A uniform rod of mass M = 2 kg and length L is suspended by two smooth hinges 1 and 2 as…
Physics · JEE Advanced · NTA Exams — Rotational Motion
A uniform rod of mass M = 2 kg and length L is suspended
by two smooth hinges 1 and 2 as shown in the figure. A force
F = 4 N is applied downward at a distance L/4 from hinge 2 .
Due to the application of force F, hinge 2 breaks. At this
instant, applied force F is also removed. The rod starts to
rotate downward about hinge 1 .
The reaction at hinge 1, before hinge 2 breaks, is
- 24 N
- 12 N
- 11 N
- 10 N
Answer
(C) 11 N
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