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Q.

A tube of length /r, which is wide enough to male surface tension effects negligible, is closed at one end. It is then lowered into a tank of mercury to a depth.h as shown in fig. (4), so that mercury rises a distance x into
the tube. If the mercury barometer stands at i also, then

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a

$(2 h - x) (h + x) = h^{2}$

b

$h (h - x) = h^{2}$

c

${(2h−x)(h−x)=h}^{2}$

d

$2 h (h - x) = h^{2}$

answer is A.

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Detailed Solution

At atmospheric pressure, the air fills the tube. Its volume is h a, where a is area of cross section of the tube. When
the tube is pushed into mercury, the air volume is (h-x) $a$ Applying Boyle's law, we have
$h \cdot ha = (2 h - x) (h - x) a or h^{2} = (2 h - x) (h - x)$

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