QJEEM23Ph6616: JEE Main 2023 Physics - Class 11 JEEM23Ph6616 Solution: Three forces are acting on a particle of mass . The forces and…

Since the particle is initially at rest, the net force acting on it is zero. This means that the forces $F_1$ , $F_2$ , and $F_3$ are balanced. Given that $F_2$ and $F_3$ are acting perpendicularly, we can represent the balance of forces as follows:

$F_1 = \sqrt{F_2^2 + F_3^2}$

We can now calculate the equivalent force resulting from $F_2$ and $F_3$ :

$F_1 = \sqrt{(8 \mathrm{~N})^2 + (6 \mathrm{~N})^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \mathrm{~N}$

Since the particle is initially at rest, when the force $F_1$ is removed, only $F_2$ and $F_3$ remain. The net force acting on the particle is the equivalent force resulting from $F_2$ and $F_3$ , which we have calculated to be $10 \mathrm{~N}$ .

Now, we can use Newton's second law to find the acceleration of the particle:

$F = ma$

Where:

• $F$ is the net force acting on the particle
• $m$ is the mass of the particle
• $a$ is the acceleration of the particle

Rearranging the equation to solve for $a$ :

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

Plugging in the values for $F$ and $m$ :

$a = \frac{10 \mathrm{~N}}{5 \mathrm{~kg}} = 2 \mathrm{~ms}^{-2}$

The acceleration of the particle when the force $F_1$ is removed is $2 \mathrm{~ms}^{-2}$ , which corresponds to Option C.