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Question

Average torque on a projectile of mass $$m$$, initial speed $$u$$ and angle of projection $$\theta$$ between initial and final positions $$P$$ and $$Q$$ about the point of projection is


A
mu2sin2θ2
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B
mu2cosθ
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C
mu2sinθ
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D
mu2cosθ2
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Solution

The correct option is A $$\dfrac {mu^{2}sin2\theta}{2}$$
Torque at a general point $$=$$ Horizontal distance covered $$\times$$ m g (as we only take r perpendicular to g)
 $$\therefore \tau = mg \times u cos \theta t$$ ......(1)
$$t =\dfrac{usin \theta}{g}$$
 Substituting t in (1), we get $$\tau = \dfrac{mu^2 \sin 2\theta}{2}$$

Physics

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