(D) In a vertical circular motion, the tension in the string is maximum at the lowest point of the path.The force equation at the lowest point is give

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(D) In a vertical circular motion, the tension in the string is maximum at the lowest point of the path.The force equation at the lowest point is given by: $T_{max} = m \omega_{max}^2 r + mg$.Given: $T_{max} = 30 \,N$, $m = 0.5 \,kg$, $r = 2 \,m$, and $g = 10 \,m/s^2$.Substituting the values into the equation:$30 = (0.5) \cdot \omega_{max}^2 \cdot (2) + (0.5) \cdot (10)$.$30 = 1 \cdot \omega_{max}^2 + 5$.$30 - 5 = \omega_{max}^2$.$25 = \omega_{max}^2$.$\omega_{max} = \sqrt{25} = 5 \,rad/s$.

Class 11 Physics Work, Energy, Power and Collision Vertical Circular Motion

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$A$ weightless string can support a tension up to $30 \,N$. $A$ stone of mass $0.5 \,kg$ is tied to its one end and is revolved in a circular path of radius $2 \,m$ in a vertical plane. Then the maximum angular velocity of the stone will be (acceleration due to gravity $g=10 \,m/s^2$)

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A
$10 \,rad/s$
B
$\sqrt{60} \,rad/s$
C
$\sqrt{30} \,rad/s$
D
$5 \,rad/s$

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$A$ weightless string can support a tension up to $30 \,N$. $A$ stone of mass $0.5 \,kg$ is tied to its one end and is revolved in a circular path of radius $2 \,m$ in a vertical plane. Then the maximum angular velocity of the stone will be (acceleration due to gravity $g=10 \,m/s^2$)

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