A heavy particle is tied to the end A of a string of length 1.6 m. Its other end O is fixed. It revolves as a conical pendulum with the string making $60^{\circ}$ with the vertical. Then $(take g = 9.8 m / s^{2})$

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Its period of revolution is $\frac{4 π}{7} \sec$

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

The tension in the string is double the weight of the particle

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

answer is A, B, C, D.

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Detailed Solution

$a) T = 2 π \sqrt{\frac{Lcos θ}{g}} = 2 π \sqrt{\frac{1.6 \times 1 / 2}{9.8}} = \frac{4 π}{7} \sec$

$b) Tcos θ = mg \Rightarrow Tcos 60 = mg \Rightarrow T = 2 mg$

$c) V = \sqrt{gr t an θ}; V = \sqrt{gLsin θ t an θ} = 2.8 \sqrt{3} m / s$

$d) a = \frac{V^{2}}{r} = \frac{(2.8 \sqrt{3})^{2}}{0.8 \sqrt{3}} = 9.8 \sqrt{3} m / s^{2}$

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