Question #4ad26

1 Answer
Aug 24, 2017

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