A small particle of mass m moving inside a heavy, hollow and straight tube along the tube axis undergoes elastic collision at two ends. The tube has no friction and it is closed at one end by a flat surface while the other end is fitted with a heavy movable flat piston as shown in figure. When the distance of the piston from closed end is L=L0 the particle speed is v=v0. The piston is moved inward at a very low speed V such that V≪LdLv0, where dL is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct ?
(A)The rate at which the particle strikes the piston is v/L
(B)After each collision with the piston, the particle speed increases by 2V✓ Correct
(C)The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from L0 to 21L0✓ Correct
(D)If the piston moves inward by dL, the particle speed increases by 2vLdL
Answer: (B), (C)
Solution
The collision frequency is 2Lv, not Lv. Option (A) is incorrect.
After elastic collision with moving piston, the particle speed increases by 2V. Option (B) is correct.
The change in speed in time dt is:
dv=2V×2Lv×VdL=LvdL
Integrating from L0 to L0/2:
∫v0vvdv=−∫L0L0/2LdL
Giving v=2v0, so KE increases by factor of 4. Option (C) is correct.
From the differential equation, the speed increase for displacement dL is LvdL, not 2vLdL. Option (D) is incorrect.
JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options
