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(B) Given mass $m = 250 \ g = 0.25 \ kg$, length of string $l = 50 \ cm = 0.5 \ m$, and maximum tension $T_{\max} = 72 \ N$. For a horizontal circular

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$24$

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(B) Given mass $m = 250 \ g = 0.25 \ kg$, length of string $l = 50 \ cm = 0.5 \ m$, and maximum tension $T_{\max} = 72 \ N$. For a horizontal circular motion, the tension in the string provides the necessary centripetal force. $T = m \omega^2 R$. Here, the radius $R$ of the circular path is equal to the length of the string $l$ (assuming the string remains horizontal). $T_{\max} = m \omega_{\max}^2 l$. Substituting the values: $72 = 0.25 	imes \omega_{\max}^2 	imes 0.5$. $72 = 0.125 	imes \omega_{\max}^2$. $\omega_{\max}^2 = \frac{72}{0.125} = 576$. $\omega_{\max} = \sqrt{576} = 24 \ rad/s$.

$A$ spherical bob of mass $250 \ g$ is attached to the end of a string having length $50 \ cm$. The bob is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is $72 \ N$. The maximum possible value of angular velocity of the bob (in $rad/s$) is

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