Verified Solution • 2025–26 Session moderate

QJEEM18Ph088b: JEE Main 2018 Physics - Class 11 JEEM18Ph088b Solution: Two particles of the same mass m are moving in circular orbits…

Two particles of the same mass m are moving in circular orbits because of force, given by F(r)=−16r−r3F\left( r \right) = {{ - 16} \over 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 :

Multiple Choice Options

  • A 6×10−26 \times {10^{ - 2}}
  • B 3×10−33 \times {10^{ - 3}}
  • C 10−1{10^{ - 1}}
  • D 6×1026 \times {10^{ 2}}

Step-by-Step Verified Solution

In circular motion the force required

∣F∣=mv2r\left| F \right| = {{m{v^2}} \over r}

∴   \therefore\,\,\, mv2r=16r+r3{{m{v^2}} \over r} = {{16} \over r} + {r^3}

⇒\Rightarrow mv2 = 16 + r4

∴   \therefore\,\,\, kinetic energy (K) = 12{1 \over 2} mv2 = 12{1 \over 2} [ 16 + r4]

∴   \therefore\,\,\, Kinetic energy of first particle (K1) = 12{1 \over 2} [16 + 1]

Kinetic energy of second particle (K2) = 12{1 \over 2} [16 + 44]

∴    \therefore\,\,\,\, K1K2{{{K_1}} \over {{K_2}}} = 16+1216+2562{{{{16 + 1} \over 2}} \over {{{16 + 256} \over 2}}} = 17272{{17} \over {272}}

⇒\Rightarrow    \,\,\, K1K2=6×10−2{{{K_1}} \over {{K_2}}} = 6 \times {10^{ - 2}}

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