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A simple pendulum of length $ℓ$ is set in motion such that the bob, of mass m, moves along a horizontal circular path, and string makes a constant angle $θ$ with the vertical. The time period of rotation of the bob is t and tension in the thread is T.

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$t = 2 π \sqrt{ℓ / g}$

$t = 2 π \sqrt{ℓ \cos θ / g}$

$T = \frac{4 π^{2} m ℓ}{t^{2}}$

The bob is in equilibrium

answer is B, C.

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Detailed Solution

$T \sin θ = m ℓ ω^{2} o r T = m ω^{2} ℓ [∵ r = ℓ \sin θ]$

$T \cos θ = m g o r m ω^{2} / ℓ \cos θ = m g$

$o r ω^{2} = \frac{g}{ℓ \cos θ}$

$t = \frac{2 π}{ω} = 2 π \sqrt{\frac{ℓ \cos θ}{g}}$

$T = m [\frac{4 π^{2}}{t^{2}}] ℓ$

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