A bob of mass $10\, kg$ is attached to wire $0.3\, m$ long. Its breaking stress is $4.8 \times 10^7 N/m^2$. The area of cross section of the wire is $10^{-6} m^2$. The maximum angular velocity with which it can be rotated in a horizontal circle ....... $rad/sec$
Medium
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(b) Centripetal force $=$ breaking force
$⇒\ m{\omega ^2}r = $ breaking stress $\times$ cross sectional area
$⇒\ m{\omega ^2}r = p \times A$
$⇒\ \omega = \sqrt {\frac{{p \times A}}{{mr}}} = \sqrt {\frac{{4.8 \times {{10}^7} \times {{10}^{ - 6}}}}{{10 \times 0.3}}} $
$\omega = 4\,rad/\sec $
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