Kinematics Question 487

Question: A string of length L it’s fixed at one end and carries a mass M at the other end.The string makes 2/p revolutions per second around the vertical axis through the fixed end , then tension in the string is [BHU 2002; DPMT 2004]

Options:

A) ML

B) 2 ML

C) 4 ML

D) 16 ML

Show Answer

Answer:

Correct Answer: D

Solution:

$ T\sin \theta =M{{\omega }^{2}}R $ -(i)

$ T\sin \theta =M{{\omega }^{2}}L\sin \theta

$ -(ii) From (i) and (ii) $ T=M{{\omega }^{2}}L $

$ =M4{{\pi }^{2}}n^{2}L $

$ =M4{{\pi }^{2}}{{( \frac{2}{\pi } )}^{2}}L $

$ =16ML $



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