A) 1 mm
B) 3 mm
C) 6 mm
D) 9 mm
Correct Answer: D
Solution :
Key Idea: The narrower the tube, the higher is the rise in water. When a glass capillary tube open at both ends is dipped in a liquid of surface tension \[T\], density \[\rho \], \[\theta \] the angle of contact, then water rises to a height \[h=\frac{2T\,\,\cos \,\,\theta }{r\rho g}\] \[\Rightarrow \] \[h\propto =\frac{1}{r}\] Where \[r\] is radius of tube. Given, \[{{h}_{1}}=3\,mm,\,\,{{r}_{2}}=\frac{{{r}_{1}}}{3}\] \[\therefore \] \[\frac{{{h}_{1}}}{{{h}_{2}}}=\frac{{{r}_{2}}}{{{r}_{1}}}\] \[\Rightarrow \] \[\frac{3}{{{h}_{2}}}=\frac{1}{3}\] \[\Rightarrow \] \[{{h}_{2}}=9\,\,mm\]
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