A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The time taken to move a distance x is proportional to

Power, $\mathrm{P}=\frac{\text{ Energy }}{\text{ time }}=\frac{\text{ Force }\times \text{ displacement }}{\text{ time }}$
Let m be the mass of the box.
$\mathrm{P}=\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}\cdot \frac{\text{ displacement }}{\text{ time }}=\mathrm{m}\frac{\mathrm{dv}}{\mathrm{dt}}\cdot \mathrm{v}$
Now, Pdt = m vdv
Also, ${\int }_0^{\mathrm{t}} \mathrm{Pdt}={\int }_0^{\mathrm{v}} \mathrm{mvdv}$ [Box starts from rest]
Since power is constant $\text{Pt=}\frac{{\mathrm{mv}}^2}{2};\mathrm{v}=\sqrt{\frac{2\mathrm{Pt}}{\mathrm{m}}}$
We know, $\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}\mathrm{so}\frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\frac{2\mathrm{Pt}}{\mathrm{m}}}$
$\int \mathrm{dx}=\sqrt{\frac{2\mathrm{P}}{\mathrm{m}}}\int \sqrt{\mathrm{t}}\mathrm{dt};\mathrm{x}=\sqrt{\frac{2\mathrm{P}}{\mathrm{m}}}\cdot \frac{3}{2}{\mathrm{t}}^{3/2}\Rightarrow \mathrm{x}\propto {\mathrm{t}}^{3/2}\Rightarrow \mathrm{t}\propto {\mathrm{x}}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$
Hence, the correct answer is option (c).