• # question_answer A spherical drop of water has radius 1 mm. If surface tension of water is $70\times {{10}^{-3}}N/m$then the difference of pressures between the inside and outside of the spherical drop is: A) $~35N/{{m}^{-2}}$                      B) $~70\text{ }N/{{m}^{2}}$C) $~140\text{ }N/{{m}^{2}}$                     D)  0

Let the excess pressure be$\Delta P$ $\Delta P=\frac{2T}{R}=\frac{2\times 70\times {{10}^{-3}}}{{{10}^{-3}}}=140\,N/{{m}^{2}}$ Hence, the correction option is [c].