Question

The surface tension and vapour pressure of water at $ 20{}^\circ C $ is $ 7.28\times {{10}^{-2}}N{{m}^{-1}} $ and $ 2.33\times {{10}^{3}}Pa $ respectively. The radius of the smallest spherical droplet which can form without evaporating is

• A. $ 6.25\times {{10}^{-5}}m $
• B. $ 6.25\times {{10}^{-3}}m $
• C. $ 7.45\times {{10}^{+3}}m $
• D. $ 8.9\times {{10}^{-4}}m $

Answer 1

Answer: (A)

Given $ T=7.28\times {{10}^{-2}}N/m $ $ p=2.33\times {{10}^{3}}Pa $ The drop will evaporate if the water pressure greater than the vapour pressure p. i.e. $ p=\frac{2T}{R}, $ where R = radius of drop $ \Rightarrow $ $ R=\frac{2T}{p}=\frac{2\times 7.28\times {{10}^{-2}}}{2.33\times {{10}^{3}}}=6.25\times {{10}^{-5}}m $