Other Physics · JEE Advanced previous year questions

2023 1 question

1. Q202023 · 4 Jun 2023 · Paper 2

   A particle of mass $m$ is moving in the $xy$-plane such that its velocity at a point $(x,y)$ is given as $\vec{v}=\alpha (y\hat{x}+2x\hat{y})$, where $\alpha$ is a non-zero constant. What is the force $\vec{F}$ acting on the particle?

   1. A.$\vec{F}=2m{\alpha }^2(x\hat{x}+y\hat{y})$
   2. B.$\vec{F}=m{\alpha }^2(y\hat{x}+2x\hat{y})$
   3. C.$\vec{F}=2m{\alpha }^2(y\hat{x}+x\hat{y})$
   4. D.$\vec{F}=m{\alpha }^2(x\hat{x}+2y\hat{y})$
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   Answer: A