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  • Answer please! - Properties of Solids and Liquids - JEE Main-2

A small soap bubble of radius 4 cm is trapped inside another bubble of radius
6 cm without any contact. Let p_{2} be the pressure inside the inner bubble and p_{0},
the pressure outside the outer bubble. Radius of another bubble with pressure
difference p_{2}-p_{0} between its inside and outside would be :

  • Option 1)

    12 cm

  • Option 2)

    2.4 cm

  • Option 3)

    6 cm

  • Option 4)

    4.8 cm

 

Answers (1)

best_answer

As we have learned

Change in Pressure of bubble in air -

\Delta P=\left ( \frac{2T}{R} \right )\times 2= \frac{4T}{R}

- wherein

T- Temperature

R- Radius

 

p_{2}-p_{1}=\frac{4s}{R}_{1}

p_{1}-p_{0}=\frac{4s}{R}_{2}

p_{2}-p_{0}={4s} \left ( \frac{1}{R_{1}}+\frac{1}{R_{2}} \right ) 

\frac{1}{R}= \frac{1}{R_{1}}+\frac{1}{R_{2}}  or R= \frac{R_{1}R_{2}}{R_{1}+R_{2}}= \frac{4*6}{4+6}cm= 2.4cm

 

 

 


Option 1)

12 cm

This is incorrect

Option 2)

2.4 cm

This is correct

Option 3)

6 cm

This is incorrect

Option 4)

4.8 cm

This is incorrect

Posted by

Avinash

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