Question

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

Easy

Solution

$P_{1} V_{1} = P_{2} V_{2}; (H + h) V_{1} = {HV}_{2}$

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