Question

An inverted bell lying at the bottom of a lake 47.6 m deep has 50 cm3 of air trapped in it. The bell is brought to the surface of the lake. The volume of the trapped air will be n100cm3, value of n is (atmospheric pressure=70 cm of hg and density of hg=13.6 g/cm3)

Answer 1

SOLUTION

Atmos pressure A=0.70m×13.6^3kg/m^3× 9.80 = 9.33^4 N/m^2

Assuming that temperature doesn't change use P1.V1 = P2.V2

P1 on lake bed

=) A+ (h.pw.g)= 9.33^4 +(47.60m× 1000kg/m^3 ×9.80) = 5.60^5N/m^2

V1= 50cm^3

P2= surface pressure= A

V2 = P1.V1/P2

=) 5.60^5 ×50/9.33^4

=) V2= 300cm^2

hope it helps ☺️⬆️

Answer 2

Answer:

According to Boyle's law PαV1

We know that,

V2=(1+P0hρωg)V1

V2=(1+70×13.6×100047.6×102×1×1000)V1

V2=(1+5)50cm3=300cm3

[P0=P2=70cmofHg=70×13.6×1000)