A) 3 cm
B) 1.5 cm
C) 4.5 cm
D) 6 cm
Correct Answer: A
Solution :
The liquid rises to such a height, that external pressure balances internal pressure. The column of liquid remaining in the tube will be held by two meniscus an upper and lower one. Therefore height of column of water left in the tube will be \[\frac{2T}{r}+\frac{2T}{r}=h\rho g\] \[\Rightarrow \] \[\frac{4T}{r}=h\rho g\] \[\Rightarrow \] \[h=\frac{4T}{r\rho g}\] Given, \[T=73.5\times {{10}^{-3}}N/m\], \[r=1\,mm=1\times {{10}^{-3}}m\] \[\rho ={{10}^{3}}kg/m\], \[g=9.8\,\,m/{{s}^{2}}\] \[\therefore \] \[h=\frac{4\times 73.5\times {{10}^{-3}}}{1\times {{10}^{-2}}\times 9.8\times 10}\] \[\Rightarrow \] \[h=3\times {{10}^{-2}}m=3\,cm\]You need to login to perform this action.
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