The correct answer is (B) T1 > T2.

Given:

L/4 and 3L/4 away from the pivoted ends.

(i) At L/4 from the pivot end:

Tension, \(T_{1} \) = centrifugal force on the rod behind this point.

\( T_{1} = {m}’\omega ^{2}r = \frac{M}{L}\frac{3}{4}L\omega ^{2}(\frac{L}{4} + \frac{3L}{8}) \) \(T_{1} = \frac{15}{32}M\omega ^{2}L \)

(ii) At 3L/4 from the pivot end:

Tension, \(T_{2} \)= centrifugal force on the rod behind this point (of L/4 length)

\(T_{2} = {m}’\omega ^{2}r = \frac{M}{L}\frac{1}{4}L\omega^{2}(\frac{3L}{4} + \frac{L}{8}) \) \(T_{2} = \frac{7}{32}M\omega^{2}L \)

Therefore, \(T_{1} > T_{2} \)

Hence, proved.

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