QJEEAdv23Ph35e5: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph35e5 Solution: A particle of mass is moving in the -plane such that its…

Given :

$\vec{v} = \alpha(y \hat{x} + 2x \hat{y})$

For the x-component, $v_x = \alpha y$ :

$\frac{dv_x}{dt} = \alpha \frac{dy}{dt} = \alpha v_y$

Given $v_y = 2\alpha x$ :

$\frac{dv_x}{dt} = \alpha (2\alpha x) = 2\alpha^2 x$

For the y-component, $v_y = 2\alpha x$ :

$\frac{dv_y}{dt} = 2\alpha \frac{dx}{dt} = 2\alpha v_x$

Given $v_x = \alpha y$ :

$\frac{dv_y}{dt} = 2\alpha (\alpha y) = 2\alpha^2 y$

Thus, combining the components :

$\vec{a} = 2\alpha^2 x \hat{x} + 2\alpha^2 y \hat{y}$

Now, the force acting on the particle is given by :

$\vec{F} = m\vec{a}$

$\vec{F} = m\alpha(2 \alpha x \hat{x} + 2 \alpha y \hat{y})$

$\vec{F} = m\alpha^2(2x \hat{x} + 2y \hat{y})$

This is equivalent to :

$\vec{F} = 2m\alpha^2(x \hat{x} + y \hat{y})$

The correct answer is Option A :

$\vec{F} = 2m\alpha^2(x \hat{x} + y \hat{y})$