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A conical pendulum of length L makes an angle $θ = 45^{0}$ w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.9m and its centre is vertically below O. The speed of the pendulum, in its circular path, will be: (Take $g = 10 m s^{- 2})$

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$2 m / s$

$0.4 m / s$

$3 m / s$

$0.2 m / s$

answer is B.

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Detailed Solution

Given, $θ = 45^{0}, r = 0.9 m, g = 10 m / s^{2}$

$T \sin θ = \frac{m v^{2}}{r} ........... (i)$

$T \cos θ = m g ........... (i i)$

From equation (i) & (ii) we have, $\tan θ = \frac{v^{2}}{r g}$

$v^{2} = r g ∵ θ = 45^{0}$

Hence, speed of the pendulum in its circular path, $v = \sqrt{r g} = \sqrt{0.9 \times 10} = 3 m / s$

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