JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
Two particles of the same mass are moving in circular orbits because of force, given by \(F(r) = \frac{{ - 16}}{r}\, - \,{r^3}\) 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
- A \(10^{-1}\)
- B \(6 \times {10^{-2}}\)
- C \(6 \times {10^2}\)
- D \(3 \times {10^{-3}}\)
Answer & Solution
Correct Answer
(B) \(6 \times {10^{-2}}\)
Step-by-step Solution
Detailed explanation
As the particles moving in circular orbits, So \(\frac{{m{v^2}}}{r} = \frac{{16}}{r} + {r^3}\) Kinetic energy, \(K{E_0} = \frac{1}{2}m{v^2} = \frac{1}{2}\left[ {16 + {r^4}} \right]\) For fist particle, \(r = 1,\) \({K_1} = \frac{1}{2}\left( {16 + 1} \right)\) Similarly, for…
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