A bob of mass `10 kg` is attached to a wire `0.3 m` long. Its breaking stress is `4.8xx10^(7) N//m^(2)`. Then area of cross-section of the wire is `10^(-6) m^(2)`. What is the maximum angular velocity with which it can be rotated in a horizontal circle?A. `8 rad//s`B. `4 rad//s`C. `2 rad//s`D. `1rad//s`

A bob of mass `10 kg` is attached to a wire `0.3 m` long. Its breaking

1.	A bob of mass `10 kg` is attached to a wire `0.3 m` long. Its breaking stress is `4.8xx10^(7) N//m^(2)`. Then area of cross-section of the wire is `10^(-6) m^(2)`. What is the maximum angular velocity with which it can be rotated in a horizontal circle?A. `8 rad//s`B. `4 rad//s`C. `2 rad//s`D. `1rad//s`
Answer» Correct Answer - B `T =mlomega^(2)` ` T/A = sigma_(max) = (mlomega^(2))/A` :. ` omega = sqrt((sigma_(max)A)/(ml)` `=sqrt((4.8xx10^(7)xx10^(-6))/(10xx10.3)` `= 4 rad//s`

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