If An Air Bubble Of Same Radius Be Formed At A Depth Of 40.0 Cm In A Soap Solution (relative Density 1.20), Then What Will Be The Pressure Inside The Air Bubble?

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Question

 

If an air bubble of same radius be formed at a depth of 40.0 cm in a soap solution (relative density 1.20), then what will be the pressure inside the air bubble?

Solution

Correct option is

 

The excess pressure in an air bubble of same radius R formed near the surface of a soap solution is

       

If P be the atmospheric pressure, then the pressure at a depth h (= 40.0 cm = 40.0  10–2) in a soap solution of density  is 

          

The relative density of soap solution (relative to density of water which is 103 kg/m3) is 1.20. Hence the density of soap solution is

        

Here h = 40.0 cm = 0.40 m. 

  

               

Putting values of P and  in eq. (i), we get  

       

            

 total pressure inside the air bubble is  

       

                  .

Testing

SIMILAR QUESTIONS

Q1

A non-viscous liquid of constant density 1000 kg/m3 flows in a streamline motion along a tube of variable cross-section. The tube is kept inclined in the vertical plane as shown in figure. The area of cross-section of the tube at two points P and Q at heights of 2 metre and 5 metre are respectively  the velocity of the liquid at point P is 1 m/s. Find the work done per unit volume by the pressure and the gravity forces as the fluid flows from point P to Q.     

                                                          

Q2

The fresh water behind a reservoir dam is 15 m deep. A horizontal pipe 4.0 cm in diameter passes through the dam 6.0 m below the water surface as shown in figure. A plug secures the pipe opening. The plug is removed. What volume of water flows out of the pipe in 3.0- hour?     

                                                          

Q3

A soap film is on a rectangular wire ring of size . If the size of the film is changed to , then calculate the work done in this process. The surface tension of soap film is 

Q4

The surface tension of a soap solution is 0.03 N/m. How much work is required to form a bubble of 1.0 cm radius from this solution? 

Q5

A mercury drop of radius 1.0 mm breaks up into 64 droplets of equal volumes. Calculate the work done in this process. (Surface tension is mercury is 0.465 N/m)  

Q6

 

A big drop is formed by coalesing 1000 small droplets of water. What will be the change in surface energy? 

What will be the ratio between the total surface energy of the droplets and the surface energy of the big drop?

Q7

 joule work is being done in breaking a big drop of water of radius R into 1000 small drops of equal size. Find out the surface tension of water.

Q8

A drop of mercury has a radius of 3.00 mm at room temperature. The surface tension of mercury at that temperature is 0.465 Nm–1. Find excess pressure inside the drop and the total pressure inside the drop. The atmospheric pressure is .

Q9

What is the excess pressure in a soap bubble of radius 5.00 mm at 20oC? The surface tension of soap solution at 20oC is .

Q10

Two spherical soap bubbles of different radii coalesce. If V is the consequent change in volume of the contained air, and S the change in the total surface area, then show that 3PV + 4ST = 0, where T is surface tension of the soap solution.