Q 11-04-199JEE MainJEE Main 2018 (16 Apr, Shift 1)Medium
Two particles of the same mass $m$ are moving in circular orbits because of force, given by $F(r) = -\dfrac{16}{r} - r^3$. The first particle is at a distance $r = 1$, and the second, at $r = 4$. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to
Answer: (C) $6\times10^{-2}$
The attractive force supplies the centripetal force:
$$\frac{mv^2}{r} = \frac{16}{r} + r^3 \quad\Rightarrow\quad mv^2 = 16 + r^4$$
So $K = \frac12mv^2 \propto 16 + r^4$:
$$\frac{K_1}{K_2} = \frac{16 + 1}{16 + 256} = \frac{17}{272} = 0.0625 \approx 6\times10^{-2}$$
Solution by Sreeraj P, M.Sc Physics