A particle of mass m is moving along a trajectory given by x=x_{0}+a \cos \\omega_{1} t y= y_{0}+b \sin \\omega_{2} t The torque,acting on the particle about the origin, at t=0 is : [10 Apr 2019 Shift 1]

Correct Answer is : +my0aω12 ^k

+m y_{0} a \\omega_{1}^{2} \stackrel{\^}{k}

m (-x_{0} b+y_{0} a ) \\omega_{1}^{2} \stackrel{\^}{k}

-m (x_{0} b w_{2}^{2}-y_{0} a w_{1}^{2} ) \stackrel{\^}{k}

JEE Main 10 Apr 2019 Paper 1 Solved Paper

Section: Physics

Question : 1 of 90

Marks: +1, -0

A particle of mass m is moving along a trajectory given by x=x0+a‌cos‌ω1‌t ‌‌y=y0+b‌sin‌ω2‌t The torque,acting on the particle about the origin, at t=0 is :

[10 Apr 2019 Shift 1]

+my0aω12
^
k
m (−x0b+y0a )ω12
^
k
−m(x0bw22−y0aw12 )
^
k
Zero

Validate

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