Average torque on a projectile of mass m, initial speed u and angle of projection θ 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 Amu2sin2θ2 Torque at a general point = Horizontal distance covered × m g (as we only take r perpendicular to g) ∴τ=mg×ucosθt ......(1) t=usinθg Substituting t in (1), we get τ=mu2sin2θ2