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Two separate bubbles (radii 0.002 m and 0.004 m) formed of the same soap solution (surface tension = 0.07 N/m) comes together to form a double bubble. The internal film common to both the bubbles has radius R. The value of R (in mm) is (Answer upto two digit after the decimal point)

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Solution


Excess pressure is always present on concave side. So interface is concave towards small bubble.
p1=po+4Tr1p2=po+4Tr2
Excess pressure=p1p2=4T(1r11r2]
This excess pressure for the interface is also equal to 4TR.
4T(1r11r2]=4TRR=r1r2r1r2=(0.002)(0.004)0.0040.002=0.004 m=4 mm


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