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 dK/dt = \gamma t, where \gamma is a positive constant of appropriate dimensions. Which of a positive constant of appropriate dimensions. Which of the following statement is (are) true?

The force applied on the particle is constant, The speed of the particle is proportional to time, The force is conservative

The force applied on the particle is constant

The speed of the particle is proportional to time

The distance of the particle from the origin increases linearly with time

Work Power and Energy PYQ — JEE Advanced 2018 · Shift 2 · Q54 with Solution

View written solutionFree

Correct answer: A, B, D

We are given $\frac{dK}{dt}=\gamma t,$ where $\gamma>0$ is constant.

Also, initially the particle is at rest at the origin, so at $t=0$ , $K=0.$

Integrating, $K=\int \gamma t\,dt=\frac{\gamma t^2}{2}+C.$ Using $K(0)=0$ , we get $C=0$ .

Hence, $K=\frac{\gamma t^2}{2}.$

Since $K=\frac{1}{2}mv^2,$ we have $\frac{1}{2}mv^2=\frac{\gamma t^2}{2}.$ So, $mv^2=\gamma t^2$ $v=t\sqrt{\frac{\gamma}{m}}.$

Since the particle moves along the positive $x$ -axis after starting from rest, speed is $v\propto t.$

Therefore, Option B is true.

From $v=t\sqrt{\frac{\gamma}{m}},$ we get $a=\frac{dv}{dt}=\sqrt{\frac{\gamma}{m}}=\text{constant}.$

Thus force is $F=ma=m\sqrt{\frac{\gamma}{m}}=\sqrt{m\gamma},$ which is a constant.

Therefore, Option A is true.

Since $v=\frac{dx}{dt}=t\sqrt{\frac{\gamma}{m}},$ integrating,

=\frac{1}{2}\sqrt{\frac{\gamma}{m}}\,t^2 + C.$$ With $x(0)=0$, $C=0$. Hence, $$x=\frac{1}{2}\sqrt{\frac{\gamma}{m}}\,t^2.$$ So distance from origin increases as $t^2$, **not linearly**. Therefore, **Option C is false**. 6. **Check whether force is conservative** The force is constant: $$F=\sqrt{m\gamma}.$$ A constant force depending only on position (in fact independent of position) is conservative. Its potential can be written as $$U(x)=-Fx + \text{constant}.$$ So mechanical work done depends only on initial and final position. Therefore, **Option D is true**. 7. **Final evaluation of options** - **A:** True - **B:** True - **C:** False - **D:** True So the correct options are $$\boxed{A,\ B,\ D}.$$