Question

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 0.465 N m⁻¹, 0.03 N m⁻¹ and 0.076 N m⁻¹ respectively.

Answer 1

Given:
$$\begin{gathered} \mathrm{Radius}\mathrm{of}\mathrm{mercury}\mathrm{drop}r=2\mathrm{mm}=2\times {10}^{-3}\mathrm{m} \\ \mathrm{Radius}\mathrm{of}\mathrm{soap}\mathrm{bubble}r=4\mathrm{mm}=4\times {10}^{-3}\mathrm{m} \\ \mathrm{Radius}\mathrm{of}\mathrm{air}\mathrm{bubble}r=4\mathrm{mm}=4\times {10}^{-3}\mathrm{m} \\ \mathrm{Surface}\mathrm{tension}\mathrm{of}\mathrm{mercury}{T}_{\mathrm{Hg}}=0.465\mathrm{N}/\mathrm{m} \\ \mathrm{Surface}\mathrm{tension}\mathrm{of}\mathrm{soap}\mathrm{solution}{T}_s=0.03\mathrm{N}/\mathrm{m} \\ \mathrm{Surface}\mathrm{tension}\mathrm{of}\mathrm{water}{T}_a=0.076\mathrm{N}/\mathrm{m} \end{gathered}$$

(a) Excess pressure inside mercury drop:
$$\begin{gathered} P=\frac{2T_{Hg}}{r} \\ =\frac{0.465\times 2}{2\times {10}^{-3}}=465\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(b) Excess pressure inside the soap bubble:
$$\begin{gathered} P=\frac{4T_s}{r} \\ =\frac{4\times 0.03}{4\times {10}^{-3}}=30\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$

(c) Excess pressure inside the air bubble:
$$\begin{gathered} P=\frac{2T_a}{r} \\ =\frac{2\times 0.076}{4\times {10}^{-3}}=38\mathrm{N}/{\mathrm{m}}^2 \end{gathered}$$