A particle of mass ‘m’ is projected with an initial velocity u at an angle ‘ $θ$ ’ to horizontal. The torque due to gravity on projectile at maximum height about the point of projection is

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$\frac{{mgu}^{2} \sin θ}{2}$

$\frac{{mgu}^{2} \sin 2 θ}{2}$

${mgu}^{2} \sin 2 θ$

$\frac{1}{2} {mu}^{2} \sin 2 θ$

answer is D.

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Detailed Solution

$\begin{array}{l} τ_{0} = \vec{r} \times \vec{F} = \frac{R}{2} \times mg \\ = \frac{(\frac{u^{2} \sin 2 θ}{g})}{2} \times mg = \frac{1}{2} {mu}^{2} \sin 2 θ \end{array}$

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