A) \[\text{44}\times \text{1}{{0}^{-\text{6}}}\text{ N}\]
B) \[\text{22}\times \text{1}{{0}^{-\text{6}}}\text{ N}\]
C) \[\text{33}\times \text{1}{{0}^{-\text{6}}}\text{ N}\]
D) None of these
Correct Answer: A
Solution :
Surface tension (T) is defined as work done (W) in increasing the surface area \[(\Delta A)\] of a film by unity at constant temperature. \[\therefore \] \[T=\frac{W}{\Delta A}\] Also, W = force (F) \[\times \] displacement \[\therefore \] \[T=\frac{FL}{2\pi rL}\] \[\Rightarrow \] \[F=2\pi rT\] Given, \[r=0.1\,mm=0.1\times {{10}^{-3}}m\], \[T=7\times {{10}^{-2}}N/m\] \[\therefore \] \[F=2\times \frac{22}{7}\times {{10}^{-4}}\times 7\times {{10}^{-2}}\] \[\Rightarrow \] \[F=44\times {{10}^{-6}}N\] Alternative The height (h) of water rise in a capillary tube is given by \[h=\frac{2T}{r\rho g}\] ... (i) Also, Weight (w) = volume (V) \[\times \] density \[(\rho )\times g\] w = Area \[\times \] height (h) \[\times \rho \times g\] ... (ii) From Eqs. (i) and (ii), we get Weight \[=Ah\rho g=\frac{2T}{r}A=\frac{2T}{r}\times \pi {{r}^{2}}=2T{{\pi }^{r}}\] Given, \[T=7\times {{10}^{-2}}N/m\], \[r=0.1\,mm=0.1\times {{10}^{-3}}m\] \[w=2\times 7\times {{10}^{-2}}\times 3.14\times {{10}^{-3}}\times 0.1\] \[\Rightarrow \] \[w=44\times {{10}^{-6}}N\]
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