Kinematics Question 770
Question: A bob of mass 10 kg is attached to wire 0.3 m long. it’s breaking stress it’s 4.8 × 107 N/m2.The area of cross section of the wire it’s 10?6 m2.The maximum angular velocity with which it can be rotated in a horizontal circle [Pb. PMT 2001]
Options:
A) 8 rad/sec
B)4 rad/sec
C) 2 rad/sec
D)1 rad/sec
Show Answer
Answer:
Correct Answer: B
Solution:
Centripetal force = breaking force therefore $ m{{\omega }^{2}}r= $ breaking stress ´ cross sectional area therefore $ m{{\omega }^{2}}r=p\times A $
therefore $ \omega =\sqrt{\frac{p\times A}{mr}}=\sqrt{\frac{4.8\times 10^{7}\times {{10}^{-6}}}{10\times 0.3}} $ $ \omega =4rad/\sec $
