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A conical pendulum of length $L$ makes an angle $θ$ with the vertical. The time period will be

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$2 π \sqrt{\frac{L}{gcosθ}}$

$2 π \sqrt{\frac{Ltanθ}{g}}$

$2 π \sqrt{\frac{L}{gtanθ}}$

$2 π \sqrt{\frac{Lcosθ}{g}}$

answer is A.

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Detailed Solution

$Tcosθ = mg, Tsinθ = {mRω}^{2}$

or $tanθ = \frac{{Rω}^{2}}{g} = \frac{R (2 π / T)^{2}}{g}$

$∴ T = 2 π \sqrt{\frac{R}{gtanθ}}$

But $R = Lsinθ$

$∴ T = 2 π \sqrt{\frac{Lcosθ}{g}}$

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