$'$To solve this problem, we need to find the force on a particle given that a machine delivers constant power $k$ watt.$\\$The particle starts from rest, so its initial velocity $v_0 = 0. \\$Let's denote:$\\$• $P = k$ as the constant power delivered to the particle.$\\$• $F$ as the force on the particle.$\\$• $v$ as the velocity of the particle at time $t. \\$The relation between power, force, and velocity is given by:$\\ P = F \cdot v \\$Since the power is constant, we have:$\\ k = F \cdot v \\$From this, we can express the force as:$\\ F = \frac{k}{v} \\$Next, we need to find the expression for velocity $v$ as a function of time $t. \\$The kinetic energy $K$ of the particle is given by:$\\ K = \frac{1}{2} m v^2 \\$The rate of change of kinetic energy is equal to the power:$\\ \frac{d}{dt}\left(\frac{1}{2} m v^2\right) = k \\$Differentiating the kinetic energy with respect to time, we get:$\\ m v \frac{dv}{dt} = k \\$Since $\frac{dv}{dt}$ is the acceleration $a,$ we have:$\\ m v a = k \\$Using $a = \frac{dv}{dt},$ we can write:$\\ m v \frac{dv}{dt} = k \\$Rearrange and integrate both sides with respect to time:$\\ m \int v \, dv = \int k \, dt \\ \frac{m}{2} v^2 = kt + C \\$Since the particle starts from rest, at $t = 0, v = 0,$ so $C = 0. \\$Thus, we have:$\\ \frac{m}{2} v^2 = kt \\ v^2 = \frac{2kt}{m} \\ v = \sqrt{\frac{2kt}{m}} \\$Substitute $v$ back into the expression for force:$\\ F = \frac{k}{v} = \frac{k}{\sqrt{\frac{2kt}{m}}} \\ F = \frac{k}{\sqrt{2k} \cdot \sqrt{\frac{t}{m}}} \\ F = \sqrt{\frac{m k}{2}} \cdot t^{-1/2} \\$Therefore, the force on the particle at time $t$ is given by Option 4:$\\ \sqrt{\frac{m k}{2}} \cdot t^{-1/2} \\$However, the correct option according to the question is Option 1:$\\ \sqrt{m k} \cdot t^{-1/2} \\$This discrepancy suggests a possible error in the problem statement or options.