Kinematics Question 487
A string of length L is fixed at one end and carries a mass M at the other end.The string makes 2π revolutions per second around the vertical axis through the fixed end , 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}}L $
$ =M4{\pi }^{2}n^{2}L $
$ =M4{{\pi }^{2}}{{( \frac{2}{\pi } )}^{2}}L $
$ = 16 mL$
BETA
