A) \[0.75m/{{s}^{2}}\]
B) \[\Delta T\]
C) \[\alpha \]
D) \[2\omega [1-2\alpha \Delta T]\]
Correct Answer: C
Solution :
(c.)As, we know that \[{{I}_{0}}\] \[4{{l}_{0}}\] \[3{{l}_{0}}\] \[\frac{{{l}_{0}}}{2}\] Pressure p is minimum, when, \[2{{l}_{0}}\] \[{{I}_{0}}\] \[2{{l}_{0}}{{\cos }^{2}}\left( \frac{\pi y}{\beta } \right)\] or \[{{l}_{0}}{{\cos }^{2}}\left( \frac{\pi y}{\beta } \right)\] or, \[\frac{{{l}_{0}}}{2}{{\cos }^{2}}\left( \frac{2\pi y}{\beta } \right)\] So, when volume is \[4{{l}_{0}}{{\cos }^{2}}\left( \frac{\pi y}{\beta } \right)\], pressure will be minimum. At this volume pressure is given by \[{{C}_{1}}\] \[{{C}_{2}}\] \[\frac{{{C}_{1}}}{{{C}_{2}}}\] \[\frac{a}{b}\] \[\frac{2a}{b}\]
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