Q.1 (v) The period of conical pendulum in terms of its length (l) , semi-vertical angle (θ) and acceleration due to gravity (g) is _____. A. 1/2π√(lcosθ/g) B. 1/2π√(lsinθ/g) C. 4π√(lcosθ/4g) D. 4π√(ltanθ/g)

Answer: C. 4π√(lcosθ/4g) Answer: (c) 4π√(lcosθ/4g) For a conical pendulum, Tcosθ = mg and Tsinθ = mω² r with r = lsinθ . Dividing, ω² = g/lcosθ , so the period is T_p = 2π√(lcosθ/g) = 4π√(lcosθ/4g) (since 4π√(x/4) = 2π√(x) ). Options (a) and (b) have the factor 1/2π in place of 2π (and (b) also has sinθ ), and (d) has the wrong trigonometric ratio.

The period of conical pendulum in terms of its length \((l)\), semi-vertical angle \((\theta)\) and acceleration due to gravity \((g)\) is _____.

Q.1 (v) The period of conical pendulum in terms of its length \((l)\), semi-vertical angle \((\theta)\) and acceleration due to gravity \((g)\) is _____.

A. \(\dfrac{1}{2\pi}\sqrt{\dfrac{l\cos\theta}{g}}\)

B. \(\dfrac{1}{2\pi}\sqrt{\dfrac{l\sin\theta}{g}}\)

C. \(4\pi\sqrt{\dfrac{l\cos\theta}{4g}}\)

D. \(4\pi\sqrt{\dfrac{l\tan\theta}{g}}\)

Doubtly Asked 27 Oct 2026 Board exam · March 2026 hsc 12th Save WhatsApp Report

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Answer: C. \(4\pi\sqrt{\dfrac{l\cos\theta}{4g}}\)

Answer: (c) \(4\pi\sqrt{\dfrac{l\cos\theta}{4g}}\)

For a conical pendulum, \(T\cos\theta = mg\) and \(T\sin\theta = m\omega^2 r\) with \(r = l\sin\theta\). Dividing, \(\omega^2 = \dfrac{g}{l\cos\theta}\), so the period is

\[T_p = 2\pi\sqrt{\frac{l\cos\theta}{g}} = 4\pi\sqrt{\frac{l\cos\theta}{4g}}\]

(since \(4\pi\sqrt{x/4} = 2\pi\sqrt{x}\)). Options (a) and (b) have the factor \(\tfrac{1}{2\pi}\) in place of \(2\pi\) (and (b) also has \(\sin\theta\)), and (d) has the wrong trigonometric ratio.

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report

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The period of conical pendulum in terms of its length \((l)\), semi-vertical angle \((\theta)\) and acceleration due to gravity \((g)\) is _____.

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