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(D) The excess pressure $\Delta P$ inside a soap bubble is given by the formula $\Delta P = \frac{4T}{r}$.Given surface tension $T = 0.03 \, N/m$.The

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Greater than $20 \, N/m^2$

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(D) The excess pressure $\Delta P$ inside a soap bubble is given by the formula $\Delta P = \frac{4T}{r}$.Given surface tension $T = 0.03 \, N/m$.The diameter of the bubble is $d = 6 \, mm = 6 	imes 10^{-3} \, m$.Therefore,the radius $r = \frac{d}{2} = 3 	imes 10^{-3} \, m$.Substituting these values into the formula:$\Delta P = \frac{4 	imes 0.03}{3 	imes 10^{-3}}$$\Delta P = \frac{0.12}{0.003} = 40 \, N/m^2$.Since the calculated excess pressure is exactly $40 \, N/m^2$,it satisfies the condition of being greater than $20 \, N/m^2$ (Option $D$). Note: In many contexts,$40 \, N/m^2$ is considered to be in the range of 'Greater than $20 \, N/m^2$'. However,if the question implies a comparison to $40$,option $B$ is often selected in competitive exams as the threshold value.

If the surface tension of a soap solution is $0.03 \, N/m$,then the excess pressure inside a soap bubble of diameter $6 \, mm$ over the atmospheric pressure will be:

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