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Q.

If an air bubble rises from the bottom of a mercury tank to the top its volume becomes $1 \frac{1}{2}$ times. When normal pressure is 76cm of Hg then the depth of the Hg tank is

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a

76 cm

b

132 cm

c

49 cm

d

38 cm

answer is A.

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Detailed Solution

$Here temperature remain constant$

$So P \propto \frac{1}{V} P_{1} V_{1} = P_{2} V_{2}$

$here P_{1} = P_{0} + H P_{2} = P_{0} and P_{0} = 76 cm of Hg$

$V_{2} = 1.5 V_{1}$

$76 \times 1.5 V_{1} = (76 + H) V_{1}$

$H = 38 cm$

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