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Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. Surface tension of mercury, soap solution and water are `0.465Nm^-1, 0.03Nm^-1 and 0.076Nm^-1` respectively.

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The correct Answer is:
B, C, D

`r=2mm`
`=2xx10^-3m`
`T_(Hg)=0.465N/m`
`T_s=0.03N/m`
`T_a=0.076N/m`
`P=(2T_(Hg))/r`
`=(0.465xx2)/(2xx10^-3)=465N/m^2`
`P=(4T_s)/r`
`=(4xx0.03)/(4xx10^-3)=30N/m^2`
`P=(2T_s)/r`
`=(92xx0.076)/(4xx10^-3)=38N/m^2`
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