Two separate air bubble (radii 0.002 m and 0.004 m) formed of the same liquid (surface tension 0.07 $N m^{- 1}$ )come together to form a double bubble. The radius and the sense of curvature of internal film surface, common to both the bubbles is:

0.002 m concave towards smaller bubble

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0.003 m concave towards smaller bubble

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0.004 m concave towards smaller bubble

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None of these

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Solution

The correct option is C 0.004 m concave towards smaller bubble
Since both are same liquid , therefore net radius will be

$\frac{4 T}{r} = \frac{4 T}{r_{1}} - \frac{4 T}{r_{2}}$

$r = \frac{r_{1} r_{2}}{r_{2} - r_{1}}$

$= \frac{0.002 \times 0.004}{0.002}$

= 0.004

Hence (c) option is correct

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Two separate air bubble (radii 0.002 m and 0.004 m) formed of the same liquid (surface tension 0.07 Nm−1)come together to form a double bubble. The radius and the sense of curvature of internal film surface, common to both the bubbles is:

The correct option is C 0.004 m concave towards smaller bubbleSince both are same liquid , therefore net radius will be4Tr=4Tr1−4Tr2r=r1r2r2−r1=0.002×0.0040.002= 0.004Hence (c) option is correct

Two separate air bubble (radii 0.002 m and 0.004 m) formed of the same liquid (surface tension 0.07 $N m^{- 1}$ )come together to form a double bubble. The radius and the sense of curvature of internal film surface, common to both the bubbles is:

The correct option is C 0.004 m concave towards smaller bubble
Since both are same liquid , therefore net radius will be

$\frac{4 T}{r} = \frac{4 T}{r_{1}} - \frac{4 T}{r_{2}}$

$r = \frac{r_{1} r_{2}}{r_{2} - r_{1}}$

$= \frac{0.002 \times 0.004}{0.002}$

= 0.004

Hence (c) option is correct