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 $



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