therefore, Option C is correct and Option B is wrong.