JEE Main On 16 April 2018 Question 4
Question: Two particles of the same mass m 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 [JEE Main 16-4-2018]
Options:
A) $ {10^{-1}} $
B) $ 6\times {10^{-2}} $
C) $ 6\times 10^{2} $
D) $ 3\times {10^{-3}} $
Show Answer
Answer:
Correct Answer: B
Solution:
The force is required for the circular motion of the body
Hence $ |F|=\frac{mv^{2}}{r} $ $ \frac{mv^{2}}{r}=\frac{16}{r}+r^{3} $
$ mv^{2}=16+r^{4} $ K.E. $ =\frac{mv^{2}}{2}=8+\frac{r^{4}}{4} $
Putting the value of $ r=1 $ and $ r=4 $
taking the ratio We get the $ \approx 6\times {10^{-2}} $
