A) 2.5 g
B) 5.0 g
C) 10 g
D) 20 g
Correct Answer: C
Solution :
When one end of capillary tube of radius r is immersed into a liquid of density p, to maintain the equilibrium the liquid level rises in the capillary tube upto height \[h=\frac{2T\cos \theta }{\rho rh}\] Mass of the water in first tube is \[m=\pi {{r}^{2}}h\rho \] \[=\pi {{r}^{2}}\left[ \frac{2T\,\cos \theta }{\rho rg} \right]\times \rho \] \[m=\frac{\pi r(2T\,\cos \theta )}{g}\] or \[m\propto r\] Similarly for second tube \[m\propto r\] or \[\frac{m}{m}=\frac{r}{r}=\frac{r}{2r}=\frac{1}{2}\] or \[m=2m=2\times 5=10\,g\]
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