When a large bubble rises from bottom of a water lake to its surface, then its radius doubles. If the atmospheric pressure is equal to the pressure of height H of a certain water column, then the depth of lake will be

Step-by-step solution

When a large bubble rises from bottom of water lake to its surface, then its radius becomes double. i.e., r 2 = 2 r 1 Since, volume V ∝ r 3 ∴ V 2 V 1 = r 2 r 1 3 = 2 r 1 r 1 = 8 ⇒ V 2 = 8 V 1 … (i) From Boyle’s law, p 1 V 1 = p 2 V 2 Where, p 2 = atmospheric pressure. p 1 V 1 = p 2 8 V 1 [from Eq. (ii)] p 2 = p 1 8 where, p 2 = atmospheric pressure and p 1 = pressure at depth d. ∴ p 1 = p a t o m s p h e r i c + p g d p 1 = p 2 + p g d … (iii) From Eqs. (ii) and (iii), we have p 2 = p 2 + p g d 8 7 p 2 = p g d 7 .