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Question

A particle of mass 10 kg is projected with a velocity 5 m/s from a point on horizontal ground, making an angle 37 with the horizontal. Find the total torque about the point of projection acting on the particle when it is at its maximum height.


A
120 Nm along ^k direction
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B
120 Nm along ^k direction
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C
60 Nm along ^k direction
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D
60 Nm along ^k direction
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Solution

The correct option is B 120 Nm along ^k direction
Given, mass of the projectile, m=10 kg
Initial velocity of projectile, u=5 m/s
Angle of projection, θ=37

Torque about point of projection due to gravity at maximum height
τ=R2^i×mg(^j)
Range (R)=u2sin2θg
τ=mg(u2sin2θ)2g
=mu2sin2θ2=mu2sinθcosθ
=10×(5)2×sin37×cos37=120
τ=120 Nm (^k)

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