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#