A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of water in the capillary tube is 4g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:
(A) 2g
(B) 2.5g
(C) 8g
(D) 16g
Solution
h=\frac{2T\;\cos\theta}{\rho rg}
h=\frac{2T\;\cos\theta}{\left({\displaystyle\frac mV}\right)rg}
h=\frac{2T\;V\:\cos\theta}{mrg}
h=\frac{2T\;\left(\pi\;r^2\;h\right)\;\cos\theta}{mrg}
1 =\frac{2T\;\left(\pi\;r\;\right)\;\cos\theta}{m\;g}
m=\left(\frac{2\pi T\cos\theta}g\right)\;r
m\;\propto\;r
\frac{m_1}{m_2}=\frac{r_1}{r_2}
\frac{4}{m_2}=\frac{r}{2r}
m_2=\;\left(\frac{2r}r\right)\ast\;4
\Rightarrow m_2=\;2\ast\;4=8\;
\Rightarrow m_2=8g\;
(C) option is correct
