Three forces $F_{1}=10 \mathrm{~N}, F_{2}=8 \mathrm{~N}, \mathrm{~F}_{3}=6 \mathrm{~N}$ are acting on a particle of mass $5 \mathrm{~kg}$. The forces $\mathrm{F}_{2}$ and $\mathrm{F}_{3}$ are applied perpendicularly so that particle remains at rest. If the force $F_{1}$ is removed, then the acceleration of the particle is:

<p>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:</p> <p>$F_1 = \sqrt{F_2^2 + F_3^2}$</p> <p>We can now calculate the equivalent force resulting from $F_2$ and $F_3$:</p> <p>$$F_1 = \sqrt{(8 \mathrm{~N})^2 + (6 \mathrm{~N})^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \mathrm{~N}$$</p> <p>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 $10 \mathrm{~N}$.</p> <p>Now, we can use Newton&#39;s second law to find the acceleration of the particle:</p> <p>$F = ma$</p> <p>Where:</p> <ul> <li>$F$ is the net force acting on the particle</li> <li>$m$ is the mass of the particle</li> <li>$a$ is the acceleration of the particle</li> </ul> <p>Rearranging the equation to solve for $a$:</p> <p>$a = \frac{F}{m}$</p> <p>Plugging in the values for $F$ and $m$:</p> <p>$a = \frac{10 \mathrm{~N}}{5 \mathrm{~kg}} = 2 \mathrm{~ms}^{-2}$</p> <p>The acceleration of the particle when the force $F_1$ is removed is $2 \mathrm{~ms}^{-2}$, which corresponds to Option C.</p>