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← JEE Advanced 2024 · Paper 1
PhysicsMultiple correct+4 / -2Question 22 of 51

JEE Advanced 2024 · Paper 1, Physics question 5

A particle of mass mm is moving in a circular orbit under the influence of the central force F(r)=−krF(r) = -kr, corresponding to the potential energy V(r)=kr2/2V(r) = kr^2/2, where kk is a positive force constant and rr is the radial distance from the origin. According to the Bohr's quantization rule, the angular momentum of the particle is given by L=nℏL = n\hbar, where ℏ=h/(2π)\hbar = h/(2\pi), hh is the Planck's constant, and nn a positive integer. If vv and EE are the speed and total energy of the particle, respectively, then which of the following expression(s) is(are) correct?

Choose one or more options
Answer and worked solution

Official answer: A, B, C

Circular orbit: mv2r=kr  ⟹  v=rk/m\dfrac{mv^2}{r} = kr \implies v = r\sqrt{k/m}.

Quantization: L=mvr=r2mk=nℏ  ⟹  r2=nℏ1mkL = mvr = r^2\sqrt{mk} = n\hbar \implies r^2 = n\hbar\sqrt{\dfrac{1}{mk}}. (A) true.

v2=kmr2=nℏkm3v^2 = \dfrac{k}{m}r^2 = n\hbar\sqrt{\dfrac{k}{m^3}}. (B) true.

Lmr2=vr=km\dfrac{L}{mr^2} = \dfrac{v}{r} = \sqrt{\dfrac{k}{m}}. (C) true.

E=12mv2+12kr2=kr2=nℏkmE = \frac12 mv^2 + \frac12 kr^2 = kr^2 = n\hbar\sqrt{\dfrac{k}{m}}, not nℏ2km\dfrac{n\hbar}{2}\sqrt{\dfrac{k}{m}}. (D) false.