The bob of a 0.2 m pendulum describes an arc of circle in a vertical plane. If the tension in the cord is 3 times the weight of the bob when the cord makes an angle 30° with the vertical, the acceleration of the bob in that position is

a=an2+a12=T-mgcos30°m2+gsin30°2=g3-322+14=g

The bob of a 0.2 m pendulum describes an arc of circle in a vertical plane. If the tension in the cord is 3 times the weight of the bob when the cord makes an angle 30° with the vertical, the acceleration of the bob in that position is

a=an2+a12=T-mgcos30°m2+gsin30°2=g3-322+14=g

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The bob of a $0.2 m$ pendulum describes an arc of circle in a vertical plane. If the tension in the cord is $\sqrt{3}$ times the weight of the bob when the cord makes an angle $30 °$ with the vertical, the acceleration of the bob in that position is

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$\frac{g}{4}$

$\frac{g}{2}$

$\frac{\sqrt{3} g}{2}$

answer is A.

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Detailed Solution

$a = \sqrt{a_{n}^{2} + a_{1}^{2}}$

$= \sqrt{{(\frac{T - mgcos 30 °}{m})}^{2} + {(gsin 30 °)}^{2}}$

$= g \sqrt{{(\sqrt{3} - \frac{\sqrt{3}}{2})}^{2} + (\frac{1}{4})} = g$

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The bob of a 0.2 m pendulum describes an arc of circle in a vertical plane. If the tension in the cord is 3 times the weight of the bob when the cord makes an angle 30° with the vertical, the acceleration of the bob in that position is

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The bob of a $0.2 m$ pendulum describes an arc of circle in a vertical plane. If the tension in the cord is $\sqrt{3}$ times the weight of the bob when the cord makes an angle $30 °$ with the vertical, the acceleration of the bob in that position is