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JEE Advanced 2023 · Paper 2

JEE Advanced 2023 Physics question on Laws of Motion

A particle of mass mm is moving in the xyxy-plane such that its velocity at a point (x,y)(x, y) is given as v⃗=α(yx^+2xy^)\vec{v} = \alpha(y\hat{x} + 2x\hat{y}), where α\alpha is a non-zero constant. What is the force F⃗\vec{F} acting on the particle?
  1. A.
    F⃗=2mα2(xx^+yy^)\vec{F} = 2m\alpha^2(x\hat{x} + y\hat{y})
  2. B.
    F⃗=mα2(yx^+2xy^)\vec{F} = m\alpha^2(y\hat{x} + 2x\hat{y})
  3. C.
    F⃗=2mα2(yx^+xy^)\vec{F} = 2m\alpha^2(y\hat{x} + x\hat{y})
  4. D.
    F⃗=mα2(xx^+2yy^)\vec{F} = m\alpha^2(x\hat{x} + 2y\hat{y})
Show answer

Correct option: A

[JEE Advanced 2023 · Paper 2]

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