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(C) Given kinetic energy $K \propto t$,so $K = \lambda t$,where $\lambda$ is a constant.Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = \lambda

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$(ii), (iv)$

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(C) Given kinetic energy $K \propto t$,so $K = \lambda t$,where $\lambda$ is a constant.Since $K = \frac{1}{2}mv^2$,we have $\frac{1}{2}mv^2 = \lambda t$,which implies $v = \sqrt{\frac{2\lambda}{m}} t^{1/2}$.The acceleration $a$ is the derivative of velocity with respect to time: $a = \frac{dv}{dt} = \sqrt{\frac{2\lambda}{m}} \cdot \frac{1}{2} t^{-1/2}$.The force $F = ma = m \cdot \sqrt{\frac{2\lambda}{m}} \cdot \frac{1}{2\sqrt{t}} = \sqrt{\frac{\lambda m}{2}} \cdot \frac{1}{\sqrt{t}}$.Thus,$F \propto \frac{1}{\sqrt{t}}$,which matches statement $(ii)$.Now,expressing $F$ in terms of speed $v$: since $t^{1/2} = v \cdot \sqrt{\frac{m}{2\lambda}}$,we substitute this into the force equation:$F = \sqrt{\frac{\lambda m}{2}} \cdot \frac{1}{v \sqrt{\frac{m}{2\lambda}}} = \frac{\lambda}{v}$.Thus,$F \propto \frac{1}{v}$,which matches statement $(iv)$.Therefore,statements $(ii)$ and $(iv)$ are correct.

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