A particle having mass m is projected with a speed v at an angle α with the horizontal ground. Find the torque of the weight of the particle about the point of projection when the particle reaches the ground.
A
2mv2sin2α
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B
mv2sin2α2
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C
mv2sin2α
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D
0
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Solution
The correct option is Cmv2sin2α
As we know, →τ=→r×→F →τ=r⊥F r⊥=Range=v2sin2αg Hence, →τ=v2sin2αg×mg →τ=mv2sin2α