Grade 12th passMechanics

Two soap bubbles of radii r1 and r2 equal to 4cm and 5cm are toiching each other over a common surface s1s2 .Its radii will be

vishal thakur

9 Years agoGrade 12th pass

1 Answer

Ishita Mandal

7 Years ago

Let the radius be R.

Let surface tension be T

Let pressure outside the bubbles be P

And pressure of the bubbles be P1 and P2 respectively.

Applying the concept of surface tension

For 1st bubble,

P1-P=4T/r1 .......(1)

For 2nd bubble,

P2-P=4T/r2 .......(2)

Equation (1)-(2) gives

P1-P2=(4T/r1)-(4T/r2) .......(3)

But we know that P1-P2=Excess pressure which is equal to 4T/R,

Where R is the radius to be calculated as per the question

So,

Replacing P1-P2=4T/R in equation (3) we get

4T/R=(4T/r1)-(4T/r2)

1/R=(1/r1)-(1/r2)

1/R=(1/4)-(1/5)

1/R=1/20

R=20cm

Its radius will be 20 cm

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Two soap bubbles of radii r1 and r2 equal to 4cm and 5cm are toiching each other over a common surface s1s2 .Its radii will be

vishal thakur

1 Answer

Ishita Mandal

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