A) 2 cm
B) 3 cm
C) 6 cm
D) 0.67 cm
Correct Answer: C
Solution :
If h is height of water raised in capillary tube of radius r, then according to \[T=\frac{rhdg}{2\cos }\] rh = constant for same surface ten sin T, same density d and same angle of contact \[\] \[\therefore \] \[{{r}_{1}}{{h}_{1}}={{r}_{2}}{{h}_{2}}\] \[\Rightarrow \] \[{{h}_{2}}=\frac{{{r}_{1}}{{h}_{1}}}{{{r}_{2}}}\] (1) Here : \[{{h}_{1}}=2\,cm,\,{{r}_{2}}=\frac{{{r}_{1}}}{3}\] \[\Rightarrow \,\,\frac{{{r}_{1}}}{{{r}_{2}}}=3\] from eq. (1) \[\therefore \] \[{{h}_{2}}=3\times 2=6\,cm\]
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