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A heavy particle is tied to the end A of string of length 1.6m. Its other end O is fixed. It revolves as a conical pendulum with the string making $60^{0}$ with vertical. $(g = 10 m / s^{2})$

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Its time period is $\frac{5 π}{7}$ sec

The tension in string is double the weight of particle

The speed of particle is $2.8 \sqrt{3} m / s$

The centripetal acceleration of particle is $9.8 \sqrt{3} m / s^{2}$

answer is B, C, D.

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Detailed Solution

$T \sin 60^{0} = \frac{m v^{2}}{\frac{\sqrt{3}}{2} l} \to (1)$

$T \cos 60^{0} = m g$

$T = 2 m g \to (2)$

from (1), (2) $\to V^{2} = \frac{3 g l}{2}$

$V = \sqrt{\frac{3 g l}{2}} = 2.8 \sqrt{3} m / s$

$a = \frac{V^{2}}{r} = \frac{\frac{3 g l}{2}}{l (\frac{\sqrt{3}}{2})} = \sqrt{3} g, T = \frac{2 π r}{V} = \frac{4 π}{7}$

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