| 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 the particle speed is \[v\,=\,{{v}_{0}}.\] The piston is moved inward at a very low speed V such that \[\text{V}\,<<\,\frac{\text{dL}}{\text{L}}{{\text{V}}_{0}}.\] Where \[dL\] the infinitesimal displacement of the piston. Which of the following statements) is/are correct? |
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A) After each collision with the piston, the particle speed increases by 2V
B) If the piston moves inward by dL, the particle speed increases by \[2v\frac{d\text{L}}{\text{L}}\]
C) The rate at which the particle strikes the piston is v/L.
D) The particle's kinetic energy increases by a factor of 4 when the piston is moved inward from \[{{\text{L}}_{0}}\] to \[\frac{1}{2}{{\text{L}}_{0}}.\]
Correct Answer: A , D
Solution :
| Initial : \[v={{v}_{0}}\] |
| Distance, \[x={{\text{L}}_{0}}\] |
| \[dt=\frac{dx}{v},\]\[dt'=\frac{2x}{v}\] |
| No. of collision \[=n=\frac{v}{2x}\] |
| Total in\[dt=\,\text{N}\,=\,\frac{v}{2x}.\frac{dx}{v}\] |
| Speed change in dx shifting \[=dv=\frac{vdx}{2xv}.2v\] |
| \[\therefore \] \[dv=\frac{vdx}{x}\] |
| \[\Rightarrow \] \[dv=v\frac{dL}{\text{L}}\] |
| \[\Rightarrow \] \[\int_{{{v}_{0}}}^{v'}{\frac{dv}{v}=\int_{l}^{1/2}{\frac{d\text{L}}{\text{L}}}}\] |
| \[\Rightarrow \] \[\text{V}'\,=\,2{{\text{V}}_{0}}\] |
| \[\therefore \] KE is 4 times. |
Solution :
Same as above
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