A capillary tube of radius 'r' is immersed in | vedclass

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Question

(A) The height of a liquid column in a capillary tube is given by $h = \frac{2T \cos 	heta}{rdg}$.For water: $h_1 = \frac{2T_1 \cos 	heta_1}{rdg}$.

Vedclass · Accepted Answer

$5 	imes 10^{-3}$

Vedclass · Answer

(A) The height of a liquid column in a capillary tube is given by $h = \frac{2T \cos 	heta}{rdg}$.For water: $h_1 = \frac{2T_1 \cos 	heta_1}{rdg}$. Given $T_1 = T$,$	heta_1 = 0^{\circ}$ (so $\cos 	heta_1 = 1$),and mass $m_1 = 5 	imes 10^{-3} \ kg$.Since $m = \pi r^2 h d$,we have $h_1 = \frac{m_1}{\pi r^2 d}$.For the second liquid: $T_2 = \sqrt{2}T$ and $	heta_2 = 45^{\circ}$.The new height $h_2 = \frac{2T_2 \cos 	heta_2}{rdg} = \frac{2(\sqrt{2}T) \cos 45^{\circ}}{rdg} = \frac{2\sqrt{2}T 	imes (1/\sqrt{2})}{rdg} = \frac{2T}{rdg} = h_1$.Since the radius '$r$' and density '$d$' (assuming same density for simplicity or as implied by the problem context) remain constant,and $h_2 = h_1$,the mass of the liquid $m_2 = \pi r^2 h_2 d = m_1 = 5 	imes 10^{-3} \ kg$.

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