Average torque on a projectile of mass m, initial speed u and angle of projection θ between initial and final positions P and Q as shown in the figure about the point of projection is:
A
mu2sin2θ2
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B
mu2cosθ
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C
μ2sinθ
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D
mu2cosθ2
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Solution
The correct option is Amu2sin2θ2 →τav⋅Δt=Δ→L |Δ→L|=|→Lf−→Li| above point of projection. =(musinθ)(Range) =(musinθ)(u2sin2θ)g=mu3sinθsin2θg |→τav|=Δ→LΔt=mu3sinθsin2θgg2usinθ =mu2sin2θ2