When a large bubble rises from the bottom of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the depth of the lake be $h$. The pressure at the bottom of the lake is $P_1 = P_{atm} + h\rho g$. Given that atmospheric pressure $P_{atm} =

Vedclass · Accepted Answer

$7H$

Vedclass · Answer

(C) Let the depth of the lake be $h$. The pressure at the bottom of the lake is $P_1 = P_{atm} + hho g$. Given that atmospheric pressure $P_{atm} = Hho g$,we have $P_1 = (H + h)ho g$.At the surface,the pressure is $P_2 = P_{atm} = Hho g$.Let the radius of the bubble at the bottom be $r$. Its volume is $V_1 = \frac{4}{3}\pi r^3$.At the surface,the radius becomes $2r$,so the volume is $V_2 = \frac{4}{3}\pi (2r)^3 = 8 	imes \frac{4}{3}\pi r^3 = 8V_1$.Assuming the temperature remains constant,we apply Boyle's Law: $P_1V_1 = P_2V_2$.Substituting the values: $(H + h)ho g 	imes V_1 = Hho g 	imes 8V_1$.Dividing both sides by $ho g V_1$,we get $H + h = 8H$.Therefore,$h = 7H$.

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

Similar Questions

Which one of the following items gives its correct total number?

Which of the following is a micronutrient for plants?

$A$ system is provided with $50 \ J$ of heat and the work done on the system is $10 \ J$. What is the change in internal energy of the system in joules?

The incorrect statement regarding the following reactions is:
$XeF_6 + \text{Excess } H_2O \rightarrow 'X' + HF$
$XeF_6 + 2 H_2O \rightarrow 'Y' + HF$

Two bodies of masses $m$ and $4m$ are placed at a distance $r$. The gravitational potential at a point on the line joining them where the gravitational field is zero is

Vedclass Test Series

Exam Paper Generator

Online Exam Module