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:

Correct option is B.
+my0aω₁²k^
F = -m(aω₁²cosω₁ti^ + bω₂²sinω₂tj^)
r→ = (x0 + acosω₁t)i^ + (y0 + bsinω₂t)j^
T→ = r→ × F→ = m(x0 + acosωt)bω₂²sinω₂tk^ + m(y0 + sinω₂t)aω₁²cosω₁tk^
at t = 0,
T→ = my0aω₁²k^