A particle of mass $m$ is initially at rest at the origin. It is subjected to a force and starts moving along the $x$-axis. Its kinetic energy $K$ changes with time as $\text{d}K/\text{d}t = \gamma t$, where $\gamma$ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?

Step 1: Find the kinetic energy as a function of time.

The rate of change of kinetic energy is given.

$$\frac{\text{d}K}{\text{d}t} = \gamma t$$

Integrate with respect to time, using the initial condition $K(0) = 0$.

$$K(t) = \int_{0}^{t} \gamma t' \, \text{d}t'$$ $$K(t) = \frac{1}{2}\gamma t^2$$

Step 2: Determine the speed of the particle.

Relate kinetic energy to speed.

$$K = \frac{1}{2}mv^2$$

Substitute the expression for $K(t)$.

$$\frac{1}{2}mv^2 = \frac{1}{2}\gamma t^2$$ $$v(t) = \sqrt{\frac{\gamma}{m}}\,t$$

Since $v \propto t$, statement (2) is true.

Step 3: Determine the force acting on the particle.

Find the acceleration by differentiating velocity.

$$a = \frac{\text{d}v}{\text{d}t}$$ $$a = \sqrt{\frac{\gamma}{m}}$$

Calculate the force using Newton's second law.

$$F = ma$$ $$F = \sqrt{\gamma m}$$

Since the force is constant, statement (1) is true.

Step 4: Determine the position of the particle.

Integrate the velocity to find the position $x(t)$, using the initial condition $x(0) = 0$.

$$x(t) = \int_{0}^{t} v(t') \, \text{d}t'$$ $$x(t) = \int_{0}^{t} \sqrt{\frac{\gamma}{m}}\,t' \, \text{d}t'$$ $$x(t) = \frac{1}{2}\sqrt{\frac{\gamma}{m}}\,t^2$$

Since $x \propto t^2$, statement (3) is false.

Step 5: Determine if the force is conservative.

The force is constant in both magnitude and direction.

$$W = Fx$$

Since the work done depends only on the displacement and is independent of the path, the force is conservative. Thus, statement (4) is true.

Answer: The true statements are (1), (2), and (4).