Question

Question #4ad26

Answer 1

Let the volume of the air bubble at the bottom of the lake be `V_b`. So its volume at the surface `V_s=2V_b`

Let the depth of the lake be `hcm`

So pressure at the bottom of the lake

`P_b="pressure of water of height h+ atmospheric prssure"`

`P_b= hxxd_wxxg+ 75xxd_(Hg)xxg`

Pressure at the surface

`P_s=75xxd_(Hg)xxg`

Now applying Boyle's law we get

`P_bxxV_b=P_sxxV_s`

`=>(hxxd_wxxg+ 75xxd_(Hg)xxg)V_b=(75xxd_(Hg)xxg)xx2V_b`

`=>h+ 75xxd_(Hg)/d_w=75xxd_(Hg)/d_wxx2`

`=>h=75xxd_(Hg)/d_w`

`=>h=75xx40/3=1000cm=10m`