JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Heat & Thermodynamics

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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 ≪ \frac{dL}{L} v_{0}

, 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 $L_{0}$ to $\frac{1}{2} L_{0}$ ✓ Correct
(D)If the piston moves inward by dL, the particle speed increases by $2 v \frac{dL}{L}$

Answer: (B), (C)

Solution

The collision frequency is

\frac{v}{2 L}

, not

\frac{v}{L}

. Option (A) is incorrect.

After elastic collision with moving piston, the particle speed increases by

2 V

. Option (B) is correct.

The change in speed in time dt is:

d v = 2 V \times \frac{v}{2 L} \times \frac{d L}{V} = \frac{v d L}{L}

Integrating from

L_{0}

L_{0} /2

\int_{v_{0}}^{v} \frac{d v}{v} = - \int_{L_{0}}^{L_{0} /2} \frac{d L}{L}

Giving

v = 2 v_{0}

, so KE increases by factor of 4. Option (C) is correct.

From the differential equation, the speed increase for displacement dL is

\frac{v d L}{L}

, not

2 v \frac{d L}{L}

. Option (D) is incorrect.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Heat & Thermodynamics

Solution

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