JEE Advanced 2019 · Paper 2 · Physics Q3

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PhysicsMultiple correct+4 / -1Question 3 of 54

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 $L = L_0$ the particle speed is $v = v_0$ . The piston is moved inward at a very low speed $V$ such that $V \ll \dfrac{dL}{L}v_0$ , where $dL$ is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct?

A horizontal tube closed at the left end and fitted with a movable piston at the right; a particle moves along the axis with speed v; the piston is a distance L from the closed end.

The rate at which the particle strikes the piston is

v/L

After each collision with the piston, the particle speed increases by

2V

If the piston moves inward by

dL

, the particle speed increases by

2v\dfrac{dL}L

The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from

L_0

\frac12L_0

Answer and worked solution

Official answer: B, D

(A) Between successive hits on the piston the particle travels $2L$ , so the rate is $\dfrac v{2L}$ , not $\dfrac vL$ . False.

(B) In the piston's frame the particle rebounds with the same speed. Back in the lab frame it gains $2V$ at each hit. True.

(C) While the piston moves in by $dL$ (time $dL/V$ ), the particle hits it $\dfrac v{2L}\cdot\dfrac{dL}V$ times. So
$dv=2V\cdot\frac{v\,dL}{2LV}=\frac{v\,dL}L,$
not $2v\frac{dL}L$ . False.

(D) From $\frac{dv}v=-\frac{dL}L$ , the product $vL$ stays constant. Halving $L$ doubles $v$ , so the kinetic energy becomes 4 times larger. True.

(B), (D)

Topics:Elastic and inelastic collisions Kinetic interpretation of temperature

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JEE Advanced 2019 · Paper 2, Physics question 3