Question

An air bubble situated at the bottom of an open kerosene tank rises to the top surface. It is observed that at the top the volume of the bubble is thrice its initial volume. If the atmospheric pressure is 72 cm of Hg, and mercury is 17 times heavier than kerosene the depth of the tank is:

• A. 2.16 m
• B. 2.88 m
• C. 12.24 m
• D. 24.48 m

Answer 1

The correct option is D 24.48 m

from boyle's law

$P_1{V}_1=P_2{V}_2$

& $V_2=3V_1$

so $P_1=3P_2$

& $P_2=P_0$

so $P_1=3P_0$

& $P_1=P_2+\rho gh$

$\rho gh=2P_0$

${\rho }_{k}g\times h=2\times 72cmofHg$

${\rho }_k\times g\times h=2\times 72\times 17\times {\rho }_{k}g$

$h=2448cm$

$h=24.48m$