JEE Main 2023 — Laws Of Motion Question with Solution

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.