Question

Radius of a bubble at the bottom of the tank at the depth of 10.336 m was found to be 1 cm. If the absolute temperature at surface is 4 times to that at the bottom. The radius (in cm) of bubble at surface is​

Answer 1

Answer:

P1 = 1 atm at surface V1 = Volume at Surface P2 = Pressure at bottom h = depth of the lake ρ = density of water If temperature is same, P1 V1 = P2 V2 1 ...

Answer 2

The radius of bubble at the surface is $0.1262m$,

The radius of bubble at the surface will be  a $0.2004m$.

Explanation:

$h=10.336$

Density of water ρ$=1000\frac{kg}{m^{2}}$

$g=9.8kg$  $P_{a}=10^{5} atm$

$p_{1}=10^{5}+(10.336*1000*9.8)$

$=20.129*10^{4}p_{a}$

Volume of the bubble at the bottom $V_{1}=\frac{4}{3}\pi r^{3}$

$=\frac{4}{3}*3.14*1^{3}$

$=4.186cm^{3}$

$\frac{p_{1}v_{1}}{t_{1}}=\frac{p_{2}v_{2}}{t_{2}}$

$r^{3}=\frac{842.599}{4.1866*10^{5}}$

$=201.260*10^{-5}$

$=0.1262m$.

If the absolute temperature at the surface is 4 times at the bottom

$T_{2}=4T_{1}$

$\frac{p_{1}v_{1}}{t_{1}}=\frac{p_{2}v_{2}}{t_{2}}$

$r^{3}=\frac{842.599*4}{4.1866*10^{5}}$

$=8.0504*10^{-3}$

$r=0.2004$.