JEE Advance - Physics (2019 - Paper 2 Offline - No. 2)
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 $$V < < {{dL} \over L}{v_0}$$, where dL is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct?
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 $$V < < {{dL} \over L}{v_0}$$, where dL is the infinitesimal displacement of the piston. Which of the following statement(s) is/are correct?
After each collision with the piston, the particle speed increases by 2V.
If the piston moved inward by dL, the particle speed increases $$2v{{dL} \over L}$$.
The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from L0 to 1/2 L0.
The rate at which the particle strikes the piston is v/L.
Explanation
Change in speed = $$(2v + {v_0} - {v_0}) = 2v$$ In every collision it acquires 2v,
$$f = {v \over {2x}}$$
$${{dv} \over {dt}} = f \times 2v$$
$$dv = {v \over {2x}}2vdt$$
$$dv = {v \over {2x}}2( - dx)$$
Integration on both sides limits v0 to v, we get
$$\int\limits_{{v_0}}^v {{{dv} \over v}} = \int\limits_l^x {{{ - dx} \over x}} $$
$$ \Rightarrow \ln {v \over {{v_0}}} = - \ln {x \over l}$$
$$ \Rightarrow v = {{{v_0}l} \over x}$$
where, $$x = {l \over 2},v = 2{v_0}$$
so, $$f = {{2{v_0}} \over {2{l \over 2}}} = {{2{v_0}} \over l}$$
$$ \therefore $$ $${{{k_f}} \over {{k_i}}} = 4$$
Comments (0)
