Physics Two Dimensional Motion Question 32
A string of length L is fixed at one end and carries a mass M at the other end.The string makes 2/π revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is
[BHU 2002; DPMT 2004]
Options:
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}}R $
$ =M,4{\pi }^{2}n^{2}L $
$ =M,4{{\pi }^{2}}{{( \frac{1}{2\pi } )}^{2}}L $
$ =16,\text{mL}$
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