Work Energy Power

JEE Main level, single correct, medium

118 of 183

Question 118 of 183

Two particles of the same mass mm are moving in circular orbits because of force, given by F(r)=−16r−r3F\left(r\right)=-\dfrac{16}{r}-{r}^{3}. The first particle is at a distance r=1r=1, and the second, at r=4r=4. The best estimate for the ratio of kinetic energies of the first and the second particle is closest to

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Answer: (C) 6×10−26\times {10}^{-2}

Centripetal force is given as F(r)=mv2rF\left(r\right)=\dfrac{m{v}^{2}}{r}.
Given in the problem, mV2r=16r+r3...(i)\dfrac{m{V}^{2}}{r}=\dfrac{16}{r}+{r}^{3}...\left(i\right)

Kinetic energy is stated as KE0=12mv2K{E}_{0}=\dfrac{1}{2}m{v}^{2}.

KE0=12[16+r4]K{E}_{0}=\dfrac{1}{2}\left[16+{r}^{4}\right] (From equation(i)\left(i\right))

The ratio of kinetic energies of first and second particle is K1K2=16+1216+2562=17272\dfrac{{K}_{1}}{{K}_{2}}=\dfrac{\dfrac{16+1}{2}}{\dfrac{16+256}{2}}=\dfrac{17}{272}

K1K2≃6×10−2\dfrac{{K}_{1}}{{K}_{2}}≃6\times {10}^{-2}.

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