Q. An air bubble of volume $1.0\ cm^{3}$ rises from the bottom of a lake $40m$ deep at a temperature of $12℃.$ The volume of the bubble when it reaches the surface, which is at a temperature of $35℃,$ will be
A) $5.4\ cm^{3}$
B) $4.9\ cm^{3}$
C) $2.0\ cm^{3}$
D) $10.0\ cm^{3}$
Explanation
(a) $\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$
$\frac{(P + h\rho g)1.0}{273 + 12} = \frac{P.V_{2}}{273 + 35}$
$V_{2} = 5.4cm^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.