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The arms of a `U` shaped tube are vertical. The arm on right side is closed and other arm is closed by light movable piston. There is mercury in the tube and initially the level of the mercury in the arms is the same. Above the mercury, there is an air column of height h in each arm, and initial pressure is the same as atmospheric pressure in both arms. Now pistion is slowly pushed down by a distance of `h//2` (see figure) Let x is the displacement of mercury level in one of the stem and `P_(1)` is pressure in left column of air after compression, `P_(2)` is pressure in right column of air after compression and `P_(0)` is atmospheric pressure thenA. `h P_(0)=((h)/(2)+x)P_(1)`B. `hP_(0)=(h-x)P_(2)`C. `P_(1)=P_(2)`D. `P_(1)-P_(2)=2xpg` where p is density of mercury |
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Answer» Correct Answer - A::B::D `p_(1)=(P_(0)h)/((h)/(2) + x )` `P_(2)= (P_(0)h)/(h-x)` ` P_(1)-P_(2)=2x pg` `(P_(0)h)/((h)/(2)+2)-(P_(0)h)/(h-x)=2xpg` `P_(0)h[h-x-(h)/(2)-x]=((h)/(2)+x)(h-x)2xpg` |
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